01 EP

02
=CVE + no of MV
03 VALENCE BOND THEORY
05
BOND ORDER
Vsepr 2 HYBRIDSATION C - C = 1 bond 2 - 1 - Diamagnetic
MV Remaining
C = C = 1 + 1π 3 - 0.5 - Paramagnetic
2,0- Linear - 180o 2- SP - Linear C -
= C = 1 + 2π 4 - 0 - Diamagnetic
Monoatomic 5 - 0.5 - Paramagnetic
3,0- Triagonal planar - 120o BP LP 3- SP2 - Triagonal planar MOLECULAR ORBITAL THEORY 6 - 1 - Diamagnetic
2,1- Bent shape 7- 0.5- Paramagnetic
EP =CVE + 0 4- SP3 - Tetrahedral Up to N2 the energy order is, 8 - 0 - Diamagnetic
4,0- Tetrahedral - 109o28| 2 5- SP3d - Triagonal pyramidal 9 - 0.5 - Paramagnetic
3,1- Pyramidal - 107o DV Remaining 1s< *1s< 2s< *2s<(π 2px=π 2py) 10 - 1 - Paramagnetic
Diatomic 6- SP3d2 - Octahedral < 2pz<(π* 2px=π*2py)< *2pz 11 - 1.5 - Paramagnetic
2,2- Bent - 104.5o 12 - 2 - Diamagnetic
LP
7- SP3d3 - Pentagonal bipyramidal 13 - 2.5 - Paramagnetic
BP
5,0- Triagonal Bipyramidal After N2 the energy order is,(for 02,F2,Ne2) 14 - 3 - Diamagnetic
4,1- See saw 15 - 2.5 - Paramagnetic
CH3-CH3 - sp 3 1s< *1s< 2s< *2s< 2pz<(π 2px=π 2py)
16 - 2 - Paramagnetic
EP =CVE + negative charge - positive charge
3,2- T shape <(π* 2px=π*2py)< *2pz 17 - 1.5 - Paramagnetic
2 CH2=CH2 - sp2 18 - 1 - Diamagnetic
2,3- Linear - 180 o
MV/DV Remaining
B2-10 Paramagnetic π2p1x= π2py1 19 - 0.5 - Paramagnetic
CH=CH - sp 20 - 0 - Diamagnetic
π*2px= π*2py
1 1
6,0- Octahedral - 90o Charged Compound O2-10 Paramagnetic
LP Odd number + 10&16 - Paramagnetic
5,1- Square pyramidal
BP =C= - sp Even number except 10&16 - Diamagnetic

4,2- Square planar LP-LP > LP-BP > BP-BP

7,0- Pentagonal bipyramidal
6,1- distorted octahedral
BOND PARAMETERS
In XeF6, oxidation state and state of
Total no. of Lone pairs in I3- hybridisation of Xe and shape of the The no. of unpaired electrons in O2

Q2 1) 3 2) 6 Q3 molecules are, respectively

1) +6, sp3d3 distorted octahedral
Q4 molecule is:
1) 0 2) 1
In which of the following pairs are 2) +4, sp3d2 square planar

Q1 3) 9 4) 12 3) 2 4) 3
the two species triagonal pyramidal ? 3) +6, sp3 , pyramidal
BOND ORDER
4) +6, sp3d2 square pyramidal
1) BrO3 and XeO3
-
2) SF4 and XeF4 1
B.O Stability Bond
2) SO 3 and NO3-
2-
4) BF3 and NF3 B.L
strength
B.O-O22+ > O2+> O2
3 2.5 3

B.L-O2>O2 >O2 + 2+

1) Max B.A-1800[sp]
1) Polar molecule, = 0 Resonance Changes B.O.
BOND 1) Hybridisation
DIPOLE Irregular geometry.
RESONANCE 1) Benzene
BOND 2) Hybridisation BOND 1) B.L with Size of atom

ENERGY sp> sp2 >sp3

MOMENT
( )
a) Different bonds
b) Lone pairs
3
ANGLE sp > sp2 > sp3
NO2+ > NO2 > NO2-
LENGTH HI > HBr > HCl > HF

2) B.E. B.O. (No. of Bonds)

c) Different surrounding atoms = 2 = 1.5 Sp Sp2 2) B.L decreases with
eg: SF4, CH3Cl 3) if LP B.A
N2> O2 >F2 2) Non polar molecule, =0 CH4 > NH3 > H2O [sp3]
multiplicity
B.L.: C2H6 > C6H6 > C2H4 > C2H2
3) B.E. E.N. difference Regular geometry 0 LP 1 LP 2 LP C-C>C=C>C=C
a) Same bonds B.O.: 1 1.5 2 3
HF> HCl > HBr > HI b) Zero lone pairs
4) If electronegativity of
2) O3 3 C.A B.A
4) Halogens c) Same surrounding atoms = 2 = 1.5 NH3 > PH3 > ASH3 > SbH3
eg: CH4, CO2
Cl2> Br2 > F2 > I2 CH3OH > CH3Cl > H2O> NH3> NF3> BF3 B.L.: H2O2 > O3 > O2 5) If electronegativity oF S.A B.A
X X X =0
BI3 > BBr3 > BCl3 > BF3
X
> B.O.: 1 1.5 2
O > O
X
O

Q9
The correct order of increasing
Q8
The correct order of bond angles:

Q5
Which of the following molecules
has the maximum bond enthalpy? Q6 Which of the following will
have maximum dipole moment?
Q7 The correct order in
which the O-O bond length 1) H2O > NH3 > CH4 > CO2
bond length of C-H,C-O,C-C & C=C is:
1) C-C < C=C < C-O < C-H
1) N2 2) CO decreases in the following is 2) H2O < NH3 < CO2 > CH4 2) C-O < C-H < C-C < C=C
1) NF3 2) NH3
1) O3 > H2O2 > O2 2) O2 > O3 > H2O2 3) H2O < NH3 > CO2 > CH4 3) C-H < C-O < C-C < C=C
3) F2 4) HF
3) CH4 4) PCl3 3) O2 > H2O2 > O3 4) H2O2 > O3 > O2 4) CO2 > CH4 > NH3 > H2O 4) C-H < C=C < C-O < C-C

Fajans Rule - Indicates Covalent character in an ionic bond.

Hydrogen bond- Formation of bond between hydrogen & most electronegetive elements like F,O,N 1) Size of the cation- Smaller the cation higher is the polarisation, so covalent character increases
Intermolecular - H – Bonding occur within one single molecule. LiCl > NaCl > KCl > RbCl > CsCl
2) Size of anion- As the size of anion increases, polarisation increases, covalent character increases
Intermolecular - H Bonding between two different molecules of same or different compounds. AgF < AgCl < AgBr < AgI
Intermolecular H bonding increases the boiling point. eg: Intermolecular H bond in p-nitrophenol 3) Charge on the cation- As the charge increases, Covalent character also increases
increases the boiling point. HF(HB) > HI > HBr > HCl. H2O(4HB) > H2Te > H2Se > H2S LiCl < BeCl2 < BCl3 < CCl4
 01 02 03 LAW OF EQUIPARTITION OF ENERGY
DEGREES OF FREEDOM SPECIFIC HEAT CAPACITY MIXING OF GASES
1 Molecule per f = - 1 KT
a) C P − C V =R n1cv1 + n2cv2 2
. For monoatomic gas, f =3 (MSH) CVmix = f
Total for a molecule = -K T
. For diatomic gas, n 1+ n 2 2
b) C P − C V =R 5 3
(a) at room temperature, f = 5 - Mono- = - Monoatomic Molecule = -K T
(GSH) M 3 n1cP1 + n2cP2 2
(b) at high temperature, R f 7 CPmix = Total for a mole= - f RT
f = 7 c) C
V
=-
-1
=-
2
R Dia = -5 n 1+ n 2 2
4 3
For triatomic gas, Tri = - Monoatomic=-R T
(a) Linear f= 5 d) C =- =1-
R
+ f 3 CP mix (mole) 2
-1
= 5 RT
P
2 mix Diatomic= -
(b) Non-linear f= 7
e) =
CP
= 1 +-
2 CV mix (mole) 2
1 Vibrational mode f = 2 Cv 3
Translatory Kinetic energy=-R T
f
(mole f= 3)
2

Ideal gas is composed of polyatomic

Q3
If CP and Cv denoted the specific heats

Q1 Q2
Consider a mixture of n moles of helium gas and

Q4
A gas mixture consists of 2 moles of 02 and 4 moles
molecule that of unit mass of nitrogen at constant pressure 2n moles of oxygen gas (molecules taken to be rigid) of Ar at temparature T. Neglecting all vibrational
has 4 vibrational modes. and volume respectively, then as an ideal gas. It‛s Cp/CV value will be: modes, the total internal energy of the system is
Total degree of freedom is R R R
a) CP-CV= b) CP-CV= c) CP-CV= d) C -CV= R a) 19/13 b) 67/45 c) 40/27 d) 23/15 a) 4RT b) 15RT c) 9RT d) 11RT
28 14 7 P
a) 12 b) 14 c) 8 d) 6

06
VELOCITY OF GAS
05
Root Mean Average Speed: Most probable FIRST LAW OF
square speed: speed: Vmps:Vavg:Vrms = 1 : 1.14 : 1.228 THERMODYNAMICS

Arithmatic mean of speed of QP= U + W

07
U= ncv T
Square root of mean of square Speed possessed by maximum
molecules of gas at given
of speed of different molecules, number of molecules of gas.
temperature.
vrms = v1 + v2 +............... + vn
W = nRT=PDV
2 2 2

2RT 2P
n
v avg = I v1 I + I v2 I + ....... + I vnI vmp = =
n Mo d
U
vrms =
3RT
=
3P
=
3kBT
v avg = 8RT = 8P MEAN FREE PATH = 1
M d m
πM πd QP
Average distance travelled by
W
molecules between two
successive collision
1 = 1- 1
1
d2 QP
λ mean = 1
2 πd n
2
Consider a gas of triatomic molecules. The molecules
are assumed to be triangular made up of massless rigid
The rms speeds of the molecules of Hydrogen,
The mean free path of molecules of gas, r2
Oxygen & Carbondioxide at the same temparature

Q5 Q6
rods whose vertices are occupied by atoms.The internal are VH, VO and Vc respectively then: (radius r) is inversely proportional to d = diameter of molecules. T
energy of a mole of the gas at temperature T is:
a) r3 b) r2 n = no. of molecules per P
a) VH >VO>VC b) VC >VO>VH
5 3 9
3RT unit volume
KINETIC THEORY OF GAS
a) RT b) RT c) RT d)
2 2 2
c) VH =VO>VC d) VH =VO=VC c) r d) r

1 PV=nRT
GAY PRESSURE 2 -
mn v 2
BOYLES CHARLES PV mn Vrms
LUSSACS 3 R=8.314 JK-1mol-1
OF GAS
P P/T
PV V V/T

LAW LAW LAW T P

V
T V

Relation between pressure

P
. PV = constant, if T =Cosntant . V α T; v = constant; P = constant.
. P α T; = constant; V = constant. and Kinetic Energy.
T T
3
. P 1V1 = P2V2 ,When gas changes it‛ s . v 1 = v 2 ,When gas change its state . P1 =
P2 , When gas change its state E = PV
state under constant T1 T2 under constant pressure. T1 T2 under constant Volume. 2
temperature.
Specific heat of Solid = 3R
WATER = 9R
 BORON FAMILY PHYSICAL PROPERTIES CHEMICAL PROPERTIES COMPOUNDS OF Al

Reducing power Al>Ga>In>Tl

AlCl3
5 Most abundant metal
Non-metal B Third most element Stability order of O.S: Al2O3 + 3C+3Cl2
heat
2AlCl3 + 3CO
(O > Si > Al > Fe) Atomic radii
B < Ga < Al < In < Tl Ionic radii Decreases Tl >In >Ga >Al >B (Inert pair effect)
+1 +1 +1 +1 +1
Aqueous solution is acidic due to the formation of HCl.
13 down the group B+3>Al+3>Ga+3>In+3>Tl+3 AlCl3 + 3H₂O Al(OH)3 + 3HCl
Ionisation
Al enthalpy Tl+3 = Strong oxidant, Ga+1 = Strong reductant
Density Increases Anhyd.AlCl3 is covalent & dimeric.
31 exists as Ga2 molecules & B > Tl > Ga > Al >In
Down the group
Maximum covalency of B is 4 (absence of d orbitals)
Ga thus low M.P (29.67o C)
-liquid up to 2000o C
B
Halides -Lewis acids (BF3 < BCl3 < BBr3) Alum
49 Used in high T thermometer Tl
Metal Ga M.p M2SO4 M|2(SO4)3 24H₂O
In decreases from B 2O 3 B(OH)3 M - Monovalent metal M| - Trivalent metal

Acidic Character
Acidic Character

Basic Character
Basic Character
Examples:
B to Ga and then Al2O3
81

Decreases
Decreases
Al(OH)3

Increases
Increases
Al Potash alum K2SO4 Al2(SO4)3 24H2O
In increases Ga2O3 Ga(OH)3
Tl B > Al > Tl > In > Ga In2O3 In(OH)3
Chrome alum K2SO4 Cr2(SO4)3 24H2O
Electronegativity Ferric alum (NH4)2 SO4 Fe2 (SO4)3 24H2O
113 B > Tl > In > Ga > Al B.p Decreases
Tl2O3
Tl(OH)3
Each cation is surrounded by 6 H2O
Nh down the group
Radioactive Oxides Hydroxides

BORIC ACID DIBORANE BORAX / TINCAL CARBON FAMILY

1) Oxidation State: +2, +4, -2 Halide
Na2B4O7 10H2O / Na2[B4O5(OH)4].8H2O
Stability of +4: C > Si > Ge > Sn > Pb
Highly reactive: it catches fire
Stability of +2: Pb > Sn > Ge > Si > C
Thermal stability order :
Aqueous solution of borax is alkaline in nature
Weak monobasic acid (Lewis acid) B2H6 + 302 B203 + 3H20, H = -ve
2) Oxides
CCl2 < SiCl2 < GeCl2 < SnCl2 < PbCl2
CCl4 > SiCl4 > GeCl4 > SnCl4 > PbCl4
H2BO3+H2O [B(OH)4] + H+ With water : B2H6(g) + 6H₂O(l) 2B(OH)3(aq) + 6H2(g)
Na2B4O7 + 7H2O -> 2NaOH + 4H3BO3 CO Neutral CO2 Acidic
SiO Neutral SiO2 Acidic CCl4 can‛t be hydrolysed due
Borax bead test (Detection of transition metal)
Reaction with ammonia :
GeO Acidic GeO2 Acidic to absence of d orbitals
Heating effect: Low temp
(-120°C) Heat Na2B4O7.10H2O Na2B4O7
740OC
2NaBO₂ + B2O3 SnO Amphoteric SnO2 Amphoteric PbI4 & PbBr4 does not exist
B2H6+ 2NH3 [BH₂ (NH3)2] (BH4] 2B3N3H6 + 12H2 Sodium
PbO2 Amphoteric due to strong oxidising
Boric
Metaborate anhydride PbO Amphoteric
H3BO3 HBO₂ H2 B 4 O 7 B 20 3
Excess NH3
273K 433K red
T.M Colour (Litharge) nature of Pb+4
Borazine or Glassy mass (Borax bead)
borazole

hot
(inorganic benzene)
Cu/Co Blue
Metaboricacid Tetraboric acid Boron trioxide

COMPOUNDS OF CARBON
(Boric anhydride) Uses : As flux
Water softening
2-3c-2e banana bond
Forms 6 H-bonds in 4 2c-2e Terminal bonds
As antiseptic
In making enamel
aqueous solution. Manufactur of glass CO CO2
There are 5 B-O-B bridge bonds H2SO4
HCOOH CO (100% pure)
Solid CO2 - Dry ice (Refrigerant)
Coal gasification
C + H2O CO + H2 CO2 in water gives carbonic
Syn gas / Water gas acid (maintain pH 7.26-7.42)
Producer gas NaOH + CO2 NaHCO3
COMPOUNDS OF SILICON CO + O2 CO + N2
Used in soft drinks
With Hb it forms 300 Used as fire extinguisher
SILICA SILICONES SILICATES ZEOLITES times stable Carboxy Hb

- SiO2-sand, quartz
(piezoelectric) - general formula (R2SiO)n - Metal derivatives\ - Sodium Aluminium silicates ALLOTROPES OF CARBON
of silicic acid H4SiO4 (Na2Al2Si2O8.XH2O)
- R2SiCl2+H2O→R2Si(OH)2
- Insoluble in H2O DIAMOND GRAPHITE FULLERENES
(Linear Chain Silicones) Basic unit is (SiO4) 4-
& inert at RT - (i) Used for Purification of H2O C-C bond lenth is 141.5pm
H2 O Tetrahedral to remove hardness of water
C-C bond length is 154 pm Graphite C -C
60 70

- Reacts with HF - RSiCl3 → RSi(OH)3 C is sp hybridised 3 Thermodynamically most (Bucky balls)

stable due to RH = 0
(Cross linked silicones) - (ii) ZSM 5 is a shape selective C is sp2 hybridised
HF + SiO2 → SiF4 catalyst to convert alcohol
Good thermal conductor C is sp2 hybridised
SiF2+2HF→ H4SiF4 - R3SiCl→stopping agent / to gasoline
H of formation is Good electric conductor C-C 143.5pm
Dimer 1.9KJ/mol C=C 138.3pm
Hexagonal ring layers
- 3D Geometry Used as abrasive for
sharpening of tools
which are 340pm apart C60 has 12 pentagons
Used as dry lubricant and 20 hexagons
in machines
ALKALI REACTION WITH OXYGEN Sodium
carbonate
METALS
METAL HYDRIDES washing soda
2M + H2 2MH Li forms only normal oxide Li2O
ATOMIC SIZE
LiH, NaH, KH, RbH, CsH Na forms normal oxide (Na2O) & Peroxide (Na2O2) Na2CO3 . 10H2O
Li < Na < K < Rb < Cs
Li <Na+ <K+< Rb+ <Cs+ THERMAL STABILITY
K forms normal oxide, Peroxide & Superoxide (KO2) METAL HYDROXIDES Solvay Process:
Rb forms normal oxide, Peroxide & Superoxide
LiH > NaH > KH > RbH > CsH Cs forms normal oxide, Peroxide & Superoxide Reactants- NH3, CO2
REDUCING CHARACTER KO2 ,RbO2 ,CsO2 (Superoxide) - Paramagnetic & coloured BASIC CHARACTER Regeneration of NH3
- Using Ca(OH)2
LiOH < NaOH < KOH < RbOH < CsOH
LiH < NaH < KH < RbH < CsH Caustic Soda Byproduct - CaCl2
Sodium bicarbonate/
NaOH we can’t obtain K2CO3 Because
Baking Soda/
KHCO3 obtained in 2nd step is
NaHCO3
(i) It is a white crystalline water soluble

IONISATION
REACTION WITH H2O solid (i) It is white crystalline solid
(ii) Deliquescent in nature (ii) It is sparingly soluble in water
ENERGY ELECTROPOSITIVE
CHARACTER / Li React with water (iii) Reaction with acidic (iii) It is not stored with strong
REACTIVITY oxide
Li > Na > K > Rb > Cs
Na React easily with water REACTION WITH NITROGEN NaOH +CO2 →Na2CO3+H2O
bases like NaOH, because
1 k it has acidic H
1E
Li < Na < K < Rb < Cs (fast reaction) Only Li react with N2 gas to form Li3N
NaHCO3+ NaOH →Na2CO3+H2O
Rb Give explosive reaction with water which on hydrolysis give NH3 gas.
Cs
6 Li(s) + N 2(g)→ 2Li3N
Na/K/Rb/Cs + N2 →X
Li3N + 3H2O → 3LiOH +NH3
MELTING HYDRATION COMPLEX
POINT ENERGY FORMATION

Li > Na > K > Rb > Cs Li+ > Na+ > K+ > Rb+ > Cs+ COMPOUNDS OF SODIUM
In alkali metals,
BOILING POINT only Li form complex
due to small size
Li > Na > K > Rb > Cs eg: LiCl 2H2O
Anomalous behaviour of Li ( Due to small size & High charge density)
Diagonal relationship with Mg
• The melting point and boiling point of lithium are higher that than other alkali metals.
Charge • Li & Mg → harder
• The hardness of lithium is higher than other metals. • Same
IONIC MOBILITY
DENSITY
Radius • Li & Mg react with cold water
• The alkali metal chlorides do not have the capability to form hydrates but lithium
I Li < K < Na < Rb < Cs chloride crystallizes to form a hydrate • Li & Mg both form nitride which on hydrolysis give NH3 gas
Li3N + 3H2O → 3LiOH + NH3
hydration >
LiCl2, H2O
Li+ < Na+ < K+ < Rb+ < Cs+ K > Vacant d orbital
> Density • Lithium nitrate decomposes to form an oxide whereas other metals on heating Mg3N2+ 6H2O →3 Mg(OH)2+ 2NH3
give nitrites.
• Compounds of lithium are partially soluble in • LiCl and MgCl2 both are deliquescent in nature.
water whereas the alkali metals are highly soluble in water.

S-BLOCK
ALKALINE
EARTH METAL HYDRIDES
REACTION WITH OXYGEN Calcium oxide/
Quick lime/ CaO
METALS ATOMIC SIZE
MgH2, CaH2, SrH2, BaH2
1) Be, Mg → Normal oxide BeO & MgO
2) Ca, Sr, Ba → Normal oxide & Peroxide.
1) CaO + H2O Ca(OH)2
Be < Mg < Ca < Sr < Ba
THERMAL STABILITY Peroxide is diamagnetic, but shows METAL HYDROXIDES Quick lime Slaked lime

BeH2 > MgH2 > CaH2 > SrH2 > BaH2 colour due to Lattice defect Uses
REDUCING CHARACTER BASIC CHARACTER Used as basic flux
BeH2 < MgH2 < CaH2 < SrH2 < BaH2 Slaked lime/Ca(OH)2 CaO +SiO2 CaSiO3
basic acidic CaSO4.2H2O
flux slag
1) Used in white wash impurity Gypsum

2) In manufacturing of
IONISATION
REACTION WITH H2O bleaching powder,
ENERGY ELECTROPOSITIVE
glass etc CaSO4.1/2H2O
Be > Mg > Ca > Sr > Ba
CHARACTER /
REACTIVITY 1) Be → Steam
1 2) Mg → Hot water
REACTION WITH NITROGEN Ca(OH)2 + Cl2 →Ca(OCl)2 Plaster of Paris
IE Bleaching powder
Ca All elements react with nitrogen to form
Be < Mg < Ca < Sr < Ba
3) Sr → React with cold water nitrates of type M3 N2 3) In removing temporary
Ba hardness of water
CaSO4
Mg3N2 + H2O → Mg(OH)2 + 2NH3
Dead burnt plaster

MELTING HYDRATION COMPLEX

POINT ENERGY FORMATION

Be > Ca > Sr > Ba > Mg Be2+ > Mg2+ > Ca2+ > Sr2+ > Ba2+ Be, Mg & Ca form complexes
BeCl2 + 2H2O BeCl2 . 2H2O
COMPOUNDS OF CALCIUM
BOILING POINT BeF2 + 2F- BeF42-
max. C.N = 4

Be > Ba > Ca > Sr > Mg Ca+EDTA Ca EDTA

Anomalous behaviour of Be ( Due to small size & High charge density)

Diagonal relationship with Al
C.N = 6

1) The coordinate number of Be is not more than 4 where as other alkali Charge
metals have coordiante number of 6 • Same
IONIC MOBILITY
DENSITY Radius
I
Ca < Mg < Be < Sr < Ba 2) The M.P and B.P are higher when compared to the other elements Be & Al are not attacked easily by acids[HNO3] forms protective layer
hydration of the group BeCl2 & AlCl3 → Lewis acid and form dimer
Be2+ < Mg2+ < Ca2+ < Sr2+ < Ba2+
3) They forms covalent bonds whereas the other members of BeCl2 in vapour as dimer and solid as polymer
the group, ionic bonds BeO, Be(OH)2 , Al2O3 , Al(OH)3 → Amphoteric
4) Be does not react with H2O like the other companions of the group Both form complexes - [BeF4]2--sp3 [Al(H2o)6]3+-sp3d2
MONODENTATE LIGAND BIDENTATE LIGAND CHELATE LIGAND GEOMETRICAL ISOMERISM Optical isomer
COO- 1) EDTA4-,en 2=d cis + l cis
X-- Halido 1) [Ma2b2] - 2 (cis + trans)
1) Oxalato COO- 2) Stability is max Total Stereoisomer
Anionic

OH- - Hydroxo 2) [Ma2bc] - 2 (cis + trans)

2) Ethane-1,2-diammine[en]
3=d cis + l cis + trans
NO2- - Nitrito COORDINATION NUMBER (CN) 3) [Mabcd] -3 (2 cis + trans)
H2N-CH2-CH2-NH2
SO42- -Sulphato 1) nx1 monodentate 4) [Ma4b2] / [Ma4bc] VBT
O2-- Oxo 2) nx2 bidentate 5) [Ma3b3]- fac & mer SPECTROCHEMICAL SERIES
S2-- Sulphido POLYDENTATE LIGAND
3) nx3 Tridentate 6) [Ma2b2c2]- 2 I-<Br-<SCN-<Cl-<F-<C2O42- <H2O
1) N(CH2CH2NH2)3- Tetradendate <NCS- < EDTA4-<NH3<en <NO2-<CN-<CO
7) [Mabcdef]- 15
Neutral

H2O- Aqua 2) Ethylene Diammine Tetra Acetate [EDTA]

NO- Nitrosyl - COO-
WERNER’S THEORY 8) [M(en)2b2] [M(en)2bc]
OOC dsp2 Square planar

NH3 - Ammine C.N=6 1) 1 Valency - O.N

0
[ML4]
CO -Carbonyl 2) 20 Valency - C.N
OPTICAL ISOMERISM
COO

}
- -
sp3 Tetrahedral
OOC
1) C.N = 6 sp3d

}
4O Atoms
Donor - [Pt(NH3)5Cl]Cl3 2) 1 bidentate
Except
[ML5] Triagonal bipyramidal
2N Atoms
Cationic

[Co en(H2O)4]
1) 10 Valency - O.N = 4 3) Cis isomer dsp3
sp3d2

}
NO2+- Nitronium 2) 20 Valency - C.N = 6
AMBIDENTATE LIGAND [M(en)2b2] [M(en)2bc]
NO+- Nitrosonium 3) AgCl Formed 3 [ML6] Octahedral

1) NO2- 2)SCN- 4) Total ions - 4 Geometrical isomer d2sp3

2= cis + trans

COORDINATION COMPOUNDS

APPLICATION OF
UNPAIRED ELECTRONS - PARAMAGNETIC CFT
OMC
3 2 ORGANO METALLIC COMPOUNDS
PAIRED ELECTRONS - DIAMAGNETIC 1) Octahedral - Ligand approaches CFSE= a x - t+ b x t + np
along the axis eg> t2g 5 5 1) Ziegler Natta Catalyst is used in
= n(n+2) 1) σ bonded - RMgX(Grignard reagent) polymerisation of alkene
eg HIGH SPIN COMPLEX LOW SPIN COMLEX
< Pairing Energy ∆0 > Pairing Energy - R2SiCl2 TiCl4 + AlEt3
1=1.73 BM 3
+ ∆0 ∆0 ∆0
π bonded - Ferrocene
5
2=2.84 BM 2) 2) Wilkinson’s Catalyst is used in
3=3.87 BM 2

CFSE
− ∆
hydrogenation
3) σ & π bonded- Metal Carbonyls
Average energy of 5 0
4=4.90 BM d-orbitals
t 2g
5=5.92 BM
Splitting of d-orbitals in
Mononuclear- [Fe(CO)6] [{(C6H5)3P}3Rh]Cl
octahedral crystal field.
[NiCl4] - sp ,Tetrahedral n=2
2- 3
[Mn(CO)6] < [Cr(CO)6] < [V(CO)6]
+ -
APPLICATION OF
[NiCO4] - sp3,Tetrahedral n=0 1) Max for CFSE
2 3 COORDINATION COMPOUNDS
[Ni(CN)4]2- - dsp2,Sq. planar n=0 CFSE= a x -
5 0+ b x
5
0 + np 2) ∆0 > ∆t ∆t = -4 ∆0 BO BL BO BL
9
3) [Fe(CN)6]4- < [Fe(CO)6] Back bonding 1) Cisplatin ([Pt(NH3)2Cl2]) is used as an
H2O act as SFL with Co3+, Cu2+
2)Tetrahedral -Ligand approaches anticancerous agent
NH3 act as SFL with Co3+, Cu2+,Co2+ b/w the axis t2g> eg 4) Jahn Teller by - d4, d7,d9 π∗
π 2) EDTA in Lead poisoning
forms d2sp3 π
Cr3+ - 3) Copper and iron poisoning-
4d and 5d metals all ligands act as SFL
t 2g
6) Color - d-d transition, E∝ m
π C≡O depencillamine & desferrioxime
2
∆ ∆t =
-4
∆
λ 4) Vitamin B-12 - Cobalt
[Pt(CN)4 ]2- [Pt(Cl)4 ]2- [Pd(Cl)4 ]2- =dsp2 5 t 9 0
[Co(en)3] < [Co(NH3)6] < [Co(H2O)6]
3+ 3+ 3+

λ π 5) Chlorophyll - Magnesium
C2O42- act as SFL with Co3+ Average energy of −
3
∆
5 t
E E λ
d-orbitals Synergic Bonding 6) Carboxy peptide- Zinc
F- act as SFL with Ni4+ d-orbitals free ion
eg

Splitting of d-orbitals in
tetrahedral crystal field.