Co - ordination Compounds MCQ Questions & Answers in Inorganic Chemistry | Chemistry

In $${\left[ {Mn{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }},Mn$$     is present as $$M{n^{2 + }}$$  or $$Mn\left( {{\text{II}}} \right),$$   so its electronic configuration $$ = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6},3{d^5}$$

In $${\left[ {Mn{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }}$$   the coordination number of $$Mn$$  is six, and in presence of weak field, ligand there will be no pairing of electrons in $$3d.$$  So, it will form high spin complex due to presence of five unpaired electrons.

Magnetic moment, $$\mu $$  is related with number of unpaired electrons as
$$\eqalign{ & \mu = \sqrt {n\left( {n + 2} \right)} \,BM \cr & {\left( {1.73} \right)^2} = n\left( {n + 2} \right) \cr} $$
On solving $$n=1$$
Thus, the complex/compound having one unpaired electron exhibit a magnetic moment of $$1.73\,BM.$$
$$\eqalign{ & \left( {\text{A}} \right){\text{In}}{\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }} \cr & C{u^{2 + }} = \left[ {Ar} \right]3{d^9} \cr} $$

( Although in the presence of strong field ligand $$N{H_3},$$  the unpaired electron gets excited to higher energy level but it still remains unpaired ).
$$\eqalign{ & \left( {\text{B}} \right){\text{In}}{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }} \cr & N{i^{2 + }} = \left[ {Ar} \right]3{d^8} \cr} $$

But $$C{N^ - }$$  being strong field ligand pair up the unpaired electrons and hence in this complex, number of unpaired electrons = 0.
$$\eqalign{ & \left( {\text{C}} \right){\text{In}}\left[ {TiC{l_4}} \right] \cr & T{i^{4 + }} = \left[ {Ar} \right] \cr} $$
no unpaired electron.
$$\eqalign{ & \left( {\text{D}} \right){\text{In}}{\left[ {CoC{l_6}} \right]^{4 - }} \cr & C{o^{2 + }} = \left[ {Ar} \right]3{d^7} \cr} $$

It contains three unpaired electrons.
Thus, $${\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}$$   is the complex that exhibits a magnetic moment $$1.73\,BM.$$

$${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }}\, + \frac{1}{2}{H_2}{O_2} + {H^ + } \to {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}\, + {H_2}O$$
$${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }} + \frac{1}{2}{H_2}{O_2} + {H^ - } \to {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }} + {H_2}O + \frac{1}{2}{O_2}$$

In $${d^3}$$  and $${d^8}$$  octahedral complexes number of unpaired $${e^ - }s$$  at central metal atom/ion never changes, therefore for such octahedral complexes terms high spin and low spin not used.

Non – superimposable mirror images are optically active, hence rotate plane polarized light.

Electronic configuration of $$N{i^{2 + }}$$  in $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$   is $$N{i^{2 + }} = 3{d^8}4{s^0}$$

$$N{i^{2 + }}$$  has $$ds{p^2}$$  hybridisation, as $$C{N^ - }$$  is a strong field ligand.

$$\therefore \,\,{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$   is diamagnetic ( because of the absence of unpaired electrons ).

$$\left[ {Co{{\left( {N{H_3}} \right)}_3}{{\left( {N{O_2}} \right)}_3}} \right]C{l_2}$$     complex exhibits geometrical isomerism $$(G.I.)$$
Geometrical isomers $$ = 2\left( {1\,cis + 1\,trans} \right)$$
Optical isomers $$= 0$$
Space isomers $$= 2$$
$$\left[ {Co{{\left( {N{H_3}} \right)}_5}\left( {N{O_2}} \right)} \right]C{l_2}$$     complex shows linkage and ionization isomerism.

Complexes of type $$M{A_4}{B_2}$$  show geometrical isomerism.

The correct IUPAC name of the given compound is tetramminenickel (Il) - tetrachloronickelate (II) thus (C) is the correct answer.

NOTE : In metal carbonyl the metal is in zero oxidation state.
In $$Ni{\left( {CO} \right)_4},O.N.{\text{of}}\,Ni = 0$$
For $$Ni\left( {Z = 28} \right)$$

In presence of $$CO\,{\text{two}}\,4s$$   electrons pair up, thus

In $$Ni{\left( {PP{h_3}} \right)_2}C{l_2},O.N.\,\,{\text{of}}\,\,Ni = + 2$$
For $$N{i^{2 + }}$$

$$PP{h_3}\,{\text{and}}\,C{l^ - }$$  can't pair up $$d - {\text{electrons}},$$   leading to $$s{p^3}$$ hybridization leading to tetrahedral geometry.