Retro (Past 13 Year) IIT-JEE Advanced (Coordination Compounds) - Class 12 MCQ

Plan This problem is based on concept of VBT and magnetic properties of coordination compound.
Draw VBT for each coordination compounds.
If unpaired electron is present then coordination compound will be paramagnetic otherwise diamagnetic.
Coordination compounds of [MA₄B₂] type show geometrical isomerism.
Molecular orbital electronic configuration (MOEC) for various coordination compounds can be drawn using VBT as

Number of unpaired electrons (n) = 3
Magnetic property = paramagnetic Geometrical isomers of [Cr(NH₃)₄CI₂]⁺ are

Magnetic property = paramagnetic Ionisation isomers of [Ti(H₂O)₅CI](NO₃)₂ are
[Ti(H₂O)₅CI](NO₃)₂ and [Ti(H₂O)₅(NO₃)]CI(NO₃)
MOEC of [Pt(en)(NH₃)CI]NO₃ is

n = 0
Magnetic property = diamagnetic Ionisation isomers are [Pt(en)(NH₃)CI]NO₃ and [Pt(en)NH₃(NO₃)]Cl


Thus, magnetic property and isomerism in given coordination compounds can be summarised as
(i) [Cr(NH₃)₄CI₂]Cl → Paramagnetic and exhibits cis-trans isomerism (r)
(ii) [Ti(H₂O)₅CI](NO₃)₂ → Paramagnetic and exhibits ionisation isomerism (p)
(iii) [Pt(en)(NH₃) CI]NO₃ → Diamagnetic and exhibits ionisation isomerism (s)
(iv) [Co(NH₃)₄(NO₃)₂]NO₃ → Diamagnetic and exhibits cis-trans isomerism (q)
(i) → (r), (ii) → (p), (iii) → (s), (iv) → (q)

Plan EDTA is a multidentate ligand as it can donate six pairs of electrons - two pair from the two nitrogen atoms and four pair from the four terminal oxygens of the —COO^- groups.

The structure of a chelate of a divalent Co²⁺ with EDTA is shown as


Each N has four N—Co—O bonds thus, total eight N—Co—O bonds.

Plan Spin only magnetic moment have the formula 

BM, where n is the number of unpaired electrons . In presence of weak ligand (as H₂O, Cl^-, F^-), there is no pairing of electrons and electrons donated by ligands are filled in outer vacant orbitals.
In presence of strong ligand (as CN^-, CO, NH₃, en), electrons are paired and electrons from ligands are filled in available inner orbitals.

Thus, order of spin only magnetic moment = Q < R < P

First of all, the compound has complex positive part [Co(H₂O)₄(NH₃)₂]³⁺ therefore , according to IUPAC conventions , positive part will be named first. Secondly , in writing name of complex , ligands are named first in alphabetical order irrespective of its charge , hence "ammine " will be written prior to “ aqua” .
Note In alphabetical order, original name of ligands are considered not the initials of prefixes. Also, special precaution should be taken in spelling name of NH₃ ligand as it is ammine. Therefore, name ofthecomplex [Co(H₂O)₄(NH₃)₂]CI₃ is Diamminetetraaquacobait? (III) chloride.

In the given complex , NiCI₂{P(C₂H₅)₂(C₆H₅)}₂ nickel is in + 2 oxidation state and the ground state electronic configuration of Ni²⁺ ion in free gaseous state is

For the given four coordinated complex to be paramagnetic , it must possess unpaired electrons in the valence shell. To satisfy this condition, four lone pairs from the four ligands occupies the four sp³ hybrid orbitals as

There fore , geometry of paramagnetic complex must be tetrahedral. On the other hand , for complex to be diamagnetic, there should not be any unpaired electrons in the valence shell. This condition can be fullfilled by pairing electrons of 3d orbitals against Hund's rule as

The above electronic arrangement gives dsp² hybridisation and therefore, square planar geometry of the complex.

Ni²⁺ + 4CN^- → [Ni(CN)₄]^2- 
Here, Ni²⁺ has d⁸ configuration with CN^- as strong ligand.

d⁸ configuration in strong ligand field gives dsp² hybridisation, hence square planar geometry.
Ni²⁺ + 4Cl^- → [NiCI₄]^2-
Here, Ni²⁺ has d⁸ configuration with Cl^- as weak ligand.



d⁸ configuration in weak ligand field gives sp³ hydridisation, hence tetrahedral geometry.
Ni²⁺ with H₂O forms [Ni(H₂O)6]²⁺ complex and H₂O is a weak ligand.


Therefore, [Ni(H₂O)₆]²⁺ has octahedral geometry.

For a diamagnetic complex , there should not be any unpaired electron in the valence shell of central metal.
In K₃[Fe(CN)₆], Fe (III) has d⁵-configuration (odd electrons), hence it is paramagnetic.

In [Co(NH₃)₆]Cl₃, Co (III) has d⁶-configuration in a strong ligand field, hence all the electrons are paired and the complex is diamagnetic.

In Na₃[Co(ox)₃], Co (III) has d⁶-configuration and oxalate being a chelating ligand, very strong ligand and all the six electrons remains paired in lower t_2g level, diamagnetic.

In [Ni(H₂O)₆]CI₂, Ni (II) has 3d⁸-configuration and H₂O is a weak ligand hence

In K₂[Pt(CN)₄], Pt(ll) has d⁸ configuration and CN^- is a strong ligand, hence all the eight electrons are spin-paired. Therefore, complex is diamagnetic.
In [Zn(H₂O)₆] (NO₃)₂ Zn (II) has 3d¹⁰ configuration with all the ten electrons spin paired, hence diamagnetic.

mmol of complex = 30 x 0.01 = 0.3
Also, 1 mole of complex [Cr(H₂O)₅CI]CI₂ gives only two moles of chloride ion when dissolved in solution.
[Cr(H₂O)₅CI]CI₂ → [Cr(H₂O)₅CI]²⁺ + 2Cl^- 
mmol of Cl^- ion produced from its 0.3 mmol = 0.6
Hence, 0.6 mmol of Ag⁺ would be required for precipitation.
0.60 mmol of Ag⁺ = 0.1 M x V(in mL)
V = 6mL

3 isomers.

Ionisation isomers are the complexes that produces different ions in solution, i.e. they have ions interchanged inside and outside the coordination sphere.
[Cr(H₂O)₄CI(NO₂)]CI and [Cr(H₂O)₄CI₂] (NO₂) have different ions inside and outside the coordinate sphere and they are isomers. Therefore, they are ionisation isomers.

Magnetic moment = 2.83 BM indicates that there is two unpaired electrons.

In [NiCl₄]^2-, Ni has d⁸ configuration and Cl^- is a weak ligand:

Both [Pt(en)₂CI₂]CI₂ and [Pt(NH₃)₂CI₂] are capable of showing geometrical isomerism.

In Cr(CO)₆, 3d⁶, has no unpaired electrons, zero magnetic moment because CO is strong field ligand.

In Ni(CO)₄, Ni is sp³ hybridised while in [Ni(CN)₄]^2-, Ni²⁺ is dsp² hybridised because CO is strong field ligand.

[Ni(NH₃)₄]²⁺ = tetraamminenickel (II)
[NiCI₄]^2- = tetrachloronickelate (II)
Cationic part is named first, hence tetraamminenickel (ll)-tetrachloronickelate(ll)

CuF₂:Cu²⁺ has 3d⁹ configuration allowed d-d transition.hence coloured.

Both statements are true. However, axis of symmetry is not a criteria of optical isomerism. Optical inactivity of the two geometrical isomers of [M(NH₃)₄CI₂] is due to the presence of plane of symmetry.

In the complex [Fe(H₂O)₅NO]SO₄, Fe is in + 1 oxidation state because NO is in +1 state. Also, NO is a strong ligand, complex has 3d⁷ configuration at Fe(l) as

(i) [Co(NH₃)₄(H₂O)₂]CI₂ : Co²⁺, 3d⁷
show geometrical isomerism, paramagnetic.
(ii) Pt(NH₃)₂CI₂ : Pt²⁺ has d⁸ configuration with all paired electrons.
Show geometrical isomerism, diamagnetic.
(iii) [Co(H₂O)₅CI]CI : Co²⁺, 3d⁷
cannot show geometrical isomerism, paramagnetic.
(iv) [Ni(H₂O)₆]CI₂ : Ni²⁺ , 3d⁸ , weak ligand, has two unpaired electrons. Paramagnetic but cannot show geometrical isomerism.
(i) → (r), (ii) → (p), (iii) → (s), (iv) → (q)

Greater the extent of dπ-pπ back bonding, smaller will be the bond order of CO bond in metal carbonyls. In Fe(CO)₅, there is maximum number of valence shell electrons (d-electrons), greatest chances of pπ-dπ back bonding, lowest bond order of CO bond.

Explanation:

1. In the molecule Fe(CO)5 there exists a synergic bond between Fe and CO.

2. A synergic bond can be defined as the movement of electrons from ligands to metal. it is also known as backbonding.

   The filled molecular orbital of Fe donates the electron pairs to the antibonding molecular orbitals of carbon monoxide CO.

3. Due to this synergic bond formation between Fe and CO, the bond order of CO decreases.
4. As bond order decreases the Bond length increases.

A = K₂[Ni(CN)₄]; B = K₂[NiCI₄]
A : Potassium tetracyanonickelate (II)
B : Potassium tetrachloronickelate (II)

A is diamagnetic, square planar complex because of strong ligand field ofCN^-.

B is paramagnetic, tetrahedral complex because of weak ligand field ofCl^-.

Described in sol.25, A has dsp² hybridisation while B has sp³ hybridisation of Ni.

[Co(NH₃)₄Br₂]CI and [Co(NH₃)₄BrCI]Br are ionisation isomers.