Chemistry MCQs for NEET — Practice Questions with Answers

The d-electron configuration of Co2+ is d7. This configuration has no unpaired electrons, making [Co(H2O)6]2+ diamagnetic. The other options have unpaired electrons, exhibiting paramagnetic behavior.

The complexes [Co(NH₃)₆] [Cr(CN)₆] and [Cr(NH₃)₆] [Co(CN)₆] are the examples of co-ordination isomerisms.This isomerism occurs only in those complexes in which both cation and anion are complex. It occurs due to exchange of ligands between cation and anion.

The existence of two different colored isomers of [Co(NH₃)₄Cl₂]⁺ is due to geometrical isomerism. One isomer has a square planar geometry, while the other has a tetrahedral geometry, leading to different d-d transitions and hence different colors.

(b) Optical isomerism is exhibited by only those complexes in which elements of symmetry are absent. Octahedral complexes of the types [M(aa)₃], [M(aa)X₂,Y₂] and [M(aa)₂X₂] have absence of elements of symmetry, thus exhibit optical isomerism. Here, aa represents bidentate ligand, x or y represents monodentate ligand and M represents central metal ion.

Hence, [Co(NH₃)₃Cl₃]⁰ due to presence of symmetry elements does not exhibit optical isomerism.

Key Idea : Greater is the number of unpaired electrons, larger is the paramagnetism.

[V(gly)₂ (OH)₂(NH₃)₂]⁺ V₂₃ = [Ar] 4s², 3d³ Oxidation state of V in [V(gly)₂ (OH)₂(NH₃)₂]⁺  is

x + (-l)x2 + (-l)x2 + (0)x2 = +1

x =+5

V^s+ =[Ar]3d° (No unpaired electron)

[Fe(en)(bpy)(NH₃)₂)²⁺  Fe₂₆ =[Ar]4s², 3d⁶

Oxidation state of Fe in [Fe(en) (bpy) (NH₃)₂]²⁺ is

x + (0) + (0) + (0) x 2 = +2

x=+2

Fe²⁺ = [Ar]3d⁶

But en, bpy and NH₃ all are strong field ligands, so pairing occurs, thus no unpaired electrons. [Co(OX)₂(OH)₂]^-

CO₂₇ = [Ar]4s², 3d⁷ 

Oxidation state of Co in[CO(OX)₂(OH)₂]^- is

x + (- 2) x 2+ (-1) x 2 = -1

x-6=-1

X =+ 5

CO⁵⁺ = [Ar]3d⁴ (4 unpaired electrons)

[Ti(NH₃)₆]³⁺

Ti₂₂ =[Ar]4s², 3d²

Oxidation state of Ti in [Ti(NH₃)₆]³⁺ is +3, thus it contains 1 unpaired electron.

Hence [Co(OX)₂(OH)₂]^- has highest paramagnetic behaviour.

(c) Key Idea: The magnitude of ${∆}_{\mathrm{oct}}$, (the orbital splitting energy) is decided by the nature of ligand. Strong field ligand has highest ${∆}_{\mathrm{oct}}$ . The increasing field strength is as

${\mathrm{I}}^{-}<{\mathrm{Br}}^{-}<{\mathrm{Cl}}^{-}<{\mathrm{F}}^{-}<{\mathrm{OH}}^{-}<{\mathrm{H}}_2\mathrm{O}<{\mathrm{C}}_2{\mathrm{O}}_4^{2-}<{\mathrm{NH}}_3<\mathrm{en}<{\mathrm{NO}}_2^{-}<{\mathrm{CN}}^{-}$

The CN is the strongest ligand among these, hence the magnitude of ${∆}_{\mathrm{oct}}$ will be maximum
in [Co(CN)₆]^3-.

Note : The magnitude of ${∆}_{\mathrm{oct}}$ is also decided by oxidation state of the metal ion. Thus, greater the ionic charge on the central metal ion, the greater the value of ${∆}_{\mathrm{oct}}$. But here the oxidatinon state of
Co metal ion is same in all complexes.

(b) Copper sulphate on reaction with KCN to give cupric cyanide precipitates firstly which reduce into Cu₂CN₂ and dissolve in excess of KCN to give soluble K₃[Cu(CN)₄] complex salt

$$\begin{gathered} \left[{\mathrm{CuSO}}_4 + 2\mathrm{KCN} \stackrel{ }{\to } \mathrm{Cu}{\left(\mathrm{CN}\right)}_2 + {\mathrm{K}}_2{\mathrm{SO}}_4\right] \times 2 \\ \mathrm{Cupric} \mathrm{cyanide} \\ 2\mathrm{Cu}{\left(\mathrm{CN}\right)}_2 \stackrel{ }{\to } {\mathrm{Cu}}_2{\left(\mathrm{CN}\right)}_2 + \mathrm{NC}-\mathrm{CN} \\ \mathrm{cyanogen} \\ {\mathrm{Cu}}_2{\left(\mathrm{CN}\right)}_2 + 6\mathrm{KCN} \stackrel{ }{\to } 2{\mathrm{K}}_3\left[\mathrm{Cu}{\left(\mathrm{CN}\right)}_4\right] \\ \mathrm{soluble} \mathrm{complex} \mathrm{salt} \\ \mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \\ 2{\mathrm{CuSO}}_4 + 10\mathrm{KCN} \stackrel{ }{\to }2{\mathrm{K}}_3\left[\mathrm{Cu}{\left(\mathrm{CN}\right)}_4+2{\mathrm{K}}_2{\mathrm{SO}}_4 +\hspace{0.17em}{\left(\mathrm{CN}\right)}_2\right] \end{gathered}$$