NCERT Revision Notes for Chapter 5 Coordination Compounds Class 12 Chemistry

Answer

When two or more than two simple salts are allowed to chemically combine in a fix ratio then addition compound is formed. On the basis of behaviour in aqueous solution they are of the following two types.

1. Double salt: - Addition compound in which simple salts don’t loss their identity and its aqueous solution give test of its all constitute ions.
• Double salts loss their identity in aqueous solution.
KCl + MgCl₂ + 6H₂O → KCl.


• All Alums are double salt.

2. Complex compound : − Addition compound in which simple salts loss their identity and it’s aqueous solution doesn’t give test of its all constitute ions.
• Complex compound don’t loss their identify in aqueous solution.
K₄[Fe(CN)₆] → 4K⁺ + [Fe(CN)₆]⁴⁻
Does not ionize to give Fe²⁺ and CN⁻ ions

Answer

Answer

Coordination sphere :-

Outside region:-
• Ionisation sphere
• Free ion
• Cation 

Naming of ligands
• —O—suffix provided to the name of anionic ligands.
• —ium suffix provided to the name of cationic ligands.
Anionic ligands ending with -ide are named by replacing -ide with suffix -o or replacing -e by -o. Ligands whose names end in -ite or -ate become -ito or -ato, i.e., by replacing the ending -e with -o.

Answer

(a) On the basis of denticity
1. Monodentate ligand
(i) Neutral ligands

• Hydrozine never acts as bidentate ligand.

(ii) Cationic ligand : -

(iii) Anionic ligands

2. Bidentate ligands
(i) Ethylene diamine (en) or (ethane 1, 2−diamine)

(ii) Propane 1, 2–diamine Or propylene diamine (Pn)

(iii) Dipyridyl (dipy)

(iv) Oxalato (OX)⁻²

(v) Glycenato (gly)⁻

(vi) Dimethyl glyoximato (dmg)⁻

3 Polydentate ligands
All bidentate and polydentate ligand are chelating ligands.
No of chelate ring = Denticity – 1
(i) Diethylenetriamine (dien)
Denticity = 3
No. of rings = 2
Neutral

(ii) (trien) – Triethylene tetramine
Denticity = 4
Chelate rings = 3
Neutral

(iii) Terpyridine
Denticity = 3
Chelate rings = 2
Neutral

(iv) Ethylenediamine tetracetato (EDTA)^–4
2N 4(O)
Denticity = 6
Chelate rings = 5

(v) Ethylenediamine triacetate (EDTA)^–3
Denticity = 5
Chelate rings = 4

4 Ambidentate ligands
• Ligands which have two different donor atoms but at time of coordination such ligand can be coordinate CMA by either of two donor atoms.

5 Flexi dentate
(a) Ligands which can change their denticity
Ex. CO₃⁻², PO₄⁻³, SO₄⁻², CH₃ – COO⁻, NO₃⁻ etc.

Ex. [Co(NH₃)₅ SO₄]Cl
x + 0 – 2 = 0
x ⇒ +3
C.N ⇒ 6
5 by NH₃ + 1 by SO₄
Monodentate

(b) On the basis of e⁻ donating and e⁻ accepting nature
(i) Normal Or classical ligands
Ligands which only donate e⁻ pair to CMA and form coordinate σ bond.
Ex. NH₂⁻, OH⁻, N⁻³, Cl⁻ , O^–2

(ii) Non classical or π acid or π acceptor ligands
Ligands which donate e- pair to CMA and form coordinate bond but simultaneously they accept e- pair from CN through back bonding (synergic bonding)
 Ex. CO,NO⁺ ,CN⁻ ,C₆H₆,C₅H₅⁻, C₂H₄

(iii) π donor and π acceptor ligand
Ligands which donate π e⁻ to CMA and also accept e⁻ density from CMA though synergic bonding.
Ex. C₆H₆, C₅H₅⁻
(a) 
(b) 

Answer

Total no of e⁻ pair accepted by CMA
1. [Ni(dmg)₂] = 4
2. [Pt(trien)]Cl₂ = 4
3. [Fe(EDTA)]⁻ = 6
4. [Co(en)₂Ox]Cl = 6

Some important points
(i). Generally C.N of monovalent cation is two and four except
Ex. 
(ii) Generally C.N or bivalent cation is four and six except.

(iii) Generally C.N of trivalent cation is six except some exceptions.
(iv) C.N of tetravalent cation is 6.
(v) C.N of CMA depends upon charge of CMA, six of CMA size of ligands and concentration of ligand.

Answer

Total number of e⁻ of CMA after accepting e⁻ pair from ligands
ENA = Z – (0.5) + 2 × C.N.
1. K₄[Fe(CN)₆] ⇒ 26 – (+2) + 2 × 6 ⇒ 36[K]r
O.S. = +2, C.N. = 6
2. K₃[Fe(CN)₆] ⇒ 26 – (+3) + 2 × 6 ⇒ 35[Kr]
O.S. = + 3, C.N. = 6
3. [Fe(η⁵ – C₅H₅)₂] ⇒ 26 – (+2) + 2× 6 ⇒ 36[Kr]
O.S. = +2, C.N.= 6

Sidgwick rule
• If EAN of CMA in metal carbonyl is equal to Atomic number of nearest inert gas then the stability of metal is high.

(C) [Fe(CO)₅] ⇒ neither oxidising nor reducing
• Sidgwick rule is applicable only for metal carbonyl.

EAN of Polynuclear Metal Carbonyl

Answer

1. Formula of a Complex

(a) In formulas of both simple and complex salts, cation precedes the anion. Nonionic compounds are written as single units.
(b) Complex ions are written inside square brackets without any space between the ions.
(c) Metal atom and ligands are written in the following order:

1. In the complex part, the metal atom is written first followed by ligands in the order, anionic → neutral → cationic.
2. If more than one ligand of one type (anionic, neutral or cationic) are present, then they are arranged in English alphabetical order, e.g. between H₂O and NH₂, H₂O should be written first. Similarly, order of NO₂⁻ , SO₃²⁻ and OH^– will be NO₂⁻

• When ligands of the same type have similar name for the first atom, then the ligand with less number of such atoms is written first. Sometimes the second atom may be used to decide the order. When number of atoms are also same e.g., Out of NO₂⁻ , NH₂⁻ will be written first. In H₃ and  will be written first as it contains only one N-atom.

1. Polyatomic ligands and abbreviations for ligands are always written in lower case letters. e.g. (en), (py), etc.
2. Charge of a complex ion is represented as over script or square bracket.

Examples:
K₄[Fe(CN)₆] — First cation and then anion
[CrCl₂(H₂O)₄] Br—Cl^– (negative ligand) before H₂O

2. Nomenclature of Coordination Compounds

Mononuclear coordination compounds are named by following these rules:
(a) In both the positively and negatively charged coordination compounds, the cation is named first followed by the anion.
(b) The name of the central atom/ion is written after the ligands are named in alphabetical order. (This procedure is reversed in writing its formula).
(c) Names of the anionic ligands end in –o.

(d) Names of neutral and cationic ligands are the same except for aqua for H₂O, ammine for NH₃, carbonyl for CO and nitrosyl for NO. These are placed within parentheses ().

(e) Positive ligands are named as :

(f) Oxidation state of the metal in a cation, anion or a neutral coordination compound is indicated by a Roman numeral in parenthesis.
(g) When the complex ion is a cation, the metal is named same as the element. For example, Co in a complex cation is called cobalt and Pt is called platinum. In an anion, Co is called cobaltate. For some metals, their Latin names are used in the complex anions, e.g. ferrate for Fe.
(h) Nomenclature of a neutral complex molecule is done in the similar way as that of a complex cation.
The following examples illustrate the nomenclature for coordination compounds:
• [Cr(NH₃)₃(H₂O)₃] Cl₃ is named as: Triamminetriaquachromium (III) chloride
• [Co(H₂NCH₂CH₂NH₂)₃]₂ SO₄ is named as: Tris (ethane –1, 2–diammine) cobalt (III) sulphate
• [Ag(NH₃)₂] [Ag(CN)₂ ] is named as: Diamminesilver (I) dicyanoargentate (I)
(i) Ligands which join two metals are known as ‘Bridge ligands’ and they are prefixed by ‘µ’ (mu).
E.g.  in this complex.

3. Nomenclature of Complexes
(a) Cationic Complex
[Cr(NH₃)₃(H₂O)₃]Cl₂
triamminetriaquachromium (III) chloride
(i) The number of the individual ligands are indicated by prefix like mono, di, tri, etc. and ligands are named in an alphabetical order.
(ii) Central metal atom and oxidation state indicated by Roman numeral in parenthesis.
(iii) Name of ionisable anion.

(b)Anionic Complex
K₃[Fe(CN)₆]
Potassium hexacyanoferrate (III)
(i) Name of ionisable metal and oxidation state
(ii) Name of ligand in an alphabetical order
(iii) Central metal atom + ate and oxidation state

(c) Neutral Complex
[Pt(NH₃)₂Cl(NO₂)]
Diammine chloronitrito–N–platinum (II)
(i) Name of ligands in an alphabetical order
(ii) Central metal atom and oxidation states.

Answer

Answer

Answer

• Central metal atom releases e^– according to its O.S.
• Central metal atom will provide vacant orbitals according to its coordination number.
• These vacant orbital undergo histrion and form coordinate s bond with donor atoms.
• Hybridisation state of central metal atom depends upon C.N. and nature of ligand M.M. dipole moment, isomerism etc. are also helpful in hybridisation prediction.
C.N = 2 sp linear
C.N = 3 sp² Trigonal planar
C.N = 4 sp³ Tetrahedral
dsp² square planar
C.N = 5 sp³d TBP
dsp³ (i) TBP
(ii) Square pyramidal (dx²y²)
C.N = 6 sp³d² Octahedral (SBP)
d²sp³ Octahedral (SBP)
• In presence of strong field ligand (SFL), the pairing of (n–1)d e^– is possible before hybridisation but this pairing is not possible in presence of WFL.

• All ligands acts as SFL with 4d and 5d series metal ions.
• P⁻ acts as SFL with Ni⁺⁴
• H₂O and C₂O₄⁻² acts as SFL with CO⁺³
• NH₃ act as WFL with Fe⁺² and Mn⁺² .
• H₂O act as SFL with Cu⁺² (C.N = 4)

(1) [Ni(CN)₄]⁻²
O.S. = +2 C.N = 4 ligand = SFL
Ni = 3d⁸4s²

(2) [Ni(Cl₄)]⁻² or [Ni(SCN)₄]⁻²
O.S = +2 C.N = 4 ligand = WFL
Ni = 3d⁸4S²

(3)[Fe(CO)₅] D.M. = O
O.S. = 0 C.N. = 5 Ligand = SFL

(4) [MnCl₄]⁻² sp³
O.S. = +2 C.N. = 4 Ligand = WFL

(5) [CuCl₅]⁻³ sp³d
O.S. = +2 C.N = 5 Ligand = WFL

(6) [Ni(CO)₄] or [Ni(CN)₄]⁻⁴ sp³
O.S. = 0 C.N = 4 Ligand = SFL

(7) [Fe(H₂O)₅(NO)]SO₄
O.S = + 1 CN = 6 Ligand = WFL

(8) [NiCl₂(PPh₃)₂] Paramagnetic
O.S = + 2 CN = 4 Ligand = WFL = No pairing

(9) [RhCl(PPh)₃],[PdCl₄]⁻², [AgF₄]⁻

(10) [NiF₆]⁻² Ni⁺⁴ → F⁻ act as SPL
O.S. = +4 CN = 6

(11) [Fe(CN)₅(O₂⁻)]⁻⁴
O.S. = +2 CN = 6 Ligand = SFL

(12) [Fe(CO)₄]⁻²
O.S. = −2 CN = 4 Ligand = SPL

Drawbacks of VBT
(1) It can’t explain stability of complex compound
(2) It can’t explain colour of complex compound
(3) It can’t explain d e- transference
(4) It can’t explain pairing of (n-1)d e^–
(5) It doesn’t give any criteria for classification of SFL and WFL.

CFT for octahedral complex
d_xy , d_yz , d_zx = t_2g
d_z² , d_x² _{– y}² = e_g

Answer

In crystal field theory bonding between metal and ligands is purely electrostatic.
Ligands are considered as negative point charges.

What Happens When Ligands Approach A Metal

Orbital Splitting In Octahedral Complexes

Orbital Splitting In Tetrahedral Complexes

Strenght Of Ligands
CO ≈ CN⁻ > PPH₃ > NO₂⁻ > NH₃ > pyridine > CH₃CH > NCG⁻ > H₂O ≈ C₂O₄²⁻ > OH⁻ > NCO⁻ > F⁻ > Cl > SCN > S²⁻ > Br⁻ > l⁻ > O₂²⁻ .

Answer

Electronic configuration of (n – 1) de⁻ of CMA (after splitting) depends upon 2 type of energies are :
(1) ∆₀₍₂₎ Pairing energy (P)
There are two conditions
(1) In presence of SFL ∆₀ > P
(2) In presence of WFL ∆₀ < P

Hybridisation State and Magnetic Nature

Configuration of metal in

Factor affecting splitting energy
(1) ∆₀ ∝ charges of central metal atom
(2) ∆₀ ∝ Z_eff of central metal atom

(3) ∆₀ ∝ strength of ligand
(4) Geometry of complex

Examples,

H₂O acts as strong ligand with Co⁺³, so splitting energy for (II) is higher than (I)
(ii) [Fe(H₂O)₆]⁺³ < [Ru(H₂O)₆]⁺³ < [Os(H₂O)₆]⁺³ ∆₀
Fe is a 3d series elements whereas Ru and Os are 4d and 5d transition elements, so H₂O is weaker ligand but it acts as strong ligand with higher d series transition element.
(iii) [CrCl₆]⁻³ <[Cr(NH₃)₆]⁺³ <[Cr(CN)₆]⁻³ ∆₀
Ligand strength for CN⁻ > NH₃ >Cl⁻

(I) is tetrahedral and (II) is octahedral complex
 ∆_t > ∆₀

Stability of complex compound

K↑, stability↑
Stability ∝ charge of CMA
∝      Z_eff of CMA
∝     Strength of ligand
∝    chelation effect

Stability order :

(1) [Fe(CN)₆]⁻⁴ <[Fe(CN)₆]⁻³
O.S = +2 O.S. = +3

(2) [Co(H₂O)₆]⁺³ <[Rh(H₂O)₆]⁺³ <[Ir(H₂O)₆]⁺³ Zeff ↑

(3) [NiCl₆]⁻⁴ <[Ni(NH₃)₆]⁺² <[Ni(CN)₆]⁻⁴ ligand strength

(6) Irving williams series
Mn⁺² +2 +2 +2 ≤Cu⁺² >Zn⁺²

Answer

Extra stabilisation (released energy) due to splitting in comparison to no splitting
[Fe(CN)₆]⁻⁴

Priority to 2P = N, New pairs
CFS = −0.4∆₀ × 6 + 0.6∆₀ × 0
= −2.4∆₀
Total pairs
CFSE = 0.4∆₀ nt_eg + 0.6 ∆₀ nt_eg + xp = new pairs

CFT for Tetrahedral complex

Answer

(a) d – d Transition
• Colour of complex compound is due to d – d transition
• Ag. Solution of Ti⁺³ is violet.

• Colour of complex compound depends upon spilitting energy.
• Complex become colourless in absence of ligand filled (no ligand, no splitting, no transition)
• Colour of f – block compounds is due to f – f transition.
Examples,
(i) [Ti(H₂O)₆]⁺ become colourless on heating due to removal of water molecules.
(ii) Anhydrous CuSO₄ is colourless but hydrated CuSO₄ is blue (test of moisture)

(b) Charger transfer
Colour of some compound is due to charge transfer (e⁻ transition) from anion to metal ion.

(c) Polarisation
• Colour of same compound can be explained on the basis of polarisation.
• Polarisation increase, possibility of finding colour increase
Examples,
(1) AgF colourless but Agl yellow.
(2) PbF₂ colourless but Pbl₂ yellow.
(3) HgF₂ colourless but Hgl₂ Red.
(4) ZnS colourless but Cds yellow. HgS Black.

Colour of Co – Ordination compounds

Why we are seeing Blue, Why Not Red or Green?
Let’s see what happens inside the solution containing co – ordination compound when we cast a white light, on it

Colours shown by metals in various oxidation states

Answer

• Same molecular formula but different structural formula
(1) Coordination Isomerism
It arises due to exchange of ligands between complex cation and complex anion.
(2) Ionisation Isomerism
• Structural isomer which give differ ions in aqueous solution
• Ionization isomerism is the result of the exchange of groups or ions between the coordinating sphere and the ionization sphere.
Examples,

(3) Hydrate Isomerism
• It is a special type of ionisation isomerism in which number of water molecules differ in coordination sphere/outside region.
Examples,

(4) Linkage Isomerism
• It arise due to presence of ambidentate ligand.
Examples,
[Co(NO₂)(NH₃)₅]Cl₂
Pentaamminenitrocobalt (III) chloride

(5) Polymerisation Isomerism
• Complexes which have differ molecular formula but have same empirical formula. It is not a true isomerism.
Example,
[PtCl₂(NH₃)₂] and [Pt(NH₃)₄][PtCl₄]

Answer

Answer

(1) Test of Ni⁺²

Square planer diamagnetic

(2) Test of Fe⁺², Fe⁺³ and Cu⁺²

(3) Test of sulphide ion

S⁻² + N⁺O → NOS
(4) Separation of hydroxides or oxides

(5) If excess amount of KCN is added into CuSO₄ solution then insoluble CuCN is formed which later into soluble complex.
(6) AgCl or AgBr ppt (not Agl) are soluble in NH₃ or NH₄OH

(7) [EDTA]⁻⁴ is used for estimation and removal of hardness of H₂O.
Ca⁺² + EDTA⁻⁴ → [Ca(EDTA)]⁻²
(8) Wilkinson catalyst [RhCl(PPh)₃] is used for hy – drogenation of alkene.
(9) Biological importance : -

1. Chlorophyll → Mg
2. Vit – B₁₂ → Co
3. Carboxypeptide → Zn
4. Plastocynin → Cu
5. Insolin → Zn
6. Haemoglobin → Fe⁺²
7. Hyoglobin → Fe⁺²

Answer

Types
(1) σ Bonded OMC
(C₂H₅)₄Pb T.E.L used as anti knocking agent [TiCl₄ + (C₂H₅)₃ Al] Ziggler natta catalyst(Heterogenous catalyst) = Used for polymerisation of alkene.
(2) π Bonded OMC
Presence of π donor ligand
[Cr(η⁶ – C₆H₆)₂], [Fe(η⁵ – C₅H₅)₂]
(3) σ and π bonded OMC
Presence of synergic bonding
[Ni(CO)₄] ∙[Fe(CO)₅]
Trans effect
K[PtCl₃(η² – C₂H₄)]

square planar diamagnetic
John Taller effect
Distortion in octahedral geometry due to unsymmetrical e⁻ cloud in eg set of orbitals.