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

Complex $$\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]Cl$$     will have four different isomers.
(i) Geometrical isomers

(ii) Optical isomers

$$\eqalign{ & {\text{Mole of}}\,AgCl = \frac{{1.435}}{{143.5}} = 0.01;\,50\,mL\,\,{\text{of}}\,\,0.2 \cr & M{\text{ complex}} \equiv 0.01mole \cr & \left[ {Co{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]Cl \to {\left[ {Co{{\left. {N{H_3}} \right)}_4}C{l_2}} \right]^ + } + C{l^ - } \cr} $$

Square planar complexes of type $$\left[ {Mabcd} \right]$$  occur in three isomeric forms.

The configuration of $$N{i^{2 + }}$$ is $$3{d^8}.$$ For the elements of the first transition series, $$C{l^ - }$$  behaves as a weak field/ high spin ligand. Hence $${Ni}$$  in $${\left[ {NiC{l_4}} \right]^{2 - }}$$  is $$s{p^3}$$ hybridised leading to tetrahedral shape.

$$\left[ {Pt{{\left( {en} \right)}_2}C{l_2}} \right]$$   is a complex of the type $$\left[ {M{{\left( {AA} \right)}_2}{B_2}} \right]$$   which is octahedral. Such compounds exhibit optical and geometrical isomerism both.

Aqueous $$CuS{O_4}$$  solution contains $${\left[ {Cu{{\left( {{H_2}O} \right)}_4}} \right]^{2 + }}$$   ions which are blue in colour. When aqueous solution of $$KF$$  is added, $${H_2}O$$  can be replaced by $${F^ - }$$ ions forming $${\left[ {Cu{F_4}} \right]^{2 - }}$$  which is green in colour.
$${\left[ {Cu{{\left( {{H_2}O} \right)}_4}} \right]^{2 + }} + \mathop {4{F^ - }}\limits_{{\text{(from KF)}}} \to $$     $$\mathop {{{\left[ {Cu{F_4}} \right]}^{2 - }}}\limits_{{\text{Green}}} + 4{H_2}O$$

According to Wemer’s theory,
$$\eqalign{ & CoC{l_3} \cdot 6N{H_3} \to {\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}3C{l^ - } \cr & CoC{l_3} \cdot 5N{H_3} \to {\left[ {Co{{\left( {N{H_3}} \right)}_5}Cl} \right]^{2 + }}2C{l^ - } \cr & CoC{l_3} \cdot 4N{H_3} \to {\left[ {Co{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]^ + }C{l^ - } \cr} $$
When $$AgN{O_3}$$   in excess is treated with these complexes then following reactions takes place :
$${\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}3C{l^ - } + \mathop {AgN{O_3}}\limits_{{\text{(Excess)}}} \to $$       $$3AgCl + {\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$$
$${\left[ {Co{{\left( {N{H_3}} \right)}_5}Cl} \right]^{2 + }}2C{l^ - } + \mathop {AgN{O_3}}\limits_{{\text{(Excess)}}} $$       $$ \to 2AgCl + {\left[ {Co{{\left( {N{H_3}} \right)}_5}Cl} \right]^{2 + }}$$
$${\left[ {Co{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]^ + }C{l^ - } + \mathop {AgN{O_3}}\limits_{{\text{(Excess)}}} $$       $$ \to AgCl + {\left[ {Co{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]^ + }$$

$$EDTA$$   has hexadentate four donor $$O$$ atoms and $$2$$ donor $$N$$ atoms and for the formation of octahedral complex one molecule is required

Thus, $${d_{xy}},{d_{yz}}$$   and $${d_{zx}}$$  orbitals have maximum electron density between the axis.

$${d_{{z^2}}}$$  and $${d_{{x^2} - {y^2}}}$$  orbitals have maximum electron density along the axes.

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