31 Year NEET Previous Questions: Coordination Compounds - Free MCQ Test

Diaminodichloroplatinum (II) commonly known as  - platin is found to have anticancer property.

The optical isomers are pair of molecules which are non super imposable mirror images of each other

The two optically active isomers are collectivity called enantiomers.

₂₈ Ni⁰ :

Remember CO is a strong ligand

Remember CN^– is also a strong ligand

Cl^– is a weak ligand, hence no pairing of electrons takes place.

CN^– is astrong fieldlig and as it is a psuedohalide ion. These ions are strong coordinating ligands and hence have the tendency to form σ-bond (from the pseudo halide to the metal) and π-bond. (from the metal to pseudo halide)

Since H₂O is a weak ligand, it will not cause pairing of electrons in the metal ion Mn²⁺.
Thus electronic configuration of the metal (Mn²⁺) in the complex will be

i.e s unpaired electrons.

Triethoxyaluminium has no Al – C linkage

Transform of [M(AA)₂ a₂]^n± does not shows optical isomerism.

The given compound may have linkage isomerism due to presence of NO₂ group which may be in the form –NO₂ or –ONO.
It may have ionisation isomerism due to presence of two ionisable group –NO₂ & – Cl. It may have geometrical isomerism in the form of form as follows :

Geometrical isomers

on solving n = 3

as per magnetic moment, it has three unpaired electron.

So      3dxy' 3dyz' 3dxz'

Non superimposable mirror images are called optical isomers and may be described as “chiral’. They are also called enantiomers and rotate plane polarised light in opposite directions.

Lesser is the number of unpaired electrons smaller will be the paramagnetic behaviour.

As Ni⁺⁺ has minimum no. of unpaired e^– thus this is least paramagnetic.

More the number of unpaired electrons present in a complex more is its paramagnetic behaviour.
To find unpaired electrons let us calculate the oxidation states of elements in each complex and then write the electronic configuration for that oxidation state to find the number of unpaired electrons in it.
We find that in their complexes V, Fe, Co and Ti are in + 3, + 2, + 5, and + 3 oxidation states respectively. In these state the number of unpaired electrons present in them are 2, 4, 4 and 1 respectively. Since the maximum number of unpaired electrons are in cobalt complex so it has highest paramagnetic behaviour i.e. choice (c) is correct answer.

In octahedral field the crystal field splitting of d- orbitals of a metal ion depends upon the field produced by the ligands.
In general ligands can be arranged in a series in the order of increasing fields and splittings which they produce around a central metal ion. A portion of the series is given below. cyanide > ethylene - diamine > ammonia > pyridine>thiocyanate > water> fluoride > oxalate > hydroxide> chloride> bromide > iodide.
Out of the given ligands water, ammonia, cyanide and oxalate, we can find from the above series of ligands that the maximum splitting will occur in case of cyanide (CN^–) i.e. the magnitude of Δ₀ will be maximum in case of [Co(CN)₆]³⁺.

The octahedral coordination compounds of the type MA₃B₃ exhibit fac-mer isomerism.

Since Cr³⁺ in the complex has unpaired electrons in the d orbital, hence will be coloured

   Number of unpaired electrons = 0

: Number of unpaired electrons = 3

: Number of unpaired electrons = 4

: Number of unpaired electrons = 2

⁴ in high spin octahedral complex

CFSE = [0.6 × 1] + [–0.4 × 3] = – 0.6 Δ₀

hybridisation, thus tetrahedral in shape. Hence the four ligands are not different to exhibit optical isomerism.

∴      Correct choice : (d)

Since, the coordination number of Ni in this complex is 4, the configuration of Ni⁺⁺ at first sight shows that the complex is paramagnetic with two unpaired electron. However, experiments show that the complex is diamagnetic. This is possible when the 3d electrons rearrange against the Hund’s rule as shown below. This is in accordance with  the fact that the ligand involved here is strong i.e., CN^– ion.

Ni⁺⁺ (after rearrangement) 

Hence, now dsp² hybridization involving one 3d, one 4s and two 4p orbitals, takes place leading to four dsp² hybrid orbitals, each of which accepts four electron pairs from CN^– ion forming [Ni (CN)₄]^2– ion.

Thus, the complex is diamagnetic as it has no unpaired electron.

Coordination isomerism occurs when cationic and anionic complexes of different metal ions are present in a salt. Interchange of ligand between the complexes give isomers e.g. [Co (NH₃)₆] [Cr (CN)₆]  is an isomer of [Co (CN)₆] [Cr (NH₃)₆]

Complexes of the type MABCD may exist in three isomeric forms.

Similarly, [Pt (Py) (NH₃) BrCl] may exist in three isomeric form in which M = Pt A = Py,  B = NH₃,  C = Br,  D = Cl.

Minimum paramagnetic behaviour = [Co (H₂O)₆]²⁺

As + ve charge on the central metal atom increases, the less readily the metal can donate electron density into the π* orbitals of CO ligand to weaken the C – O bond.
Hence, the C – O bond would be strongest in Mn(CO)₆⁺.

(a) [Ti(NH₃)₆]³⁺ : 3d¹ configuration and thus has one unpaired electron. (b) [Cr(NH₃)₆]³⁺ : In this complex Cr is in +3 oxidation state.

Thus, the complex is paramagnetic.

(c) [Co(NH₃)₆]³⁺ : In this complex are cobalt ion is in + 3 oxidation state with 3d⁶ configuration.
Co³⁺,[Ar]3d⁶

(inner orbital or d²sp³ hybrid orbital low spin complex) diamagnetic

(d) In this complex Zn exists as

Zn⁺⁺ ion

Zn⁺⁺ ion : 3d¹⁰ 4s⁰

Due to presence of paired electrons complex is diamagnetic in nature.

[Ni(NH₃)₆]²⁺

 Ni²⁺ = 3d⁸, according to therefore, hybridisation is sp³d² & complex is paramagnetic.

Nickel ions are frequently detected by the formation of red precipitate of the complex of nickel dimethylglyoxime, when heated with dimethylglyoxime.

Dimethylglyoxime

Nickel dimethylglyoxime

C.F.S.E  =  – 0.4 × 6Δ₀ + 3P

[Cu(NH₃)₄]²⁺ hybridisation 2 Cu⁺² – 3 ⁹ has one unpaired e^-

So magnetic moment