s-Block Elements

Section (A) : General facts about elements

The elements in which the last electron enters the outermost s-orbital are called s-block elements.
As the s-orbital can accommodate only two electrons, two groups 1 and 2 belong to the s-block.
Flame Test
The alkali metals and alkaline earth metals and their salts impart characteristic colour to an oxidizing
flame.
Reason: This is because the heat from the flame excites the outermost orbital electron to a higher
energy level. When they drop back to the ground state, there is emission of radiation in the visible
region.
The electrons in beryllium and magnesium are too strongly bound (due to small size) to get excited by
flame. Hence, these elements do not impart any colour to the flame.
Metal Li Na K Rb Cs
Colour Crimson red Yellow Violet / Lilac Red violet Blue
Metal Be Mg Ca Sr Ba
Colour No colour No colour Brick red Crimson red Apple green

Higher oxidation zone

(melting zone) 1550ºC

Higher reduction zone 1560ºC

520ºC
Lower oxidation zone
Lower reduction zone 1450ºC

300ºC

Section (B) Based on Periodic trends

Group – 1st(IA) Elements : (Alkali Metals)
Atomic and Physical properties of the Alkali metals
Lithium Sodium Potassium Rubidium Caesium Francium
Property
Li Na K Rb Cs Fr
Atomic number 3 11 19 37 55 87
Atomic mass (g mol ) –1 6.94 22.99 39.10 85.47 132.91 (223)
Electronic configuration [He] 2s 1 [Ne] 3s 1 [Ar] 4s 1 [Kr] 5s1 [Xe] 6s1 [Rn] 7s1
Ionization enthalpy / kJ mol –1 520 496 419 403 376 ~375
Hydration enthalpy/kJ mol –1 –506 –406 –330 –310 –276 –
Metallic radius / pm 152 186 227 248 265 –
+
Ionic radius M / pm 76 102 138 152 167 (180)
m.p. / K 454 371 336 312 302 –
b.p / K 1615 1156 1032 961 944 –
Density / g cm–3 0.53 0.97 0.86 1.53 1.90 –
Standard potentials E/ V for (M+/ M) –3.04 –2.714 –2.925 –2.930 –2.927 –
Occurrence in lithosphere † 18* 2.27** 1.84** 78-12* 2-6* ~ 10–18*
*ppm (part per million), ** Percentage by weight
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s-Block Elements
Group IIA Elements (Alkaline Earth Metals)
Beryllium Magnesium Calcium Strontium Barium Radium
Property
Be Mg Ca Sr Ba Ra
Atomic number 4 12 20 38 56 88
Atomic mass (g mol–1) 9.01 24.31 40.08 87.62 137.33 226.03
Electron configuration [He] 2s2 [Ne] 3s2 [Ar] 4s2 [Kr] 5s2 [Xe] 6s2 [Rn] 7s2
Ionization enthalpy (I) / kJ mol–1 899 737 590 549 503 509
Ionization enthalpy (II) /kJ mol–1 1757 1450 1145 1064 965 979
Hydration enthalpy (kJ/mol) – 2494 – 1921 –1577 – 1443 – 1305 –
Metallic radius / pm 112 160 197 215 222 –
Ionic radius M2+/ pm 31 72 100 118 135 148
m.p. / K 1560 924 1124 1062 1002 973
b.p / K 2745 1363 1767 1655 2078 (1973)
Density / g cm–3 1.84 1.74 1.55 2.63 3.59 (5.5)
Standard potential E/ V for (M2+/ M) –1.97 –2.36 –2.84 –2.89 – 2.92 –2.92
Occurrence in lithosphere 2* 2.76** 4.6** 384* 390* 10–6*

Section (C) & (D) : Based on Chemical Bonding, Properties of elements

Properties of Alkali and Alkaline earth metals
S.No. Atomic Properties Alkali metals Alkaline earth metals
1. Outer Electronic ns1 ns2
configuration
2. Oxidation number (i)These elements easily form univalent The IP1 of these metals are much lower than IP2 and
and valency +ve ion by losing loosely solitary ns1 thus it appears that
electron due to low IP value. these metals should form univalent ion rather than
divalent ions but in actual practice, all these give
bivalent ion.
3. Atomic and Ionic Increase down the group, because value The atomic and ionic radii of the alkali earth metal
radii of n (principal quantum number) are smaller than corresponding alkali metals.
increases. Order = Li < Na < K < Rb < Cs. Reason
higher nuclear charge (Zeff)
On moving down the group size increase, as value
of n increases.
Be < Mg < Ca < Sr < Ba
4. Ionisation Energy As size increases, I.E. decreases down Down the group IE decreases due to increase in
the group (so Cs have lowest I.P.) size. Be > Mg > Ca > Sr > Ba
Order = Li > Na > K > Rb > Cs IE1 of Alkali metal < IE1 of Alkaline earth metal
IE2 of Alkali metal > IE2 of Alkaline earth metal
Reason
IE1 of Alkaline earth metal is large due to increased
nuclear charge in Alkaline earth metal as compared
to Alkali metal but IE2 of Alkali metal is large
because second electron in Alkali metal is to be
removed from cation which has already acquired
noble gas configuration
5. Electropositive Alkali metals are strongly electropositive Due to low IE they are strong electropositive but not
character or metallic and metallic. Down the group as strong as Alkali metal because of comparatively
character electropositive nature increase so metallic high IE. The electropositive character increase down
nature also increases. the group.
i.e. M  M+ + e– Order = Be < Mg < Ca < Sr < Ba
Metallic Nature: Electropositive character
1 /I.P.
Order = Li < Na < K < Rb < Cs.

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s-Block Elements
6. Hydration of ions (i) Hydration represents for the Hydration energy = Be2+ > Mg2+ > Ca2+ > Sr2+ >Ba2+
dissolution of a substance in water to get
absorb water molecule by weak valency
forces Hydration of ions in the process
when ions on dissolution in water get
hydrated.
(ii) Hydration energy  charge density on
ion
Degree of hydration  1/Cation size 
charge  1/ionic mobility  1/conductivity
Hydration energy = Li+ > Na+ > K+ > Rb+ >
Cs+
(iii) Li+ being smallest in size has
maximum degree of hydration and that is
why lithium salts are mostly hydrated and
moves very slowly under the influence of
electric field. e.g : LiCl.2H2O.
7. Photoelectric effect The phenomenon of emission of
electrons when electromagnetic rays
strikes
against them is called photoelectric effect;
Alkali metal have low I.P. so show
photoelectric effect. Cs and K are used in
Photoelectric cells.
8. Electronegativity (i)These metals are highly electopositive (i) Their electronegativities are also small but are
and there by possess low values of higher than that of alkali metals
electro negativities. (ii) Electronegativity decrease from Be to Ba
(ii)Electronegativity of alkali metals
decreases down the group.
Order = Li > Na > K > Rb > Cs

S.No. Physical Property Alkali metals Alkaline earth metals

1. Density (i)All are light metals. (i) Heavier than alkali metals.
(ii) Density increase down the group but K is lighter (ii) Density decrease slightly up to Ca after
than Na. which it increases.
Order = Li < K < Na < Rb < Cs (iii) Density of Mg is greater than Ca.
2. Hardness (i) All are silvery white metals. Relatively soft but harder than Alkali
(ii) Light soft, malleable and ductile metals with metals.
metallic luster.
(iii) Diamagnetic and colour less in form of ions.
(iv) These metals are very soft and can be cut with a
knife. Lithium is harder than any other alkali metal.
The cutting of sodium The hardness depends upon cohensive energy.
metal
Cohensive energy  Force of attraction between
atoms.
3. Melting points/ Boiling (i) Lattice energy decreases from Li to Cs and thus They have low Melting points and Boiling
points Melting points and Boiling points also decrease from points but are higher than corresponding
Li to Cs. value of group I.
M.P. = Li > Na > K > Rb > Cs Reason
B.P. = Li > Na > K > Rb > Cs They have two valence electrons which
may participate in metallic bonding
compared with only one electron in Alkali
metal. Consequently group II elements are
harder and have higher cohesive energy
and so, have much higher Melting points /
Boiling points than Alkali metal .
M.P. = Be > Ca > Sr >Ba > Mg ,B.P. = Be
> Ba > Ca > Sr > Mg
4. Specific heat It decreases from Li to Cs. Li > Na > K > Rb > Cs values are lesser than that of alkali metals,
(*need not to memorise) decreases down the group.
*need not memorise.

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s-Block Elements
S.No. Chemical Alkali metals Alkaline earth metals
Property
1. (i) They generally form oxides and peroxides. (i) Be and Mg are kinetically inert towards oxygen
M+O2   M2O (Oxide) 
O2
M2O2 (Peroxide) becasue of formation of a film of oxide on their
The alkali metals tarnish in dry air due to the formation surface. However powdered Be burn brilliantly.

of their oxides on their surface. 2Be + O2 (air) 
 2BeO(amphoteric) ;
4M + O2 — 2M2O 
3Be + N2 (air)   Be3N2
They react vigorously in oxygen forming following
oxides. (ii) Mg is more electropositive and burns with
4 Li + O2 — 2 Li2O (Monoxide) dazzling brilliance in air give MgO and Mg3N2.

2 Na + O2 — Na2O2 (Peroxide) Mg + O2(air) 
 MgO ;
M + O2 — MO2 ( Superoxide) where M = K, Rb, Cs Mg + N2(air)  
 Mg3N2
Principal Combustion Product (Minor Product) Peroxides are coloured due to lattice defect.
Metal Oxide Peroxide Superoxide (Similar property with Li because both shows
Li Li2O (Li2O2) diagonal relation.)
Na (Na2O) Na2O2
KO2(Orange/Yellow
(iii) Ba gives BaO2 not BaO.
K (iv) Calcium, strontium and barium are readily
Crystalline)
Action with attacked by air to form the oxide and nitride. They
RbO2 (Orange/Yellow
O2 and N2 Rb
Crystalline) also react water with increasing vigour even in cold
Cs
CsO2 (Orange/Yellow to form hydroxides.
Crystalline) (v) BeO, MgO are used as refractory, because they
The oxides and peroxides are colourless when pure. have high M.P.
(ii) All super oxide are paramagnetic and peroxides (vi) Other metals (Ba or Sr form peroxide)
are diamagnetic in nature. 
(iii) The increasing stability of the peroxide or M + O2 
 MO2
superoxide as the size of the metal ion increases is
due to the stabilisation of large anions by larger
cations through lattice energy effect.
(iv) Since all the alkali metals are highly reactive
towards air ; they are kept in kerosene oil. Reactivity
increases from Li to Cs.
(v) Only Lithium reacts with N2 (at room temperature)
to form ionic lithium nitride Li3N because Li being
strongest reducing agent converts N2 into N3–.
3Li + 1/2N2 — Li3N
2. (i) Alkali metals decompose water to form the (i) Ca, Sr, Ba and Ra decompose cold water readily
hydroxides having the formula MOH and dihydrogen. with evolution of hydrogen.
2M + 2H2O — 2MOH(aq.) + H2(g) (M = An alkali M + 2H2O   M(OH) 2 + H2
metal).
(ii) Li decompose water slowly, sodium reacts with (ii) Magnesium decomposes boiling water but
beryllium is not attacked by water even at high
water quickly K, Rb and Cs react with water
temperatures as its oxidation potential is lower
vigorously.
(iii) It may be noted that although lithium has most than the other members
negative E value (In below table), its reaction with
water is less vigorous than that of sodium which has
the least negative E value among the alkali metals.
This behaviour of lithium is attributed to its small size
and very high hydration energy. It’s explanation lies in
Kinetics, released energy in case of K, Rb, Cs is
Action with sufficient to melt or even vapourise and so more
water surface area is exposed to the water and kinetically
reaction is faster than lithium. Other metals of the
group react explosively with water.
Alkali metals
Property
Li Na K Rb Cs Fr

Standard
– 2.714

– 2.925

– 2.930

– 2.927
– 3.04

potentials
–

E/V for
(M+/M)

(iv) They also react with proton donors such as

alcohol, gaseous ammonia and terminal alkynes
evolution of hydrogen.
2M + 2C2H5OH — 2C2H5OM + H2
Ethyl alcohol Metal ethoxide
3. (i)They react with H2 forming metal hydride with (i) Except Be, all alkaline earth metals form hydrides
formula MH which are of ionic nature. Stability of (MH2) on heating directly with H2.
hydride decreases down the group. Since the (ii)The stability of hydrides decreases from Be to Ra.
Hydrides
electropositive character decreases from Cs to Li. (iii) BeH2 is prepared by the action of LiAlH4 on
2M + H2 —2MH BeCl2.

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s-Block Elements
(iii) The metal hydrides react with water to give MOH BeCl2 +LiAlH4 — 2BeH2 + LiCl+ AlCl3
and H2. (act as reducing agent) BeH2 & MgH2 is covalent and polymeric but other are
MH + H2O — MOH + H2 ionic.
H H H
Be Be
H H H

(iv) The ionic hydrides of Ca, Sr, Ba liberate H2 at

anode and metal at cathode.
4. (i)The alkali metals react vigorously with halogens to (i)The alkaline earth metals directly combine with
form ionic halides M+X–. halogens on heating to give metal halides MX2
2M + X2 2 M+X– (X=F,Cl,Br,I)
(ii) Alkali metals halides (Cl2, Br2, I2) formation is (ii) Thermal decomposition of (NH4)2BeF4 is the best
increases form Li to Cs due to increase in route for the preparation of BeF2, and BeCl2 is
electropositive character. conveniently made from the oxide.
Order of reactivity towards F2
Li > Na > K > Rb > Cs BeO + C + Cl BeCl2 + CO
(iii) LiX have more covalent character (It is because of Anhydrous beryllium halide can not be obtained from
the high polarisation capability of Lithium ion (fajan’s materials made in aqueous solution because the
rules)). hydrated ions [Be(H2O)4]2+ is formed. i.e.
(iv)Halides having ionic nature have high melting point [Be(H2O)4]Cl2
and are good conductor of current in fused state. On dehydration, hydrolysis takes place.
These are readily soluble in water. heat
(v) Halides of potassium, rubidium and ceasium have [Be(H2O)4]Cl2   Be(OH)2 + 2HCl
property of combining with extra halogen atoms (iii) Except for beryllium halides, all other halides of
forming polyhalides. alkaline earth metals are ionic in nature. Beryllium
KI + I2 — KI3 halides are essentially covalent and soluble in
organic solvents. Beryllium chloride has a chain
structure in the solid state as shown below :
Cl–Be–Cl

Cl Cl
Halides Be Be Be
Cl
Cl
In the vapour phase BeCl2 tends to form a chloro-
bridged dimer which dissociates into the linear
monomer at high temperatures of the order of 1200
K.
(iv)The ionic character of halides increases from Be
to Ra.
(v)Beryllium halides have covalent character due to
small size and high effective nuclear charge and thus
do not conduct electricity in molten state.
(vi) The fluorides are relatively less soluble than the
chlorides owing to their high lattice energies.
(vii)The decreases in solubility of halides down the
group is due to decrease in hydration energy
because of increasing size of metal cation .
(viii) The tendency to form halide hydrates gradually
decreases (for example, MgCl2·6H2O, CaCl2·6H2O,
SrCl2·6H2O and BaCl2·2H2O) down the group. The
dehydration of hydrated chlorides, bromides and
iodides of Ca, Sr and Ba can be achieved on heating;
however, the corresponding hydrated halides of Be
and Mg on heating suffer hydrolysis.
(ix) CaCl2 has strong affinity with water and is used
as dehydrating agent.
5. Property
Alkali metals The alkaline earth metals are strong reducing agents.
Li Na K Rb Cs Fr This is indicated by large negative values of their
Standard reduction potentials (below table). However their
– 2.714

– 2.925

– 2.930

– 2.927
– 3.04

Reducing potentials reducing power is less than those of their

–

nature E/V for corresponding alkali metals. Beryllium has less

(M+/M) negative value compared to other alkaline earth
(*need not
to (i) Reducing agent is electron donor. The alkali metals metals.
memorise) are strong reducing agents, lithium being the most and However, its reducing nature is due to large
sodium the least powerful (above table). The standard hydration energy associated with the small size of
electrode potential (E) which measures the reducing Be2+ ion and relatively large value of the atomization
power represents the overall change : enthalpy of the metal.
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s-Block Elements
M(s)  M(g) Sublimation enthalpy Alkaline earth metals
Property
M(g) M+(g) + e– Ionization enthalpy Be Mg Ca Sr Ba Ra
M+(g) + H2O  M+ (aq) Hydration enthalpy Standard

– 1.97

– 2.36

– 2.84

– 2.89

– 2.92

– 2.92
(ii) Lithium is expected to be least reducing agent due potentials
to it's very high I.E. However, lithium has the highest E/V for
hydration enthalpy which accounts for its high negative (M+/M)
E value and its high reducing power.
Reducing Nature in gas phase
= Li < Na < K < Rb < Cs.
Reducing Nature in aqueous condition
= Li > Cs > Rb > K > Na.

6. (i) These oxides are easily hydrolysed by water to form Basic/thermal stability
the hydroxides.
Thus M2O (oxide) + H2O  MOH = Be(OH)2<Mg(OH) 2<Ca(OH) 2<Sr(OH)2 <Ba(OH) 2
M2O2 (peroxide) + H2O  2 MOH +H2O2
Basic MO2 (superoxide) + H2O  2 MOH +H2O2+O2
nature of (ii) The Hydroxide which are obtained by the reaction
hydroxide of the oxide. With water all are white crystalline solids.
The alkali metal hydroxides are the strongest of all
bases and dissolve freely in water with evolution of
much heat an account of intense hydration.
Basic nature/Solubility in water/Thermal stability
= LiOH < NaOH < KOH < RbOH < CsOH
7. (i) The carbonates (M2CO3) and bicarbonates (i) All these metal carbonates MCO3 are insoluble in
(MHCO3) are highly stable to heat, where M as alkali neutral medium but soluble in acids and decompose
metals. on red heating.
(ii) Group 1 metals are so strongly basic, they (except (ii) The stability of carbonates increases with
lithium) also form solid bicarbonates. No other metals increase in electropositive character of metal.
form solid bicarbonates. Lithium carbonate is not so BeCO3 < MgCO3 < CaCO3 < SrCO3 < BaCO3
stable to heat. Its hydrogencarbonate does not exist
(iii) Bicarbonates of alkaline earth metals do not exist
as a solid. Although NH4HCO3 also exists as a solid. in solid state but are known in solution only on
(iii) The stability of these salts increases with the heating their solution bicarbonates decomposed to
increasing electropositive character from Li to Cs. It is
liberate CO2 .
therefore Li2CO3 decompose on heating. 
Thermal stability/Solubility in water. M(HCO3)2  MCO3+CO2 + H2O
Li2CO3 < Na2CO3 < K2CO3 < Rb2CO3 < Cs2CO3 (Solution)
LiHCO3 does not exist in solid form due to high (iv)Solubility of carbonates decrease on moving
polarizing power of Li+ and uncomparable size of Li+ down the group.
cation and HCO3– anion. BeCO3 > MgCO3 > CaCO3 > SrCO3 > BaCO3

Li2CO3  Li2O + CO2
(iv) Bicarbonates are decomposed at relatively low
Carbonates temperature.
2MHCO3   M2CO3 + H2O+ CO2
and 300ºC
bicarbonates
(v) Hydrolysis of carbonate
Na2CO3 + 2H2O —2NaOH + H2CO3
Li2CO3 + 2H2O — sparingly soluble
(vi) The crystal structures of NaHCO3 and KHCO3 both
show hydrogen bonding, but are different.
(a) In NaHCO3, the HCO3– ions are linked into an
infinite chain.
(b) in KHCO3, RbHCO3, CsHCO3, HCO3– forms a
dimeric anion.
Solubility in water NaHCO3 < KHCO3 < RbHCO3 <
CsHCO3

(a)

(b)

8. (i) A metal shows complex formation only when it has Be2+ on account of smaller size forms many
following characteristics. complexes such as [Be F3]–, [BeF4]2–
(a) Small size, (b) High nuclear charge, (c) Presence Chlorophyll contains Mg2+ [Photosynthetic pigment in
Complex
of empty orbitals in order to accept electron pair plants] (C.No.= 4)
ion
from ligand (electron pair donor species). [Be(H2O)4]2+ + H2O — [Be(H2O)3OH]+ + H3O+
formation
(ii) Due to small size only Lithium in alkali metals,
forms a few complex ions. Rest all alkali metals do not
possess the tendency to form complex ion.

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s-Block Elements
Reacts vigorously with acids The alkaline earth metals readily react with acids
Reaction 2M + H2SO4 —M2SO4 + H2 liberated dihydrogen.
9.
with acids
M + 2HCl — MCl2 + H2
(i) Alkali metals get dissolved in mercury to form Alkaline earth metals get dissolved in mercury to
Formation amalgams with evolution of heat and the form amalgams with evolution of heat and the
10. of amalgamation is highly exothermic. amalgamation is highly exothermic.
amalgams (ii) Alkali metals form alloys themselves as well as with
other metals.
(i) MSO4 type sulphates are formed
11. Sulphates (i) All these form sulphates of type M2SO4. (ii)The solubility of sulphates decreases on moving
(ii) Except Li2SO4 rest all are soluble in water. down the group. The sulphates of the alkaline earth
Thermal stability /solubility in water metals are all white solids and stable to heat. BeSO4,
Li2SO4 < Na2SO4 < K2SO4 < Rb2SO4 < Cs2SO4 and MgSO4 are readily soluble in water; the solubility
(iii)These sulphates on fusing with carbon form decreases from CaSO4 to BaSO4. The greater
sulphides. hydration enthalpies of Be2+ and Mg2+ ions overcome
M2SO4 + 4C — M2S + 4CO the lattice enthalpy factor and therefore their
sulphates are soluble in water.
Thermal stability
BeSO4 < MgSO4 < CaSO4 < SrSO4 < BaSO4
Solubility in water
BeSO4 > MgSO4 > CaSO4 > SrSO4 > BaSO4

12. Sulphides All metals react with S forming sulphides such as (iii) MSO4 + 2C — MS + 2CO2
Na2S and Na2Sn (n = 2, 3, 4, 5 or 6). The polysulphide M2+ + S2– — MS
ions are made from zig-zag chains of sulphur atoms.
13. (i) Nitrates of both are soluble in water and On heating they decompose into their corresponding
decompose on heating. oxides with evolution of a mixture of nitrogen dioxide
(ii) LiNO3 decomposes to give NO2 and O2 and rest all and oxygen.
give nitrites and oxygen.
M(NO3)2   MO + 2NO2 + ½O2
2MNO3 —2MNO2 +O2 (except Li)
(M = Be, Mg , Cr, Sr, Ba)
4LiNO3 — 2Li2O + 4NO2 + O2
Nitrates 
2NaNO3  2NaNO2 + O2
500ºC

2NaNO3  Na2O + N2 + O2
800ºC
Na
2NaNO3   Na2O + N2 + O2
Li3N + 3H2O — 3LiOH + NH3  Be3N2 + 6H2O  3Be(OH)2 + 2NH3 
14. Nitride
Mg3N2 + 6H2O  3Mg(OH)2 + 2NH3 
When Li is heated with carbon, an ionic carbide Li2C2 The binary compounds of carbon with other elements
15. Carbide is formed. (less electronegative or of similar electronegativity)
2Li + 2C — Li2C2 are called carbides. They are classified into following
Other metals do not react with carbon directly but form 3 categories :
carbides when heated with ethyne, or when ethyne is (i) Ionic (ii) Covalent (iii) Interstitial (or metallic)
passed through a solution of metal in liquid ammonia. (i) Ionic carbides (or salt like carbides) : Generally
Na + C2H2 — NaH + C2 — Na2C2 formed by the most electropositive elements such as
[CC–H]– [CC]2– alkali and alkaline earth metals and aluminium
Na2C2 + 2H2O —NaOH + C2H2 (Boron is exception). Based on the product obtained
on hydrolysis, they are further sub-classified into
three types.
(a) Methanides : These give CH4 on reaction with
H2O.
Al4C3 + 12H2O —4Al(OH)3 + 3CH4 ;
Be2C + 4H2O —2Be(OH)2 + CH4
These carbides contain C4– ions in their constitution.
(b) Acetylides : These give C2H2 on reaction with
H2O.
CaC2 + 2H2O —Ca(OH) 2 + C2H2
Al2 (C2)3 + 6H2O —2Al(OH)3 + 3C2H2
SrC2 + 2H2O —Sr (OH) 2 + C2H2
Such compounds contain C22– ions.
(c) Allylides : These give 1-propyne on reaction
with H2O.
Mg2C3 + 4H2O —2Mg(OH)2 + CH3–CCH
..
Such compounds contain C34– [: C – C  C :]4  ions.
..

(ii) Covalent carbides : Molecules like SiC and B4C are

also examples of covalent carbides.

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s-Block Elements
(iii) Interstitial or metallic carbides
Such carbides are formed by transition metals and
some of the lanthanides and actinides. Interstitial
carbides retain many of the properties of metals.
They conduct electricity by metallic conduction and
have properties of metals (a lusture like a metal). In
these compounds carbon atoms occupy octahedral
holes in the closed packed metal lattice. These are
generally very hard and have very high melting point
(e.g. WC). Carbides of Cr, Mn, Fe, Co and Ni are
hydrolysed by water or dilute acids.

Lattice Energy: Energy change when one mole of crystalline lattice is formed from gaseous ions
eg. 2Al3+ + 3O2– —Al2O3 + L.E.

Hydration Energy: It is the energy change when gaseous ions form aqueous ions.
eg. Na+ + aq. —Na+ + H.E. of Na+
SO42- + aq. —SO42- + H.E. of SO42–
Solutions in liquid NH3
Alkali metals dissolve in liquid ammonia (high conc. 3 M) and give blue solution which is conducting,
reducing and paramagnetic in nature.
Reason
On dissolving Metal in NH3
M(s) + NH3()  M+(NH3) + e–(NH3)
M+ + x (NH3)  [M(NH3)x]+  Ammoniated cation
e– + y (NH3)  [e(NH3)y]–  Ammoniated electron
The blue colour is due to  Ammoniated electron
The paramagnetic nature is due to  Ammoniated electron
The conducting nature is due to  Ammoniated M+ + Ammoniated electron
On standing the colour fades due to formation of amide after liberating hydrogen.
M+ + e– + NH3  MNH2(amide) + H2(g)
In the absence of impurities like. Fe, Pt, Zn etc, the solutions are stable.
In concentrated solution, the blue colour changes to bronze colour and diamagnetic due to the
formation of metal clusters and ammoniated electrons also associate to form electron pairs
2e– (NH3)y  [ e–(NH3)y]2
Solutions are of much lower density than the pure solvent, i.e., they occupy for greater volume than that
expected from the sum of the volumes of metal and solvent
Peroxide and superoxides of Na & K are widely used as oxidising agent and air purifiers in space
capsules, submarines and breathing mask.

Alkaline metal in liq. NH3

Like alkali metals, the alkaline earth metals dissolve in liquid ammonia to give deep blue black solutions forming
ammoniated ions
M + (x + y)NH3 — [M(NH3)x]2+ + 2[e(NH3)y]–
From these solutions, the hexa-ammoniates [M(NH3)6]2+ can be recovered.

Uses of alkali metal

(1) Lithium metal is used to make useful alloys,
 with lead to make ‘white metal’ bearings for motor engines.
 with aluminium to make aircraft parts.
 with magnesium to make armour plates.
(2) It is used in thermonuclear reactions.
(3) Lithium is also used to make electrochemical cells.
(4) Sodium is used to make a Na/Pb alloy needed to make PbEt4 and PbMe4. These organolead
compounds were earlier used as anti-knock additives to petrol, but nowadays vehicles use
lead-free petrol.
(5) Liquid sodium metal is used as a coolant in fast breeder nuclear reactors.
(6) Potassium chloride is used as a fertilizer.
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s-Block Elements
(7) Potassium hydroxide is used in the manufacture of soft soap. It is also used as an excellent
absorbent of carbon dioxide.
(8) Caesium is used in devising photoelectric cells.

Uses of alkaline metal

(1) Beryllium is used in the manufacture of alloys.
(2) Copper-beryllium alloys are used in the preparation of high strength springs.
(3) Metallic beryllium is used for making windows of X-ray tubes.
(4) Magnesium forms alloys with aluminium, zinc, manganese and tin. Magnesium-aluminium alloys
being light in mass are used in air-craft construction.
(5) Magnesium (powder and ribbon) is used in flash powders and bulbs, incendiary bombs and signals.
(6) A suspension of magnesium hydroxide in water (called milk of magnesia) is used as antacid in
medicine.
(7) Magnesium carbonate is an ingredient of toothpaste.
(8) Calcium is used in the extraction of metals from oxides which are difficult to reduce with carbon.
(9) Calcium and barium metals, owing to their reactivity with oxygen and nitrogen at elevated
temperatures, have often been used to remove air from vacuum tubes.
(10) Radium salts are used in radiotherapy, for example, in the treatment of cancer.
Biological Importance Of Sodium And Potassium:
 Sodium ions are found primarily on the outside of cells, being located in blood plasma and in the
interstitial fluid which surrounds the cells. These ions participiate in the transmission of nerve
signals, in regulating the flow of water across cell membranes and in the tranasport of sugars
and amino acids into cells. Sodium and potassium, although so similar chemically, differ
quantitatively in their ability to penetrate cell membranes, in their transport mechanisms and in
their efficiency to activate enzymes. Thus, potassium ions are the most aundant cations within
cell fluids, where they activate many enzymes, participate in the oxidation of glucose to produce
ATP and, with sodium, are responsible for the transmission of nerve signals
A typical 70 kg man contains about 90 g of Na and 170 g of K compared with only 5 g of iron and
0.06 g of copper.
Biological Importance of Magnesium and Calcium :
 Monovalent sodium and potassium ions and divalent magnesium and calcium ions are found in
large proportions in biological fluids. These ions perform important biological functions such
as maintenance of ion balance and nerve impulse conduction.
 All enzymes that utilise ATP in phosphate transfer require magnesium as the cofactor. The main
pigment for the absorption of light in plants is chlorophyll which contains magnesium. About 99
% of body calcium is present in bones and teeth. It also plays important roles in neuromuscular
function, interneuronal transmission, cell membrane integrity and blood coagulation.
 The calcium concentration in plasma is regulated at about 100 mgL–1. It is maintained by two
hormones : calcitonin and parathyroid hormone. Do you know that bone is not an inert and
unchanging substance but is continuously being solubilised and redeposited to the extent of 400
mg per day in man? All this calcium passes through the plasma.
An adult body contains about 25 g of Mg and 1200 g of Ca compared with only 5 g of iron and 0.06 g
of copper. The daily requirement in the human body has been estimated to be 200–300 mg.
ANOMALOUS PROPERTIES OF LITHIUM
The anomalous behavior of lithium is due to the :
(i) Exceptionally small size of its atom and ion,
(ii) High polarising power (i.e., charge/ radius ratio).
As a result, there is increased covalent character of lithium compound which is responsible for their
solubility in organic solvent. Further, lithium shows diagonal relationship to magnesium.
S.No. Property Li
1. Hardness Li is much harder.
2. M.P and B.P Higher M.P and B.P
3. Reactivity Less reactive
4. Reducing agent Strong
5. Combustion in air Li form monoxide (Li2O) and nitride (Li3N) ; not for other.
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s-Block Elements
Favored for Li; not for other.
6. Hydration of ion Li has maximum degree of hydration for this reason. Lithium salts are mostly
hydrated. E.g LiCl.2H2O.
Hydrogen Li is not obtained in the solid form while all other elements form solid hydrogen
7.
Carbonate carbonates.
8. Ethynide Favored for Li ; not for other.
4LiNO

3  2Li2O + 4NO2 + O2
Lithium Oxide
9. Lithium nitrate
Where as other alkali metal nitrates decompose to give the corresponding nitrite.

2NaNO3  2NaNO2 + O2
Sodium nitrite
These are much less soluble in water. Solubility in water is less than the
10. LiF and Li2O
corresponding compounds of other alkali metal.
11. Carbide Li reacts directly with carbon to form anionic carbide.
Lithium hydroxide is less basic Li 2CO3, LiNO3 and LiOH all form the oxides on gentle
12. Hydroxide
heating.
13. Carbonate Less stable.
14. Nitrite Less stable.
Lithium forms a bicarbonates in solution it does not form a solid bicarbonate.
15. Bicarbonate
Where as the other all forms stable solid bicarbonates.
Complex ion Lithium has a great tendency to form. Complexes not for other. Due to small size of
16.
formation Lithium.
17. Reaction with NH3 Li when heated in NH3 imide (Li2NH) while other alkali metals form amides (MNH2)

Points of Similarities between Lithium and Magnesium

The similarity between lithium and magnesium is particularly striking and arises because of their
similar size: atomic radii, Li = 152 pm, Mg = 160 pm; ionic radii : Li+ = 76 pm, Mg2+ = 72 pm. The main
points of similarity are :
S.No. Properties Li and Mg
1. Hardness Li and Mg are much harder.
2. Density These are lighter than other elements in the respective group.
3. Reaction with water Both react slowly with water.
4. Solubility of hydroxide and oxide Less soluble and their hydroxides decompose in acid on heating.
5. Reaction with N2 By direct combination with nitrogen both form a nitride Li 3N and Mg3N2.
The oxides Li2O and MgO donot combine with excess oxygen to give
6. Oxides
any superoxide.
Carbonates of both decompose easily on heating to form the oxides
7. Carbonates
and CO2. Solid hydrogen carbonates are not formed by Li and Mg.
8. Solubility of halides in ethanol Both LiCl and MgCl2 are soluble.
Both LiCl and MgCl2 are deliquescent and crystallise from aqueous
9. Hydration of ion
solution as hydrates, LiCl.H2O and MgCl2.6H2O.

Anamolous Behaviour of Beryllium

The properties of beryllium the first member of the alkaline earth metal, differ from the rest of the
member. Its is mainly because of
(i) Its small size and high polarizing power.
(ii) Relatively high electro negativity and ionization energy as compared to other members.
(iii) Absence of vacant d–orbitals in its valence shell.
Some important points of difference between beryllium and other members (especially magnesium) are
given below.

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s-Block Elements
S.No. Properties
1. Hardness Be is harder than other members of its group
2. Density Be is lighter than Mg
3. M.P. and B.P. Higher then other members of its group.
4. Reaction with water Be does not react with water while Mg reacts with boiling water.
5. Nature of oxides BeO is amphoteric while MgO is weakly basic.
6. Nature of compounds Be forms covalent compounds whereas other members form ionic compounds.
Beryllium carbide reacts with water to give methane whereas carbides of other
alkaline earth metals gives acetylene gas.
7. Carbide Be2C + 4H2O → 2Be (OH)2 + CH4
MgC2 + 2H2O → Mg (OH)2 + C2H2
CaC2 + 2H2O → Ca (OH)2 + C2H2
The beryllium hydride is electron deficient and polymeric, with muti center
8. Hydride
bonding like aluminium hydride.
Beryllium does not exhibit coordination number more than four as it has four
9. Co-ordination number orbitals in the valence shell. The other members of this group has coordination
number 6.
Be dissolves in alkalies with evolution of hydrogen
Be + 2NaOH +2H2O→ Na2BeO2.2H2O + H2
10. Reaction with Alkali
(sodium beryllate
Other alkaline earth metals don't react with alkalies.
Resemblance of Beryllium with Aluminium (Diagonal relationship)
The following points illustrate the anomalous behaviour of Be and its resemblance with Al.
S.No. Properties Be and Al
1. Nature of compounds Unlike groups-2 elements but like aluminium, beryllium forms covalent compounds.
2. Nature of hydroxide The hydroxides of Be, [Be(OH)2] and aluminium [Al(OH)3] are amphoteric in
nature, whereas those of other elements of group – 2 are basic in nature.
3. Nature of oxide The oxides of both Be and Al i.e. BeO and Al2O3 are high melting insoluble solids.
4. Polymeric structure BeCl2 and AlCl3 have bridged chloride polymeric structure.

5. Salts The salts of beryllium as well as aluminium are extensively hydrolysed.

6. Carbides Carbides of both the metal reacts with water liberating methane gas.
Be2C + 4H2O → 2Be (OH)2 + CH4
Al4C3 + 12H2O → 4Al(OH)3 + 3CH4
7. Oxides and The oxides and hydroxides of both Be and Al are amphoteric and dissolve in
hydroxides sodium hydroxide as well as in hydrochloric acid.
BeO + 2HCl → BeCl2 + H2O
BeO + 2NaOH → Na2BeO2 + H2O
Al2O3 + 6HCl → 2AlCl3 + H2O
Al2O3 + 2NaOH → 2NaAlO2 + H2O
8. Reaction with acids Like Al, Be is not readily attacked by acids because of the presence of an oxide
film.

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s-Block Elements
Section (E) : Oxides, Peroxides, Super Oxides, Hydroxides
GROUP -I & II OXIDES
1. Sodium Oxide (Na2O)
Preparation:
Name of compound Name and Brief about the process Related chemical reaction
(1) By burning sodium at 180°C in a limited supply of air or 1
2Na + O2 180º
  Na2O
oxygen and distilling off the excess of sodium in vacuum. 2
Na2O2 + 2Na —2Na2O
Sodium Oxide (2) By heating sodium peroxide, nitrate or nitrite with
(Na2O) sodium. 2NaNO3 + 10Na —6Na2O + N2
2NaNO2 + 6Na —4Na2O + N2
(3) Sodium oxide is formed when the mixture of sodium 3NaN3 + NaNO2  2Na2O + 5N2
azide and sodium nitrite is heated.

Chemical Properties:
(1) It is white amorphous substance.
(2) It dissolve violently in water, yielding caustic soda (NaOH) and evolving a large amount of heat.
Na2O + H2O  2NaOH
Uses : It is used as dehydrating and polymerising agent in organic chemistry.
2. Sodium Peroxide (Na2O2)
Preparation
Name of compound Name and Brief about the process Related chemical reaction
(1) By heating the metal in excess of air or 2Na + O2 (excess) 
300ºC
 Na2O2
oxygen at 300°, which is free from
Sodium Peroxides moisture and CO2.
(Na2O2) (2) Industrial method : 2Na + O2 — Na2O
It is a two stage reaction in the presence
Na2O + O2 — Na2O2
of excess air.
Properties:
(1) It is a pale yellow solid (when impure), becoming white in air from the formation of a film of NaOH and
Na2CO3.
(2) In cold water (~0°C) produces H2O2 but at room temperature produces O2. In ice-cold mineral acids
also produces H2O2.
Na2O2 + 2H2O ~0ºC
 2NaOH + H2O2
2Na2O2 + 2H2O 
25ºC
 4NaOH + O2
Na2O2 + H2SO4 
~0ºC
 Na2SO4 + H2O2
2Na2O2 + H2SO4  25ºC
 2Na2SO4 + 2H2O + O2
(3) It reacts with CO2, giving sodium carbonate and oxygen and hence its use for purifying air in a confined
space e.g. submarine, ill-ventilated room.
2Na2O2 + 2CO2 — 2Na2CO3 + O2
Na2O2 + CO — Na2CO3
(4) It is an oxidising agent and oxidises charcoal, CO, NH 3, SO2.
3Na2O2 + 2C — 2Na2CO3 + 2Na [deposition of metallic Na]
CO + Na2O2 — Na2CO3
SO2 + Na2O2 — Na2SO4
2NH3 + 3Na2O2 — 6NaOH + N2
(5) Sulphides are oxidised to corresponding sulphates
Na2O2 — Na2O + [O] ; Na2S + 4[O] —Na2SO4
(6) Na2O2 — Na2O + [O]; 2Al + 3[O] —Al2O3; Al2O3 + Na2O — 2NaAlO2.
Uses :
(1) For preparing H2O2, O2.
(2) Oxygenating the air in submarines.
(3) Oxidising agent in the laboratory.
Oxides of Potassium K2O K2O2 K2O3* KO2 KO3
Colours White White Red Bright Yellow Orange Red Solid
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s-Block Elements
3. Potassium sesquioxide (need not memorize).
Preparation:
Name of compound Name and Brief about the process Related chemical reaction
2KNO3 + 10K h  6K2O + N2
eatin g

K2O
(Potassium oxide)
By heating
potassium.
potassium nitrate with
K2O  K2O
h eatin g

(White) (Yellow)
K2O + H2O — 2KOH

Name of compound Name and Brief about the process Related chemical reaction
K2O2 By burning potassium at 300°C in a 2K + O2   K2O2
Co n tro l l e d
(Potassium peroxide) limited supply of air or oxygen. a i ra t 3 0 0ºC

Name of compound Name and Brief about the Related chemical reaction
process
Pas O2
(i) Passage of O2 through a blue K in liq. NH3  K2O2 — K2O3 — KO2
solution of K in liquid NH3 yields white red yellow
0C
oxides K2O2 (white), K2O3 (red) 2KO2 + 2H2O ~  2KOH + H2O2 + O2
KO2 and KO2 (deep yellow) i.e KO2
(Potassium superoxide) reacts with H2O and produces
H2O2 and O2 both.
(ii) It is prepared by burning K + O2 —KO2
potassium in excess of oxygen
free from moisture.

Name of compound Name and Brief about the process Related chemical reaction
It is obtained when oxygen is
K2O3 3O2
passed through liquid ammonia 4K (dissolved in liquid NH3)   2K2O3
(Potassium sesquioxide)
containing potassium.

Name and Brief

Name of compound Related chemical reaction
About the process
10 to15C
KO3
From KOH KOH + O3 (ozonised oxygen)  KO3
(Potassium ozonide) (Dry powdered) (orange solid)

Properties of Potassium superoxide (KO2)

It is a orange coloured (chrome yellow) powder and reacts with water according to following reaction.
2KO2 + 2H2O — 2KOH + H2O2 + O2
It reacts directly with CO and CO2.
2KO2 + CO — K2CO3 + O2 ; 2KO2 + CO2 —K2CO3 + O2
If more CO2, in presence of moisture is present; then
4KO2 + 4CO2 + 2H2O — 4KHCO3 + 3O2
On heating with sulphur, it forms potassium sulphate
2KO2 + S — K2SO4
Uses : It is used as an oxidising agent and air purifier in space capsules, submarine and breathing mask as it
produces O2 and removes CO2.

4. Magnesium Oxide (MgO):

Name of compound Name and Brief about the process Related chemical reaction
It is also called magnesia and obtained by
Magnesium Oxide (MgO) MgCO3 —MgO + CO2
heating natural magnesite.

Properties :
(1) It is white powder.
(2) It's m.p. is 2850°C. Hence used in manufacture of refractory bricks for furances. And it is acts as basic
flux and facilitates the removal of acidic impurities of Si, P and S from steel through slag formation.
(3) It is very slightly soluble in water imparting alkaline reaction.

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s-Block Elements
5. Calcium Oxide (CaO):
Preparation
Name of compound Name and Brief about the process Related chemical reaction
It is commonly called as quick lime or lime
Calcium Oxide
and made by decomposing lime stone at a CaCO3 CaO + CO2
(CaO)
high temperature about 1000°C.
* The Carbon dioxide is removed as soon as it is produced to enable the reaction to proceed to completion.

Chemical Properties :
(1) It is white amorphous powder of m.p. 2570°C. On exposure to atmosphere; it absorbs moisture and
carbondioxide.
CaO + H2O — Ca(OH) 2 ; CaO + CO2 — CaCO3
acidic oxide
(2) It emits intense light (lime light), when heated in oxygen-hydrogen flame.
(3) It combines with limited amount of water to produce slaked lime. This process is called slaking of lime.
Quick lime slaked with soda gives solid sodalime (CaO). Being a basic oxide.
CaO + H2O — Ca(OH) 2
(4) Soda lime (basic oxide) combines with some acidic oxides at high temperature.
CaO + SiO2 — CaSiO3
6CaO + P4O10 — 2Ca3(PO4)2
Uses :
(i) It is an important primary material for manufacturing cement and is the cheapest form of alkali.
(ii) It is used in the manufacture of sodium carbonate from caustic soda.
(iii) It is employed in the purification of sugar and in the manufacture of dye stuffs.
Magnesium Peroxide (MgO2) and Calcium Peroxide (CaO2)
These are obtained by passing H2O2 in a suspension of Mg(OH) 2 and Ca(OH) 2.
Uses : MgO2 is used as an antiseptic in tooth paste and as a bleaching agent.
HYDROXIDES
1. Sodium Hydroxides(Caustic Soda) NaOH (White) :
Preparation :
Name of Name and Brief about the
Related chemical reaction
compound process
Cathode:Na++ e– 
Hg
 Na-amalgam
1
Anode : Cl–  Cl2 + e–
2
2Na-amalgam + 2H2O 2NaOH + 2Hg + H2
(1) Electrolysis of Brine :
Sodium hydroxide is prepared by
the electrolysis of sodium
chloride in Castner-Kellner cell. A
brine solution is electrolysed
using a mercury cathode and a
carbon anode. Sodium metal
Sodium discharged at the cathode
Hydroxides combines with mercury to form
(NaOH) sodium amalgam. Chlorine gas is
evolved at the anode.
The amalgam is treated with
water to give sodium hydroxide
and hydrogen gas.

Anode Cathode

(2) By Diaphragm cell

Na2CO3 + Ca(OH)2  2NaOH + CaCO3

(3) Caustication of Na2CO3
(suspension)
(Gossage's method)
Since the Ksp(CaCO3) < Ksp(Ca(OH)2), the reaction shifts towards right.
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s-Block Elements
Properties:
(1) Sodium hydroxide is a white, translucent solid. It melts at 591 K. It is readily soluble in water to give a
strong alkaline solution. Crystals of sodium hydroxide are deliquescent. The sodium hydroxide solution
at the surface reacts with the CO2 in the atmosphere to form Na2CO3.
(2) It is white crystalline, deliquescent, highly corrosive solid.
(3) It is stable towards heat.
(4) It's aqueous solution alkaline in nature and soapy in touch.
(5) NH4Cl + NaOH  NaCl + NH3  + H2O
FeCl3 + 3NaOH  Fe(OH)3  + 3NaCl
Brown ppt
ZnCl2 + 2NaOH  Zn(OH)2 + 2NaCl
Zn(OH)2 + 2NaOH Na2ZnO2 + 2H2O [Same with AlCl3, SnCl2, PbCl2]
soluble
(6) Acidic and amphoteric oxides gets dissolved easily e.g.
CO2 + 2NaOH  Na2CO3 + H2O
Al2O3 + 2NaOH  2NaAlO2 + H2O
(7) Aluminium and Zn metal gives H2 from NaOH.
2Al + 2NaOH + 2H2O  3H2 + 2NaAlO2
(8) Several non metals such as P, S, Cl etc. yield a hydride instead of hydrogen.e.g.
4P + 3NaOH + 3H2O  PH3 + 3NaH2PO2 (Disproportionation reaction)
(9) NaOH is stable towards heat but reduced to metal when heated with carbon.
2NaOH + C  2Na + 2 CO + H2
+2
Na2MO2
+3
(10). NaOH + Metal Oxide (M) Na3MO3
+4
Na2MO3
Above are general reactions of NaOH with metal oxides having metal's Oxidation number +2, +3 & +4
respectively.
Uses : It is used in
(i) The manufacture of soap, paper, artificial silk and a number of chemicals.
(ii) In petroleum refining.
(iii) In the purification of bauxite.
(iv) In the textile industries for mercerising cotton fabrics.
(v) For the preparation of pure fats and oils .
(vi) As a laboratory reagent.
2. Potassium Hydroxide (KOH):
Preparation:
(1) It is prepared by electrolysis of KCl solution.
(2) KOH resembles NaOH in all its reactions. However KOH is much more soluble in alcohol. This
accounts for the use of alcoholic KOH in organic chemistry.
(3) KOH is called caustic potash, because of their corrosive properties (for example on glass or on skin)
and its aqueous solution is known as potash lye.
2KOH + 4NO  2KNO2 + N2O + H2O
4KOH + 6NO  4KNO2 + N2 + 2H2O
(4) It is used for the absorption of gases like CO2, SO2, etc. It is used for making soft soaps.
Properties: Same as NaOH
(1) It is stronger base compared to NaOH.
(2) Solubility in water is more compared to NaOH.
(3) In alcohol, NaOH is sparingly soluble but KOH is highly soluble.
(4) As a reagent KOH is less frequently used but in absorption of CO 2, KOH is preferably used compared
to NaOH. Because KHCO3 formed is soluble whereas NaHCO3 is insoluble and may therefore choke
the tubes of apparatus used.

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s-Block Elements

+Na2SO4

Na2PbO2

Na2PbO3

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s-Block Elements
3. Magnesium Hydroxide (Mg(OH)2):
It occurs in nature as the mineral brucite.
Preparation:
Name and Brief about the
Name of compound Related chemical reaction
process
It can be prepared by adding MgSO4 + 2NaOH  Mg(OH)2 + Na2SO4
Magnesium
caustic soda solution to a MgCl2 + 2NaOH  Mg(OH)2 + 2NaCl
Hydroxide
(Mg(OH)2)
solution of Magnesium sulphate MgCl2 + Ca(OH)2  Mg(OH)2 + CaCl2
or chloride solution. MgO + H2O  Mg(OH)2

Chemical Properties:
(1) It can be dried at temperature upto 100°C only otherwise it breaks into its oxide at higher temperature.
Mg(OH)2  MgO + H2O
(2) It is slightly soluble in water imparting alkalinity.
(3) It dissolves in NH4Cl solution.
Mg(OH)2 + 2NH4Cl MgCl2 + 2NH3.H2O
Thus, Mg(OH)2 is not therefore precipitated from a solution of Mg2+ ions by NH3.H2O. in presence of
excess of NH4Cl.
Uses : A suspension of Mg(OH)2 in water is used in medicine as an antacid (An antacid is substance which
neutralizes stomach acidity) under the name, milk of magnesia.
4. Calcium Hydroxide (Ca(OH)2) (White Powder):
Preparation :
Name of compound Name and Brief about the process Related chemical reaction
Calcium Hydroxide (Ca(OH)2) By spraying water on quicklime. CaO + H2O  Ca(OH)2

Properties:
(1) It is a white amorphous powder.
(2) It is sparingly soluble in water.
(3) It's solubility in hot water is less than that of cold water. Hence solubility decreases with increase in
temperature.
(4) The aqueous solution is known as lime water and a suspension of slaked lime in water is known as milk
of lime.
(5) When carbon dioxide is passed through lime water it turns milky due to the formation of calcium
carbonate.
Ca(OH)2 + CO2  CaCO3 + H2O
On passing excess of carbon dioxide, the precipitate dissolves to form calcium hydrogen carbonate.
CaCO3 + CO2 + H2O  Ca(HCO3)2
Milk of lime reacts with chlorine to form hypochlorite, a constituent of bleaching powder.
2Ca(OH) 2 + 2Cl2  CaCl2 + Ca(OCl)2 + H2O
Bleaching powder
Uses:
(i) It is used in the preparation of mortar, a building material.
(ii) It is used in white wash due to its disinfectant nature.
(iii) It is used in glass making, in tanning industry, for the preparation of bleaching powder and for
purification of sugar.

Section (F) : Carbonates, Bicarbonates

CARBONATES
1. Sodium Carbonate (Washing soda) Na2CO3.10H2O (White Solid) :

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s-Block Elements
Preparation:
1.
Name of
Name and Brief about the process Related chemical reaction
compound
NaCl + H2SO4(conc.)   NaHSO4 + HCl
m i l dheati ng

NaCl + NaHSO4    Na2SO4 + HCl

S t ro n g l y
h e a te d
(1) Leblanc Process
(Salt Cake)
Na2SO4 + 4C  Na2S + 4CO 
Na2S + CaCO3 Na2CO3 + CaS
(2) Solvay Process
Step-1 (In ammonia absorber) 2NH3 + CO2 + H2O  (NH4)2CO3
(i)Saturation of brine with ammonia and CO2 CaCl2 + (NH4)2CO3  CaCO3 + 2NH4Cl
(ii) Ammoniated brine is filtered to remove MgCl2 + (NH4)2CO3  MgCO3 + 2NH4Cl
calcium and magnesium impurities as their
Sodium insoluble carbonates.
Carbonate Step-2 (In carbonation tower) :
(Washing soda) (i) Formation of insoluble NaHCO3
Na2CO3.10H2O (ii) Reaction is exothermic and hence there is a
cooling arrangement. NH3 + CO2 + H2O  NH4HCO3 ;
(iii) NaHCO3 is insoluble in
cold brine solution because NH4HCO3 + NaCl  NaHCO3 + NH4Cl
30ºC

of the common ion effect. It is separated by

filtration and the filtered is used for recovering
NH3 & CO2.
Step-3 (Calcination to get sodium carbonate) : 2 NaHCO3  Na2CO3 + CO2 + H2O
150ºC

 / Steam
Step - 4 (In recovery tower) : NH4 HCO3  NH2 + CO2 + H2O
Recovery of ammonia and carbondioxide.  / Steam
2NH4 Cl +Ca(OH)2   2NH3+ 2H2O+CaCl2
CaCl2 is obtained as by product.
* advantage is taken of low solubility of NaHCO3, it gets precipitated in the reaction of NaCl + NH4HCO3.
2. Naturally from trona
heat
2(Na2CO3.NaHCO3.2H2O) 
 3Na2CO3 + CO2 + 5H2O

Properties
(1) Anhydrous Na2CO3 is called as soda ash, which does not decompose on heating but melts at 852°C.
(2) Sodium carbonate is a white crystalline solid which exists as a decahydrate, Na 2CO3·10H2O. This is
also called washing soda. It is readily soluble in water. On heating, the decahydrate loses its water of
crystallisation to form monohydrate. Above 373K, the monohydrate becomes completely anhydrous and
changes to a white powder called soda ash.
Na2CO3·10H2O   Na2CO3·H2O + 9H2O
375K

373K
Na2CO3·H2O   Na2CO3 + H2O
(soda ash)
Carbonate part of sodium carbonate gets hydrolysed by water to form an alkaline solution.
Na2CO3 + H2O  H2CO3 (weak acid) + NaOH (strong)
(3) Na2CO3 absorbs CO2 yielding sparingly soluble sodium bicarbonate which can be calcined at 250° to
get pure sodium carbonate.
Na2CO3 + H2O + CO2 2NaHCO3 (solid)
(4) It dissolved in acid with effervescence of CO2 and causticised by lime to give caustic soda.
Na2CO3 + HCl  2NaCl + H2O + CO2
Na2CO3 + Ca(OH)2  2NaOH + CaCO3
Uses :
(i) It is used in water softening, laundering and cleaning.
(ii) It is used in the manufacture of glass, soap, borax and caustic soda.
(iii) It is used in paper, paints and textile industries.
(iv) It is an important laboratory reagent both in qualitative and quantitative analysis.

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s-Block Elements
2. Potassium Carbonate (K2CO3)
Name of compound Name and Brief about the process Related chemical reaction

KCl + H2SO4 (conc.)   KHSO4 + HCl

m i l dheati ng
By leblance process, it can be
Potassium KCl + KHSO4    K2SO4 + HCl
S t ro n g l y
prepared but by solvay process it
Carbonate h e a te d
cannot be prepared because KHCO3
(K2CO3)
is soluble in water. K2SO4 + 4C  K2S + 4CO 
K2S + CaCO3 K2CO3 + CaS

Properties:
It resembles with Na2CO3, m.p. is 900°C but a mixture of Na2CO3 and K2CO3 melts at 712°C.
Uses It is used in glass manufacturing.
* need not memories.

Note : Calcium carbonate and Magnesium carbonate found in nature.

Calcium bicarbonate and Magnesium bicarbonate are present in temporary hardness of water.
Unstable and unimportant. Same for KHCO3.

Section (G) : Chlorides, Sulphates

CHLORIDES
Sodium Chloride (NaCl) and Potassium Chloride, Calcium Chloride
Preparation:
NaCl : Found in nature as rock salt or in sea water.
KCl : Found in nature as sylvine (KCl) or carnallite (2KCl.MgCl2.6H2O)
CaCl2 : Obtained as byproduct in Solvay’s process.
Properties of NaCl :
(1) It is nonhygroscopic but the presence of MgCl2 in common salt renders it hygroscopic.
(2) It is used to prepare freezing mixture in laboratory [Ice-common salt mixture is called freezing mixture
and temperature goes down to –23°C.]
(3) For melting ice and snow on road.
Uses of NaCl:
(i) It is used as a common salt or table salt for domestic purpose.
(ii) It is used for the preparation of Na2O2, NaOH and Na2CO3.
Magnesium Chloride (MgCl2)
It occurs in nature as mineral carnallite, KCl.MgCl2.6H2O.
Preparation : By Dow’s Processes (Natural Brine process and Dolomite process). See Metallurgy,
stdXII.
Properties:
(1) It crystallises as hexahydrate. MgCl2. 6H2O
(2) It is deliquescent solid.
(3) This hydrate undergoes hydrolysis as follows:
MgCl2·6H2O  Mg(OH)Cl + HCl + 5H2O
Mg(OH)Cl  MgO + HCl
Hence, Anh. MgCl2 cannot be prepared by heating this hydrate. Because of this formation of HCl. Sea
water cannot be used in marine boilers which corrodes the iron body.
(4) Anhydrous MgCl2 can be prepared by heating a double salt like. MgCl2.NH4Cl.6H2O as follows:
–H O
MgCl2 . NH4Cl . 6H2O  2 MgCl2. NH4Cl   MgCl2 + NH3 + HCl
s t ro n g
 

(5) It is a colourless crystalline solid, highly deliquescent and highly soluble in water.
(6) Sorel Cement is a mixture of MgO and MgCl2 (paste like) which set to hard mass on standing. This is
used in dental filling, flooring etc.
(7) Anh. CaCl2 is used in drying gases and organic compounds but not NH 3 or alcohol due to the formation
of CaCl2.8NH3 and CaCl2.4C2H5OH.

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s-Block Elements

SULPHATES
1. Potassium Sulphate (K2SO4)
It occurs in stassfurt potash beds as schonite K2SO4.MgSO4.6H2O and Kainite, KCl.MgSO4.3H2O from
which it is obtained by solution in water and crystallisation. It separates from the solution as anh,
crystals whereas Na2SO4 comes as decahydrate.
Preparation:
(1) It is prepared by the reaction of potassium chloride or hydroxide with concentrated. H 2SO4.
2KCl + H2SO4  K2SO4 + 2HCl ; 2KOH + H2SO4  K2SO4 + 2H2O
(2) K2SO4.MgSO4.6H2O + 2KCl  2K2SO4 + MgCl2+ 6H2O
Uses : It is used to prepare alum.
It is a white crystalline solid and soluble in water.
It is used as a fertilizer for tobacco and wheat.


KCl + H2 SO4

Reactions Charts
2. Magnesium Sulphate (MgSO4):
It occurs in nature as minerals kiesserite (MgSO4.H2O), epsom salt (MgSO4.7H2O) and kainite
(KCl.MgSO4.3H2O).
Preparation:
(1) It is obtained by dissolving kieserite. MgSO4.H2O in boiling water and then crystallising the solution as a
hepta hydrate. i.e. MgSO4.7H2O. It is called as Epsom salt.
(2) It is also obtained by dissolving magnesite in hot dil. H 2SO4.
MgCO3 + H2SO4  MgSO4 + H2O + CO2
(3) By dissolving dolomite (CaCO3.MgCO3) in hot dil. H2SO4 and removing the insoluble CaSO4 by
filtration.
CaCO3.MgCO3 (dolomite) + 2H2SO4  MgSO4 + CaSO4 + 2CO2 + 2H2O
(4) It is isomorphous with FeSO4.7H2O, ZnSO4.7H2O.
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s-Block Elements
Chemical Properties:
Heating effect:
(1) When heated to 150°C, it changes to monohydrate. On further heating, it becomes anhydrous at
200°C. On strong heating, it decomposes into MgO.
strong
MgSO4.7H2O
150C
 MgSO4.H2O
200C
 MgSO4  MgO + SO2 + O2.
heating
(2) Magnesium sulphate when heated with lamp black at 800°C produces SO2 and CO2 gases.
2MgSO4 + C  2MgO + 2SO2 + CO2
(3) It forms double salts with alkali metal sulphates, e.g., K2SO4.MgSO4.6H2O.
3. Calcium Sulphate (Plaster of paris) CaSO4.½ H2O
It occurs as anhydrite CaSO4, hemihydrate CaSO4.½H2O and as the dihydrate (CaSO4.2H2O) gypsum,
alabaster or satin-spar.
Preparation:
(1) It is a hemihydrate of calcium sulphate. It is obtained when gypsum, CaSO 4·2H2O, is heated to 393 K.
2(CaSO4.2H2O)   2(CaSO4).H2O + 3H2O
393 K

Above 393 K, no water of crystallisation is left and anhydrous calcium sulphate, CaSO4 is formed. This
is known as ‘dead burnt plaster’.
It has a remarkable property of setting with water. On mixing with an adequate quantity of water it forms
a plastic mass that gets into a hard solid in 5 to 15 minutes.
(2) It can be prepared by reacting any calcium salt with either sulphuric acid or a soluble sulphate.
CaCl2 + H2SO4  CaSO4 + 2HCl ; CaCl2 + Na2SO4  CaSO4 + 2NaCl
Properties:
It is a white crystalline solid. It is sparingly soluble in water and solubility decreases as temperature
increases.
It dissolves in dilute acids. It also dissolves in ammonium sulphate due to the formation of double
sulphate, (NH4)2SO4.CaSO4.H2O.
The setting process is exothermic. The process of setting takes place in stages. In the first stage, there
is conversion of Plaster of Pairs into orthorhombic form of gypsum (setting step) and in the second
stage orthorhombic form changes into monoclinic form (hardening step).
The setting of Plaster of Paris may be catalysed by sodium chloride while it is retarded by borax or
alum. Addition of alum to Plaster of Paris makes the setting very hard. The mixture is known as
Keene’s cement.

Dead plaster has no setting property as it takes up water only very slowly.
A suspension of gypsum when saturated with ammonia and carbon dioxide forms ammonium sulphate,
a nitrogenous fertilizer.
2NH3 + CaSO4 + CO2 + H2O  (NH4)2 SO4 + CaCO3
When strongly heated with carbon, it forms calcium sulphide.
CaSO4 + 4C CaS + 4CO
Uses: For preparing blackboard chalk.
In anhydrous form as drying agent.

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s-Block Elements
Fertilizer
1. Cynamide: It is an organic compound with the formula CN2H2. This white solid is widely used in
agriculture and the production of pharmaceuticals and other organic compounds.
Cyanamide is produced by hydrolysis of calcium cyanamide, which in turn is prepared from calcium
carbide via the frank-Caro process.
CaC2 + N2  CaCN2 + C ; frank-Caro process
CaCN2 + H2O + CO2  CaCO3 + H2NCN (Cynamide)
The main reaction exhibited by cyanamide involves additions of compounds containing an acidic
proton. Water, hydrogen sulfide, and hydrogen selenide react with cyanmide to give urea, thiourea, and
selenourea, respectively :
H2NCN + H2E H2NC (E) NH2 ; (E= O, S, Se)
2. Fluorapatite: It is a phosphate mineral with the formula Ca5 (PO4)3.
Cement
Cement is a product obtained by combining a material rich in lime, CaO with other material such as clay
which contains silica, SiO2 along with the oxides of aluminium, iron and magnesium.
The raw materials for the manufacture of cement are limestone and clay. When clay and lime are
strongly heated together they fuse and react to form cement clinker. This clinker is mixed with 2-3% by
weight of gypsum (CaSO4.2H2O) to form cement. Thus important ingredients present in Portland
cement are dicalcium silicate (Ca2SiO4) 26%, tricalcium silicate (Ca3SiO=5) 51% and tricalcium
aluminate (Ca3Al2O6) 11%.
Setting of cement : When mixed with water, the setting of cement takes place to give a hard mass.
This is due to the hydration of the molecules of the constituents and their rearrangement. The purpose
of adding gypsum is only to slow down the process of setting of the cement so that it gets sufficiently
hardened.
Uses : Cement has become a commodity of national necessity for any country next to iron and steel. It
is used in concrete and reinforced concrete, in plastering and in the construction of bridges, dams and
buildings.

Common Names
The names marked with asterisk (*) should be memorized with formulae. Others are given only for
reference. You need not memorize them.
Metal Ore name Formula
Lithium (Li) Spodumene LiAl(SiO3)2
Lepidolite KLi2Al(Al,Si)3O10(F,OH)2
Petalite LiAl(Si2O5)2
Sodium (Na) *Washing soda Na2CO3.10H2O
*Baking soda NaHCO3
*Sodium carbonate
Na2CO3
(soda ash/ washing soda)
*Sodium chloride
NaCl
(rock salt or halite)
*Sodium nitrate (Chile saltpeter) NaNO3
Salt cake Na2SO4
Fusion mixture Na2CO3 + K2CO3 (eq. molar mix.)
Sodium sesquicarbonate (trona) Na2CO3.NaHCO3.2H2O ( it is a double salt )
Na(NH4)HPO4.4H2O (it is obtained by mixing
*Microcosmic salt solutions of sodium phosphate and ammonium
phosphate or chloride)
Soda feldspar or sodium feldspar
Na2O. Al2O3. 6SiO2
(albite)
Potash feldspars or orthoclase or
K2O. Al2O3.6SiO2
microcline or Potassium feldspars
*Hypo Na2S2O3.5H2O
*Sodium aluminium fluoride
Na3AlF6
(cryolite)
*Borax (Tincal) Na2B4O7.10H2O
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s-Block Elements
Na2SO4.10H2O
(Sodium sulfate is the sodium salt of sulfuric
acid. When anhydrous, it is a white crystalline
*Sodium sulphate (glauber’s salt)
solid of formula Na2SO4 known as the mineral
thenardite; the decahydrate Na2SO4·10H2O is
known as Glauber's salt)
Sodium aluminium silicate
NaAlSi3O8
(Soda Feldspar)
Potassium (K) Sylvite KCl
Schonite K2SO4.MgSO4.6H2O
Kainite MgSO4.KCl.3H2O
*Carnallite MgCl2.KCl.6H2O
KNO3 (used especially as a fertilizer and
*Indian saltpetre (Nitre)
explosive)
Pearl ash K2CO3
Schonite K2SO4.MgSO4.6H2O( it is a double salt)
Langbeinite K2SO4.2MgSO4
Polyhalite K2SO4.MgSO4.2CaSO4.2H2O
*Potassium Alum K2SO4. Al2 (SO4)3. 24H2O
Alunite or Alumstone K2SO4. Al2 (SO4)3. 4Al(OH)3
Mica K2O. 3Al2O3. 6SiO2.2H2O
Feldspar KAlSi3O8(K2O.Al2O3.6SiO2)
Beryllium (Be) Beryl 3BeO. Al2O3 6SiO2
Chrysoberyl BeO.Al2O3
Phenacite BeSiO4
Bromalite BeO
*Baryta Ba(OH)2
Magnesium
*Magnesite MgCO3
(Mg)
*Dolomite MgCO3.CaCO3
*Epsom salt MgSO4.7H2O
Kieserite MgSO4.H2O
Asbestos CaMg3 (SiO3)4
Talc Mg(Si2O5)2 Mg (OH)2
Brucite Mg(OH)2
*Magnesia MgO
Artinite MgCO3.Mg(OH)2 .3H2O
*Sorel cement (magnesia cement) Mg4Cl2(OH)6(H2O)8
Calcium (Ca) *Quick lime CaO
*Slaked lime Ca(OH)2
*Hydrolith CaH2
*Calcium cynamide CaCN2 OR CaNCN
*Limestone (Marble / Whiting) CaCO3
Anhydrite CaSO4
*Gypsum CaSO4.2H2O
*Fluorspar or Fluorite CaF2
Phosphorite Ca3 (PO4)2
*Fluorapatite 3Ca3 (PO4)2.CaF2 OR Ca5(PO4)3F
*Plaster of paris CaSO2.½H2O
*Bleaching powder CaOCl2
*Rock phosphate Ca3 (PO4) 2
Wollastonite CaSiO2
Colmanite 2CaO.3Ba2O3.5H2O
Strontium(Sr) Strontianite SrCO3
Celestite SrSO4
Barytes or Heavy spar BaSO4
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s-Block Elements
Periodic Properties of s-Block
Properties Order

Thermal stability LiH > NaH > KH > RbH > CsH

Basic strength BeO < MgO < CaO < SrO

Basic Strength or Solubility in water or thermal
LiOH < NaOH < KOH < RbOH < CsOH
stability
Basic Strength or Solubility in water Be(OH)2<Mg(OH)2<Ca(OH)2<Ba(OH)2

Thermal stability Be(OH)2<Mg(OH)2<Ca(OH)2<Sr(OH)2< Ba(OH)2

Solubility in water or thermal stability Li2CO3 < Na2CO3 < K2CO3 < Rb2CO3 < Cs2CO3

Solubility in water BaCO3 < CaCO3 < MgCO3 < BeCO3

Thermal stability BeCO3 < MgCO3 < CaCO3 < BaCO3

Solubility in water BaSO4 < SrSO4 < CaSO4 < MgSO4 < BeSO4

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 p-Block Elements (B & C family)

Section (A), (B) & (C) : General facts about elements, Based on Periodic trends &
Based on Chemical Bonding
Introduction :
Group 13 to 18 of the periodic table of elements constitute the p–block. The p–block contains metals,
metalloids as well as non–metals.
Configuration ns2np1 ns2np2 ns2np3 ns2np4 ns2np5 ns2np6
He
B C N O F Ne
Al Si P S Cl Ar
Ga Ge As Se Br Kr
In Su Sb Te I Xe
Tl Pb Bi Po At Rn
Group Oxidation State +3 +4 +5 +6 +7 +8
Other Oxidation State +1 +2, –4 +3, –3 +4, +2, –2 +5, +3, +1, –1 +6, +4, +2
The p-block elements have general valence shell electronic configuration ns 2 np1–6.
The first member of each group from 13–17 of the p–block elements differ in many respects from the
other members of their respective groups because of small size, high electronegativity and absence of
d–orbitals.
The first member of a group also has greater ability to form p–p multiple bonds to itself (e.g. C=C,
CC, NN) and to element of second row (e.g C=O, C=N, CN, N=O) compared to the other members
of the same group.
The highest oxidation state of p–block element is equal to the group number –10. Moving down the
group, the oxidation state two less than the highest group oxidation state and becomes more stable in
groups 13 to 16 due to inert pair effect (reluctance of s-subshell electrons to participate in chemical
bonding)
Ge liquid expands when it forms the solid. This property is unique to Ga, Ge and Bi.

Catenation :
Carbon atoms have the tendency to link with one another through covalent bonds to form chains and
rings. This property is called catenation. This is because C–C bonds are very strong. Down the group
the size increases tendency to show catenation decreases. This can be clearly seen from bond
enthalpies values. The order of catenation is C >> Si > Ge  Sn. Lead does not show catenation. Due
to the property of catenation and p-p bonds formation, carbon is able to show allotropic forms.
Bond Bond enthalpy (kJ mol–1)
C—C 348
Si––Si 297
Ge—Ge 260
Sn—Sn 240

ANOMALOUS BEHAVIOUR OF CARBON :

Like first member of other groups, carbon also differs from rest of the members of its group. It is due to
its smaller size, higher electronegativity, higher ionisation enthalpy and unavailability of d orbitals.
Carbon accommodate only four pairs of electrons around it and thus this would limit the maximum
covalence to four whereas other members can expand their covalence due to the presence of d
orbitals, Carbon also has unique ability to form p-p multiple bonds with itself and with other atoms of
small size and high electronegativity. Few examples of multiple bonding are C=C, CC, C=O, C=S and
CN. Heavier elements do not form p-p bonds because their atomic orbital are too large and diffuse
to have effective overlapping.

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p-Block Elements (B & C family)
IMPORTANT TRENDS AND ANOMALOUS PROPERTIES OF BORON
The tri-chlorides, bromides and iodides of all these elements being covalent in nature are hydrolysed in
water. Species like tetrahedral [M(OH)4]– and octahedral [M(H2O)6]3+, except in boron, exist in aqueous
medium. It is due to the absence of d orbitals that the maximum covalence of boron is 4. Since the d-
orbitals are available with Al and other elements, the maximum covalence can be expected beyond 4.
GROUP 13 : BORON FAMILY & GROUP 14 : CARBON FAMILY
Occurrence :
Element Abundance Source Element Abundance Source
 Coal
 Natural gas
 oil
 Borax : (hydrocarbon)
Na2[B4O5(OH)4].8H2O  Natural
9 ppm  Colemnite : Graphite
B (Rare Ca2B6O11.5H2O C  Natural
180 ppm
(Boron) element)  Kernite : (Carbon) diamond
Na2[B4O5(OH)4].2H2O  Carbonates:
 Boric acid :  Calcite(CaCO3)
H3BO3  Magnesite
(MgCO3)
 Dolomite(MgC
O3.CaCO3]
83000 ppm
 Bauxite :
(Most
Al2O3.H2O–Al2O3.3H2O  Silica (sand &
abundant
Al  Aluminosilicate Si quartz) SiO2
metal, 272000 ppm
(Aluminium) rocks (feldspars, (Silicon)  Silicate
3rd most
mica) minerals
Abundant
element)  Cryolite : Na3AlF6
Ga Ge  Silver and Zinc
19 ppm  Ores of Al, Zn, Ge 1.5 ppm
(Gallium) (Germanium) ores, coal
In Sn  Cassiterite
0.24 ppm  ZnS & PbS ores 2.1 ppm
(Indium) (Tin) (SnO2)
Tl Pb
0.5 ppm  ZnS & PbS ores 13 ppm  Galena (PbS)
(Thallium) (Lead)

Allotropy :
Elements Allotropes Elements Allotropes
Crystalline Diamond,
5 crystalline forms :
hexagonal diamond
B -rhombohedral C
-graphite, -graphite
2 amorphous forms fullerenes
Amorphous - brown powder
Al No allotrope Si
Crystalline - greyish metallic
-Germanium
Ge
Ga No allotrope -Germanium
-Germanium
-tin (grey tin)
Sn
In No allotrope -tin (white tin)
13.2ºC
-Sn(diamond structure) -Sn(metallic)
Tl -Thallium -lead (Pb-I)
Pb
-Thallium -lead (Pb-II)

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 p-Block Elements (B & C family)
Allotropy of carbon family
 Silicon is a very high melting-point solid with the same structure as diamond. The non-existence of
an allotrope with the graphite structure clearly shows the inability of silicon atoms to multiple bond with
themselves.
Allotropes of Carbon:
Diamond Graphite Fullerene

Structure

z
Hybridisation sp3 sp2 sp2
Density
3.51 2.22 1.65
(g/cm3)
Hf (KJ/mol) 1.9 0 38.1
Bond length 154 pm 141.5 pm 143.5 pm & 138.3 pm
 Crystalline lattice.  Layered structure  Cage like
 3-D network: each C-  Interlayer force-Vanderwaal’s forces molecules.
atoms is linked to four  Each carbon atom is linked to three  C60–Soccer ball
other C-atoms in other carbon atoms, fourth electron shape–Buck
tetrahedral manner. forms a  bond. minsterfullerene
 One of the hardest –  Good conductor along the sheet  20- six membered
next to boron nitride and semi-conductor perpendicular rings and 12- five
(only at certain to the sheet. membered rings.
conditions)  Inter layer distance 340 pm so  Six membered
 Uses: sharpening cleavage between layers is easy. ring is fused with
hard tools, cutter tools; Soft and slippery– lubricant at high six or five
as a gem. temperature. membered ring
 Natural graphite is found as a  Five membered
mixture with mica, quartz & ring is fused only
silicates. with six
3C+SiO2  
 SiC + 2CO  2500ºC
 membered ring.
C(graphite) + Si gas  Heating of
 Thermodynamically most stable graphite in an
among allotropes. electrical arc in
the presence of
Graphite  1600ºC
 synthetic diamond
50000–60000atm inert gases such
as helium or
argon can result
into fullerene

Ex-1. Thermodynamically graphite is more stable than diamond but diamond does not transform into graphite
on its own. Why?
Sol. This conversion is not favoured by kinetic factors (the activation energy for this is very high).

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p-Block Elements (B & C family)
Atomic & Physical Properties
Boron is non-metallic in nature. It is extremely hard and black coloured solid. It exists in many allotropic
forms. Due to very strong crystalline lattice, boron has unusually high melting point. Rest of the
member are soft metals with low melting point and high electrical conductivity. Gallium with low melting
point (303 K), could exist in liquid state during summer. Its high boiling point (2676 K) makes it a useful
material for measuring high temperatures. Aluminium is a good conductor of heat and electricity. It is
malleable and ductile. Density of the elements increases down the group from boron to thallium.
Boron family:
Property B Al Ga In Tl
Character Metalloid Metallic Metallic Metallic Metallic
Atomic Number 5 13 31 49 81
Atomic Mass/g mol–1 10.81 26.98 69.72 114.82 204.38
[Xe] 4f14
Electronic configuration [He] 2s2 [Ne] 3s2 [Ar] 3d10 [Kr] 4d10
5d10
General electronic configuration=(ns2 np1) 2p1 3p1 4s24p1 5s2 5p1
6s26p1
Covalent Radius / pm 85 143 135 167 170
(B < Ga < Al < In < Tl) In Ga, poor shielding of 10 d-electrons

Ionic Radius X– / pm
27 53.5 62 80 88.5
(B < Al < Ga < In < Tl)

Ionization enthalpy iH1  801 577 579 558 589

(kJ mol–1)
(B > Al < Ga > In < Tl) iH2  2427 1816 1979 1820 1971
2877
iH3  3659 2744 2962 2704
Poor shielding of d-orbital and f-orbital in Ga & Tl respectively
Electronegativity 2.0 1.5 1.6 1.7 1.8
(B > Al < Ga < In < Tl)
Marginal increase after Al
Melting point / K 2453 933 303 430 576
Boiling point / K
3923 2740 2676 2353 1730
(B > Al > Ga > In > Tl)
Density/[g cm–3 ( at 293 K)
2.35 2.70 5.90 7.31 11.85
(B < Al < Ga < In < Tl)
+3 +3 +3, +1 +3, +1 +3, +1
+1 oxidation state arises due to inert pair effect.
Oxidation State
Stability of Oxidation state: +1: Ga < In < Tl
+3 : Al > Ga > In > Tl

Carbon family :
Property C Si Ge Sn Pb
Non Non
Character Metalloid Metallic Metallic
Metallic Metallic
Atomic Number 6 14 32 50 82
Atomic Mass/g mol–1 12.01 28.09 72.60 118.71 207.2
Electronic configuration [Xe] 4f14
[He] 2s2 [Ne] 3s2 [Ar] 3d10 4s2 [Kr] 4d10
5d10 6s2
General electronic configuration=(ns2 np2) 2p2 3p2 4p2 5s2 5p2
6p2

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p-Block Elements (B & C family)
Covalent Radius / pm
77 118 122 140 146
(C < Si < Ge < Sn < Pb)
Ionic Radius M+4 / pm
– 40 53 69 78
(Si < Ge < Sn < Pb)

Ionization enthalpy iH1  801 577 579 558 589

(kJ mol–1)
(C > Si > Ge > Sn < Pb) iH2  2427 1816 1979 1820 1971
iH3  3659 2744 2962 2704 2877
Poor shielding f-orbital
Electronegativity
2.5 1.8 1.8 1.8 1.9
(C > Si  Ge  Sn  Pb)
Melting point / K 4373 1693 1218 505 600
Boiling point / K – 3550 3123 2896 2024
Density/[g cm–3 ( at 293 K)
3.51 2.34 5.32 7.26 11.34
(C (diamond) > Si < Ge < Sn < Pb)
+4 +4 +4, +2 +4, +2 +4, +2
+2 oxidation state arises due to inert pair effect.
Oxidation State
Stability of Oxidation state: +2: Ge < Sn < Pb
+4: Ge > Sn > Pb

The Elements
Preparation of elements :

BORON :
Source Process Comments
 low purity (95-98%) boron
(black)
 The product thus obtained
 Na2B4O7+2HCl+5H2O   4H3BO3+2NaCl
is boiled with HCl and
From Borax  filtered when Na2O or
 2H3BO3  B2O3 + 3H2O
(Na+)2B4O72- .10H2O MgO dissolves leaving
 B2O3+3Mg/Na 
High temp.
2B+3MgO/Na2O behind elemental boron. It
is thoroughly washed to
remove HCl and then
dried finally.
 2BX3+3H2 
red hot W
 2B(crystalline) + 6HX Problem in obtaining high
or Tantalum
purity boron:
(X = Cl or Br) 99.9% pure
From BX3  High melting point
 2BCI3 + 3Zn   3ZnCl2 + 2B
(2180ºC)
 2BI3 
red hot W
 2B + 3I2(Van Arkel method)  Liquid gets corroded
or Tantalum

From Diborane   Thermal decomposition of

 B2H6   2B(crystalline) + 3H2
(B2H6) diborane.
 By heating it with
From potassium  potassium metal.
 KBF4 + 3K   4KF + B
fluoroborate  It is then treated with dilute
(KBF4) HCl to remove KF and B is
then washed and dried.

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p-Block Elements (B & C family)
ALUMINIUM :
Method Process Comments
 The purified Al2O3 is mixed with
Na3AlF6 (cryolite) or CaF2
(fluorspar) which lowers the
melting point of the mixture and
brings conductivity. The fused
The electrolytic reactions are: matrix is electrolysed. Steel
cathode and graphite anode
Cathode : Al3+(melt) + 3e–   Al(l) are used. The graphite anode
Electrolytic 2–
Anode : C(s) + O (melt)  – is useful here for reduction to
 CO(g) + 2e
reduction 2–
C(s) + 2O (melt)   CO2 (g) + 4e
– the metal.
(Hall-Heroult The overall reaction may be taken as:  The electolysis of the molten
process) mass is carried out in an
2Al2O3 + 3C   4Al + 3CO2
electrolytic cell using carbon
electrodes. The oxygen
liberated at anode reacts with
the carbon of anode producing
CO and CO2. This way for
each kg of aluminium
Several other extraction processes will be studied in produced, about 0.5 kg of
detail in Metallurgy. carbon anode is burnt away.

CARBON :
Preparation of Process

Carbon black (soot) By incomplete combustion of hydrocarbon.


Graphite 3C + SiO2   SiC + 2CO 
2500ºC
 C (graphite) + Si gas
 Natural diamond can be extracted from mines.
Diamond  Synthetic diamond
1600ºC
Graphite 
50000  60000 atm
 synthetic diamond.

SILICON :
From Process Comments
SiO2 (excess) + 2C   Si (pure) + 2CO
Reducing SiO2 with high purity coke in an
SiO2 Si (pure) + 2Cl2  SiCl4
electric furnace
SiCl4 + 2Mg   Si (highly pure) + MgCl2
Sodium Reduction of Na2[SiF6] Zone refining is used to get ultra pure
Na2[SiF6] silicon from highly pure silicon, which can
Na2 [ SiF6] + 4Na  Si (ultra pure) + 6NaF
be used in semi-conductor industry.

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 p-Block Elements (B & C family)
Section (D) : Properties of Elements
Chemical Properties :
Reaction
Boron Family Carbon Family
with

E + O2  EO2

4E + 3O2  2E2O3 (diox ide
)
1 
E+ O2  EO
2 (m onoox ide)
O2 B2O3 – acidic CO2 – acidic CO – neutral
Al2O3 – amphoteric SiO2 – acidic SiO – unstable
Ga2O3 – amphoteric GeO2 – acidic GeO – acidic
In2O3 – basic SnO2 – amphoteric SnO – amphoteric
Tl2O3 – basic PbO2 – amphoteric PbO – amphoteric

2E + N2  2EN 1273 K
N2 EN + H2O 
E(OH)3 + NH3 2C(s)+O2(g)+4N2(g)  2CO(g)  4N2 (g)
(Produc er
gas )
E = B or Al
2E + 3X2  2EX3 ; (X=F, Cl, Br, I) E + 2X2  EX4 ; (X = F, Cl, Br, I)
BX3 – covalent  All members form MX4 ; Ge & Pb form MX2
(BI3 cannot be formed directly)  PbI4 does not exist
X2 – Ionic
(Halogen)
AlF3  Stability of EX4 decreases down the group
GaF3 – Ionic  Stability of EX2 increases down the group
InF3 – Ionic  Stability :
TlI3 – exist as Tl+I3– & T l I also forms GeX4 > GeX2 ; PbX2 > PbX4
B – does not react with water C(s) + H2O (steam) red
 CO(g)  H2 (g)
25ºC heat (water gas)
Al(OH)3
+H2O red
Al E(s) + H2O(steam) 
 EO2+H2;
(Si,Ge,Sn) heat
>480ºC
Al2O3  Pb is unafftected by water, probably
H2O +H2O
Ga because of a protective oxide layer.
Not attacked by cold & hot water  C, Si, Ge – not attacked by cold water.
unless oxygen is present.
In
Tl – oxidises in moist air & decomposes
steam at red heat.
+H2SO4 C
H3BO3 + SO2 +HCl
(Hot & Conc.)
Si No reaction
B dilute
+HNO3 Ge
H3BO3 + NO2 Sn dissolves in HCl (dil. & conc) but Pb in only
(Hot & Conc.)
(Boron reacts with only oxidising acids) dil.HCl.
HCl
PbCl2 + H2
+HCl (conc.) (coating)
AlCl3 + H2 Pb
(does not H2SO4
Al dissolve in) PbSO4 + H2
Acids +HNO3 (coating)
(Conc.)
Do not react
because it forms +HNO3
passive oxide layer.
(Hot & conc.)
Ga, In, T can also react with dilute mineral
acids. C (graphite) (Mellitic acid)
+HF/HNO3
Graphite oxide
(mixture)
Si + 6HF  H2SiF6 + 2H2
E + dil. HNO3 E(NO3)2
(Sn,Pb )

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 p-Block Elements (B & C family)
2B + 2NaOH + 2H2O   2NaBO2 + 3H2 C – unaffected by alkalies
2B + 6NaOH fused
 2Na3BO3 + 3H2 Si + 2NaOH + H2O   Na2SiO3 + 2H2

NaOH + H2O
NaAlO2.2H2O + H2
NaOH or
+ —
Na [Al(OH)4] 
Al E + NaOH   Na2EO3 + H2
(Sn,Pb ) or
NaOH
Na3AlO3 + H2 Na2 [E(OH)6 ]
(Al & Ga readily dissolves in alkalies).
In & Tl do not react with alkali.
3Mg +2B  Mg3B2 
2Mg +Si   Mg2Si (magnesium silicide)
Metal
3Ca +2B  Ca3B2
SiO2
B2O3 + Si

B
CO2
Reducing  B2O3 + C
Property MnO2
Al2O3 + Mn

Al
Cr2O3
 Al2O3 + Cr

Section (E) : Oxides, Hydroxides, Oxyacids, Borax

Oxides, oxy acids and hydroxide
Boron trioxide (B2O3) :
Preparation :

 Properties :
It is a acidic oxide and is anhydride of boric acid and it reacts with alkalies or bases to form borates.
3Na2O + B2O3   2Na3BO3 (sodium orthoborate).
It reacts with water slowly to form orthoboric acid.
H2O + B2O3   2HBO2 ; HBO2 + H2O   H3BO3
When heated with transition metal salts, it forms coloured compounds.

3B2O3 + Cr2(SO4)3   3SO3  + 2Cr(BO2)3(green)

2B2O3 + 2Cu(NO3)2  4NO2  + O2  + 2Cu(BO2)2 (blue)
It also shows weakly basic properties according to the following reaction.
B2O3 + P2O5 2BPO4
It reacts with hydrogen fluoride in presence of H2SO4 forming BF3 .
B2O3 + 6HF + 3H2SO4   2BF3 + 3H2SO4.H2O.

Boric acid (H3BO3) :

Preparation :
(i) It is precipitated by treating a concentrated solution of borax with sulphuric acid.
Na2B4O7 + H2SO4 + 5H2O   Na2SO4 + 4H3BO3 
(ii) From Colemanite: Powdered colemanite is suspended in water and excess SO 2 is passed through it.
On filtering and cooling the filtrate, white crystals of H3BO3 are obtained.
Ca2B6O11 + 4SO2 + 11H2O   2Ca(HSO3)2 + 6H3 BO3
 Properties:
It is a weak monobasic acid, soluble in water and in aqueous solution the boron atom completes its
octet by accepting OH– from water molecules:
B(OH)3(aq) + 2H2O() [B(OH)4]– (aq) + H3O+(aq). pK = 9.25.
It, therefore, functions as a Lewis acid and not as a proton donor like most acids.

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 p-Block Elements (B & C family)
Since B(OH)3 only partially reacts with water to form H3O+ and [B(OH)4]–, it behaves as a weak acid.
Thus H3BO3 cannot be titrated satisfactorily with NaOH as a sharp end point is not obtained. If certain
organic polyhydroxy compounds such as glycerol, mannitol or sugars are added to the titration mixture,
then B(OH)3 behaves as a strong monobasic acid and it can be now titrated with NaOH and the end
point is detected using phenolphthalein as indicator (pH = 8.3 - 10.0).
The added compound must be a cis-diol, to enhance the acid properties. The cis-diol forms very stable
complex with the [B(OH)4]–, thus removing it from solution. The reaction is reversible and thus removal
of one of the products shifts the equilibrium in the forward direction and thus all the B(OH) 3 reacts with
NaOH; in effect it acts as a strong acid in the presence of the cis-diol.
2B(OH)3 + 2NaOH 
 Na[B(OH)4] + NaBO2 + 2H2O

HB(OH)4 + 2 + H+ + 4H2O

Ethanol does not form similar complex but catechol, salicylic acids, mannitol form similar complexes.

When heated it first forms metaboric acid (HBO2) and then boron trioxide.

Orthoboric acid is greasy to touch less soluble in cold water but more soluble in hot water. In the solid
state, the B(OH)3 units are hydrogen bonded together in to two dimensional sheets with almost
hexagonal symmetry. The layered are quite a large distance apart (3.18 Å) and thus the crystal breaks
quite easily into very fine particles.

Figure : 1
 Polymeric metaborate species are formed at higher concentration, for example,
3B(OH)3 H3O+ + [B3O3(OH)4]– + H2O, pK = 6.84


 Boric acid dissolves in aqueous HF forming HBF4 (fluoroboric acid).
B(OH)3 + 4HF H3O+ + BF4–– + 2H2O


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 p-Block Elements (B & C family)
 Test for Borate radical :
When boric acid is heated with ethyl alcohol, the evolved gas is burned forming a green edged flame.
H3BO3 + 3C2H5OH   B(OC2H5)3 + 3H2O
ethylborate (volatile)
 Uses :
(i) It is an antiseptic and its water solution is used as an eyewash.
(ii) It is also used in glass, enamel and pottery industry.

Ex-2. It has been observed that BF3 does not hydrolyses completely whereas BCl3 or BBr3 get easily
hydrolysed to form B(OH)3 and HX ? Explain.
Sol. The greater stability of B–F bond as compared to B–Cl and B–Br bonds is due to additional –bonding
in B–F bonds of BF3 molecules. The B–Cl and B–Br bonds are relatively weak and are easily cleaved
by water forming strong B–OH bonds instead of stable addition product (BF3.OH2) formed by BF3.

Borax (Na2B4O7.10H2O) :
 Preparation :
It is found in nature but can also be prepared by the following methods.
(i) From Colemanite.
When colemanite powder is heated with Na2CO3 solution, the following reaction occurs with the
precipitation of CaCO3.
Ca2B6O11 + 2Na2CO3   2CaCO3  + Na2B4O7 + 2NaBO2
The filtrate is cooled when white crystals of borax are precipitated. The mother liquor on treatment with
CO2 converts NaBO2 to Na2B4O7 which precipitates out on crystallization.
4NaBO2 + CO2   Na2B4O7 + Na2CO3
(ii) From orthoboric acid.
Borax is obtained by the action of Na2CO3 on orthoboric acid.
4H3BO3 + Na2CO3   Na2B4O7 + 6H2O + CO2



 Properties :
(i) Borax is a white powder, less soluble in cold water, more soluble in hot water.
(ii) Its aqueous solution is alkaline because of its hydrolysis to weak acid H 3BO3 and strong alkali NaOH.
Na2B4O7 + 7H2O   4H3BO3 + 2NaOH
(iii) Action of heat.
When borax powder is heated, it first swells due to loss of water in the form of steam but at 740oC it
becomes converted into colourless transparent borax bead.

Na2B4O7.10H2O   Na2B4O7 + 10 H2O
740ºC
Na2B4O7   2NaBO2 + B2O3 (borax bead)
(iv) Oxidation of boric acid or sodium metaborate with H2O2.

Na2B4O7   2NaBO2 + 2H2O2 + 6H2O   Na2 [(OH)2B (O—O)2B(OH)2].6H2O
Sodium per oxoborate is used as a brightner in washing powder. In very hot water (over 80ºC) the
peroxide linkages —O—O— break down to give H2O2.
(v) It is a useful primary standard for titrating against acids. One mole of it reacts with two moles of acid.
This is because when borax is dissolved in water both B(OH) 3 and [B(OH)4]– are formed, but only the
[B(OH)4]– reacts with HCl.
[B4O5(OH)4]2– + 5H2O 2B(OH)3 (weak acid) + 2[B(OH)4]– (salt)
2[B(OH)4]2– + 2H3O+   2B(OH)3 + 4H2O
On cooling, the white flakes of boric acid are obtained
 Borax is also used as a buffer since its aqueous solution contain equal amounts of weak acid and its
salt.
H O H SO
(vi) Na2[B4O5(OH)4] + 12HF  2
 [Na2O(BF3)4]   4BF3 + 2NaHSO4 + H2O
2 4

 Correct formula of borax is Na2[B4O5(OH)4] . 8H2O. It contains boron in both planar BO3 and tetrahedral
BO4 units. It contains five B—O—B linkages.

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 p-Block Elements (B & C family)
 Borax-bead test :
Borax reacts with certain metal salts such as, Ni2+, Co2+, Cr3+, Cu2+, Mn2+ etc. to form coloured
metaborates. The colour of the metaborates can be used to identify the metallic ions (cations) in salts.

 2NaBO 2  B 2 O3 ; CuO + B2O3 
740ºC
Na2B4O7·10H2O  Na2B4O7   Cu(BO2)2 (blue bead)
10H2 O glassy mass

 Uses :
(i) In borax bead test. (ii) In purifying gold.
(iii) As flux during welding of metals. (iv) In production of glass.

 (A) 
(i) C H OH
Ex-.3 (a) Na2B4O7 + concentrated H2SO4 + H2O  2 5
(B)
(ii) ignite

(B) is identified by the characteristic colour of the flame. Identify (A) and (B).
(b) Complete the following reaction and identify the products formed.


Na2B4O7  (A)
 (B )

Sol. (a) Na2B4O7 + concentrated H2SO4 + 5H2O   Na2SO4 + 4H3BO3

H3BO3 + 3C2H5OH   B(OC2H5)3-volatile (burn with green edged flame) + 3H2O


(b) Na2B4O7   B2O3
 (NaBO 2 )

Aluminium Oxide (Al2O3) :

 It is also called alumina. It occurs in nature in the form of bauxite and corundum. It is also found in the
form of gems. Some important aluminium oxide gems are :
(A) Oriental Topaz-yellow (Fe3+), (B) Sapphire-blue (Fe2+ / 3+ / Ti4+),
(C) Ruby-red (Cr3+), (D) Oriental Emerald-green (Cr3+ / V3+)


 Preparation :
Pure Al2O3 is obtained by igniting Al2(SO4)3, Al(OH)3 or ammonium alum.
 
Al2(SO4)3   Al2O3 + 3SO3 ; 2Al(OH)3  Al2O3 + 3H2O

(NH4)2SO4.Al2(SO4)3.24H2O   2NH3 + Al2O3 + 4SO2  + 25H2O

 Properties :
It is a white amorphous powder insoluble in water but soluble in acids (forming eg., AlCl 3) as well as
alkalies (forming e.g., NaAlO2), Thus amphoteric in nature. It is a polar covalent compound. Exists in
two forms -Al2O3 or corundum and -Al2O3.
Addition of Cr2O3 or Fe2O3 makes alumina coloured.
-Al2O3 
 -Al2O3
1000ºC

 Uses :
(i) It is used for the extraction of aluminium.
(ii) It is used for making artificial gems.
(iii) It is used for the preparation of compounds of aluminium.
(iv) -Al2O3 is used in making furnace linings. It is a refractory material.
(v) It is used as a catalyst in organic reactions.
(vi) Corundum is extremely hard and is used as ‘Jewellers rouge’ to polish glass.
(vii) -Al2O3 dissolves in acids absorbs moisture and is used in chromatography.

Ex-4 What will happen if aluminium is heated with coke in an atmosphere of nitrogen ?

Sol. Al2O3 + N2 + 3C   2AlN + 3CO

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p-Block Elements (B & C family)
Carbonmonoxide (CO) :
Preparation :
(i) It is formed together with CO2, when carbon or carbonaceous matter is oxidized by air or oxygen. It is
also produced when CO2 is reduced by red- hot carbon; this reaction is of importance in metal
extractions.
C(s) + CO2(g)   2CO(g)
(ii) In the laboratory it can be prepared by dehydrating methanoic acid with concentrated sulphuric acid.
373K
HCOOH (liq)   CO(g) + H2O
conc.H2SO4

(iii) If oxalic acid is dehydrated in the same way, CO 2 is formed as well.

conc. H2SO4 , 
H2C2O4   CO + CO2
–H2O

(iv) On commercial scale it is prepared by the passage of steam over hot coke. The mixture of CO and
H2 thus produced is known as water gas or synthesis gas.
4731273K
C (s) + H2O (g)   CO (g) + H2(g) (water gas).
When air is used instead of steam, a mixture of CO and N2 is produced, which is called producer gas.
1273K
2 C (s) + O2 (g) + 4 N2 (g)   2 CO (g) + 4 N2 (g) (Producer gas).
Water gas and producer gas are very important industrial fuels. Carbon monoxide in water gas or
producer gas can undergo further combustion forming carbon dioxide with the liberation of heat.
(v) CO2 + H2   CO + H2O

(vi) K4Fe(CN)6 + 6H2SO4 (concentrated) + 6H2O   2K2SO4 + FeSO4 + 3(NH4)2SO4 + 6CO
(vii) HCN + 2H2O   HCOOH + 2NH3 (absorbed by H2SO4)

HCOOH   2O + CO
H
(viii) Also obtained as by-product when carbon is used in reduction processes such as, of phosphite
rock to give phosphorus.
 Properties :
(i) Carbon monoxide is a colourless, odourless gas which burns in air with a blue flame, forming CO 2. It
is sparingly soluble in water and is a neutral oxide. CO is toxic, because it forms a complex with
haemoglobin in the blood and this complex is more stable than oxy-haemoglobin. This prevents the
haemoglobin in the red blood corpuscles from carrying oxygen round the body. This causes oxygen
deficiency, leading to unconsciousness and then death.
Hb—O2 + CO   Hb—CO + O2
Ordinary gas masks are no protection against the gas, since it is not readily adsorbed on active
charcoal. In the presence of air, a mixture of manganese (IV) oxide and copper(II) oxide catalytically
oxidizes it to CO2, and this mixed catalyst is used in the breathing apparatus worn by rescue teams in
mine disasters.
(ii) Carbon monoxide is a powerful reducing agent, being employed industrially in the extraction of iron
and nickel .
Fe2O3(s) + 3CO(g)   2Fe(s) + 2CO2(g) ; NiO(s) + CO(g)   Ni(s) + CO2(g)
(iii) It reacts with many transition metals, forming volatile carbonyls; the formation of nickel carbonyl
followed by its decomposition is the basis of the Mond’s process for obtaining very pure nickel .
28ºC 180ºC
Ni(s) + 4CO(g)   Ni(CO)4(liq)   Ni(s) + 4CO(g)
(iv) In addition to reacting with oxygen, carbon monoxide combines with sulphur to give carbonyl
sulphide and with chlorine in the presence of light to give carbonyl chloride (phosgene), used in the
production of polyurethane foam plastics. Phosgene is an exceedingly poisonous gas.
CO(g) + S(s)   COS(s) (carbonyl sulphide) ;
CO(g) + Cl2(g)   COCl2(g) (carbonyl chloride)
(v) Although carbon monoxide is not a true acid anhydride since it does not react with water to produce
an acid, it reacts under pressure with fused sodium hydroxide to give sodium methanoate :
dil. HCl
NaOH(liq) + CO(g)   HCOONa(s)   HCOOH(aq)

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 p-Block Elements (B & C family)
(vi) With hydrogen under pressure and in the presence of zinc oxide or chromium (III) oxide catalyst it
reacts to give methanol; this reaction is of industrial importance.
CO(g) + 2H2(g)   CH3OH(liq)
(vii) CO is readily absorbed by an ammonical solution of copper (I) chloride to give CuCl.CO.2H 2O. It
reduces an ammonical solution of silver nitrate to silver (black) and, in the absence of other gaseous
reducing agents, this serves as a test for the gas. It can be estimated by reaction with iodine pentoxide,
the iodine which is produced quantitatively being titrated with standard sodium thiosulphate solution.
5CO(g) + 2O5(s)  2(s) + 5CO2(g)
(viii) It reduces an aqueous PdCl2 solution to metallic Pd.

Carbon dioxide (CO2) :

 Preparation :
(i) In the laboratory it can be conveniently made by the action of dilute hydrochloric acid on marble
chips:
CO32-(aq) + 2H+(aq)   CO2(g) + H2O()

(ii) Industrially it is produced as a by-product during the manufacture of quicklime and in fermentation
processes:
CaCO3(s)   CaO(s) + CO2(g) ; C6H12O6(aq){glucose}   2C2H5OH(aq) + 2CO2(g)
 Properties :
(i) It is a colourless, odourless and heavy gas which dissolves in its own volume of water at ordinary
temperature and pressure. Like all gases, it dissolves much more readily in water when the pressure is
increased and this principle is used in the manufacture of soda water and fizzy drinks.
(ii) CO2 is easily liquefied (critical temperature = 31.1ºC) and a cylinder of the gas under pressure is a
convenient fire extinguisher. When the highly compressed gas is allowed to expand rapidly solid carbon
dioxide (‘dry ice’) is formed. Solid carbon dioxide sublimes at –78ºC and, since no massy liquid is
produced, it is a convenient means of producing low temperatures.
(iii) Carbon dioxide is the acid anhydride of carbonic acid, which is a weak dibasic acid and ionises in
two steps as follows :
H2CO3(aq) + H2O () HCO3– (aq) + H3O+ (aq)
HCO3– (aq) + H2O () CO32– (aq) + H3O+ (aq)
–
H2CO3 / HCO3 buffer system helps to maintain pH of blood between 7.26 to 7.42.
A solution of carbonic acid in water will slowly turn blue litmus red and when the solution is boiled, all
the CO2 is evolved.
(iv) Carbon dioxide readily reacts with alkalies forming the carbonate and, if CO 2 is in excess, the
hydrogen carbonate. This is the basis of the lime-water test for CO2 gas.
Ca(OH)2(aq) + CO2(g)  CaCO3(s) + H2O(liq); CaCO3(s) + H2O(liq) + CO2(g)  Ca(HCO3)2(aq)
The above reaction accounts for the formation of temporarily hard water.
(v) Carbon dioxide, which is normally present to the extent of ~ 0.03% by volume in the atmosphere, is
removed from it by the process known as photosynthesis. It is the process by which green plants
convert atmospheric CO2 into carbohydrates such as glucose. The overall chemical change can be
expressed as:
hv
6 CO2 + 12 H2O  
 C6H12O6 + 6 O2 + 6 H2O
Chlorphyll
By this process plants make food for themselves as well as for animals and human beings. But the
increase in combustion of fossil fuels and decomposition of limestone for cement manufacture in recent
years seem to increase the CO2 content of the atmosphere. This may lead to increase in green house
effect and thus, raise the temperature of the atmosphere which might have serious consequences.
(vi) Gaseous CO2 is extensively used to carbonate soft drinks. Being heavy and non–supporter of
combustion it is used as fire extinguisher. A substantial amount of CO 2 is used to manufacture urea.
 Recovery of CO2 :
(a) Na2CO3 + CO2 + H2O 2NaHCO3
(b) Girbotol process : 2HOCH2CH2NH2 + CO2 + H2O (HOCH2CH2NH3)2CO3


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 p-Block Elements (B & C family)


Ex-5. H2C2O4   gas (A) + gas (B) + liquid (C)
oxalic acid
Gas (A) burns with blue flame and is oxidised to gas (B). Gas (B) turns lime water milky.
NH3 ,  NH3 , 
Gas (A) + Cl2   (D)   (E)   (B)
Identify (A) to (E) and explain reactions involved.

Sol. H2C2O4   CO + CO2 + H2O
NH3 ,  NH3 , 
CO + Cl2   COCl2   NH2CONH2   CO2


Carbon suboxide (C3O2) :

This is an evil-smelling gas and can be made by dehydrating propanedioic acid (malonic acid), of which
it is the anhydride, with phosphorus pentoxide :
3 CH2(COOH)2 + P4O10   3C3O2 + 4H3PO4
o
When heated to about 200 C, it decomposes into CO2 and C:
C3O2(g)   CO2(g) + 2C(s)
The molecule is thought to have a linear structure: O=C=C=C=O.

Silicon Dioxide (SiO2) :

Silicon dioxide, commonly known as silica, occurs in several crystallographic forms. Quartz, cristobalite
and tridymite are some of the crystalline forms of silica, and they are interconvertable at suitable
temperature. Silicon dioxide is a covalent, three-dimensional network solid in which each silicon atom is
covalently bonded in a tetrahedral manner to four oxygen atoms. Each oxygen atom in turn covalently
bonded to another silicon atoms. Each corner is shared with another tetrahedron. The entire crystal
may be considered as giant molecule in which eight membered rings are formed with alternates silicon
and oxygen atoms. Silica in its normal form is almost non-reactive because of very high Si–O bond
enthalpy. It resists the attack by halogens, dihydrogen and most of the acids and metals even at
elevated temperatures. However, it is attacked by HF and NaOH.
SiO2 + 2 NaOH   Na2SiO3 + H2O ; SiO2 + 4 HF   SiF4 + 2 H2O
Quartz is extensively used as a piezoelectric material; it has made possible to develop extremely
accurate clocks, modern radio and television broadcasting and mobile radio communications. Silica gel
used as a drying agent and as a support for chromatographic materials and catalysts. Kieselghur, an
amorphous form of silica is used in filtration plants.

Stannous Oxide (SnO) :

 Preparation :
By heating stannous hydroxide, Sn(OH)2, in absence of air.
Sn(OH)2  SnO + H2O
 Properties :
SnO is an amphoteric dark grey or black solid oxide, insoluble in water. It dissolves in acids to form
stannous salts.
SnO (basic) + 2H+  Sn2+ + H2O ;
SnO (acidic) + 4OH– + H2O  [Sn(OH)6]4– or SnO22– (stannite)
 Stannites are only known in aqueous solutions. Stannites absorb oxygen from air and are oxidised to
stannate which are stable in nature.
2 Na2SnO2 + O2  2 Na2SnO3


 Uses :
For the preparation of stannous chloride and stannous sulphate.

Stannic Oxide (SnO2) :

 Preparation :
By heating tin with concentrated HNO3.

Sn + 4HNO3   H2SnO3 + 4NO2 + H2O ; H2SnO3   H2O + SnO2

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p-Block Elements (B & C family)
Properties :
(i) It is a white solid insoluble in water.
(ii) It is amphoteric in nature.
(iii) It dissolves in concentrated H2SO4 to form stannic sulphate.
SnO2 + 2H+   Sn+4 + 2H2O
(iv) It also dissolves in concentrated alkalies to form alkali metal stannate solution.
SnO2 + 6OH–   [Sn(OH)6]2– or SnO32– (stannate)

Litharge (PbO) :
PbO is prepared by heating Pb at 180oC. It is a volatile yellow organic solid.

2Pb + O2   2PbO
It is an amphoteric oxide and dissolves in acids as well as in alkalies.
It is used in rubber industry and in the manufacture of flint glasses, enamels, and storage batteries.

Lead Dioxide (PbO2) :

 Preparation :

(i) PbO + NaOCl   PbO2 (insoluble) + NaCl

(ii) Pb3O4 + 4HNO3 (dilute) 

 2Pb(NO3)2 + PbO2 + 2H2O

 Properties :
It is a chocolate/dark brown coloured insoluble solid.
(i) On heating at 440oC it gives the monoxide.
440ºC
2PbO2   2PbO + O2

(ii) PbO2 is an oxidising agent and reduced to PbO since stability of Pb(II) > Pb(IV) based on inert pair
effect.
(a) It oxidizes HCl to Cl2.
PbO2 + 4HCl   PbCl2 + 2H2O + Cl2
(b) It oxidises Mn salt to permanganic acid.
2MnSO4 + 5PbO2 + 6HNO3   2HMnO4 + 2PbSO4 + 3Pb(NO3)2 + 2H2O

(c) It reacts with SO2 at red heat to form lead sulphate.


PbO2 + SO2   PbSO4
(iii) It dissolves in concentrated NaOH solution.
PbO2 + 2OH– + 2H2O   [Pb(OH)6]2– (plumbate)
(iv) It reacts with concentrated HNO3 to evolve oxygen gas.
PbO2 + 2HNO3   Pb(NO3)2 + 1/2O2 + H2O
PbO2 + H2SO4   PbSO4 + 2H2O + O2

 Uses :
It is used in match industry for making ignition surface of match boxes, in the preparation of KMnO 4 and
in explosives.

Red Lead (Pb3O4) :

 Preparation :
It is prepared by heating PbO at 450oC for a long time.
450ºC
6PbO + O2   2Pb3O4

 Properties :
(i) It is a red powder insoluble in water but when heated with concentrated HNO 3 it gives a red
precipitate of PbO2.
Pb3O4 + 4HNO3  2Pb(NO3)2 + PbO2 + 6H2O
(ii) When heated above 550oC, it decomposes into PbO.

Pb3O4   6PbO + O2

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 p-Block Elements (B & C family)
(iii) It oxidizes concentrated HCl to chlorine.
Pb3O4 + 8HCl  3PbCl2 + 4H2O + Cl2
(iv) When heated with concentrated H2SO4 it evolves oxygen.
2Pb3O4 + 6H2SO4  6PbSO4 + 6H2O + O2
 Uses :
It is used as an oxidizing agent, for making metal protecting paints like red oxide paint, for making
special lead cement and for making flint glass.

Section (F) : Hydrides

Compounds
Hydrides
Boranes
Binary compounds of B with H are called boron hydrides or boranes. These compounds form following
two types of series :
BnHn+4 - B2H6, B5H9, B6H10, B10H14
BnHn+6 - B4H10, B5H11, B6H12, B9H15
The chemistry of diborane has aroused considerable interest because of its usefulness in many
synthetic reactions and also because the elucidation of its structure helped to clarify the basic concepts
about the structure of electron deficient compounds.
 Preparation of Diborane (B2H6) :
ether
(i) 4BF3 + 3LiAlH4   2B2H6 + 3Li [AlF4]
silent electric
(ii) 2BCl3 + 6H2 (excess)   B2H6 + 6HCl
discharge
ether
(iii) 8BF3 + 6LiH  B2H6 + 6LiBF4
(iv) 2NaBH4 + 2 
ether
 B2H6 + 2Na + H2
ether
(v) 3NaBH4 + 4BF3  3NaBF4 + 2B2H6
450K
(vi) It can also be prepared by treating NaBH4 with concentrated H2SO4 or H3PO4.
2NaBH4 + H2SO4   B2H6 + 2H2 + Na2SO4 ; 2NaBH4 + 2H3PO4   B2H6 + 2H2 + 2NaH2PO4
453 K
(vii) 2BF3 + 6NaH   B2H6 + 6NaF (Industrial method)
750 atm
(viii) B2O3 + 3H2 + 2Al   B2H6 + Al2O3
150ºC

(ix) Mg3B2 + H3PO4 
 mixture of boranes mainly, B4H10  B2H6.

 Properties :
(i) B2H6 is colourless gas and highly reactive (boiling point 183 K).
(ii) Controlled pyrolysis of diborane leads to most of the higher boranes.
It catches fire spontaneously in air and explodes with O 2.Reaction with oxygen is extremely exothermic.
B2H6 + 3O2  B2O3 + 3H2O  H = – 2160 kJ mol–1
 Mixtures of diborane with air or oxygen inflame spontaneously producing large amount of heat.
Diborane has a higher heat of combustion per unit weight of fuel than most other fuels. It is
therefore used as a rocket fuel.
  At red-heat the boranes decomposes to boron and hydrogen.
(iii) Reaction with water is instantaneous.
B2H6 + 6H2O  2B(OH)3 + 6H2
Dibroane is also hydrolysed by weaker acids (e.g. alcohols) or aqueous alkali.
B2H6 + 6ROH  2B(OR)3 + 6H2
B2H6 + 2KOH + 2H2O  2KBO2 + 6H2
(iv) Reaction with HCl replaces a terminal H with Cl.
B2H6 + HCl  B2H5Cl + H2
(v) Reaction with chlorine gives the trichloride.
B2H6 + 6Cl2  2BCl3 + 6HCl
(vi) The electron deficient 3c-2e BHB bridges are sites of nucleophilic attack.

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 p-Block Elements (B & C family)
  Small amines such as NH3, CH3NH2 and (CH3)2NH give unsymmetrical cleavage of diborane.
B2H6 + 2NH3  [H2B (NH3)2]+ + [BH4]–
  Large amines such as (CH3)3N and pyridine give symmetrical cleavage of diborane.
2(CH3)3N + B2H6  2H3B  N(CH3)3
  B2H6 + 2Me3P  2Me3PBH3
  200ºC, 20 atm
B2H6 + 2CO   2BH3CO (borane carbonyl)
  The boronium ion products [H2BL2]+, are tetrahedral and can undergo substitution by other
bases
[H2B(NH3)2]+ + 2PR3   [H2B(PR3)2]+ + 2NH3

  The reaction with ammonia depends on conditions.

 B2H6.2NH3 or [H2B(NH3)2]+ [BH4]– (ionic compound).
Excess NH3
B2H6 + NH3 
low temperature
Excess NH3
  (BN)x boron nitride.
higher temperature ( 200ºC)
Ratio 2NH3 : 1 B2H6
 
 B3N3H6 borazine.
higher temperature (200ºC)
Borazine is much more reactive than benzene. Borazine readily undergoes addition reactions which do
not occur with benzene. Borazine also decomposes slowly and may be hydrolysed to NH 3 and boric
acid at elevated temperature. If heated with water, B3N3H6 hydrolyses slowly.
B3N3H6 + 9H2O  3NH3 + 3H3BO3 + 3H2O
(vii) Reduction of diborane can be accomplished with sodium or with sodium borohydride.
  2B2H6 + 2Na  NaBH4 + NaB3H8
  B2H6 + NaBH4  NaB3H8 + H2.
  Reductions of diborane with NaBH4 can also lead to higher borane anions.
  2NaBH4 + 5B2H6  Na2B12H12
(viii) B2H6 + 2LiH  2LiBH4

Ex-6. Complete the following reactions and identify the products formed.
NaBH4
 (A)   (B)
140ºC
(a) BCl3 + NH4Cl 
C6H5Cl
17001800ºC
(b) BCl3 + H2 + Cfibre   product(s)
NaBH4
 B3N3H3Cl3   B3N3H6 (borazine)
140ºC
Sol. (a) 3BCl3 + 3NH4Cl 
C6H5Cl
17001800ºC
(b) 4BCl3 + 6H2 + Cfibre   B4C(fibre) + 12 HCl

Aluminium Hydride (AlH3) :

Aluminium hydride is obtained by interaction of LiAlH4 with100% H2SO4 in THF :
2LiAlH4 + H2SO4   2AlH3 + 2H2 + Li2SO4
The white hydride is thermally unstable. With donor ligands however, a range of molecular complexes
AlH3L & AlH3L2 are formed indicative of the lewis acidic behaviour of AlH3.
Hydrides of carbon :
Carbon forms a vast number of chain and ring compounds including :
 The alkanes (Paraffins) CnH2n+2  The alkenes (olefines) CnH2n
 The alkynes (acetylenes) CnH2n–2  Aromatic compounds
Silanes
SiH4 (monosilane)
160 ºC
(1) small scale preparation : SiO2 + LiAlH4   SiH4
(2) Hydrolysis of Magnesium silicide
Mg2Si + H2O  mixture of silanes
(3) Reduction of chlorosilanes by LiAlH4 to produce silane
(4) Photolysis of SiH4–H2 mixture can make higher silanes
(5) Among silanes only SiH4 & Si2H6 are indefinitely stable at 25°C.

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 p-Block Elements (B & C family)
Section (G) : Halides
Halides, Alums and Other metal salts :
Boron Halide :
Boron trihalides are strictly monomeric, trigonal planar molecules. This difference from aluminium is
attributed to back-donation from “nonbonding” electrons on halogen atoms into the “empty” p z orbital of
boron which lends some double bond character to B–X bonding (manifested by bond shortening) and
stabilizes the monomer.
Aluminium Chloride (AlCl3.6H2O) :
It is a colourless crystalline solid, soluble in water. It is covalent. Anhydrous AlCl 3 is a deliquescent
white solid.
 Preparation :
(i) By dissolving aluminium, Al2O3, or Al(OH)3 in dilute HCl :
2Al + 6HCl  2AlCl3 + 3H2 Al2O3 + 6HCl  2AlCl3 + 3H2O; Al(OH)3 + 3HCl  AlCl3 + 3H2O
The solution obtained is filtered and crystallized when the crystals of AlCl 3.6H2O are obtained.
(ii) Anhydrous AlCl3 is obtained by the action of Cl2 on heated aluminium.
(iii) By heating a mixture of Al2O3 and coke and passing chlorine over it.
Al2O3 + 3C + 3Cl2  2AlCl3 (anhydrous) + 3CO
 Properties :
(i) Action of heat : Hydrated salt when heated strongly is converted to Al 2O3.

2AlCl3.6H2O   Al2O3 + 6HCl + 3H2O
(ii) Action of moisture on anhydrous AlCl3: When exposed to air, anhydrous AlCl3 produces white
fumes of HCl.
AlCl3 + 3H2O Al(OH)3 + 3HCl
(iii) Action of NH3 : Anhydrous AlCl3 absorbs NH3 since the former is a Lewis acid.
AlCl3 + 6NH3  AlCl3.6NH3 (white solid)
(iv) Action of NaOH solution: When NaOH solution is added dropwise to an aqueous AlCl3 solution, a
gelatinous precipitate of Al(OH)3 is first formed which dissolves in excess of NaOH solution to give a
colourless solution of sodium meta-aluminate.
AlCl3 + 3NaOH  Al(OH)3 + 3NaCl; Al(OH)3 + NaOH  NaAlO2 + 2H2O
This reaction is important as a test to distinguish between an aluminium salt from salts of Mg, Ca, Sr,
and Ba. (When NaOH solution is added to their salt solutions, a white precipitate of hydroxide forms
which does not dissolve in excess of NaOH).
(v) Action of NH4OH solution: When NH4OH solution is added to a solution of AlCl 3, a white
precipitate of Al(OH)3 is formed which does not dissolve in excess of NH4OH.
AlCl3 + 3NH4OH  Al(OH)3(white gelatinous) + 3NH4Cl
This reaction is important as a test to distinguish an Al salt from a Zn salt. (With a Zn salt a white
precipitate of Zn(OH)2 is formed which dissolves in excess of NH4OH solution).
(vi) Hydrolysis with water: When AlCl3 is dissolved in water, it undergoes hydrolysis rapidly to
produce Al(OH)3 which is a weak base and HCl which is a strong acid. Hence the solution is acidic to
litmus.
[Al(H2O)6]3+ [Al(H2O)5OH]+2 + H+
The complex cation has a high tendency to get dimerised.
2[Al(H2O)5OH]2+  [(H2O)4 (H2O)4 ]+4 + 2H2O
(vii) 4LiH + AlCl3  LiAlH4 + 3LiCl
 Uses :
(i) As catalyst for cracking of petroleum.
(ii) As catalyst in Friedel-Crafts reactions.
(iii) For preparing aluminium compounds.

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 p-Block Elements (B & C family)

Ex-7. What happens when :

(write chemical equations only)
(a) Anhydrous AlCl3 is exposed to atmosphere.
(b) AlCl3 dissolves in aqueous sodium hydroxide (excess)

Sol. (a) AlCl3 + 3H2O  Al(OH)3 + 3HCl (HCl fumes in air).

(b) AlCl3 + 3NaOH  Al(OH)3+ 3NaCl
Al(OH)3 + OH–  [Al(OH)4]–
Ex-8. Anhydrous AlCl3 cannot be prepared by heating hydrated salt. Why ?
Sol. Gets hydrolysed forming Al2O3.

2AlCl3.6H2O   Al2O3 + 6HCl + 3H2O

Alums ; M2SO4. M2 (SO4)3. 24H2O or MM (SO4)2. 12H2O

Alums are transparent crystalline solids having the above general formula where M is almost any
univalent positive cation (except Li+ because this ion is too small to meet the structural requirements of
the crystal) and M’ is a trivalent positive cation (Al 3+, Ti3+, V3+, Cr3+, Fe3+, Mn3+, Co3+, Ga3+ etc.). Alums
contain the ions [M(H2O)6]+, [M’(H2O)6]3+ and SO42– in the ratio 1 : 1 : 2. Some important alums are :
(i) Potash alum K2SO4.Al2(SO4)3.24H2O (ii) Chrome alum K2SO4.Cr2(SO4)3.24H2O
(iii) Ferric alum K2SO4.Fe2(SO4)3.24H2O (iv) Ammonium alum (NH4)2SO4.Al2(SO4)3.24H2O
Alums are double salts which when dissolved in water produce metal ions (or ammonium ions) and the
sulphate ions.
 Preparation :
A mixture containing solutions of M2SO4 and M’2(SO4)3 in 1 : 1 molar ratio is fused & then the resulting
mass is dissolved into water. From the solution thus obtained, alums are crystallised.
 Uses :
(i) As a mordant in dye industry. The fabric which is to be dyed is dipped in a solution of the alum and
heated with steam. Al(OH)3 obtained as hydrolysis product of [Al(H2O)6]3+ deposits into the fibres and
then the dye is absorbed on Al(OH)3.
(ii) As a germicide for water purification
(iii) As a coagulating agent for precipitating colloidal impurities from water.

Ex-9. List the cations which are capable of replacing aluminium in alums ?
Sol. Cations of about the same size as that of Al3+ such as Ti3+, Cr+3, Mn+3, Fe3+ and Co3+ are capable of
replacing aluminium in alums.
Carbon Halides :
CF4 (Carbon tetrafluoride)
It is an extraordinarily stable compound.

 Lab preparation : SiC + F2   SiF4 + CF4
 No hydrolysis is possible.
CCl4 (Carbon tetrachloride)
Common solvent, fairly readily decomposed photo chemically.
 Although it is thermodynamically unstable with respect to hydrolysis, the observe of acceptor orbitals
carbon makes the attack very difficult.
CBr4 (Carbon tetrabromide)
Pale yellow solid at room temperature.
 Insoluble in water, soluble in non-polar solvent.
CI4 (Carbon tetraiodide)
Bright red, crystalline material.
 odor like that of iodine.

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p-Block Elements (B & C family)
Stannous Chloride (SnCl2·2H2O) :
 Preparation :
(i) Sn + 2HCl (concentrated)  SnCl2(aq) + H2
(ii) SnO + 2HCl  SnCl2(aq) + H2O
The solution on crystallization gives colourless crystals of SnCl 2·2H2O.
 Properties :
(i) It is a colourless solid soluble in water. It is soluble in alcohol and ether also.
(ii) It is a stronger reducing agent.
(a) Reaction with Hg2Cl2 solution: When SnCl2 solution is added to an aqueous solution of mercuric
chloride, a silky white precipitate of mercurous chloride, Hg2Cl2 is formed which turns black due to
further reduction of Hg2Cl2 to black mercury.
2HgCl2 + SnCl2  Hg2Cl2 + SnCl4 ; Hg2Cl2 + SnCl2  2Hg + SnCl4
(b) It reduces ferric chloride, FeCl3 to ferrous chloride, FeCl2.
2FeCl3 + SnCl2  2FeCl2 + SnCl4
(c) It also reduces CuCl2 to CuCl (white).
(iii) SnCl2 partially hydrolyses in water forming the basic chloride, Sn(OH)Cl.
SnCl2 + H2O  Sn(OH)Cl (white) + HCl
As it produces a weak base and strong acid its aqueous solution is acidic. Its hydrolysis can be
prevented by adding concentrated HCl to it during the process of its preparation.
 Uses :
(i) In dye industry as a reducing agent.
(ii) For the test of mercuric salt.
(iii) For the preparation of other stannous compounds.
Stannic Chloride (SnCl4) :
 Preparation :
(i) By the action of Cl2 gas on heated Sn, Sn + 2Cl2  SnCl4
(ii) By the action of Cl2 on stannous chloride, SnCl2 + Cl2  SnCl4
 Properties :
(i) It is a colourless fuming liquid ; boiling point is 114ºC.
(ii) Action of moisture: It absorbs moisture and becomes converted into hydrated stannic chlorides,
SnCl4·3H2O, SnCl4·5H2O, SnCl4·6H2O and SnCl4·8H2O.SnCl4. 5 H2O is known as “butter of tin” or
“oxymercurate of tin”.
(iii) Hydrolysis with water : It hydrolyses in dilute solution but it is incomplete and can be repressed in
presence of halogen acid.
SnCl4 + 4H2O  Sn(OH)4 + 4HCl ; Sn(OH)4 + 4HCl  SnCl4 + 4H2O ;
SnCl4 + 2HCl  H2SnCl6 (stannic acid)
(v) In presence of ammonium chloride, it forms ammonium salt of H2SnCl6 (stannic acid).
SnCl4 + 2 NH4Cl  (NH4)2SnCl6
 Uses :
For the preparation of stannic compounds.

Lead Chloride (PbCl2) :

 Preparation:
Pb(OH)2·PbCO3 (basic lead carbonate) + 4HCl  2PbCl2 + CO2 + 3H2O
 Properties :
It is a white crystalline solid, insoluble in cold water but soluble in boiling water. It dissolves in
concentrated HCl forming a complex ion.
2 HCl + PbCl2 H2PbCl4 (chloroplumbous acid)
 Uses :
It is used for making pigments for paints.

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 p-Block Elements (B & C family)
Lead Tetrachloride (PbCl4) :
 Preparation :
It is prepared by the following methods:
(i) By dissolving PbO2 in cold concentrated HCl
PbO2 + 4HCl  PbCl4 + 2H2O
PbCl4 dissolves in excess of HCl to form a stable solution of H2PbCl6.
PbCl4 + 2HCl  H2PbCl6
 When NH4Cl is added to a solution of chloroplumbic acid, a yellow precipitate of ammonium
chloroplumbate is formed.
H2PbCl6 + 2NH4Cl  (NH4)2PbCl6 + 2HCl
 When crystals of ammonium chloroplumbate is added to ice cold concentrated H 2SO4, lead
tetrachloride is formed and separates as a yellow oily liquid.
(NH4)2PbCl6 + H2SO4   PbCl4 + (NH4)2SO4 + 2HCl
(ii) By the action of Cl2 on a solution of PbCl2 in concentrated HCl
PbCl2 + Cl2   PbCl4

 Properties :
(i) It is a yellow oily liquid which solidifies at –10oC and is soluble in organic solvents like ethanol and
benzene.
(ii) Rapid hydrolysis with water forms PbO2 precipitate
PbCl4 + 2H2O   PbO2 + 4HCl
 Uses :
It is used for making stannic compounds.

Carbonates (CO32–) and Bicarbonates (HCO3–)

Carbonic acid is a dibasic acids giving rise to two series of salts, carbonates (normal salts) and
bicarbonates (acid salts) due to successive removal of the replaceable hydrogens from H 2CO3.
H2CO3 + NaOH   NaHCO3 + H2O ; NaHCO3 + NaOH   Na2CO3 + H2O

 Preparation :
(i) With NaOH : 2NaOH + CO2   Na2CO3 ; Na2CO3 + H2O + CO2 
 2NaHCO3
(ii) By precipitation : BaCl2 + Na2CO3  BaCO3  + 2NaCl

SiCl4, Silicones, Silicates & Zeolites :

Silicones :
Silicones are synthetic organosilicon compounds having repeated R 2SiO units held by Si–O–Si
linkages. These compounds have the general formula (R2SiO)n where R = alkyl or aryl group.
The silicones are formed by the hydrolysis of alkyl or aryl substituted chlorosilanes and their
subsequent polymerisation. The alkyl or aryl substituted chlorosilanes are prepared by the following
reactions.
Cu
(a) RCl + Si   R3SiCl + R2SiCl2 + RSiCl3
300ºC

(b) RMgCl + SiCl4  RSiCl3 + MgCl2

2RMgCl + SiCl4  R2SiCl2 + 2MgCl2
3RMgCl + SiCl4  R3SiCl + 3MgCl2

After fractional distillation, the silane derivatives are hydrolysed and the ‘hydroxides’ immediately
condense by intermolecular elimination of water. The final product depends upon the number of
hydroxyl groups originally bonded to the silicon atom:

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 p-Block Elements (B & C family)

In this manner several molecules may combine to form a long chain polymer whose both the ends will
be occupied by –OH groups. Such compounds are generally represented from the following formula.

The polymer chain depicted above is terminated by incorporating a small quantity of the
monochlorosilane derivative into the hydrolysis mixture.
 Silicones can be prepared from the following types of compounds only.
(i) R3SiCl (ii) R2SiCl2 (iii) RSiCl3
 Silicones from the hydrolysis of (CH3)3 SiCl
H2O
(CH3)3SiCl   (CH3)3Si(OH)

 Silicones from the hydrolysis of a mixture of (CH3)3 SiCl & (CH3)2 SiCl2
The dichloro derivative will form a long chain polymer as usual. But the growth of this polymer can be
blocked at any stage by the hydrolysis product of mono-chloro derivative.

 Silicones from the hydrolysis of trichloro derivative.

When a compound like CH3SiCl3 undergoes hydrolysis, a complex cross-linked polymer is obtained as
chain can grow in three places as

 The hydrocarbon layer along the silicon-oxygen chain makes silicones water-repellent.
 Silicones find a variety of applications because of their chemical inertness, water repelling nature,
heat resistance and good electrical insulation property.
Products having the physical properties of oils, rubbers and resins can be produced using silicones.
Silicone varnishes are such excellent insulators and so heat-resistance that insulating wiring with them
enabled motors to work over-loads that would have set fire to the insulation formerly used. Silicone
fluids are used as hydraulic systems of planes as they are thermally stable and their viscosity alters
very little with temperature. Silicone rubbers are used in placed of ordinary rubber as they retain their
elasticity at much lower temperature than ordinary rubber.

Ex-10. Complete the following reactions

ZnO  Cu
(a) CO + H2   .....................................
420670k, 300atm
(b) R3SiOH + OHSiR3 ......................... + ..............................
(c) Na2CO3 + Si  .......................... + ..............................
ZnO  Cu
Sol. (a) CO + 2H2   CH3OH
420670k, 300atm
(b) R3SiOH + OHSiR3  R3Si–O–SiR3 + H2O
(c) Na2CO3 + Si  Na2SiO3 + C

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 p-Block Elements (B & C family)
Section (H) : Miscellaneous (Silicones, Silicates, Zeolites & Alums)
Silicates :
Binary compounds of silicon with oxygen are called silicates but they contain other metals also in their
structures.
(i) Since the electronegativity difference between O & Si is about 1.7, so Si–O bond can be considered
50% ionic & 50% covalent.
rSi4 
(ii) If we calculate the radius ratio then, = 0.29
rO2 
It suggests that the coordination number of silicon must be 4 and from VBT point of view we can say
that Si is sp3 hybridized. Therefore silicate structures must be based upon SiO 44– tetrahedral units.
(iii) SiO44– tetrahedral units may exist as discrete units or may polymerise into larger units by sharing
corners.
Classification of Silicates :
(A) Orthosilicates :
These contain discrete [SiO4]4– units i.e., there is no sharing of corners with one another as shown is
figure.

Figure : 2
e.g. Zircon (ZrSiO4), Forsterite of Olivine (Mg2SiO4), Willemite (Zn2SiO4)
(B) Pyrosilicate :
In these silicates two tetrahedral units are joined by sharing oxygen at one corner thereby giving
[Si2O7]6– units.

Figure : 3
e.g. Thorteveitite (Sc2Si2O7), Hemimorphite (Zn3(Si2O7) Zn(OH)2H2O)
  (–) charge will be present on the oxygen atoms which is bonded with one Si atom.
(C) Cyclic silicates :
If two oxygen atoms per tetrahedron are shared to form closed rings such that the structure with
general formula (SiO32–)n or (SiO3)n2n– is obtained, the silicates containing these anions are called cyclic
silicates. Si3O96– and Si6O1812– anions are the typical examples of cyclic silicates.
–
O

–
O
– – –
O O O
O O O
O – O –

–
O
O O O O

O– O– O– O
–

– – – –
O O O O O O O
O–

–
6–
O
Si3O9 Si6O1812–
Figure : 4 Figure : 5
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 p-Block Elements (B & C family)
(D) Chain silicates :
Chain silicates may be further classified into simple chain & double chain compounds.
In case of simple chains two corners of each tetrahedron are shared & they form a long chain of
tetrahedron. Their general formula is also same as the cyclic silicates i.e. (SiO 3)n2n–

Figure : 6

Similarly, double chain silicates can be drawn in which two simple chains are joined together by shared
oxygen. Such compounds are also known as amphiboles. The asbestos mineral is a well known
example of double chain silicates. The anions of double chain silicates have general formula (Si4O11)n6n
– –
O O

O– O–
O O
O O O O O
O– O– O–

O O O

O– O– O–
O O O O O O
O– O––

O– O–
Figure : 7
e.g., Synthetic silicates (Li2SiO3, Na2SiO3), Spondumene (LiAl(SiO3)2),
Enstatite (MgSiO3), Diopside (CaMg(SiO3)2 ), Tremolite (Ca2Mg5(Si4O11)2 (OH)2 ), etc.
(E) Two dimensional sheet silicates :
In such silicates, three oxygen atoms of each tetrahedral are shared with adjacent SiO 44– tetrahedrals.
Such sharing forms two dimension sheet structure with general formula (Si2O5)n2n–
e.g. Talc (Mg(Si2O5)2 Mg(OH)2 , Kaolin Al2(OH)4 (Si2O5)
(F) Three dimensional silicates :
These silicates involve all four oxygen atom in sharing with adjacent SiO 44– tetrahedral units.
e.g. Quartz, Tridymite, Crystobalite, Feldspar, Zeolite and Ultramarines.

Ex-11. Draw the structure of cyclic silicate containing Si6O1812– ion.

Sol.

Si6O1812–

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 p-Block Elements (B & C family)
Zeolites :
If aluminium atoms replace few silicon atoms in three-dimensional network of silicon dioxide, overall
structure known as aluminosilicate, acquires a negative charge. Cations such as Na +, K+ or Ca2+
balance the negative charge. Examples are feldspar and zeolites. Zeolites are widely used as a catalyst
in petrochemical industries for cracking of hydrocarbons and isomerisation, e.g., ZSM-5 (A type of
zeolite) used to convert alcohols directly into gasoline. Hydrated zeolites are used as ion exchangers in
softening of “hard” water.

Carborundum (SiC) :
 Preparation : electric furnace
SiO2 + 3C  SiC + 2CO
2000ºC
 Properties :
(i) It is a very hard substance (Hardness = 9.5 Moh)
(ii) On heating it does not melt rather decomposes into elements.
(iii) Not attacked by acids. However, it gives the following two reactions at high temperature.
 
SiC + 2NaOH + 2O2   Na2SiO3 + CO2 + H2O ; SiC + 4Cl2   SiCl4 + CCl4

  It has a diamond like structure in which each atom is sp 3 hybridized. Therefore, each atom is
tetrahedrally surrounded by 4 atoms of other type.

Ex-12. Write the chemical equations involved in the preparation of elemental boron from mineral colemanite.
Sol. Ca2B6O11 + 4SO2 + 11H2O   2Ca(HSO3)2 + 6H3BO3

2H3BO3  B2O3 + 3H2O ; B2O3 + 2Al   Al2O3 + 2B

Uses of boron :
(i) Boron is used in the construction of high impact-resistant steel and, since it absorbs neutrons, in
reactor rods for controlling atomic reactions.
(ii) Boron carbide is used as an abrasive.

Uses of Al :
It is extensively used :
(i) for manufacture of cooking and household utensils.
(ii) as aluminium plating for tanks, pipes, iron bars and other steel objects to prevent corrosion.
(iii) for manufacture of aluminium cables.
(iv) for making precision instruments, surgical apparatus, aircraft bodies, rail coaches, motorboats, car.
(v) Aluminates are important constituents of portland cement.

Uses of carbon : Graphite fibres embedded in plastic material form high strength, lightweight
composites. The composites are used in products such as tennis rackets, fishing rods, aircraft and
canoes. Being good conductor, graphite is used for electrodes in batteries and industrial electrolysis.
Crucibles made from graphite are inert to dilute acids and alkalies. Being highly porous and having
enormous surface area activated charcoal is used in adsorbing poisonous gases; also used in water
filters to remove organic contaminators and in air conditioning system to control odour. Carbon black is
used as black pigment in black ink and as filler in automobile tyres. Coke is used as a fuel and largely
as a reducing agent in metallurgy. Diamond is a precious stone and used in jewellery. It is measured in
carats (1 carat = 200 mg.).

Uses of silicon :
(i) Ultrapure form of germanium and silicon are used to make transistors and semiconductor devices.
(ii) Silicon is a very important component of ceramics, glass and cement.

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 p-Block Elements (B & C family)
MISCELLANEOUS SOLVED PROBLEMS (MSPS)
1. Al and Ga are trivalent in their compounds but monovalent compounds are the most stable down the
13th group. Why ?
Sol. Down the group (13th), the stability of +3 state decreases and that of +1 state increases due to the
prominent "inert pair" effect.
Al3+ > Ga3+ > ln3+ > Tl3+
Most stable  least stable
Tl+ > ln+ > Ga+ > Al+
2. If you have a mixture of CO and CO2, how would you know about the relative proportions of the two
gases in the given mixture ?
Sol. (i) Pass mixture through the Ca(OH)2 solution; CO2 is absorbed by Ca(OH)2. The residual volume
will be that of CO
Ca(OH)2 + CO2  CaCO3 + H2O
(ii) Pass mixture through I2O5 ; CO reduces I2O5 to I2.
5CO + I2O5  I2 + 5CO2
I2 thus liberated is determined by titration with Na 2S2O3.
2Na2S2O3 + I2  2NaI + Na2S4O6
This is the quantitative method of estimation of CO.
3. What will happen if borontrifluoride is kept in moist air ?
(A) It will strongly fume. (B) It will partially hydrolyse.
(C) It will completely hydrolyse. (D) None of these
Ans. (A)
Sol. In moist air it strongly fume : but it is partially hydrolysed by excess of water.
4BF3 + 6H2O  3H3O+ + 3BF4– + B(OH)3
BF3 is a colourless gas.
4. What happens when : (write only chemical reactions)
(a) iodine is treated with SnCl2.
(b) carbondioxide is passed through a concentrated aqueous solution of sodium chloride saturated with
ammonia.
(c) red lead is treated with nitric acid.
(d) dilute nitric acid is slowly reacted with tin.
Sol. (a) 2SnCl2 + I2  2SnCl2I2  SnCl4 + SnI4
(b) NaCl + NH4OH + CO2  NaHCO3 + NH4Cl
(c) Pb3O4 + 4HNO3  2Pb(NO3)2 + PbO2 + 2H2O
(d) Sn + 10HNO3(dilute)  4Sn(NO3)2 + NH4NO3 + 3H2O
5. True / False
(a) BCl3 in aqueous solution exists as B3+ and Cl–.
(b) Pure crystalline boron is very unreactive and it is attacked only at high temperatures by strong
oxidising agents such as a mixture of hot concentrated H2SO4 and HNO3.
(c) AlX3 (X = Cl, Br) exists as dimer and retains dimer formula in non-polar solvents like ether, benzene
etc.
(d) Be2C is called acetylide because it reacts with water yielding ethyne.
(e) Pb3O4 a double oxide, is obtained by heating lead (II) oxide in air.
Ans. (a) False (b) True (c) True (d) False (e) True
Sol. (a) Statement is incorrect. BCl3 hydrolyses in aqueous solution to give boric acid. Because it has large
ionisation energies and to make the enthalpy of solution of BCl 3 negative, the enthalpy of hydration of
B3+ should be very high (~ 600 g kJ) which is unlikely for the small B3+ cation.
(b) 2B + 6HNO3 (aq.)  2H3BO3 (aq.) + 6NO2 (g)
(c) Statement is correct and its dimer structure is as follows. It acquires this structure for attaining an
octet of electrons. Dimer formula retains in non-polar solvent like ether, benzene

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 p-Block Elements (B & C family)
(d) Statement is incorrect as it is methanide because it gives methane on reaction with water.
Be2C + 4H2O  2Be(OH)2 + CH4

(e) 3PbO + O2   Pb3O4

6. Statement-1 : PbO2 is an oxidising agent and reduced to PbO.

 Statement-2 : Stability of Pb(II) > Pb(IV) on account of inert pair effect.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(C) Statement-1 is True, Statement-2 is False.
(D) Statement-1 is False, Statement-2 is True.
Ans. (A)
Sol. Both are correct statements and statement-2 is the correct explanation of statement-1.
7. Write the chemical equations to represent the following reactions.
(a) The oxidation of HCl (aq) to Cl2 (g) by PbO2.
(b) The disproportionation of SnO to Sn and SnO2.
Sol. (a) PbO2 + 4HCl   PbCl2 + 2H2O + Cl2 (b) 2SnO   Sn + SnO2

8. What will happen if we take Si (CH3) Cl3 as a starting material for the preparation of commercial silicon
polymer ?
Sol. With Si(CH3)Cl3 the chain will grow in three places and we will get cross-linked silicon polymer as
shown below :
CH3 CH3
| |
— O — Si — O — Si — O —
| |
O O
| |
— O — Si — O — Si — O —
| |
CH3 CH3

9. Give three properties of diamond.

Sol. Diamond is very hard, high melting solid. It is an electrical insulator.
10. The silicate anion in the mineral kionite is a chain of three SiO 4 tetrahedra that share corners with
adjacent tetrahedra.The mineral also contains Ca2+ ions, Cu2+ ions, and water molecules in a 1:1:1
ratio.
(a) Give the formula and charge of the silicate anion.
(b) Given the complete formula for the mineral.
Sol. (a) The silicate anion has three SiO4 tetrahedra that share corners with adjacent tetrahedra thus silicate
is Si3O10, hence it can be represented as with charge as = 3 × 4n + 10 × (–2) = –8
8–
 O O O 
 | | | 
O — Si — O — Si — O — Si — O
 | | | 
 O O O 

(b) Ca2+, Cu2+ and H2O are in the ratio of 1 : 1 : 1 and to balance (–8) charge of silicate as ion, (+8)
charge is required thus there are two units each of Ca2+, Cu2+ and H2O thus, kinoite has formula
Ca2Cu2Si3O10.2H2O.

11. In what respect the reaction of N2 with (i) CaC2 (calcium carbide) & (ii) BaC2 (barium carbide) differ from
each other.
Sol. (i) CaC2 reacts with N2 to form calcium cyanamide.
1373K
CaC2(s) + N2(g)   CaCN2(s) + C(s)
Calcium cyanamide
(ii) BaC2 reacts with N2 to form barium cyanide
Heating
BaC2(s) + N2(g)   Ba(CN)2 (s)
Barium cyanide

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 p-Block Elements (B & C family)
12. Statement-1 : The thermal stability of hydrides of carbon family is in order :
CH4 > SiH4 > GeH4 > SnH4 > PbH4
 Statement-2 : E—H bond dissociation enthalpies of the hydrides of carbon family decrease down the
group with increasing atomic size.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(C) Statement-1 is True, Statement-2 is False.
(D) Statement-1 is False, Statement-2 is True.
Ans. (A)
Sol. Both are correct statements and statement-2 is the correct explanation of statement-1. Down the group
the size of atom increases and thus bond length increases.
13. Which one of the following element does not dissolve in fused or aqueous alkalies?
(A) Boron (B) Silicon (C) Aluminium (D) None of these
Ans. (D)
Sol. Boron dissolved in fused alkalies according to the following reaction.
fused
2B + 6NaOH   2Na3BO3 + 3H2
Silicon and aluminium dissolved in both fused and aqueous alkalies.
14. What happens when CO2 (g) is passed through sodium meta borate solution ?
Sol. 4NaBO2 + CO2   Na2B4O7 + Na2CO3

15. Which of the following statement(s) is/are correct ?

(A) B2O3 and SiO2 are acidic in nature and are important constituents of glass.
(B) Borides and silicide are hydrolysed by water forming boranes and silanes respectively.
(C) Diborane on reaction with chlorine (g) forms B2H5Cl.
(D) SiO44– gets hydrolysed by acid or water and form Si2O76–.
Ans. (A), (B) and (D)
Sol. (A), (B) and (D) are correct statements but (C) is incorrect.
B2H6 + 6Cl2   2BCl3 + 6HCl
16. Match the following :
Column-I Column-II
(A) Boron (p) Forms acidic oxides.
(B) Carbon (q) Pure crystalline form is obtained by Van Arkel method.
(C) Tin (r) Exists in allotropic forms.
(D) Aluminium (s) Hydroxide is amphoteric in nature.
Ans. (A - p,q,r); (B - p,r) ; (C - r,s) ; (D - s)
Sol. (A) Exists in various allotropic forms and its oxide, B2O3 is acidic in nature.
red hot W
2BI3   2B + 3I2
Van Arkel method
 (B) Exists in various allotropic forms like diamond, graphite etc. and its oxide CO 2 is acidic in nature.
 (C) Exists in allotropic forms like grey tin (-Sn) and white tin (-Sn). Hydroxide is amphoteric in nature.
Sn(OH)4 + 2OH–   [Sn(OH)6]2–
Sn(OH)4 + 4H+   Sn4+ + 4H2O
 (D) Hydroxide is amphoteric in nature.
Al(OH)3 + OH–   [Al(OH)4]–
Al(OH)3 + 3H+   Al3+ + 3H2O

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