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

Geometrical isomers of following type of square planar complexes is possible. $$M{a_2}{b_2}$$  type, $$M{a_2}bc$$   type and $$Mabcd$$   type.

The $$CFSE$$   for octahedral complex is given by
$$\eqalign{ & CFSE = \left[ { - 0.4\,{t_{2g}}{e^ - } + 0.6\,{e_g}{e^ - }} \right] \cr & {\text{For}}\,\,M{n^{3 + }},\left[ {3{d^4}} \right] \to t_{2g}^3e_g^1 \cr & \therefore \,\,CFSE = \left[ {\left( { - 0.4 \times 3} \right) + \left( {0.6 \times 1} \right)} \right] = - 0.6 \cr & {\text{For}}\,\,F{e^{3 + }},\left[ {3{d^5}} \right] \to t_{2g}^3e_g^2 \cr & CFSE = \left[ { - \left( {0.4 \times 3} \right) + \left( {0.6 \times 2} \right)} \right] = 0 \cr & {\text{For}}\,\,{\text{C}}{{\text{o}}^{2 + }},\left[ {3{d^7}} \right] \to t_{2g}^5e_g^2 \cr & CFSE = \left[ {\left( { - 0.4 \times 5} \right) + \left( {2 \times 0.6} \right)} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 0.8 \cr & {\text{For}}\,\,C{o^{3 + }},\left[ {3{d^6}} \right] \to t_{2g}^4e_g^2 \cr & CFSE = \left[ {\left( { - 0.4 \times 4} \right) + \left( {2 \times 0.6} \right)} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 0.4 \cr} $$

Enantiomorphs or Enantiomers A pair of molecules related to each other as an object and its mirror images are known as enantiomorphs or enantiomers. These are not superimposable on its mirror image.
The example is $${\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]^ + }$$
Dichlorobis ( ethylene diamine ) cobalt (III)

$$Fe\left( {{\text{II}}} \right)$$  sate $$ - 3{d^6};$$  due to strong $$C{N^ - }$$  ligand spin paired complex $$\left( {{d^2}s{p^3}} \right)$$  will be formed .
Hence $$n = 0,\mu = 0\,B.M$$
$$Fe\left( {{\text{III}}} \right)$$   state $$3{d^5};{d^2}s{p^3}$$   complex, $$n = 1,$$
$$\mu = 1.73\,B.M.$$

$$\eqalign{ & {\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }} \cr & C{o^{3 + }} = 1{s^2}\,2{s^2}\,2{p^6}\,3{s^2}\,3{p^6}\,3{d^6} \cr} $$
$$C{N^ - }$$  is a strong field ligand and as it approaches the metal ion, the electrons must pair up.
The splitting of the $$d$$ - orbitals into two sets of orbitals in an octahedral $${\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }}$$   may be represented as

Here, for $${d^6}\,ions,$$   three electrons first enter orbitals with parallel spin put the remaining may pair up in $${t_{2g}}$$  orbital giving rise to low spin complex ( strong ligand ) field.
$$\therefore \,\,{\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }}$$    has no unpaired electron and will be in a low spin configuration.

Key Idea Jahn-Teller distortion is observed in those octahedral complexes in which $$d$$-electrons are filled unsymmetrically.

Except $${d^8},$$  all are unsymmetrically filled, thus $${d^8}$$ complex will not show Jahn-Teller distortion.

(A) Geometry of complexes $$ \to $$  Both square planar
(B) Hybridisation of central metal cation $$ \to ds{p^2}$$
(C) Magnetic behaviour $${\left[ {Cu{{\left( {en} \right)}_2}} \right]^2} \to $$    Paramagnetic; $$\left[ {Ni{{\left( {dmg} \right)}_2}} \right] \to $$    Diamagnetic
(D) Number of stereoisomers $$= 0$$

$${\left[ {Fe{{\left( {{C_2}{O_4}} \right)}_3}} \right]^{3 - }}.$$    The iron is present in the highest oxidation state $$F{e^{3 + }}$$  and $${C_2}O_4^{2 - }$$  is a chelating ligand. Chelates are always form more stable complexes

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Complexes of the type $${M_{ABCD}}$$  may exist in three isomeric form. Similarily $$\left[ {Pt\left( {N{O_2}} \right)\left( {Py} \right)\left( {N{H_3}} \right){{\left( {N{H_2}OH} \right)}^ + }} \right]$$       may exist in three isomeric form.