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CHEMISTRY
FOR JEE MAIN & ADVANCED
SECOND
EDITION

Exhaustive Theory
(Now Revised)

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d and f-block Elements
PlancEssential
Questions recommended for revision
 31. d A N D f- B L O C K
ELEMENTS

d-BLOCK ELEMENTS

1. INTRODUCTION
The elements in which the last electron enters (n – 1)d orbitals of the atom are called d-block elements. Also, these
elements lie in between s and p block elements in the long form of the periodic table. So, they are also called
transition elements.

2. ELECTRONIC CONFIGURATION AND IRREGULARITIES

The valence shell configurations of these elements can be represented by (n – 1)d1–10ns0,1,2. All the
d-block elements are classified into four series viz 3d, 4d, 5d and 6d orbitals of (n – 1)th main shell. Each series
has 10 elements. Cr(3d5, 4s1), Cu(3d10, 4s1), Mo(4d5, 5s1), Pd(4d10, 5s0), Ag(4d10, 5s1) and Au(5d10, 6s1) clearly show
irregularities in the configurations. These are explained on the basis of the concept that half-filled and completely
filled d-orbitals are relatively more stable than other d-orbitals.

PLANCESS CONCEPTS

It should be noted here that when atoms of these elements form cations, electrons are removed from
the outermost s-subshell instead of the penultimate d-subshell, although the former was filled earlier.

25
Mn : [Ar] 3d5, 4s2 Mn2+ : [Ar]3d5

26
Fe : [Ar]3d6, 4s2 Fe2+ : [Ar] 3d6
Vaibhav Krishnan (JEE 2009 AIR 22)

Illustration 1: To what extent do the electronic configurations decide the stability of oxidation states in the first
series of the transition elements? Illustrate your answer with example.  (JEE MAIN)

Sol: Empty, Half-filled and completely filled orbitals have extra stability
E.g. Mn2+ = [Ar]3d5 , Sc3+ = [Ar]3d0 , Zn2+ = [Ar] 3d10
3 1 . 2 | d and f-block Elements

Illustration 2: What may be the stable oxidation state of the transition element with the following d-electron
configurations in the ground state of their atoms: 3d3, 3d5, 3d8 and 3d4?  (JEE ADVANCED)

Sol: Ground state configuration Stable oxidation state

3d3 +5
3d +2, + 7
5

3d8 +2
3d4 3d4 does not exist

3. GENERAL PROPERTIES OF THE TRANSITION METALS

3.1 Atomic and Ionic Radii
(a) The atomic and ionic radii for transition elements are smaller than their corresponding s-block elements and
are greater than their corresponding p-block elements.
(b) The atomic and ionic radii for transition elements for a given series show a decreasing trend for the first five
elements and then becomes almost constant for next five elements of the series. For example, in 3d-series
atomic radius decreases from 21Sc to 25Mn and then becomes constant for next five i.e. 26Fe to 30Zn
Explanation: This is due to the combined effect of the increasing 17
effective nuclear charge, (ENC) and increasing screening effect along
16
the period. An increase in ENC favors a decrease in atomic radii, whereas
15
increase in number of d-elements increases the screening effect and
thus increases the atomic radii. Thus both ENC and screening effect act 14
opposite to each other and therefore the atomic size is governed by the 13
net influence of these two. 12
11
(c) The atomic and ionic radii of the elements of 4d-series are higher than Sc Ti V Cr Mn Fe Co Ni Cu Zn
3d-series as the number of shells increases down the group. However, Figure 31.1: Atomic radii of element
the elements of 4d-series and 5d-series on moving down the group of 3d-series
reveal almost constant value. For example, Zirconium and Hafnium, the
members of 4d and 5d-series, respectively have the almost same size i.e. 145 pm. Similarly, Zr4+ and Hf4+ have
their atomic radii as 80 pm and 81 pm respectively. This is due to the Lanthanoid contraction.
(d) The ionic radii decreases as charge on the cation increases (i.e., higher oxidation state). e.g.
Ti2+ > Ti3+ > Ti4+
Cr2+ > Cr3+ > Cr5+ > Cr6+
Fe2+ > Fe3+
(e) For ions having same oxidation states, the ionic radii decreases with increase in atomic number e.g.
For 3d-series
Sc2+ > Ti2+ > V2+ > Cr2– > Mn2+ > Fe2+ > Co2+ > Ni2+ > Cu2+
Ionic radii (in Å) 0.95 0.90 0.88 0.84 0.80 0.76 0.74 0.72 0.69

Illustration 3: In a transition series, with an increase in atomic number, the atomic radius does not change very
much. Why is it so?  (JEE MAIN)

Sol: With increase in atomic number along a transition series, the nuclear charge increases which tends to
decrease the size of the atom. But, the addition of electrons in the d-subshell increases the screening effect which
counterbalances the increased nuclear charge. Hence, along a transition series the atomic radius does not change
very much.
 Chem i str y | 31.3

3.2 Atomic Volume and Density

(a) The size decreases along the period and, therefore, atomic volume also decreases along the period.
(b) Atomic volumes are smaller than group 1 and 2 members i.e. s-block elements.
(c) The density, however, increases along the period.

3.3 Melting and Boiling Points

(a) All the transition elements have a higher melting point as compared to s-block elements due to strong
metallic bonding as well as unpaired d-electrons leading to covalence.
(b) It is evident from that the melting point of transition metals or a given series increases on moving left to right
in a period and attains a maximum value and after that the m.p. goes on decreasing towards the end of period.
This is due to the fact that the strength of inter particle bonds in transition elements is also directly related to the
number of half-filled d-orbitals. In the beginning, the number of unpaired electrons in d-orbitals increases till
the middle of the period (d1 to d5). After this, the pairing of electrons occurs in d-orbitals (d6 to d10). An increase
in inter atomic bonds due to the increase in number of unpaired electron results in higher m.p.

W
3500
Re
Ta
3000 Os
Mo
Ru
lr
2500 Nb
Tc
Melting point (K)

Cr Rh
Ti Zr Pt
2000
Y Fe
Hf Pd
(4d-series) V
1500 Co
Sc Ni
(3d-series) Mn Cu
Au
1000 Ag
La
(5d-series) Zn
500 Cd

Hg

1 2 3 4 6 5
7 8 9 10 11 12
Groups
Figure 31.2: Trends in melting points of different group

3.4 Metallic Character

(a) All the transition metals possess one or two electrons in their outermost shell and thus exhibit metallic nature.
(b) All are hard, ductile and malleable solids with strong metallic bonding (except mercury which is liquid) and
possess hcp, bcc or ccp crystal lattices.
(c) Transition metals show a gradual decrease in electropositive character on moving along the period.
(d) Strong metallic bonding in transition metals is due to greater effective nuclear charge and a large number of
valence electrons (inner d-subshell and outermost s-subshell).
(e) Due to strong metallic bonding, transition metals are hard, possess high densities and high enthalpies for
atomization.
3 1 . 4 | d and f-block Elements

(f) Due to metallic bonding, these are good conductors of heat and electricity.
(g) Transition metals form numerous useful alloys with other metals.

3.5 Enthalpies of Atomisation

Transition elements have a high enthalpy of atomization due to strong interatomic attraction. Greater the number
of valency electrons, stronger is the resultant bonding and higher is the enthalpy of atomization. The members
of 4d and 5d-series have greater enthalpy of atomization than those of 3d-series. Thus, they form metal-metal
bonding frequently in their compounds.

3.6 Ionization Energy

(a) The ionization energy (IE) of d-block elements lies in between s-and p-block elements showing less
electropositive character than s-block
(b) Smaller atomic size and a fairly high IE is noticed for transition metals.
(c) IE values first increases up to 25Mn and then becomes irregular or constant values due to irregular trend of atomic
size after 25Mn in 3d-series. Similar trend is noticed in 5d-and 6d-series.
(d) The magnitude of ionization energies give an idea about the relative stabilities of various oxidation states of
transition elements.

lr
Pt Au(5d)
9 Os
IE1 (x 10-2 kj mol-1)

Pd
8 W Re Fe Co
Hf Ta
Cu(3d)
Mn
Ni
7 Mo Rh Ag(4d)
Zr Nb Tc Ru
Ti
V Cr
6 Sc
Transition elements
Figure 31.3: Ionization energies of transition elements

PLANCESS CONCEPTS

The ionization energy of 5-d series are higher than that of 3-d and 4-d series due to the poor shielding
effect of 4-f electrons present in 3-d series
Ionization energy of Zn, Cd and Hg are abnormally higher on account of greater stability of s-subshell

Nikhil Khandelwal (JEE 2010 AIR 443)

Illustration 4: In the series Sc(Z = 21) to Zn(Z = 30), the enthalpy of atomization of zinc is the lowest, i.e. 125 kJ mol–1.
Why?  (JEE ADVANCED)

Sol: In the formation of metallic bonds, no electrons from 3d-orbitals are involved in case of zinc, while in all other
metals of the 3d series, electrons from the d-orbitals are always involved in the formation of metallic bond. This is
why the enthalpy of atomization of zinc is the lowest in the series.
 Chem i str y | 31.5

3.7 Reactivity
(a) Transition metals are less reactive than s-block elements. Their low reactivity is due to:
(i) High ionization energy.
(ii) Existence of metallic bonding among atoms which gives rise to higher heat of sublimation.
M(s) → M(g) ∆H = Heat of sublimation (∆Hs)

M(g) → M(g)

+
+ e- ∆H = Heat of ionization or ionization energy (IE)
+
+
M(g) → M(aq) ∆H = – Heat of hydration (– ∆Hh)

+
M(s) + H2O → M(aq) + e- ∆H = ∆Hs + IE – ∆Hh

+
 ore negative the value of ∆H for the change, lesser is the energy level for M(aq)
M and greater will be the stability
of that oxidation state in aqueous solution.

3.8 Variable Valency and Oxidation State

(a) Most of the transition elements show variable valencies or different oxidation states because of incomplete
d-subshell. The variable oxidation states of transition elements are due to the participation of ns and (n – 1)
d electrons in bonding.
(b) It is thus evident that for the first five transition elements, the minimum oxidation state is given by the
electrons in outermost s-subshell and the maximum oxidation state by the total number of ns and (n – 1)
d-subshell electrons.
(i) 21
Sc shows +2 and +3 (due to 4s2 and 3d1 electrons)
(ii) 22
Ti shows + 2, +3 and +4 (due to 4s2 and 3d2 electrons)
(iii) 23
V shows +2, +3, +4, +5 (due to 4s2 and 3d3 electrons)
(iv) 24
Cr shows +2, +3, +4, +5, +6 (due to 4s1 and 3d5 electrons)
(v) 25
Mn shows +2 to +7 (due to 4s2 and 3d5 electrons)

PLANCESS CONCEPTS

•• The transition elements in their lower oxidation states (+2 and +3) usually form ionic compounds. In
higher oxidation state, compounds are normally covalent. For example, Mn in MnCl2 has Mn2+ ion whereas
in KMnO4, Mn exists in +7 state of covalent nature.
•• Some transition metals also exhibit zero oxidation state in their compounds such as carbonyls. Ni and Fe
in Ni(CO)4 and Fe(CO)5 have zero oxidation state.
•• The highest oxidation states of transition metals are found in their compounds with fluorine and oxygen.
This is due to higher electronegativity and small atomic size of fluorine and oxygen. Eg.MnO4-
Neeraj Toshniwal (JEE 2009 AIR 21)

Illustration 5: Which metal in the first series of transition metal exhibits +1 oxidation state most frequently and
why?  (JEE ADVANCED)

Sol: Copper, because it will achieve a completely filled d-orbital and a stable configuration on losing an electron.
3 1 . 6 | d and f-block Elements

3.9 Colour
Substances appear coloured when they absorb light of a particular wavelength in the visible region of the spectrum
and transmit light of other wavelengths. The colour which we see is the colour of the transmitted wavelengths. In
other words, the colour of the compound observed by us is the complementary colour of the colour absorbed by
the compound.
In the s-and p-block elements, there cannot be any d-d transistions and the energy needed to promote the s or p
electrons to a higher level is much greater and may correspond to ultraviolet region, in which case the compound
will not appear coloured to the eye.
Table 31.1: Color spectrum

Wavelength absorbed in nm Colour absorbed Colour observed

< 400 UV region White/colourless
400–435 Violet Yellow-green
435–480 Indigo Yellow
480–490 Green-blue Orange
490–500 Blue-green Red
500–560 Green Purple
560–580 Yellow-green Violet
580–595 Yellow Indigo
595–605 Orange Green-blue
605–750 Red Blue-green
> 750 Infra-red White/colourless

Illustration 6: Transition metal ions like Cu+, Ag+, Zn2+, Hg2+ and Cd2+ are colourless. Explain.  (JEE ADVANCED)

Sol: Due to forbidden transition some metal ions are colourless. All the ions reported above have no unpaired
electrons in them and the d-orbital [(n – 1)d10] is also completely filled. Thus, due to d-d transition above mentioned
metal ions are colourless.

3.10 Magnetic Properties

Magnetic Properties: When a substance is placed in a magnetic field of strength H, the intensity of the magnetic
field on the substance may be greater than or less than H.
Diamagnetic: The substances which are weakly repelled by a magnetic field; absence of unpaired electrons.
Paramagnetic: The substances which are weakly attracted by the magnetic field and lose their magnetism when
removed from the field; presence of unpaired electrons.
Paramagnetism is expressed by magnetic moment,
µ= n(n + 2) B.M.
n = Number of unpaired electrons
B.M. = Bohr Magneton, unit of magnetic moment
 Chem i str y | 31.7

Illustration 7: The paramagnetic character in 3d transition series increases up to Cr and then decreases. Explain
 (JEE MAIN)
Sol: As number of unpaired electron increases paramagnetic nature increases.The number of unpaired electrons
increases from 21Sc: [Ar]3d1,4s2 to 24Cr : [Ar] 3d5, 4s1 and after chromium, the pairing of electrons takes place and
thus number of unpaired electrons goes on decreasing continuously to 30Zn: [Ar] 3d10, 4s2.

PLANCESS CONCEPTS

The magnetic properties of d-block elements are due to the only spin value of the unpaired electrons
present in d-orbital while in the case of f-block elements, it is due to both the orbital motion as well as
spin contribution.
Magnetic moment for d-block elements: µ = n(n + 2) B.M. (where n is the number of unpaired electrons)

Magnetic moment of f-block elements:

= µ 4s(s + 1) + ( + 1) . (where s is sum of spin quantum number
and  , the angular momentum quantum number)

Aman Gour (JEE 2012 AIR 230)

4. SOME IMPORTANT COMPOUNDS OF d-BLOCK ELEMENTS

4.1 Chromate and Dichromate
(a) Preparation: 4FeO.Cr2O3 + 8Na2CO3 + 7O2 
Roasting
→ 8Na2CrO4 + 2Fe2O3 + 8CO2
inair
The roasted mass is extracted with water when Na2CrO4 goes into the solution leaving behind insoluble Fe2O3.
The solution is treated with calculated amount of H2SO4.
2Na2CrO4 + H2SO4 → Na2Cr2O7 + Na2SO4 + H2O
The solution is concentrated when less soluble Na2SO4 crystallizes out. The solution is further concentrated
when crystals of Na2Cr2O7 are obtained. Then a hot saturated solution of Na2Cr2O7 is treated with KCl, then
reddish orange crystals of K2Cr2O7 are obtained on crystallization
(b) K2Cr2O7 is preferred to Na2Cr2O7 because Na2Cr2O7 is hygroscopic but K2Cr2O7 is not.
(c) Similarities between hexavalent Cr & S-compounds:
(i) SO3 & CrO3 → Both acidic
(ii) CrO −42 & SO24− Isomorphous
OH −
(iii) SO2Cl2 & CrO2Cl2  → SO24− & CrO24− respectively
OH −
(iv) SO3Cl– + CrO3Cl–  → SO24− & CrO24−
O O O
(v) CrO3 & B(SO3) has same structure Cr O Cr O Cr

O O O
(vi) Potassium dichromate reacts with hydrochloric acid and evolves chlorine.
K2Cr2O7 + 14HCl → 2KCl + 2CrCl3 + 7H2O + 3Cl2

(vii) It acts as a powerful oxidizing agent in an acidic medium (dilute H2SO4)

Cr2O27− + 14H+ + 6e– → 2Cr3+ + 7H2O (Eº = 1.33V)
The oxidation state of Cr changes from +6 to +3
3 1 . 8 | d and f-block Elements

(d) Uses: As a volumetric reagent in the estimation of reducing agents such as oxalic acid, ferrous ions, iodide
ions, etc.
(i) For the preparation of several chromium compounds such as chrome alum, chrome yellow, chrome red,
zinc yellow, etc.
(ii) In dyeing chrome tanning, calico printing, photography etc.
(iii) Chromic acid as a cleansing agent for glass ware

4.2 Manganate and Permanganate

Preparation: This is the most important and well known salt of permanganic acid. It is prepared from the pyrolusite
ore. It is prepared by fusing pyrolusite ore either with KOH or K2CO3 in the presence of atmospheric oxygen or any
other oxidizing agent such as KNO3. The mass turns green with the formation of potassium magnate, K2MnO4
2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
2MnO2 + 2K2CO3 + O3 → 2K2MnO4 + 2CO2
The fused mass is extracted with water. The solution is now treated with a current of chlorine or ozone or carbon
dioxide to convert magnate into permanganate.
2K2MnO4 + Cl2 → 2KMnO4 + 2KCl
2K2MnO4 + H2O + O3 → 2KMnO4 + 2KOH + O2
3K2MnO4 + 2CO2 → 2KMnO4 + MnO2 + 2K2CO3

Another Method of Preparation:

3K2MnO4 + 2H2SO4 → 2KMnO4 + MnO2 ↓ + 2K2SO4 + 2H2O
or 3K2MnO4 + 2H2O + 4CO2 → 2KMnO4 + MnO2 + 4KHCO3
1
But, in the above method
of Mn is lost as MnO2 but when oxidized either by Cl2 or by O3,
3
2K2MnO4 + Cl2 → 2KMnO4 + 2KCl [Unwanted MnO2 does not form]
OR
2K2MnO4 + O3 + H2O → 2KMnO4 + 2KOH + O2
∆
Heating effect: 2KMnO 4 → K 2MnO 4 + MnO2 + O2
green 200ºC Black
at red
2K2MnO4 
→ 2K2MnO3 + O2
hot

Oxidising Property of KMnO4: (In acidic medium)

(i) MnO −4 + Fe+2 + H+ → Mn+2 + O2 + H2O +Fe3+

(ii) MnO −4 + H2O2 + H+ → Mn+2 + O2 + H2O

(iii) MnO −4 + H2S → Mn2+ + S ↓ + H2O

In alkaline solution: KMnO4 is first reduced to manganate and then to insoluble manganese dioxide. Colour
changes first from purple to green and finally becomes colourless. However, a brownish precipitate is formed
2KMnO4 + 2KOH → 2K2MnO4 + H2O + O

2K2MnO4 + 2H2O → 2MnO2 + 4KOH + 2O

alkaline
2KMnO4 + H2O  → 2MnO2 + 2KOH + 3[O]

or 2MnO −4 + H2O → 2MnO2 + 2OH– + 3[O]

 Chem i str y | 31.9

In neutral or weakly acidic solution:

in presence Zn+2or ZnO
(i) 2KMnO4 + 3MnSO4 + 2H2O 
→ 5MnO2 + K2SO4 + 2H2SO4

Conversion of Mn+2 to MnO −4 :

(i) PbO2 (ii) Pb3O4 + HNO3 (iii) Pb2O3 + HNO3

(iv) NaBiO3/H+ (v) (NH4)2S2O8/H+ (vi) KIO4/H+

PLANCESS CONCEPTS

In the oxidation reactions of KMnO4 in acidic medium, only H2SO4 is used to produce an acidic medium
and not HCl or HNO3 because HCl reacts with KMnO4 and produce Cl2 while HNO3, itself acts as an
oxidising agent.
B Rajiv Reddy (JEE 2012 AIR 11)

Illustration 8: Complete the equation of following chemical reactions:  (JEE MAIN)

(i) MnO −4 (aq) + S2O32− (aq) + H2O(  ) →
(ii) CrO7− (aq) + H2S(g) + H+(aq) →

Sol:
(i) In neutral or faintly alkaline solutions
MnO −4 + 2H2O + 3e– → MnO2 + 4OH– × 8

S2O32− + 10OH– → 2SO24− + 5H2O + 8e– × 3

8MnO32− + 10OH– + H2O → 8MnO2 + 6SO24− + 2OH–

(ii) In acidic solutions

Cr2O27− + 14H+ + 6e– → 2Cr3+ + 7H2O

H2S → S + 2H+ + 2e– × 3

CrO27− + 3H2S + 8H+ → 2Cr3+ + 3S + 7H2O

Illustration 9: Write steps involved in the preparation of  (JEE ADVANCED)

(i) Mn2CrO4 from chromite ore and
(ii) K2MnO4 from pyrolusite ore.

Sol: (i) Chromite ore is fused with sodium carbonate in excess of air.
4FeCr2O4 + 8Na2CO + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2
Chromite ion Sod. Chromites
(ii) K2MnO4 from pyrolusite ore
2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O

2MnO2 + 2K2CO3 + O3 → 2K2MnO4 + 2CO2

3 1 . 1 0 | d and f-block Elements

4.3 Silver Nitrate

Properties:
(i) It is called lunar caustic because in contact with skin it produces a burning sensation that of caustic soda with
the formation of finely divided silver (black colour).
(ii) Thermal decomposition
1
AgNO3 → AgNO2 + O2
2
(iii) 6AgNO3 + 3I2 + 3H2O → 5AgI + AgIO3 + 6HNO3 (excess)
∆
(iv) Ag2SO4 → 2Ag + SO2 + O2
B
(v) A - (AgNO3) → white ppt appears quickly
added

A
B - (Na2S2O3) → It takes time to give white ppt.
added
∆
(vi) Ag2S2O3 + H2O  → Ag2S + H2SO4

AgCl, AgBr, AgI (but not Ag2S) are soluble in Na2S2O3 forming [Ag(S2O3)2]–3 complexes

(vii) AgBr+AgNO3 

KBr
→ AgBr ↓ + KNO3
Pale yellow ppt.

212ºC
Heating effect: 2AgNO3  → 2AgNO2 + O2

2AgNO3 
300ºC
→ 2Ag + 2NO + O2
(viii)
Aqua regia
Insoluble
Zn/HCl
Ag NO3 Dil.HCl AgCl [H] Ag + HCl
Na Na2CO3
OH NaOH Ag [4AgCl + 2Na2CO3 4Ag + 4NaCl + 2CO2 + O2]
(conc.)
K2S2O8
Ag2O[2AgCl + 2NaOH Ag2O + 2NaCl + H2O]
2
O

AgO Black Glucose

2
H

Ag [Ag2O + C6H12O6 2Ag + C5H11CO2H]

Ag
Gluconic acid

Ag2O + H2O2 → 2Ag + H2O + O2

K2S2O8 + 2AgNO3 + 2H2O → 2AgO + 2KHSO4 + 2HNO3

 Chem i str y | 31.11

PLANCESS CONCEPTS

AgO is supposed to be paramagnetic due to d9 configuration. But actually it is diamagnetic and exists
as AgI[AgIIIO2]
Silvering of mirror: The process of depositing a thin and uniform layer of silver on a clean glass surface is
known as silvering of mirrors. It is employed for making looking glasses, concave mirrors and reflecting
surfaces. The process is based on the reduction of ammoniacal silver nitrate solution by some reducing
agent like formaldehyde, glucose, etc. The silver film deposited on the glass is first coated with a varnish
and finally painted with red lead to prevent its loss due to scrap.
Rohit Kumar (JEE 2012 AIR 79)

4.4 Zinc Compounds

1. Zinc oxide, ZnO (Chinese white or philosopher’s wool)
It found in nature as zincite or red zinc ore.
(a) Preparation:
(i) 2Zn + O2 → 2ZnO
∆
(ii) ZnCO3  → ZnO + CO2
∆
(iii) 2Zn(NO3)2  → 2ZnO + 4NO2 + O2
(iv) Zn(OH)2 
∆
→ ZnO + H2O

(b) Physical Properties: It is a white powder, which becomes yellow on heating and again turns white on cooling,
is insoluble in water, and sublimes at 400ºC.
(c) Chemical Properties:
(i) ZnO + H2SO4 → ZnSO4 + H2O
(ii) ZnO + 2NaOH → Na2ZnO2 + H2O
∆
(iii) ZnO + H2 → Zn + H2O
400ºC
(iv) ZnO + C → Zn + CO

2. ZnCl2 (Zinc Chloride)

(a) Preparation:

ZnO + 2HCl → ZnCl2 + H2O 


ZnCO3 + 2HCl → ZnCl2 + H2O + CO2  It crystallises as ZnCl2 . 2H2O

Zn(OH)2 + 2HCl → ZnCl2 + 2H2O 

Anhydrous ZnCl2 cannot be made by heating ZnCl2.2H2O because
∆
ZnCl2. 2H2O  → Zn(OH)Cl + HCl + H2O
∆
Zn(OH)Cl  → ZnO + HCl
To get anh. ZnCl2 :

Zn + Cl2 → ZnCl2

Zn + 2HCl(dry) → ZnCl2 + H2
Or Zn + HgCl2 → ZnCl2 + Hg
3 1 . 1 2 | d and f-block Elements

(b) Properties:
(i) It is deliquescent white solid (when anhydrous)
(ii) ZnCl2 + H2S → ZnS
excess
(iii) ZnCl2 + NaOH → Zn(OH)2  → Na2[Zn(OH)4]
excess
(iv) ZnCl2+ NH4OH → Zn(OH)2  → [Zn(NH3)4]2+
(c) Uses:
(i) Used for impregnating timber to prevent destruction by insects
(ii) As a dehydrating agent when anhydrous
(iii) ZnO. ZnCl2 used in dental filling

3. ZnSO4 (Zinc Sulphate)

(a) Preparation:
Zn + dil H2SO4 → ZnSO4 + H2

ZnO + dil H2SO4 → ZnSO4 + H2O

ZnCO3 + dil H2SO4 → ZnSO4 + H2O + CO2

ZnS + 2O2 → ZnSO 4 


3  Parallel reaction
ZnS + O2 → ZnO + SO2 
2 

ZnS + 4O3 → ZnSO4 + 4O2

(b) Properties

39-70°C >70°C >280°C

ZnSO4 .7H2O ZnSO4 .6H2O ZnSO4 .H2O ZnSO4

>800°C
1
O + SO2 + ZnO
2 2
(c) Uses:
(i) In eye lotion
(ii) Lithopone making (ZnS + BaSO4) as white pigment

Illustration 10: The addition of NH4OH to ZnSO4 solution produces white precipitate but no precipitate is formed
if it contains NH4Cl. Why?  (JEE ADVANCED)

Sol: NH4OH is a weak hydroxide. It ionizes slightly, furnishing OH– ions. However, the OH– ions are sufficient to
cause the precipitation of Zn(OH)2 as its ionic product exceeds the Ksp.
ZnSO4 + 2NH4OH → Zn(OH)2 + (NH4)2SO4]

White ppt
In the presence of NH4Cl, the ionization of NH4OH is further suppressed and sufficient OH– ions are not available
to cause precipitation as the ionic product does not exceed the Ksp.
 Chem i str y | 31.13

PLANCESS CONCEPTS

Zinc oxide is white at room temperature but turns yellow on heating

Krishan Mittal (JEE 2012, AIR 199)

4.5 Copper Compounds

1. CuO:
(a) Preparation:
∆
(i) CuCO3.Cu(OH)2  → 2CuO + H2O + CO2 (Commercial process)
Malachite green
(Native Cu-carbonate)
(ii) 2Cu + O2 → 2CuO & Cu2O + 1 O2 → 2CuO
2
∆
(iii) Cu(OH)2  → CuO + H2O
250ºC
(iv) 2Cu(NO3)2  → 2CuO + 4NO2 + O2

(b) Properties:
(i) CuO is insoluble in water
(ii) Readily dissolves in dil. acids
CuO + H2SO4 → CuSO4 + H2O

CuO + HCl → CuCl2

CuO + HNO3 → Cu(NO3)2

(iii) It decomposes when, heated above 1100ºC
4CuO → 2Cu2O + O2
(iv) CuO is reduced to Cu by H2 or C under hot condition
CuO + C → Cu + CO ↑

CuO + H2 → Cu + H2O↑

2. CuCl2:
(a) Preparation:
CuO + 2HCl(conc.) → CuCl2 + H2O
Cu(OH)2.CuCO3 + 4HCl → 2CuCl2 + 3H2O + CO2

(b) Properties:
(i) It is crystallized as CuCl2.2H2O of Emerald green colour
(ii) Dilute solution in water is blue in colour due to the formation of [Cu(H2O)4]2+ complex.
(iii) conc. HCl or KCl added to dil. solution of CuCl2, the colour changes into yellow, owing to the formation
of [CuCl4]2–
3 1 . 1 4 | d and f-block Elements

(iv) The conc. aq. solution is green in colour having the two complex ions in equilibrium
2[Cu(H2O)4]Cl2 → [Cu(H2O)4]2+ + [CuCl4]2– + 4H2O
(v) CuCl2 → CuCl
∆
•• CuCl2 + Cu-turning  → 2CuCl
•• 2CuCl2 + H2SO3 + H2O → 2CuCl + 2HCl + 2H2SO4
•• 2CuCl2 + Zn/HCl → 2CuCl + ZnCl2
•• CuCl2 + SnCl2 → CuCl + SnCl4

CuF2 .2H2O → light blue Anh. CuCl is dark brown mass obtained
2

CuCl2 .2H2O → green  by heating CuCl2 .2H2O at150º C in presence

CuBr → almost black  of HCl vap.
2

CuI2 does not exist

150ºC
CuCl2.2H2O  → CuCl2 + 2H2O
HCl gas

3. CuSO4:
(a) Preparation:
CuO + H2SO4(dil) → CuSO4 + H2O.
Cu(OH)2 +H2SO4 (dil) → CuSO4 + 2H2O.Cu(OH)2.
CuCO3 +H2SO4(dil) → CuSO4 + 3H2O + CO2
1
Cu + H2SO4 + O2 → CuSO4 + H2O [Commercial scale]
2
(Scrap)
Cu + dil. H2SO4 → no reaction {Cu is below H in electrochemical series}

(b) Properties:
(i) It is crystallized as CuSO4.5H2O
On exposure 100°C
(ii) CuSO4.5H2O CuSO4.3H2O CuSO4.H2O
Effloroscence Pale blue Bluish white
Take place
230°C

CuSO4(anh.)
800°C white
750°C
1
CuO + SO2 + O2
2 CuO + SO3

PLANCESS CONCEPTS

Anhydrous copper sulphate (white) regains its blue colour when moistened with a drop of water (it is a
test of water).
T P Varun (JEE 2012, AIR 640)
 Chem i str y | 31.15

Illustration 11: Blue copper sulphate turns white on heating. Why ?  (JEE MAIN)
Sol: Hydrated copper sulphate (CuSO4.5H2O) is blue which on heating loses its water of crystallization to form
Heat
anhydrous CuSO4 a white compound. CuSO4.5H2O  → CuSO4 + 5H2O

Illustration 12: Cu+ ion is not stable in an aqueous solution. Why ?  (JEE ADVANCED)
Sol: Cu (aq) is much more stable than Cu (aq). This is because although the second ionization enthalpy of copper
2+ +

is large but Dhyd H for Cu2+(aq) is much more negative than that for Cu+(aq) and therefore it more than compensates
for the second ionization enthalpy of copper. Thus many copper (I) compounds are unstable in aqueous solution
and undergoes disproportion as follows 2Cu+(aq) → Cu2+(aq) + Cu(s)

4.6 Iron Compounds

1. FeSO4.7H2O:

(a) Preparation:
(i) Scrap Fe + H2SO4 → FeSO4 + H2↑
(dil.)
(ii) From Kipp’s waste
FeS + H2SO4(dil) → FeSO4 + H2S↑
7
(iii) FeS2 + 2H2O + O → FeSO4 + H2SO4
2 2
(b) Properties:
(i) It undergoes aerial oxidation forming basic ferric sulphate

4FeSO4 + H2O + O2 → 4Fe(OH)SO4

300ºC high
(ii) FeSO4.7H2O  → FeSO 4  → Fe2O3 + SO2 + SO3
anh. white temp.

(iii) Aq. Solution is acidic due to hydrolysis

FeSO4 + 2H2O Fe(OH)2 + H2SO4

Weak base
(iv) It is a reducing agent
•• Fe2+ + MnO −4 + H+ → Fe3+ + Mn2+ + H2O
•• Fe2+ + Cr O2− + H+ → Fe3+ + Cr3+ + H O
2 7 2

•• Au + Fe
3+ 2+
→ Au + Fe 3+

•• Fe2+ + HgCl2 → Hg2Cl2 ↓ + Fe3+

White ppt.
(v) It forms double salt. Example (NH4)2SO4.FeSO4.6H2O

2. FeO (Black):

(a) Preparation: FeC2O4 →

∆
FeO + CO + CO2
in absence of air

(b) Properties: It is stable at high temperature and on cooling slowly disintegrates into Fe3O4 and iron
4FeO → Fe3O4 + Fe
3 1 . 1 6 | d and f-block Elements

3. FeCl2:
heating in current of HCl
(a) Preparation: Fe + 2HCl  → FeCl2 + H2
OR
∆
2FeCl3 + H2 → 2FeCl2 + 2HCl

(b) Properties:
(i) It is deliquescent in air like FeCl3
(ii) It is soluble in water, alcohol and ether also because it is sufficiently covalent in nature
(iii) It volatilizes at about 1000ºC and vapour density indicates the presence of Fe2Cl4. Above 1300ºC density
becomes normal
(iv) It oxidizes on heating in air
12FeCl2 + 3O2 → 2Fe2O3 + 8FeCl3
(v) H2 evolves on heating in steam 3FeCl2 + 4H2O → Fe3O4 + 6HCl + H2
(vi) It can exist as a different hydrated form
FeCl2.2H2O → colorless
FeCl2.4H2O → pale green
FeCl2.6H2O → green
4. FeCl3:
(a) Preparation: 2 Fe(s) + 3 Cl2(g) → 2 FeCl3(s)
(b) Properties: Anhydrous ferric chloride is prepared by heating metallic iron in a stream of dry chlorine gas.
(i) FeCl3 solid is almost black. It sublimes at about 300ºC, giving a dimeric gas.
(ii) FeCl3 dissolves in both ether and water, giving solvated monomeric species.
(iii) Iron (III) chloride is usually obtained as yellow-brown lumps of the hydrate FeCl3.6H2O.
(iv) This is very soluble in water and is used both as an oxidizing agent, and as a mordant in dyeing.

PLANCESS CONCEPTS

Anhydrous ferric chloride is soluble in non-polar solvents like ether, alcohol, etc as it possesses covalent
bonds and has a chlorine bridge structure. Cl Cl Cl
Fe Fe
Cl Cl Cl

Saurabh Chaterjee (JEE Advanced 2013, AIR)

Illustration 13: FeCl3(aq) gives CO2 with NaHCO3(aq) Explain.  (JEE ADVANCED)

Sol: Fe3+ ions hydrolyse to form alkaline solution which reacts with NaHCO3 to liberate CO2.
Fe3+ + 3H2O → Fe(OH)3 + 3OH–
OH– + 2NaHCO3 → Na2CO3 + CO2 + H2O
 Chem i str y | 31.17

f-BLOCK ELEMENTS

1. INTRODUCTION

1.1 Inner transition elements/f-block elements

The elements in which the additional electrons enters (n – 2)f orbitals are called inner transition elements. The
valence shell electronic configuration of these elements can be represented as (n-2) f0.2….14(n –1)d0,1,2ns2. These are
also called as f-block elements because the extra electrons go to f-orbitals which belongs to (n–2) th main shell.
4f-block elements are also called Lanthanides or rare earths. Similarly, 5f-block elements are called actinides or
actinones. The name Lanthanides and Actinide have been given due to close resemblance with Lanthanum and
Actinium respectively. Lanthanides constitutes the first inner transition series while actinides constitutes second
inner transition series.

General Characteristics:
1. Electronic Configuration : [Xe] 4fn+15d06s2 or [Xe] 4fn5d16s2
2. Oxidation state: They readily form M+3 ions. Some of them also exhibit oxidation state of +2 and +4.
3. Colouration: Ions of Lanthanides and Actinides are coloured in the solid state as well as in aqueous solution
because of absorption of light due to f-f-transition, since they have partly filled f-orbitals.

Magnetic properties: La3+(4f0) and Lu3+(4f14) having no unpaired electrons do not show paramagnetism while all
other tri-positive ions of lanthanides are paramagnetic

Illustration 14: What is the basic difference between the electronic configuration of transition and inner transition
elements.  (JEE MAIN)

Sol: General electronic configuration of transition elements = [Noble gas] (n – 1) d1–10ns1–2 and for inner transition
elements = (n – 2) f1–14(n – 1)d0–1ns0–2. Thus, in transition elements, last electron enters d-orbitals of the penultimate
shell while in inner transition elements, it enters f-orbital of the penultimate shell.

Illustration 15: What are inner transition elements ? Decide which of the following atomic numbers are the atomic
numbers of the inner transition elements: 29, 59, 74, 95, 102, 104.  (JEE MAIN)

Sol: Inner transition elements are those which have incomplete 4f of 5f orbitals. Thus 59, 95 and 102 are inner
transition elements.

2. LANTHANIDES AND THEIR PROPERTIES

The lanthanide series of chemical elements comprises the fifteen metallic chemical elements with atomic numbers
57 through 71, from lanthanum through lutetium. These fifteen lanthanide elements, along with the chemically
similar elements, scandium and yttrium, are often collectively known as the rare earth elements.

2.1 Lanthanide Contraction

In lanthanides, the additional electron enters the 4f-sub shell but not in the valence-shell i.e. sixth shell. The
shielding effect of one electron in 4f-sub-shell by another in the same sub-shell is very little, being even smaller
than that of d-electrons, because the shape of f-sub shell is very much diffused, while there is no comparable
increase in the mutual shielding effect of 4f-electrons. This results in outermost shell electrons in the experience
increasing nuclear attraction from the growing nucleus. Consequently, the atomic and ionic radii go on decreasing
as we move from La57 to Lu71.
3 1 . 1 8 | d and f-block Elements

Lanthanides in the periodic table

Figure 31.4: Position of lanthanides in the periodic table

Consequence of Lanthanide contraction

1. Atomic and ionic radii of post-Lanthanide elements: The atomic radii of second row transition elements
are almost similar to those of the third row transition elements because the increase in size on moving down
the group from second to third transition elements is cancelled by the decrease in size due to lanthanide
contraction.
2. High density of post Lanthanide elements: It is because of a very small size due to lanthanide contraction
and increase in molar mass.
3. Basic strength of oxides and hydroxides: Due to lanthanide contraction, the decrease in size of lanthanides
ions, from La3+ to Lu3+ increases the covalent character (i.e. decreases the ionic character) between Ln+3 and
OH– ions in Ln(III) hydroxides (Fajan’s rules). Thus La(OH)3 is the most basic while Lu(OH)3 is the least basic.
Similarly, there is a decrease in the basic strength of the oxides.
4. Separation of Lanthanides: Due to the similar size (lanthanide contraction) of the lanthanides, it is difficult
to separate them. But a slight variation in their properties is utilized to separate.

PLANCESS CONCEPTS

The existence of lanthanoids in oxidation state of +2 and +4 is due to the fact that empty, half-filled or
completely filled f-subshells provide lower energy levels and the ions get stabilized. For example, Ce and
Tb show +4 oxidation state by attaining stable f0 and f7 configuration respectively whereas Eu and Yb
show +2 oxidation state by attaining stable f7 and f14 configuration, respectively.

Mredul Sharda (JEE Advanced 2013)

Illustration 16: Why is the separation of lanthanoids difficult? Explain. (JEE MAIN)

Sol: All the Lanthanoid ions are of almost the same size, so they have almost similar chemical and physical properties
and thus their separation becomes difficult.

Illustration 17: Name the members of the lanthanoid series which exhibit +4 oxidation states and those which
exhibit +2 oxidation states. Try to correlate this type of behavior with the electronic configuration of these
elements.  (JEE MAIN)

Sol: +4 = Ce, Pr, Nd, Tb, Dy. +2 = Eu, Yb. These states are accounted by the extra stability of half-filled and
completely filled f-orbitals.

2.2 Chemical Reactivity

These are very reactive metals like alkaline earth metals, however, they show very little difference in their chemical
reactivity. On strong heating with H2 and carbon, these form salt like non-stoichiometic hydrides and carbides. They
burn in oxygen to give sesquioxides M2O3 (except Ce which gives CeO3). Their ionic oxides react with water to form
 Chem i str y | 31.19

insoluble hydroxides. The oxides and hydroxides being strong base react with CO2 to form carbonates (M2CO3). On
burning in sulphur these form sulphides (M2S3).

With sulphate
From sulphide, M2S3
With acid
Liberate hydrogen
With halogens
Produce halides, MX3
With water
Lanthanoid metal From hydroxides, M(OH)3+H2
With C, 2770 K From carbides, MC2
With N2
From nitrides, MN
With O2
From oxides, M2O3

Figure 31.5: Chemical reactivity of lanthanoid metal

2.3 Uses of Lanthanoids

(i) Pyrophoric alloys, known as Misch metals, contain lanthanoids about 90-95% (Ce 40.5%, lanthanum and
neodymium 44%), iron 4.5%, calcium, carbon and silicon about 10.5% are used in cigarette and gas lighters,
flame throwing tanks, toys, tank and tracer bullets as well as in shells.
(ii) Any alloy containing 30% Misch metals and 1% Zr are used for making parts of jet engines.
(iii) Cerium salts are commonly used as catalyst in petroleum cracking (cerium phosphate), volumetric analysis
and as oxidizing agent (ceric sulphate), in dying cotton, in lead accumulators etc,
(iv) Oxides of praseodymium(Pr2O3) and neodymium (Nd2O4) are used in the preparation of coloured glasses and
standard filters.
(v) Oxides of cerium and thorium are used in the preparation of gas lamp mantles.
(vi) Cerium oxide is used to prepare sunglasses as cerium cuts off heat and ultraviolet light.

3. ACTINIDES AND THEIR PROPERTIES

(i) The differentiating electron occupies 5f-subshell and thus these elements also have three outermost shells
not filled to their capacity . These are called actinoids or actinones.
Ac: …………..5f0, 6s2 6p6 6d1, 7s2
89

90
Th: …………..5f1, 6s2 6p6 6d1, 7s2

103
Lw: …………..5f0, 6s2 6p6 6d1, 7s2
(ii) The electronic configuration of actinoids is [Rn] 5f0–14, 6d0–2, 7s2 where [Rn] stands for radon core. Like
lanthanoids, they are placed together because of similar chemical nature.
(iii) Like lanthanoids contraction, these too show actinoid contraction due to poor shielding effect of 5f-subshells.
Thus, atomic size of actinoids too decreases gradually from Ac to Lw.
(iv) Actinoids show a range of oxidation states, which is due to comparable energies of 5f, 6d and 7s-orbitals. Tl
general oxidation state of actinoids is +3; the elements in the first half of the series show higher oxidation
states.
(v) All these elements are strong reducing agents and are very reactive metals. Actinoids are radioactive and,
therefore, it is difficult to study their chemical nature. However, relatively more stable isotopes of these
elements beyond uranium have been discovered and the chemistry of these elements has been studied to an
extent by using radiotracer techniques.
Like lanthanoids, they react with oxygen, halogens, hydrogens, sulphur and acids.
3 1 . 2 0 | d and f-block Elements

Uses of Actinoids: Only U, Th have found applications in nuclear reactions undergoing nuclear fission to produce
nuclear power and nuclear bombs.

PLANCESS CONCEPTS

The 5f orbitals extend in space beyond 6s and 6p orbitals and participate in bonding.

Vaibhav Krishnan (JEE 2009, AIR 22)

Illustration 18: The chemistry of the actinoid elements is not so smooth as that of the lanthanoids.  (JEE MAIN)

Sol: This difference is due to occurrence of a wide range of oxidation states in actinoids. Also, their radioactivity
causes a hindrance in their study.

POINTS TO REMEMBER

•• General electronic configuration of d-block elements is (n – 1)d1–10ns0,1,2 and that of f-block element is
(n-2) f0.2….14(n –1)d0,1,2ns2

•• Their melting and boiling points are high which are attributed to the involvement of (n–1)d electrons resulting
in strong metallic bonds.

•• Successive ionisation enthalpies do not increase as steeply as in the main group elements with increasing
atomic number. Hence, the loss of variable number of electrons from (n –1) d orbitals is not energetically
unfavourable.

•• Ionisation energies where the electron is removed from half-filled or completely filled orbiatls are especially
large. Hence, Zn3+ is not formed.

•• The metals, in addition to variable oxidation states, they exhibit paramagnetic behaviour, catalytic properties
and tendency for the formation of coloured ions, interstitial compounds and complexes.

•• Reactivity of these elements is calculated as a sum of heat of sublimation, ionization enthalpy as well as heat
of hydration.

•• The transition elements are sufficiently electropositive to dissolve in mineral acids. Of the first series, with the
exception of copper, all the metals are relatively reactive.

•• The transition metals react with a number of non-metals like oxygen, nitrogen, sulphur and halogens to form
binary compounds. The first series transition metal oxides are generally formed from the reaction of metals
with oxygen at high temperatures.

•• These oxides dissolve in acids and bases to form oxometallic salts.

•• The two series of inner transition elements, lanthanoids and actinoids constitute the f-block of the periodic
table. With the successive filling of the inner orbitals, 4f, there is a gradual decrease in the atomic and ionic
sizes of these metals along the series (lanthanoid contraction). This has far reaching consequences in the
chemistry of the elements succeeding them.

•• Lanthanum and all the lanthanoids are rather soft white metals. They react easily with water to give solutions
giving +3 ions. The principal oxidation state is +3, although +4 and +2 oxidation states are also exhibited by
some occasionally.
 Chem i str y | 31.21

Solved Examples

JEE Main/Boards AgNO3 decomposes on exposure to light and is thus

stored in brown bottles to prevent the action of light.
Example 1: On what basis can you say that scandium 2AgNO3 → 2Ag + 2NO2 + O2
(Z = 21) is a transition elements since Zinc (Z = 30) is not?
Example 7: The species [CuCl4]2– exist but [Cul4]2– does
Sol: On the basis of incompletely filled 3d-orbitals. In
not.
case of scandium atom in its ground state (3d1), it is
regarded as a transition element. On the other hand, Sol: This can be explain by considering the reducing
zinc atom has completely filled d-orbitals (3d10) in its properties of the two ion. I– ion is a stronger reducing
ground state as well as in its oxidized state, hence it is agent than Cl– ion. It reduces Cu2+ ion into Cu+ ion.
not regarded as transition element. Hence, cupric iodide is converted into cuprous iodide.
Thus, the species [Cul4]2– does not exist.
Example 2:Explain briefly how +2 becomes more and
more stable in the first half of the first row transition Example 8: Which of the two compounds Lu(OH)3 and
elements with increasing atomic number? La(OH)3 is more basic and why?
Sol: Elements of 1st half 3d-series are: Sc, Ti, V, Cr, Mn. Sol: Large is cation more is covalent character–Fajan’s
Their electronic configuration are Sc = [Ar]4s2,3d1; Ti = rule. More the ionic character more is the basicity. Due
[Ar]4s2, 3d2; V = [Ar]4s2, 3d2; Cr = [Ar] 4s1, 3d5 and Mn = to lanthanoid contraction, the size of Lu3+ is increased
[Ar]4s2, 3d5. As we move from Sc to Mn, the number of and therefore Lu(OH)3 shows more covalent character.
empty d-orbitals reduced in +2 ion thereby imparting Thus, La(OH)3 is more ionic and thus more basic.
more stability.
Example 9: Why is europium (II) more stable than
Example 3: Give reasons for the following. Transition cerium (II)?
metals have high enthalpies of atomization.
Sol: In Eu2+, f-subshell is half-filled and d-subshell is
Sol: Due to strong interatomic interaction between completely filled and thus more stable.
valence electrons.
Example 10: In d-block metal ions, the colour of the
Example 4: What are interstitial compounds? Why are complex changes with the ligand used while in f-block
such compounds well known for transition metals? metal ion complexes, the colour remains almost constant
and does not depend upon the nature of ligand, why?
Sol: Interstitial compounds are formed when small
non-metallic atoms like H and C are trapped inside the Sol: Colour of a metal ion depends upon energy
crystal lattice of metals. difference of two energy levels involved in transition.
Splitting of degenerate d-orbitals into t2g and eg
Example 5: Why is Eº value for the M3+/M2+ couple orbitals is greatly affected by nature of ligand. i.e., why
much more positive than that of Cr3+/Cr2+ or Fe3+/Fe2+? [Ni(NH3)6]2+ is blue, [Ni(H2O)6]2+ is green and [Ni(NO2)6]4–
Explain. is brown. Color of f block metal is due to f-f transition.
4f-orbitals in lanthanoids are deeply embedded in the
Sol: Much larger third ionization energy of Mn (where atom and are deeply shielded bu 5s and 5p orbitals. Thus
the required change is d5 to d4) is mainly responsible for they remain almost unaffected by complex formation.
this. This also explains why the +3 state of Mn is of little Because of this color does not depend upon nature of
importance. ligand.

Example 6: Why is AgNO3 kept in brown coloured

bottles ?

Sol: Due to decomposition of AgNO3 in presence of light.

3 1 . 2 2 | d and f-block Elements

JEE Advanced/Boards as indicated by E° value. The EoOP for H/H+, Mn2+/Mn3+,

Cr2+/Cr3+ and Fe2+/Fe3+ are 0, – 1.5, + 0.4 and – 0.8 V.
Example 1: What is the effect of increasing pH of Thus, stability of Fe3+ ion in acid medium is less than Cr3+
K2Cr2O7 solution? but more than Mn3+.

Sol: In aqueous solution, we have Example 5: The yellow colour aqueous solution of
Cr2O27− + H2O 2CrO24− + 2H+ Na2CrO4 changes to orange red on passing CO2 gas.
Explain.
When pH < 4 (acidic medium), it exists as Cr2O27− and
the colour is orange. When pH > 7 (basic medium), it Sol: CO2 on dissolution in water produces acidic
exists as CrO −4 and the colour is yellow. medium.
CO2 + H2O → H2CO3 2H+ + CO32−
Example 2: Why does Mn(II) ion show the maximum
paramagnetic behavior among bivalent ions of the first Na2CrO4 changes to Na2Cr2O7 (orange-red) in acidic
transition series ? medium.
2CrO24− + 2H+ → Cr2O27− + H2O
Sol: The electronic configurations of Mn and Mn (II) ion
are
Example 6: An aqueous solution of inorganic compound
25
Mn : 1s22s22p63s23p63d54s2 (X) gives following reactions:
Mn+2: 1s2,2s22p63s23p63d54s0 (i) With an aqueous solution of barium chloride a
The Mn+2 ion has five unpaired electrons in its 3d subshell precipitate insoluble in dil. HCl is obtained.
which is the maximum value for a transition metal (ii) Addition of excess of KI gives a brown ppt. which
ion. Hence, Mn(II) shows the maximum paramagnetic turns white on addition of excess of hypo.
behavior (due to unpaired electrons) among bivalent
ions of the first transition series. (iii) With an aqueous solution of K4Fe(CN)6 a chocolate
coloured precipitate is obtained. Identify (X) and give
equations for the reaction for (i), (ii) and (iii) observations.
Example 3: How can AgNO3 be determined
volumetrically? Sol: (i) Step (i) suggests that compound (X) contains
SO24− radical.
Sol: AgNO3 is titrated using NH4SCN as an intermediate
solution with ferric alum as indicator. When precipitation (ii) Step (iii) suggests that the compound (X) contains
is completed, an additional drop of NH4SCN produces Cu2+ radical.
red colour with the indicator of ferric alum. (iii) Hence, the salt is CuSO4
AgNO3 + NH4SCN → AgSCN + NH4 NO3 Reactions:
White ppt (i) CuSO 4 + BaCl2 → BaSO 4 + CuCl2
Fe3+ + 3SCN– → Fe(SCN) 3 (X) White ppt.
  Red colour (Insoluble in HCl)
(ii) 2CuSO 4 + 4KI → 2Cu I2 + 2K 2SO 4
Example 4: For M2+/M and M3+/M2+ systems the E°
values for some metals are as follows: (X)
2CuI2 → Cu2 I2 + I2
Cr2+/Cr – 0.9V/Cr3+/Cr2+ –0.4V
Unstable
Mn2+/Mn – 1.2V/Mn3+/Mn2+ +1.5V
I2 + 2Na2S2O3 → Na2S 4 O6 + 2NaI
Fe2+/Fe – 0.4V/Fe3+/Fe2+ +0.8V Colourless
Use this data to comment upon:
(iii)
The stability of Fe3+ in acid solution as compared to that
of Cr3+ or Mn3+. 2CuSO 4 + K 4 Fe ( CN)  → Cu2 Fe ( CN)  + 2K 2SO 4
 6  6

( )
X Chocolate coloured ppt.
Sol: Fe3+ converts more easily to Fe2+ and Mn3+ do not
 Chem i str y | 31.23

Example 7: Give complete and balanced chemical AgCl + 2NH4OH → Ag(NH3)2Cl + 2H2O
equations for the following:
(D) Soluble
(i) Mercurous nitrate reacts with excess of KI solution.
Ag2NO3 + Na2S2O3 → Ag2S2O3 + 2NaNO3
(ii) Sodium chromite solution reacts with H2O2 in
presence of NaOH. (B) (E) White

(iii) Nickel sulphate reacts with dimethyl glyoxime Ag2S2O3 → Ag2S + SO3
reagent in ammoniacal solution. (E)   Black

Sol: (i) Hg2(NO3)2 + 2KI(excess) → Hg2I2 + 2KNO3

Example 10: Pyrolusite on heating with KOH in the
Hg2I2 + 2KI → K2HgI4 + Hg presence of air gives a dark green compound (A). The
(ii) Na2CrO3 + H2O2 → Na2CrO4 + H2O solution of (A) on treatment with H2SO4 gives a purple
coloured compound (B), which gives following reactions:
CH3–C=NOH (i) KI on reaction with alkaline solution of (B) changes
(iii) NiSO4+2 +2NH4OH →
CH3–C=NOH into a compound (C)
(ii) The colour of the compound (B) disappears on
OH O
treatment with the acidic solution of FeSO4.
CH3–C=N N=C–CH3
Ni +(NH4)2SO4 +2H2O (iii) With conc. H2SO4, compound (B) gives (D) which
CH3–C=N N=C–CH3 can decompose to yield (E) and oxygen. Identify (A)
to (E) and write balanced chemical equations for the
O OH formation of (A) and (B) and for the steps (i) to (iii).

Example 8: Out of cobalt and zinc salts, which is [O]

Sol: MnO2 + 2KOH  → K2MnO4 + H2O
attracted in a magnetic field?
Pyrolusite Dark green (A)
Sol: Out of cobalt and zinc salts, the cobalt salts are Dil. H2SO 4
3K2MnO4 + 2H2O 
→ 2KMnO4 + MnO2 + 4KOH
attracted in a magnetic field, because the cobalt ion
containing unpaired electron is characterized by a Purple (B)
permanent magnetic moment. Zn2+ ion contains 3d10 (i) 2KMnO4 + KI + H2O → KIO3 + 2MnO2 + 2KOH
configuration, i.e, no unpaired electrons, so zinc salts
are not attracted in magnetic field. (C)
(ii) 2KMnO4 + 8H2SO4 + 10FeSO4
Example 9: A certain metal (A) is boiled in dilute nitric → 2MnSO4 + K2SO4 + 5Fe2(SO4)3 + 8H2O
acid to give a salt (B) and an oxide of nitrogen (C). An
aqueous solution of (B) with brine gives a precipitate (iii) 2KMnO4 + H2SO4(conc) → K2SO4 + Mn2O7 + H2O
(D) which is soluble in NH4OH. On adding aqueous
(D)
solution of (B) to hypo solution a white precipitate (E) is
2Mn2O7 → 4MnO2 + 3O2
obtained. (E) turns black on standing. Identify (A) to (E).
(D) (E)
Sol: (i) The compound (B) reacts with NaCl (brine) to
give white precipitate (D) soluble in NH4OH, so (D) is
AgCl.
(ii) Thus, (B) must contain Ag+ ion.
(iii) (B) is obtained from (A) and dil. HNO3, so (B) is
AgNO3 and (A) is Ag.

Reactions:
3Ag + 4HNO3 → 3AgNO3 + NO + 2H2O
(A)  
(B)  
(C)
AgNO3 + NaCl → AgCl + NaNO3
(B) (D)
3 1 . 2 4 | d and f-block Elements

JEE Main/Boards

Exercise 1 Q.11 Explain the following observations:

(a) The elements of the d-series exhibit a larger number
Q.1 Why are Mn2+ compound more stable than Fe2+ of oxidation states than the elements of f-series.
towards oxidation in their + 3 state?
(b) The Cu+ salts are colorless while Cu2+ salts are
coloured.
Q.2 Write complete equation for oxidation of Fe2+ by
Cr2O27− in acidic medium.
Q.12 Mention the direct consequence of the following
factors on the chemical behavior of the transition
Q.3 Answer the following equations: elements:
(i) Which element in the first series of transition elements (i) They have incomplete d-orbitals in the ground state
does not exhibit variable oxidation state and why ? or in one of the oxidized states of their atoms.
(ii) Why do actinides, in general exhibit a greater range (ii) They contribute more valence electrons per atom in
of oxidation states than the lanthanides ? the formation of metallic bonds.

Q.4 Explain briefly how +2 state becomes more and Q.13 What are the characteristics of the transition
more stable in the first half of the first row transition elements and why are they called transition elements?
elements with increasing atomic number ? Which of the d-block elements may not be regarded as
the transition elements?
Q.5 Write chemical equations for the reactions involved
in the manufacture of potassium permanganate from Q.14 How would you account for the following
pyrolusite ore. situations ?
(i) The transition metals generally form coloured
Q.6 What is misch metal ? Mention its two important uses. compounds.
(ii) With 3d4 configuration, Cu2+ acts as a reducing agent
Q.7 To what extent do the electronic configurations but Mn3+ acts as oxidizing agent
decide the stability of oxidation states in the first series
(iii) The actinides exhibit a larger number of oxidation
of the transition elements? Illustrate your answer with
states than the corresponding lanthanides.
examples.

Q.15 How would you account for the following

Q.8 (a) Name two properties of the central metal ion
which enable it to form stable complex entities. (i) The transition elements have high enthalpies of
atomization.
(b) Account for the following: Zinc salts are white, Cu2+
salts are coloured. (ii) The transition metals and their compounds are
found to be good catalysts in many processes.
Q.9 How do you account for the following ?
Q.16 Explain giving reasons:
(a) All scandium salts are white ?
(i) Transition metals and many of their compounds
(b) The first ionization energies of the 5d transition show paramagnetic behavior.
elements are higher than those of the 3d and 4d
(ii) The enthalpies of atomization of the transition
transition elements in respective groups ?
metals are high.

Q.10 What may be the stable oxidation state of the (iii) The transition metals generally form coloured
transition element with the following d electron compounds.
configurations in the ground state of their atoms: 3d3, (iv) Transition metals and their many compounds act as
3d5, 3d8 and 3d4? good catalyst
 Chem i str y | 31.25

Q.17 Give reason for each of the following: Q.25 Describe the preparation of potassium dichromate
from iron chromite ore. What is the effect of increasing
(i) Size of trivalent lanthanide cations decreases with
pH on a solution of potassium dichromate ?
increase in the atomic number
(ii) Transition metal fluorides are ionic in nature, whereas
Q.26 Assign reason for the following:
bromides and chlorides are usually covalent in nature.
(i) From element to element, the actinides contraction
(iii) Chemistry of all the lanthanides is quite similar
is greater than lanthanide contraction.
(ii) The Eº value for Mn3+/Mn2+ couple is much more
Q.18 Discuss the general characteristics of the 3d series
positive than that for Cr3+/Cr2+.
of the transition elements with special reference to their.
(iii) Scandium (Z = 21) does not exhibit variable oxidation
(i) Atomic sizes
states and yet it is regarded as transition element.
(ii) Enthalpies of atomization
(iii) Tendency for complex formation Q.27 Describe the oxidizing action of potassium
dichromate and write the ionic equations for its reaction
Q.19 Predict which of the following will be coloured with:
in aqueous solution. Ti3+, V3+, Cu+, Sc3+, Mn3+, Fe3+ and (i) Iodide (ii) iron (iii) H2S
Co3+. Give reasons for each.
Q.28 (a) Describe the general trends in the following
Q.20 Write down electronic configuration of the properties of the first series of the transition elements:
following:
(i) Stability of +2 oxidation state
(i) La3+ (ii) Gd3+ (iii) Eu2+
(ii) Formation of oxometal ions.
(iv) Zn4+ (v) Ru2+ (vi) Ce4+
(b) Assign reason for each of the following:

Q.21 What are alloys ? Name an important alloy which (i) Transition elements exhibit variable oxidation states.
contains some of the lanthanide metals. Mention its (ii) Transition metal ions are usually coloured
uses.
Q.29 (a) Write the steps involved in the preparation of:
Q.22 (a) Give one example each of amphoteric and
(i) K2Cr2O7 from Na2CrO4
acidic oxides of transition metals
(ii) KMnO4 from K2MnO4
(b) Describe the trends in the following cases:
(b) What is meant by lanthanide contraction?
(i) Melting points of elements in the 3d transition series.
What effect does it have on the chemistry of the
(ii) Atomic sizes of elements in the 4f transition series.
elements which follow lanthanides?

Q.23 How would you account for the following ?

(i) Sc, the first member of first transition series does not Exercise 2
exhibit variable oxidation state.
Single Correct Choice Type
(ii) K2PtCl6 is well-known compound whereas
corresponding compound of nickel is not known. Q.1 The number of moles of acidified KMnO4 required
(iii) Only transition metals form complex compounds to convert one mole of sulphite ion into sulphate ion is
with ligands such as CO. (A) 2/5 (B) 3/5 (C) 4/5 (D) 1

Q.24 Use Hund’s rule to derive the electronic Q.2 N2(g) + 3H2(g)
Fe + Mo
2NH3(g); Haber’s process,
configuration of Ce3+ ion, and calculate its magnetic
Mo is used as
moment on the basis of ‘spin-only’ formula.
(A) A catalyst (B) A catalytic promoter
(C) An oxidizing agent (D) None of these
3 1 . 2 6 | d and f-block Elements

Q.3 An ornament of gold is made up of 75% of gold, it Q.10 Coinage metals show the properties of
is of……….carat.
(A) Typical elements
(A) 18 (B) 16 (C) 24 (D) 20
(B) Normal elements
(C) Inner-transition elements
Q.4 Solution of MnO −4 is purple-coloured due to
(D) None of these
(A) d-d transition
(B) Charge transfer from O to Mn
Q.11 Bayer’s reagent used to detect olefinic double
(C) Due to both d-d-transition and charge transfer bond is
(D) None of these (A) Acidified KMnO4
(B) Aqueous KMnO4
Q.5 Transition elements having more tendency to form
(C) 1% alkaline KMnO4 solution
complex than representative elements (s and p-block
elements) due to (D) KMnO4 in benzene
(A) Availability of d-orbitals for bonding
Q.12 The transition metal used in X-rays tube is
(B) Variable oxidation states are not shown by transiton
elements (A) Mo (B) Ta (C) Pb (D) Tc
(C) All electrons are paired in d-orbitals
Q.13 The higher oxidation states of transition elements
(D) f-orbitals are available for bonding are found to be in the combination with A and B, which
are
Q.6 During estimation of oxalic acid Vs KMnO4 self-
(A) F, O (B) O, N (C) Cl,O (D) S, F
indicator is
(A) KMnO4 (B) Oxalic acid Q.14 1 mole of Fe2+ ions are oxidised to Fe3+ ions with
(C) K2SO4 (D) MnSO4 help of (in acidic medium)
(A) 1/5 moles of KMnO4 (B) 5/3 moles of KMnO4
Q.7 A compound of mercury used in cosmetics, in
(C) 2/5 moles of KMnO4 (D) 5/2 moles of KMnO4
ayurvedic and yunani medicines and known as Vermilon
is
Q.15 The metals present in insulin and haemoglobin
(A) HgCl2 (B) HgS (C) Hg2Cl2 (D) HgI are respectively
(A) Zn Hg (B) Zn Fe (C) Cu, Hg (D) Cu, Fe
Q.8 Acidified chromic acid + H2O2
Org. solvent
→ X + Y, X and Y are Q.16 Solid CuSO4.5H2O having covalent, ionic as well
(A) CrO5 and H2O (B) Cr2O3 and H2O as co-ordinate bonds. Copper atom/ion forms……….co-
ordinate bonds with water.
(C) CrO2 and H2O (D) CrO and H2O
(A) 1 (B) 2 (C) 3 (D) 5

Q.9 Transition element are usually characterised by

Q.17 Purple of cassius is:
variable oxidation states by Zn does not show this
property because of (A) Pure gold (B) Colloidal solution of gold
(A) Completion of np-orbitals (C) Gold (I) hydroxide (D) Gold (III) chloride
(B) Completion of (n – 1)d orbitals
Q.18 Number of moles of SnCl2 required for the
(C) Completion of nd orbitals reduction of 1 mole of K2Cr2O7 into Cr2O3 is (in acidic
(D) None of these medium)
(A) 3 (B) 2 (C) 1 (D) 1/3
 Chem i str y | 31.27

Q.19 The aqueous solution of CuCrO4 is green because Q.3 Potassium manganate (K2MnO4) is formed when
it contains.  (1988)
(A) Green Cu2+ ions (A) Chlorine is passed into aqueous KMnO4 solution
(B) Manganese dioxide is fused with potassium
(B) Green CrO24− ions
hydroxide in air
(C) Blue Cu2+ ions and green CrO24− ions (C) Formaldehyde reacts with potassium permanganate
(D) Blue Cu2+ ions and yellow CrO24− ions in presence of a strong alkali
(D) Potassium permanganate reacts with conc. sulphuric
Q.20 Manganese steel is used for making railway tracks acid
because
Q.4 The aqueous solution of the following salts will be
(A) It is hard with high percentage of Mn
coloured in the case of  (1990)
(B) It is soft with high percentage of Mn
(A) Zn(NO3)2 (B) LiNO3
(C) It is hard with small concentration of manganese (C) Co(NO3)2 (D) CrCl3 (E) Potash alum
with impurities
(D) It is soft with small concentration of manganese Q.5Which of the following alloys contains Cu and Sn
with impurities  (1993)
(A) Bronze (B) Brass (C) Gun metal (D) Type metal
Q.21 Transition elements in lower oxidation states act
as Lewis acid because Q.6 Which of the following statement(s) is (are) correct.
When a mixture of NaCl and K2Cr2O7 is gently warmed
(A) They form complexes
with conc. H2SO4  (1998)
(B) They are oxidizing agents
(A) A deep red vapours is evolved
(C) They donate electrons (B) The vapours when passed into NaOH solution gives
(D) They do not show catalytic properties yellow solution of Na2CrO4
(C) Chlorine gas is evolved
Q.22 The Ziegler-Natta catalyst used for polymerization (D) Chromyl chloride is formed
of ethane and styrene is TiCl4 + (C2H5)3Al, the catalyzing
species (active species) involved in the polymerization is Q.7 Addition of high proportions of manganese
(A) TiCl4 (B) TiCl3 (C) TiCl2 (D) TiCl makes steel useful in making rails or railroads, because
manganese  (1998)
(A) Gives hardness to steel
Previous Years’ Questions (B) Helps the formation of oxide of iron

Q.1 In nitroprusside ion, Iron and NO exist as FeII and (C) Can remove oxygen and sulphur
NO+ rather than FeIII and NO. These forms can be (D) Can show highest oxidation state of +7
differentiated by  (1998)
(A) Estimating the concentration of iron Q.8 Assertion: Rusting of an iron is an example of
(B) Measuring the concentration of CN– corrosion. (2008)
(C) Measuring the solid state magnetic moment Reason: Rusting of iron is decreased by acids and
(D) Thermally decomposing the compound electrolytes

Q.2 Among the following the compound that is both Q.9 Assertion: AgBr is used in photography (1996)
paramagnetic and coloured is  (1997)
Reason : AgBr undergoes photochemical reaction.
(A) K2Cr2O7 (B) (NH4)2(TiCl6)
(C) VOSO4 (D) K3[Cu(CN4)] Q.10 Assertion: Tungsten filament is used in electric
bulbs. (1994)
Reason : Tungsten is a metal of high melting point.
3 1 . 2 8 | d and f-block Elements

Q.11 Assertion: In transition elements ns orbital is Q.17 In context of the lanthanoids, which of the
filled up first and (n – 1)d afterwards, during ionization following statements is not correct? (2012)
ns electrons are lost prior to (n – 1)d electrons. (1995)
(A) All the members exhibit +3 oxidation state
Reason: The effective nuclear charge felt by (n – 1)d
(B) Because of similar properties the separation of
electrons is higher as compared to that of ns electrons.
lanthanoids is not easy.
(C) Availability of 4f electrons results in the formation
Q.12 Assertion: The degree of complex
of compounds in +4 state for all the members of the
formation in actinides decreases in the order
series.
M4 + > MO22+ > M3+ > MO2+ (1997)

(D) There is a gradual decrease in the radii of the
Reason: Acitnides form complex with π-bonding members with increasing atomic number in the series.
ligands such as alkyl phosphines and thioethers.
Q.18 Which of the following arrangements does not
Q.13 Larger number of oxidation states are exhibited represent the correct order of the property stated
by the actinoids than those by the lanthanoids, the against it? (2013)
main reason being  (2008)
(A) V 2+ < Cr 2+ < Mn2+ < Fe2+ : Paramagnetic behaviour
(A) 4f orbitals more diffused than the 5f orbitals
(B) Ni2+ < Co2+ < Fe2+ < Mn2+ : Ionic size
(B) Lesser energy difference between 5f and 6d than
(C) Co3+ < Fe3+ < Cr3+ < Sc3+ : Stability in aqueous
between 4f and 5d orbitals
solution
(C) More energy difference between 5f and 6d than
(D) Sc < Ti < Cr < Mn: Number of oxidation states
between 4f and 5d orbitals
(D) More reactive nature of the actinoids than the
Q.19 The pair having the same magnetic moment is:
lanthanoids
 (2016)

Q.14 Knowing that the Chemistry of lanthanoids (Ln) [At. No.: Cr = 24, Mn = 25, Fe = 26, Co = 27]
is dominated by its +3 oxidation state, which of the 2+ 2+
following statements in incorrect?  (2009) (A) Cr (H2O )  and Fe (H2O ) 
 6  6
2+ 2+
(B) Mn (H2O )  and Cr (H2O ) 
(A) Because of the large size of the Ln (III) ions the  6  6
2+
and Fe (H2O ) 
2−
bonding in its compounds is predominantly ionic in (C) CoCl4 
 6
character. 2+
(D) Cr (H2O ) 
2−
and CoCl4 
(B) The ionic sizes of Ln (III) decrease in general with  6
increasing atomic number.
(C) Ln (III) compounds are generally colourless.
(D) Ln (III) hydroxides are mainly basic in character.

Q.15 The correct order of E 2+ values with negative

M /M
sign for the four successive elements Cr, Mn, Fe and Co
is (2010)
(A) Mn > Cr > Fe > Co (B) Cr > Fe > Mn > Co
(C) Fe > Mn > Cr > Co (D) Cr > Mn > Fe > Co

2−
Q.16 The magnetic moment (spin only) of NiCl4  is
 (2011)
(A) 5.46 BM (B) 2.83 BM
(C) 1.41 BM (D) 1.82 BM
 Chem i str y | 31.29

JEE Advanced/Boards

Exercise 1 Q.10 (i) K2MnO4 in acidic medium changes to MnO2

and KMnO4. What would be the equivalent weight of
Q.1 Explain why the greenish solution of potassium K2MnO4.
manganate turns purple when CO2 is bubbled in the (ii) Draw the structures of MnO −4 and CrO24−
solution.
Q.11 A complex has the formula PtCl4.2KCl. The
Q.2 Explain why mercurous compounds are formed electrical conductance of the compound shows that
when mercury is oxidized in a limited amount of an each formula unit had 3 ions. AgNO3 on treatment with
oxidizing agent whereas with an excess of oxidizing the complex does not give a precipitate of AgCl. What
agent mercuric compounds are formed. should be the correct formula of the complex?

Q.3 Explain why [Co(NH3)6]3+ is diamagnetic and [CoF6]3– Q.12 [NiCl4]2– and [Ni(CO)4] both are tetrahedral in
is strongly paramagnetic. shape but [NiCl4]2– is paramagnetic whereas [Ni(CO)4]
is diamagnetic. Explain in the difference in magnetic
Q.4 What happens when NaOH or NH4OH are added in behavior both the complexes.
excess to AlCl3 and ZnCl2 ?
Q.13 FeSO4 solution is mixed with (NH4)2SO4 in the molar
Q.5 Why is zinc oxide used in paints instead of lead ratio 1 : 1. It gives test of Fe2+. When CuSO4 is mixed
salts? with liquid ammonia (in the ratio 1 : 4) the mixture does
not give test of Cu2+. Explain the difference.

Q.6 Identify from [A] to [E].

Q.14 (A), (B) and (C) are three complexes of Cr(III).
Its formula is H12O6Cl3Cr. All the three complexes
Colorless salt [A] NaOH [B] (White precipitate)
have water and chloride ions as ligands. (A) does
+AgNO3 Dissolves in NaOH not react with conc. H2SO4 whereas (B) and (C)
loss 6.75% and 13.5% of their original weight
[C]
[E] White respectively on treatment with conc. H2SO4. Find
precipitate Pass H2S [A], [B] and [C]
[D] (White precipitate)
Q.15 A metal complex having composition Cr(NH3)4Cl2.
Q.7 Why are melting and boiling points of zinc,
Br has been isolated in two forms (A) and (B). (A) reacts
cadmium, and mercury is lower than those of other
with AgNO3 to give a white precipitate soluble in dilute
transition metals?
ammonia while (B) gives a pale yellow precipitate
soluble in concentrated ammonia. Write the formulae
Q.8 Why is HCl not used to acidify KMnO4 for volumetric of (A) and (B) and hybridization state of Cr in the
estimations? compounds.

?
Q.9 Colourless salt (A)  → (B) + (C) gas. Q.16 A monomeric compound of cobalt gives the
(B) Dissolves both in acid and alkali solution. following data on quantitative analysis: Co3+: 21.24%;
NH3: 24.77%; Cl–: 12.81%; SO34− : 34.65%; H2O: 6.53%
(C) Turns lime water milky and acidified K2Cr2O7 solution Deduce the empirical formula of the complex and the
green. possible isomers.
(A) Gives white precipitate (D) with H2S when the
solution is alkaline. Identify [A] to [D].
3 1 . 3 0 | d and f-block Elements

Exercise 2 Q.9 Pick out the incorrect statements:

(A) MnO2 dissolves in conc. HCl, but does not form
Single Correct Choice Type Mn4+ ions
Q.1 Cr2O27− 2CrO24− , X and Y are respectively (B) MnO2 oxidizes hot concentrated H2SO4 liberating
oxygen
(A) X = OH–, Y = H+ (B) X = H+, Y = OH– (C) K2MnO4 is formed when MnO2 in fused KOH is
(C) X = OH , Y = H2O2
–
(D) X = H2O2, Y = OH – oxidized by air, KNO3, PbO2 or NaBiO3
(D) Decomposition of acidic KMnO4 is not catalysed by
Q.2 CrO3 dissolves in aqueous NaOH to give sunlight
(A) Cr2O27− (B) CrO24−
Q.10 The rusting of iron is formulated as Fe2O3.xH2O
(C) Cr(OH)3 (D) Cr(OH)2 which involves the formation of
(A) Fe2O3 (B) Fe(OH)3
Q.3 (NH4)2Cr2O7 (Ammonium dichromate) is used in fire
works. The green coloured powder blown in air is (C) Fe(OH)2 (D) Fe2O3 + Fe(OH)3

(A) Cr2O3 (B) CrO2 (C) Cr2O4 (D) CrO3

Q.11 Metre scales are made-up of alloy

Q.4 The d-block elements which is liquid at room (A) Invar (B) Stainless steel
temperature, having high specific heat, less reactivity (C) Electron (D) Magnalium
than hydrogen and its chloride (MX2) is volatile on
heating is Q.12 A metal M which is not affected by strong acids
(A) Cu (B) Hg (C) Ce (D) Pm like conc. HNO3, conc. H2SO4 and conc. solution of
alkalies like NaOH, KOH forms MCl3 which finds use for
toning in photography. The metal M is-
Q.5 Iron becomes passive by………………due to formation
of…………… (A) Ag (B) Hg (C) Au (D) Cu
(A) Dil. HCl, Fe2O3
Q.13 KMnO4 + HCl → H2O + X(g), X is a (acidified)
(B) 80% conc. HNO3, Fe3O4
(A) Red liquid (B) Violet gas
(C) Conc. H2SO4, Fe3O4
(C) Greenish yellow (D) Yellow-brown gas
(D) Conc. HCl, Fe3O4
Q.14 In nitroprusside ion, the iron exists as Fe2+ and
Q.6 Cu + conc. HNO3 → Cu(NO3)2 + X (oxide of NO as NO+ rather than Fe3+ and NO respectively. These
nitrogen); then X is- forms of ions are established with the help of

(A) N2O (B) NO2 (C) NO (D) N2O3 (A) Magnetic moment in solid state
(B) Thermal decomposition method
Q.7 When KMnO4 solution is added to hot oxalic acid (C) By reaction with KCN
solution, the decolourisation is slow in the beginning
but becomes instantaneous after some time. This is (D) By action with K2SO4
because
(A) Mn2+ acts as auto catalyst Multiple Correct Choice Type

(B) CO2 is formed Q.15 The metal(s) which does/do not form amalgam
(C) Reaction is exothermic is/are

(D) MnO −4 catalyses the reaction (A) Fe (B) Pt (C) Zn (D) Ag

Q.8 CuSO4 solution reacts with excess KCN to give Q.16 The highest oxidation state among transition
element is
(A) Cu(CN)2 (B) CuCN
(A) + 7 by Mn (B) + 8 by Os
(C) K2[Cu(CN)2] (D) K3[Cu(CN)2]
(C) +8 by Ru (D) + 7 by Fe
 Chem i str y | 31.31

Q.17 Amphoteric oxide(s) is/are Q.25 Addition of non-metals like B and C to the
interstitial sites of a transition metal results the metal-
(A) Al2O3 (B) SnO (C) ZnO (D) Fe2O3
(A) Of more ductability (B) Of less ductability
Q.18 Interstitial compounds are formed by- (C) Less malleable (D) Of more hardness
(A) Co (B) Ni (C) Fe (D) Ca
Q.26 Mercury is a liquid at 0ºC because of-
Q.19 To an acidified dichromate solution, a pinch of (A) Very high ionization energy
Na2O2 is added as shaken. What is observed:
(B) Weak metallic bonds
(A) Blue colour
(C) High heat of hydration
(B) Orange colour changing to green
(D) High heat of sublimation
(C) Copious evolution of oxygen
(D) Bluish-green precipitate Q.27 The ionization energies of transition elements are-
(A) Less than p-block elements
Q.20 Amongst CuF2, CuCl2 and CuBr2
(B) More than s-block elements
(A) Only CuF2 is ionic
(C) Less than s-block elements
(B) Both CuCl2 and CuBr2 are covalent
(D) More than p-block elements
(C) CuF2 and CuCl2 are ionic but CuBr2 is covalent
(D) CuF2, CuCl2as well as CuBr2 are ionic Q.28 The catalytic activity of transition elements is
related to their-
Q.21 CuSO4(aq) + 4NH3 → X, then X (A) Variable oxidation states
(A) [Cu(NH3)4] 2+
(B) Surface area
(B) Paramagnetic (C) Complex formation ability
(C) Coloured (D) Magnetic moment
(D) Has magnetic moment 1.73 BM
Q.29 In the equation: M + 8CN– + 2H2O + O2 →
Q.22 Amphoteric oxide(s) of Mn is/are 4[M(CN)2]– + 4OH–, metal M is-

(A) MnO2 (B) Mn3O4 (A) Ag (B) Au (C) Cu (D) Hg

(C) Mn2O7 (D) MnO

Q.30 An elements of 3d-transtion series shows two
oxidation states x and y, differ by two units then-
Q.23 The lanthanide contraction is responsible for the
(A) Compounds in oxidation state x and ionic if x > y
fact that
(B) Compounds in oxidation state x are ionic if x < y
(A) Zr and Hf have same atomic sizes
(C) Compound in oxidation state y are covalent if x < y
(B) Zr and Hf have same properties
(D) Compounds in oxidation state y are covalent if x > y
(C) Zr and Hf have different atomic sizes
(D) Zr and Hf have different properties
Q.31 Ion(s) having non zero magnetic moment (spin
only) is/are
Q.24 Potash alum is a double salt, its aqueous solution
(A) Sc3+ (B) Ti3+ (C) Cu2+ (D) Zn2+
shows the characteristic of-
(A) Al3+ ions (B) K+ ions
(C) SO24− ions (D) Al3+, ions but not K+ ions
3 1 . 3 2 | d and f-block Elements

Assertion Reasoning Type Q.3 Which of the following pair is expected to exhibit
same colour in solution?  (2005)
Questions given below consist of two statements each
printed as Assertion (A) and Reason (R); while answering (A) VOCl2; FeCl2 (B) CuCl2; VOCl2
these questions you are required to choose any one of (C) MnCl2; FeCl2 (D) FeCl2; CuCl2
the following four responses:
(A) if both assertion and reason are true and reason is Q.4 Which of the following will not be oxidized by O3?
the correct explanation of assertion  (2005)
(B) if both assertion and reason are true but reason is (A) KI (B) FeSO4 (C) KMnO4 (D) K2MnO4
not correct explanation of assertion
(C) if assertion is true but reason is false Q.5 Which of the following alloys contains Cu and Zn ?
 (1993)
(D) if assertion is false and reason is true
(A) Bronze (B) Brass (C) Gun metal (D) Type metal
Q.32 Statement-I: Equivalent mass of KMnO4 is equal
to one-third of its molecular mass when it acts as a Q.6 Which of the following statement(s) is/are correct
oxidizing agent in an alkaline medium. when a mixture of NaCl and K2Cr2O7 is gently warmed
with conc. H2SO4? (1998)
Statement-II: Oxidation number of Mn is + 7 in KMnO4.
(A) A deep red vapours is formed
Q.33 Statement-I: K2CrO4 has yellow colour due to (B) Vapours when passed into NaOH solution gives a
charge transfer. yellow solution of Na2CrO4
Statement-II: CrO24− ion is tetrahedral in shape. (C) Chlorine gas is evolved
(D) Chromyl chloride is formed
Q.34 Statement-I: The highest oxidation state of
chromium in its compounds is +6. Q.7 Which of the following statement(s) is/are correct?
Statement-II: Chromium atom has only six electrons in  (1998)
ns and (n – 1) d orbitals. (A)The electronic configuration of Cr is [Ar] 3d54s1
(Atomic number of Cr = 24)
Q.35 Statement-I: Tungsten has a very high melting (B) The magnetic quantum number may have a negative
point. value
Statement-II: Tungsten is a covalent (C) In silver atom 23 electrons have a spin of one type and
24 of the opposite type (Atomic number of Ag = 47)
Q.36 Statement-I: Cu+ ion is colourless. (D) The oxidation state of nitrogen in HN3 is – 3
Statement-II: Four water molecules are coordinated to
Cu+ ion. Q.8 Reduction of the metal centre in aqueous
permanganate ion involves  (2011)
(A) Three electrons in neutral medium
Previous Years’ Questions (B) Five electrons in neutral medium
(C) Three electrons in alkaline medium
Q.1 The pair of compounds having metals in their
(D) Five electrons in acidic medium
highest oxidation state is  (2004)
(A) MnO2, FeCl3 (B) [MnO4]–,CrO2Cl2
Q.9 A compound of vanadium has a magnetic moment
(C) [Fe(CN)6]3–, [Co(CN)3] (D) [NiCl4]2–, [CoCl4]– of 1.73 BM. Work out the electronic configuration of
the vanadium ion of the compound.  (1997)
Q.2 When I– is oxidized by MnO −4 in alkaline medium,
I– converts into- (2004) Q.10 Give reasons: CrO3 is an acid anhydride.  (1999)
(A) IO3− (B) I2 (C) IO −4 (D) IO–
 Chem i str y | 31.33

Q.11
Moist air Zn Q.17 Among the following complexes (K − P ) (2011)
B (MCl4) A 
K 3 Fe ( CN)  (K ) , Co (KNH3 )  CI3 , (L ) ,

{
White fumes Purple
with pungent (M=Transition color  6  6
element-colorless)
smell
Na3 Co ( oxalate )  (M) , Ni (H2O )  CI2 (N) ,
 3  6
Identify the metal M and hence MCl4. Explain the
difference in colours of MCl4 and M. (2005) K 2 Pt ( CN)  ( O ) and  Zn (H2O )  (NO3 ) (P )
 4  6 2

Q.12 Among the following metal carbonyls, the C − O (A) K, L, M N (B) K, M, O, P

bond order is lowest in  (2007)
(C) L, M, O, P (D) L, M, N, O
+
(A) Mn ( CO )  (B) Fe ( CO ) 
 6  5
Q.18 The colour of light absorbed by an aqueous
(C) Cr ( CO )  (D)  V ( CO )  solution of CuSO 4 is  (2012)
 6
 6
(A) Orange-red (B) Blue-green
Q.13 Among the following the coloured compound is (C) Yellow (D) Violet
 (2008)
(A) CuCI (B) K 3 Cu ( CN) 
 4 Q.19 Consider the following complex ions, P, Q and R
(D) Cu ( CH3CN)  BF4
2+ 3+
(C) CuF2 = FeF6  , Q  V (H2O ) =
P =
3−
 and R Fe (H O ) 
 4  6  2 6

The correct order of the complex ions, according to

Q.14 The spin only magnetic moment value (in Bohr
their spin–only magnetic moment values (in B.M.) is 
magnetron units) of Cr(CO)6 s (2009)
 (2013)
(A) 0 (B) 2.84 (C) 4.90 (D) 5.92 (A) R < Q < P (B) Q < P < Q
(C) R < P < Q (D) Q < P < R
Q.15 Match each of the reactions given in column I
with the corresponding product(s) given in column II.
Q.20 For the octahedral complexes Fe3+ in SCN‒
 (2009)
(thiocyanato-S) and in CN‒ ligand environments, the
Column I Column II difference between the spin-only magnetic moments in
Bohr magnetrons (When approximated to the nearest
(A) Cu + dilHNO3 (p) NO integer) is [Atomic number of Fe = 26]  (2015)
(B) Cu + concHNO3 (q) NO2
(C) Zn + dil HNO3 (r) N2O Q.21 Among [Ni(CO)4], [NiCl4]2‒ [Co(NH3)4 Cl2], Cl,
Na3 [CoFe6], Na2O2 the total number of paramagnetic
(D) Zn + conc HNO3 (s) Cu(NO3)2
compounds is (2016)
(t) Zn(NO3)2
(A) 2 (B) 3 (C) 4 (D) 5

Q.16 The complex showing a spin-only magnetic

Q.22 The geometries of the ammonia complexes
moment of 2.82 B.M. is  (2010)
Ni2+ ,Pt2+ and Zn2+ respectively, are  (2016)
(A) Ni ( CO )
2−
(B) NiCI 4  (A) Octahedral, square planar and tetrahedral
4
2− (B) Square planar, octahedral and tetrahedral
(C) Ni (PPh2 ) (D) Ni ( CN) 
4  4
(C) Tetrahedral, square planar and octahedral
(D) Octahedral, tetrahedral and square planar
3 1 . 3 4 | d and f-block Elements

PlancEssential Questions
JEE Main/Boards JEE Advanced/Boards

Exercise 1 Exercise 1
Q.3 Q.5 Q.6 Q.5 Q.7 Q.9

Q.9 Q.12 Q.14 (ii) Q.11 Q.13

Q.17 Q.19 Q.22 (a)

Exercise 2
Q.23 (ii) Q.29
Q.4 Q.8 Q.11
Exercise 2 Q.14 Q.19 Q.20

Q.3 Q.4 Q.8 Q.24 Q.26 Q.29

Q.11 Q.19 Q.21

Previous Years’ Questions
Previous Years’ Questions Q.8 Q.11

Q.1 Q.2 Q.6

Q.9

Answer Key

JEE Main/Boards
Exercise 2
Single Correct Choice Type

Q.1 A Q.2 B Q.3.A Q.4 B Q.5 A Q.6 A

Q.7 B Q.8 A Q.9 B Q.10 D Q.11 C Q.12 A

Q.13 A Q.14 A Q.15 B Q.16 D Q.17 B Q.18 A

Q.19 D Q.20 A Q.21 A Q.22 B

Previous Years’ Questions

Q.1 C Q.2 C Q.3 B, C Q.4 C, D Q.5 A, B Q.6 A, B, D Q.7 A, C

Q.8 C Q.9 B Q.10 A Q.11 A Q.12 B Q.13 B Q.14 C

Q.15 A Q.16 B Q.17 C Q.18 A Q.19 A

 Chem i str y | 31.35

JEE Advanced/Boards Q.8 See text for the reaction of HCl and KMnO4.

Q.9 [(A) = ZnSO3; (B) = ZnO; (C) = SO2; (D) = ZnS]

Exercise 1
Q.10 3/2 M Q.11 K2[PtCl6]
Q.1 Formation of MnO −4
Q.12 Cl– is a weak ligand, so no pairing of electrons. CO
Q.2 Due to disproportionation of Hg22+ is a strong field ligand and, therefore, causes pairing of
electrons.
Q.3 NH3-strong ligand causes pairing of electrons. So
complex is diamagnetic. F– is a weak ligand and pairing Q.13 In the first case, a double salt is formed. In the
is not possible. second case, a complex is formed. So, no test.
Q.4 [Al(OH)3] and NaAlO2; Zn(OH)2 and Na2ZnO2 also Q.14 (A) = [Cr(H2O)6]Cl3; (B) = [Cr(H2O)5]Cl2H2O
[Zn(NH3)4]2+
(C) = [CrCl2(H2O)4]Cl2.2H2O
Q.5 [ZnS is white but PbS is black]
Q.15 (A) = [Cr(NH3)4ClBr]+Cl– and (B) = [Cr(NH3)4Cl2]+Br–.
Q.6 [A] = ZnCl2; [B] = Zn(OH)2; [C] = Na2ZnO2; [D] = ZnS;
[E] = AgCl Q.16 [Co(NH3)4Cl(H2O)]SO4; [Co(NH3)4(SO4)(H2O)]
Cl[Co(NH3)4(SO4)(Cl)]H2O
Q.7 All the three elements have d10

Exercise 2

Single Correct Choice Type

Q.1 A Q.2 B Q.3 A Q.4 B Q.5 B Q.6 B

Q.7 A Q.8 D Q.9 D Q.10 D Q.11 A Q.12 C
Q.13 C Q.14 A

Multiple Correct Choice Type

Q.15 A, B Q.16 B, C Q.17 A, B, C Q.18 A, B, C Q.19 A, C Q.20 A, B

Q.21 A, B, C, D Q.22 A, B Q.23 A, B Q.24 A, B, C Q.25 B, C, D Q.26 A, B

Q.27 A, B Q.28 A, B, C Q.29 A, B Q.30 B, C Q.31 B, C

Assertion Reasoning Type

Q.32 A Q.33 B Q.34 A Q.35 C Q.36 C

Previous Years’ Questions

Q.1 B Q.2 A Q.3 B Q.4 C Q.5 A, B, C Q.6 A, B, C, D

Q.7 A, B, C Q.8 A Q.12 B Q.13 C Q.14 A

Q.15 A → p, s; B → q, s; C → r, t; D → q, t Q.16 B Q.17 C Q 18 A

Q.19 B Q.20 4 Q.21 B Q.22 A

3 1 . 3 6 | d and f-block Elements

Solutions

JEE Main/Boards Sol 7: Electronic configurations play an important role

in deciding the stability of an ion. Generally half-filled &
full filled states are more stable. It also depends on no.
Exercise 1 of unpaired e–, symmetry, exchange pairs, etc.

Sol 1: 2+
Mn
4s 3d stability
2+
Mn 2+
Cr
4s 3d stable
2+ → Not so stable.
Fe

Mn2+ has half-filled 3d orbitals. Sol 8: (a) To form stable complexes, the metal most
have
So, its stable.
(i) Electronic configuration (ii) Hybridisation

Sol 2: Fe+2 → Fe+3 + e– 4s

4s 3d
3d
(b)
Cr2O27− + 14H+ + 6e– → 2Cr+3 + 7H2O
+2
Zn
Zn
+2

unpaired ee--
no unpaired
→ nonounpaired e–.
Sol 3: (i) Zn doesn’t show variable O. S. because of 4s 3d
stable pseudo inert gas configuration of Zn+2 4s
+2
3d
Cu
Cu
+2

Zn2+ → [Ar] 3d10

unpaired ee--
11 unpaired
(ii) Actinides show greater range of O. S. because of → 1 unpaired e–.
comparable energies of 5f, 6d, 7s-orbitals. So, Cu+2 shows colour but not Zn2+.

Sol 4: In the first half of 3d elements, Mn2+ implies Sol 9: (a) All scandium salts have Sc+3
that there are unpaired e– in 3d shell as we go from Sc
to Mn, the exchange pairs increases as Mn2+ is nearly Sc+3 → [Be] 3s2 3p6
symmetrical. There are no unpaired e– to show colour.
(b) Due to lanthanide contraction, the sizes of 5d
Sol 5: Pyrolusite → MnO2 (impure) elements are smaller compared to expected size. But
fusion nuclear charge will be high.
2MnO + 4KOH +MnO
Pyrolusite
2
O2  → 2K2MnO4 + 2H2O
2 (impure)
2MnO2 + 4KOH + O2 fusion fusion
3MnO2 + 6KOH + KClO3  → 3K2MnO4 + KCl + Sol 10: 3d3 → V → +3
2K2MnO4 + 2H2O
3H2O 3s5 → Mn → +7, +2
3MnO2 + 6KOH + KClO2 fusion
K2MnO4 
electrolysis 3K KMnO
→ 2MnO4 4+ KCl + 3H2O 3d8 → Ni ; → +2
electrolysis
K2MnO4 KMnO4 3d4 → Cr → +6, +3
+ -
H2O H + OH
Sol 11: (a) Most of the f-block elements show +3
2- - -
MnO4 MnO4 + e
+
2H + 2e
-
H2 oxidation state (few have +2, +4). It is because of the
fact that its outer shell has 2s electron & penultimate
Sol 6: Misch metal is an alloy of rare-earth elements shell has 1d e–. So, stability of there O. S has less
in various naturally occurring proportion. Generally it dependence on f e–.
contains 50% cerium, 25% lanthanum & small amounts (b) Cu+ → 3d10 → No. unpaired e– → White
of other. It is used preparation of most of the rare earth Cu2+ → 3d9 → 1 unpaired e– → Colour
elements. When alloyed with iron, its was is lighters.
 Chem i str y | 31.37

Sol 12: (i) They can exhibit multiple oxidation sates. They Sol 19: Ti+3, V+3, Mn3+, Fe3+, Co3+ have unpaired e–. So,
can accept e– pairs from ligands to form complexes. they are colored. Cu+, Sc3+ have no unpaired e–. So, they
(ii) Metallic bond strength is very high. So, they are are not colored.
strong metals.
Sol 20:
Sol 13: They are called transition elements because (a) La3+ → [Xe]
their position in the periodic table is between s & p
blocks their properties are transitional between highly (b) Cd3+ → [Xe] 4f7
reactive s-metals & constant bond forming p elements. (c) Eu2+ → [Xe] 4f7
Zn, Cd, Hg has ns2 (n – 1)d10 configuration, so, they are
not included in transition elements. (d) Zn4+ → [Ar] 3d10
(e) Ru2+ → [Kr] 5s2 4d4
Sol 14: (i) Most of the transition metals form coloured (f) Ce4+ → [Xe]
compounds due to the presence of unpaired e– or
complex formation.
Sol 21: Alloy is a mixture a solid solution compared
(ii) Mn+2 has half-filled orbitals, So Mn+3 acts as oxidizing of a metal and another element. Misch metal contains
agent. Cr+ has full-filled d-orbitals, So Cr2+ acts as Lanthanoids. It is used in preparation of f-block
reducing agent. elements. When mixed with iron, its used in lighters.
(iii) The energy difference between 7s, 7d, 5f is very less.
So, they can show multiple oxidation states. Sol 22: (a) ZnO → Amphoteric.
Chromium oxide → Acidic
Sol 15: (i) They have high enthalpies of atomisation
because of the strong metallic bonds. Each element in (b) (i) M. P’s first increases to the right till chromium
d-block provides more electrons for metallic bonding group, then decreases.
compared to s-block metals. (ii) The size decreases as we go right in the series. It
(ii) They are good catalysts, because of the presence decreases due to poor-shielding effect of e–.
of more than 1 stable O. S So, they can be used for
oxidation & reductions. Sol 23: (i) Configuration of Sc → 4s2 3d1
It has only 1 stable O. S. i. e. +3 because Sc+, Sc+2 have
Sol 16: (i) Paramagnetic is due to the presence of
no stabilising factors & Sc+3 is inert gas configuration.
unpaired e– many transition metals have unpaired e– in
d-orbitals. (ii) K2NiCl6 doesn’t exist because of the small size of Ni
compared to Pt. It cannot hold 6 Cl– ions around it.
Sol 17: (i) d-orbitals have poor shielding effect. As (iii) Co is a strong ligand and metals need vacant
atomic number increases, there is increase in nuclear d-orbitals for forming bonds.
charge also. So, the size decreases.
(ii) The electronegativity difference between transition Sol 24: Ce → [Xe] 4f1 5d1 4f1
metals & F is high but not so for Cl, Br.
Ce+3 → [Xe] 4f1
(iii) Most of them have similar sizes, they show a
common stable oxidation state (+3). Magnetic moment = n(n + 2) BM
and n = 1 i. e. no of unpaired e–
Sol 18: (i) The size of atoms decrease as we go from
∴ MM = 3 = 1. 73 BM
left to right in the series because poor screening effect
of d-electrons.
(ii) There is not a good trend for enthalpies of
atomisation for 3d-elements. As we go from left to
right, if first increases and then decreases with a sudden
dip at the centre.
(iii) Tendency of complex formation increases as the
size decreases.
3 1 . 3 8 | d and f-block Elements

red hot Sol 29: (a) (i) 2Na2CrO4 + H2SO4 → Na2SO4 + Na2Cr2O7
Sol 25: 4FeCr O +8Na2CO + 7O2 
red hot →
2 4O + 8Na
4FeCr 2
3 + 7O
CO 4 2 3 2 air
air + H2O
double
8 Na2CrO4 + 2Fe2O3 + 8CO2 Na2Cr2O7 + KCl  → K2Cr2O7 + 2NaCl
decomposition

(ii) 3K2MnO4 + 2H2O  2KMnO 4 + MnO2 ↓ + 4KOH

Dissolve Purple dark brown
in water
(b) Lanthanide contraction is used to describe the
greater than expected decrease in ionic radii of the
elements in the lanthanide series. It is due to the poor
Fillerate Residue shielding-effect of 4f electrons. The atomic radius of Hf
Na2CrO4 Fe2O3 (6th period) is less than that of Zr(5th period).

2Na2CrO
2Na+2CrO
H2SO 2SONa
4 + H→ SO
Na2+SONa
4 + Cr O + H O

Exercise 2
4
4 4 2 4 2 2 7 2
Na2Cr2O7 + H2O
H2- +
Cr2O27−Cr → CrO 2− +
H 2-
2O7 3Cr2O3
Single Correct Choice Type

Sol 26: (i) The actinoid contraction is high because of Sol 1: (A) KMnO4 → Mn+2, Z = 5
poor shielding effect of earlier 4f e– and nearly coming SO32− → SO24− , Z = 2
5f e–.
2
∴ n=
(ii)
2+
Mn
Mn
2+ 5

→ Stability
Sol 2: (B) Fe is catalyst
2+
CrCr
2+ Mo is promoter

Mg 3
→ Not so stable Sol 3: (A) Carats = 24 × = 24 × = 18
Mm 4
So E for Mn3+/Mn2+ is much positive than Cr+3/Cr+2
(iii) It is regarded as transition metal because if has Sol 4: (B) MnO −4 , Cr2O27− show colour due to charge
incomplete d-orbitals and its last electron enters into transfer from O to metal.
3d orbitals.
Sol 5: (A) It is due to availability of vacant d-orbitals
Sol 27: In all the reaction for bonding.
K2Cr2O7 → Cr+3
Sol 6: (A) KMnO4 is the self indicator as it’s colour
(i) I– → I2
changes from purple (+7) during it’s reduction.
(ii) Fe2+ → Fe3+
(iii) H2S → S Sol 7: (B) Vermilon is from the ore cinnabar (HgS).

H +
Sol 28: (a) (i) Mn, Zn have stable +2 O. S. because of Sol 8: (A) Chromic acid + H2O2 → CrO5
Org Solvent
presence of half or full filled orbitals. Fe, Ni, Co also Chromium
peroxide (Blue )
have stable +2 O. S. whereas others don’t show +2 in
most of the compounds. + H2O

(ii) Most of the transition metals don’t form oxometal

ions (Cr, Mn are exceptions. ) They form MO. Acidity Sol 9: (B) Zn has completed (n – 1) d orbitals. So, it’s
increases, ionic nature decreases. not included in transition elements.
(b) (i) They exhibits variable O. S. because of the
presence of unpaired e– in d-orbitals. (ii) Same is the Sol 10: (D) Coinage metals → Cu, Au, Ag i. e. transition
reason for paramagnetism & colour of compounds. metal.
 Chem i str y | 31.39

Sol 11: (C) Bayer’s reagent is alkaline KMnO4 Previous Years’ Questions

Sol 12: (A) Molybdenum (MO) is used in X-rays tube. Sol 1: (C) The existence of Fe2+ and NO+ in nitroprusside
ion [Fe(CN)5NO]2– can be established by measuring
the magnetic momentum of the solid compound
Sol 13: (A) F, O because of high electro-negativity. which should correspond to (Fe2+ = 3d6) four unpaired
electrons.
Sol 14: (A) Fe+2 → Fe+3, Z = 1
Sol 2: (C) V+4 → 3d14s0 1 unpaired electrons. Hence it is
KMnO4 → Mn2+, Z = 5
paramagnetic and coloured compound.
∴ 1/5 moles of KMnO4
Sol 3: (B, C) 2KOH + MnO2 + O → K 2MnO 4
Manganese Potassium manganate
Sol 15: (B) Haemoglobin → Fe dioxide

Insulin → Zn +H2O

HCHO + 2KMnO4 + 2KOH → K2MnO4 + H2O + HCOOH
Sol 16: (D) Cu → 5H2O as ligands
H2O will form H-bonding with SO24− ion. Sol 4: (C, D) Co(NO3)2 and CrCl3 has unpaired electron,
hence they are coloured; while Zn(NO3)2, LiNO3 and
Sol 17: (B) It’s colloidal solution of gold. potash alum have no unpaired electron hence they are
colourless.
It’s formed by the reaction of gold salts with Tin(II)
chloride.
Sol 5: (A, B) Gun metal contain Cu and Sn while type
metal contains Pb, Sn and Sb.
Sol 18: (A) SnCl2 → Sn4+, Z = 2
K2Cr2O7 → Cr2O3, Z = 6 Sol 6: (A, B, D) 4NaCl + K2Cr2O7 + 6H2SO4 → 2CrO2Cl2
6 + 4NaHSO4 + 2KHSO4 + 3H2O
n= = 3.
2 CrO2Cl2 + 4NaOH → Na2CrO4 + 2NaCl + 2H2O
Chromyl chloride Yellow solution
Sol 19: (D) Cu2+ → blue
CrO24− → yellow. Sol 7: (A, C) Manganese (Mn) imparts hardness to steel
as well as removes oxygen and sulphur from steel by
Sol 20: (A) It is very hard. It has high proportions of forming slag as MnSiO3
manganese. Fe2O3 + 3Mn → 3MnO + 2Fe
MnO + SiO2 → MnSiO3
Sol 21: (A) They can form complexes by taking e– pairs
from ligands. (Slag)

Sol 8: (C) Rusting involves reduction of absorbed oxygen

Sol 22: (B)TiCl3 is an intermediate during catalysis.
to OH– ions and oxidation of iron to Fe2+ ions. The two
TiCl4 → TiCl3 + Cl. ions combine to yield Fe(OH)2 which gets oxidized to give
Fe2O3.nH2O (rust). The presence of acid helps dissolution
of pure iron to ferrous ions while electrolytes increases
the conductivity and assist the cell action.

Sol 9: (B) AgBr is the most sensitive silver halide to

photo reduction. Hence it is used as the light sensitive
material in photographic films. The unchanged AgBr
is dissolved in hypo solution to cast an image on
photographic plate.
hv
2AgBr → 2Ag + Br2
3 1 . 4 0 | d and f-block Elements

Sol 10: (A) Tungsten is a metal of high melting point Sc3+ is highly stable (it does not show + 2)
and its filament gives brilliant light on passing electric
(D) The oxidation states increases as we go from group
current.
3 to group 7 in same period.

Sol 11: (A) In transition elements ns orbital is filled 2+ 2+

Sol 19 : (A) Each Cr (H2O )  and Fe (H2O ) 
up first and (n – 1)d afterwards, during ionization ns  6  6
electrons are lost prior to (n – 1)d electrons because Contain 4 unpaired electron.
the effective nuclear charge felt by (n – 1)d electrons is
higher as compared to that of ns electrons.
JEE Advanced/Boards
Sol 12: (B) The higher the charge on metal ion, smaller
is the ionic size and more is the complex forming Exercise 1
decreases in the order M4 + > MO22+ > M3+ > MO2+ . The
higher tendency of complex formation of MO22+ of Sol 1:
charge on metal atom M in MO22+
K 2MnO 4 + CO2 + H2O → KMnO 4 + MnO2 + KHCO3
Purple
Sol 13: (B): Being lesser energy difference between 5f
and 6d than 4f and 5d orbitals. Sol 2: Hg22+ → Hg + Hg2+

Sol 14: (C) Ln+3 compounds are mostly coloured. Sol 3: No. of e– in Co3+ → 24
NH3 is a strong ligand
Sol 15: (A) The correct order of E° 2+ values with
M /M
4s 3d
negative sign for the four successive elements Cr, Mn,
Fe and Co is Mn > Cr > Fe > Co
-
F is a weak ligand
2− 4s 3d
Sol 16: (B) In NiCl4  ,n=2

=
µ n (n + 2 ) BM there are unpaired e-

= 2 ( 2 + 2 )= 2.82 BM Sol 4: For NaOH we get MaAlO2, Na2ZnO2 .

For NH4OH, we get [Zn(NH3)4]2+.
Sol 17: (C) The general O.S of lanthanides is +3, only
few elements exhibit +4 O.S. But it doesn’t form complexes.

Sol 18: (A) Sol 5: PbS is black solid where as ZnS is white.
(A) V 2+ = 3 unpaired electrons
Sol 6: A → ZnCl2 because Zn(OH)2 & Al(OH)3 dissolve
Cr 2+ = 4 unpaired electrons in excess NaOH but Al2S3 is gray solid.
Mn2+ = 5 unpaired electrons
NaOH
2+ ZnCl2 Zn(OH2) (White ppt)
Fe = 4 unpaired electrons
Hence the order of paramagnetic behaviour should be AgNO3 Excess NaOH
2+ 2+ 2+ 2+
V < Cr = Fe < Mn
AgCl Na2ZnO2
(B) Ionic size decrease from left to right in same period
White ppt
(C) As per data from NCERT. H2S
Co3+ / Co2+ = 1.97 ; Fe3+ / Fe2+ = 0.77 ; Cr3+ / Cr 2+ = −0.41
ZnS (White ppt)
 Chem i str y | 31.41

Sol 7: They are not transition metals. They have (x – 1) In [NiCl4]2–, there is magnetic moment but not in
d10 configuration. [Ni(CO)4].

Sol 8: KMnO4 will oxidize Cl– to Cl2 Sol 13: Fe+2 doesn’t form complex with NH3 so, there
are Fe2+, NH+4 , SO24− ions in the solution.
?
Sol 9: Colourless salt (A) → (B)+(C) gas. In 2nd case, [Cu(NH3)4]2+ will be formed & there will no
B must be Zn or Al salt. Cu2+ Ions.

ZnS is white
Sol 14: In A, all H2O are ligands to Cr.
⇒ B = ZnO
In B, some H2O is evaporated So, that is water of
C is a reducing agent & turns lime water milky crystalisation.
⇒ C = SO2 6.75
WH O = Mcomplex = 18 g.
2 100
⇒ A = ZnSO4 & D = ZnS
So, there is 1 water of crystallisation.

Sol 10: MnO24− → MnO −4 + 1e−  ……(i) In (C)

13.5
MnO24− → 4H+ + 2e− → MnO2 + 2H2O  ……(ii) WH = M = 36 gm.
2O 100
2 × (i) A → [Cr(H2O)6] Cl3
3MnO24− + 4H+ → 2MnO −4 + MnO2 + 2H2O B → [Cr(H2O)5Cl]Cl2 H2O
So 3 moles of MnO24− need C → [Cr(H2O)4Cl2]Cl. 2H2O
2
2e– ⇒ Z =
3 Sol 15: AgBr is not soluble in dilute NH3
M 3M
∴E= = So, there is Cl– in A Cr3+ & strong
2 2
AgBr is pale yellow
(b) MnO-4 CrO -4
2

So, there is Br– in B

O O
A → [Cr(NH3)4Cl Br]Cl → d2sp3
Mn Cs
O O O- O O- O- B → [Cr(NH3)4Cl2]Br → d2sp3

Sol 16: Ratio of moles of Co3+ :

Sol 11: K2[PtCl6] → It has 3 ions.
NH3 : Cl– : SO24− : H2O
We don’t get : AgCl precipitate because there are no.
21.24 24.77 12.81 34.65 6.53
Cl– ions in the solution. There are only K+ and [PtCl6]2– : : : :
ions. 57 17 35 96 18
i. e. 1 : 3 : 1 : 1 : 1
Sol 12: Ni2+ with weak ligands is paramagnetic So, complex can be
4s 3d 4p → [Co(NH3)3 ClSO4] . H2O

[Co(NH3)3 SO4(H2O)]Cl

There are unpaired e- [Co(NH3)3 (H2O)Cl] SO4

Ni with strong ligands diamagnetic

4s 3d 4p
3 1 . 4 2 | d and f-block Elements

Exercise 2 Sol 12: (C) Ag, Pt, etc, are inert to conc. acids & alkalis.
AuCl3 is used for toning in photography.
Single Correct Choice Type
X Sol 13: (C) Cl– will be oxidised to Cl2 (greenish-yellow
Sol 1: (A) Cr2O27− 2CrO27−
Y gas).
y is H ,
+
Cr2O27− is stable in acidic medium.
Sol 14: (A) NO has unpaired e–. NO+ is stable taking
x is OH–
magnetic moment into consideration.
NaOH
→ CrO24−
Sol 2: (B) CrO3 
Multiple Correct Choice Type
Sol 3: (A) (NH4)2 Cr2O7 
∆
→ N2+Cr2O3+4H2O Sol 15: (A,B) Pt doesn’t form any alloys. Fe doesn’t
  ↓ form amalgam.
Green
Sol 16: (B, C) Os, Ru show +8 O. S.
Sol 4: (B) Mercury is ligand in room temperature (Hg). There are in same group as Iron (Fe)

Sol 5: (B) 80% conc HNO3 is an oxidising agent Sol 17: (A, B, C) Al2O3, SnO, ZnO are amphoteric
So Fe → Fe3O4 Fe2O3 → Basic.

Sol 6: (B) Cu + Conc HNO3 → Cu(NO3)2 + NO2 + H2O Sol 18: (A, B, C) Transition metals from interstitial
compounds. So, Co, Ni, Fe.
Sol 7: (A) Mn+7 
reduction
→ Mn+2

Acts as catalyst and increases further rate of reaction. Sol 19: (A, C)
Cr2O27− + 3H2O2 + 8H+ → 2Cr3+ + 3O2 ↑ + 2H2O
Blue colour
Sol 8: (D) CuSO4 + KCN → Cu(CN)2
(Excess)

Sol 20: (A, B) CuF2 is ionic CuCl2, CuBr2 are covalent

CuSO4 + KCN Cu(CN)2
(Excess) (exist as polymers).

Sol 21: (A, B, C, D) CuSO4(aq)+4NH3→ [Cu(NH3)4]2+ SO24−

Cu (CN) +
1
(CN)2 Cu2+
Cu
2+
& NH
& NH3 isis aa strong
strongligand
ligand
3
2
3s 3d
KCN

K3[Cu(CN)4] It’s paramagnetic, so coloured

MM
= n(n + 2) BM & n = 1 = 3 BM.
Sol 9: (D) MnO4 + 4HCl → Cl2 + MnCl2 + xH2O
Acidic KMnO4 decomposition can be catalysed by Sol 22: (A, B) MnO2, Mn3O4 → Amphoteric
sunlight. MnO → Basic
Mn2O7 → Acidic
Sol 10: (D) Fe2O3.xH2O & FeO(OH).Fe(OH)3 constitute
the rust.
Sol 23: (A,B) Zr, Hf have similar sizes & properties due
to lanthanide contraction.
Sol 11: (A) Invar is used to make meter scales. Invar
consists of Nickel, Iron & small quantities of other
metals.
 Chem i str y | 31.43

Sol 24: (A, B, C) Potash alum → KAl (SO4)2 Sol 33: (B) The yellow colour of CrO24− is due to charge
transfer from O to metal.
SO, we get K+, Al3+, SO24− .
O
Sol 25: (B, C, D) They will get harder due to the filling of
interval spaces. So, they will be less ductile & malleable.
2-
CrO4 Cr i.e. tetrahedral

O O-
Sol 26: (A, B) Metallic bonding is very weak in Hg. O-

Sol 34: (A) Maximum O. S. of Cr is +6. Because it has

Sol 27: (A, B) Transition elements have high IP only 6e– in 4s & 3d orbitals.
than s-block & lower IP than p-block. They are like
connecting blocks between s & p-blocks. So, they are
Sol 35: (C) Tungsten has very high MP due to strong
called transition elements.
metals bondings. It’s a metal, not a covalent compound.

Sol 28: (A, B, C) Catalytic activity depends on

Sol 36: (C) Cu+ is colourless because of no unpaired
Variable OS e–. All of Cu+ salts need not have 4 H2O molecules
Surface area statement-2 is false.
Reactivity (Complex forming ability)
It doesn’t depend on magnetic moment. Previous Years’ Questions

Sol 29: (A, B) Ag, Au form such complexes Sol 1: (B) In MnO −4 , Mn7+ is in highest oxidation state
possible for Mn. In CrO2Cl2, Cr6+ is in highest oxidation
Cu generally forms [Cu(CN)4]3–.
state possible for Cr.

Sol 30: (B, C) If x > y then x will be covalent Sol 2: (A) MnO −4 + I– + OH– → MnO24− + IO3−
Y will be Ionic
Sol 3: (B) In CuCl2, Cu2+ had d9 configuration, exhibit
Ex. KMnO4 → MnO −4 is covalent
d-d- transition and show colour. Similarly in VOCl2, V4+
MnCl2 → Ionic. had d1 configuration, can exhibit d-d transition and
show colour.
Sol 31: (B, C) Sc3+ → no unpaired e–
Sc
3+
no unpaired e
-
Sol 4: (C) KMnO4 is itself a very strong oxidising agent,
Ti
3+
1 unpaired e- O3 cannot oxidise it.

Sol 5: (A,B,C) Brass = Cu and Zn

Gun metal = Cu, Sn, Zn
2+
Cu 1 unpaired e-

Bronze = Cu, Zn and Sn

Zn
2+
0 unpaired e- Type metal = Pb, Sn, Sb

Sol 6: (A,B,C,D) 4NaCl + K2Cr2O7 + 6H2SO4 → 2CrO2Cl2

3+ 2+
So, Ti , Cu
Chromyl chloride
(Red vapour)
Assertion Reasoning Type + 4NaHSO3 + 2KHSO4 + 3H2O

Sol 32: (A) In alkaline medium

mol. wt mol. wt
Mn+7 → Mn+4, Z = 3 = =
2 3
Z is different for different conditions
3 1 . 4 4 | d and f-block Elements

Sol 7: (A,B,C) Cr : [Ar]3d54s1 Since (CO) is strong ligand, in Cr ( CO ) no unpaired

6
electron is present. So ‘spin only’ magnetic moment is
Magnetic quantum number: –  ………0…….+ 
zero.
Ag(4d105s1): All paired electrons have opposite spin.
The last one has unpaired spin.
Sol 15: A → p, s; B → q, s; C → r, t; D → q, t

Sol 8: (A) In neutral medium: 3Cu + dil 8HNO3 → 3Cu (NO3 ) + 4H2O + 2NO
2

MnO −4 → MnO2(Mn7+ + 3e– → Mn4+) Cu + conc. 4HNO3 → 4Zn (NO3 ) + 5H2O + N2O
2

In alkaline medium: 4Zn + dil. 10HNO3 → 4Zn (NO3 ) + 5H2O + N2O

2

MnO −4 → MnO2 (Mn7+ + 3e– → Mn4+) Zn + conc.4HNO3 → Zn (NO3 ) + 2H2O + 2NO2

2

In acidic medium: 2−
Sol 16: (B) Ni ( CN)  ,O.S. of Ni = +2
MnO −4 → Mn (Mn + 5e → Mn )
2+ 7+ – 2+
 4

Ni ( 28 ) = 3d8 4s2
Sol 9: µ = n(n + 2) BM where ‘n’ is number of unpaired
Ni+2 = 3d8
electrons.
No. of unpaired electrons = 2
1.73 = n(n + 2)
Magnetic moment µ =2.82 BM
n = 1; V4+ 3d1

Sol 17: (C) Following compounds are diamagnetic.

Sol 10: CrO3 is anhydride of chromic acid:
L : Co (NH3 )  CI3
CrO3 + H2O → H2CrO4  6

Chromic acid M : Na3 Co ( Ox ) 


 3

O : K 2 Pt ( CN) 
 4
Sol 11: A = [Ti(H2O)6]3+ and M = Ti, B = TiO2, Ti(IV) has
no electron in 3d orbital, no d-d transition is possible, P :  Zn (H2O )  (NO3 )
 6 2
therefore MCl4 is colourless. In A, there is one electron
in 3d orbital and its d-d transition is responsible for
Sol 18: (A) Aqueous solution of copper sulphate absorbs
colour.
orange red light and appears blue (complementary
colour).
Sol 12: (B) B
V G
(A) Mn+ = 3d5 4s1 in presence of CO effective configu- R Y
ration = 3d6 4s0. Three lone pair for back bonding with O
vacant orbital of C in CO Sol 19: (B)
(B) Fe0 = 3d6 4s2 in presence of CO effective configura-
P = Fe+3 (no. of unpaired e– = 5)
tion = 3d8 four lone pair for back bonding with CO.
(C) Cr0 = 3d5 45s1 Effective configuration = 3d6 three Q = V+2 (no. of unpaired e– = 3)
lone pair for back bonding with CO. R = Fe+2 (no. of unpaired e– = 4)
(D) V– = 3d4 4s2 Effective configuration = 3d6 three pair
As all ligands are weak field, hence the no. of unpaired
for back bonding with co. Maximum back bonding is
electrons remains same in the complex ion.
present in Fe(CO)s there for CO bond order is lowest
here. =µ n (n + 2 )B.M
Hence (B) is correct.
Sol 13: (C) In the crystalline form CuF2 is blue coloured.
3− 3−
Sol 20: Fe ( SCN)  and Fe ( CN) 
Sol 14: (A) Cr ( CO )   6  6
 6
In both the cases the electronic configuration of Fe3+
Cr ( 24 ) =  Ar  3d5 4s1
will be 1s2 ,2s2 ,2p6 ,3s2 ,3p6 3d5
 Chem i str y | 31.45

Since SCN‒ is a weak field ligand and CN– is a strong

field ligand, the pairing will occur only in case of
3−
Fe ( CN) 
 6

5
(high spin)
Case -1 3d (no paining)

(low spin)
Case - 2 (pairing)

Case-1 µ= n ( n + 2 )= 5 (5 + 2 ) = 35= 5.91 BM

Case-2 µ = n (n + 2 ) = 1 (1 + 2 ) = 3 = 1.73 BM

Difference in spin only magnetic moment

= 5.91 − 1.73 = 4.18
≈4

Sol 21: (B) Ni ( CO )  − sp3− - Diamagnetic

 4
−2
NiCI 4  sp3 − Paramagnetic

Co (NH ) CI  CI − d2sp3 − Diamagnetic

 3 4 2

Na3 CoF6  − sp3d2 − Paramagnetic

Na2O2i.e. O12− − Diamagnetic

CsO2 i.e O −2 − Paramagnetic

2+
Sol 22: (A) Ni (NH3 )  = Octahedral
 6
+2
Pt (NH )  = Square planar
 3 4

+2
 Zn (NH )  = Tetrahedral
 3 4