Question

$$\left[ {Co{{\left( {N{H_3}} \right)}_4}{{\left( {N{O_2}} \right)}_2}} \right]Cl$$      exhibits

A. linkage isomerism, geometrical isomerism and optical isomerism
B. linkage isomerism, ionisation isomerism and optical isomerism
C. linkage isomerism, ionisation isomerism and geometrical isomerism  
D. ionisation isomerism, geometrical isomerism and optical isomerism
Answer :   linkage isomerism, ionisation isomerism and geometrical isomerism
Solution :
The compound $$\left[ {Co{{\left( {N{H_3}} \right)}_4}{{\left( {N{O_2}} \right)}_2}} \right]Cl$$      exhibits linkage, ionisation and geometrical isomerism.
Hence,
(i) its linkage isomers are $$\left[ {Co{{\left( {N{H_3}} \right)}_4}{{\left( {N{O_2}} \right)}_2}} \right]Cl$$      and $$\left[ {Co{{\left( {N{H_3}} \right)}_4}{{\left( {ON{O_2}} \right)}_2}} \right]Cl$$
(ii) its ionisation isomers are $$\left[ {Co{{\left( {N{H_3}} \right)}_4}\left( {N{O_2}} \right)Cl} \right]N{O_2}$$      and $$\left[ {Co{{\left( {N{H_3}} \right)}_4}{{\left( {N{O_2}} \right)}_2}} \right]Cl$$
(iii) its geometrical isomers are
Co-ordination Compounds mcq solution image

Releted MCQ Question on
Inorganic Chemistry >> Co - ordination Compounds

Releted Question 1

Amongst $$Ni{\left( {CO} \right)_4},{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}{\text{and}}\,NiCl_4^{2 - }$$

A. $$Ni{\left( {CO} \right)_4}\,{\text{and}}\,NiCl_4^{2 - }$$     are diamagnetic and $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$   is paramagnetic
B. $$NiCl_4^{2 - }\,{\text{and}}\,{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$     are diamagnetic and $$Ni{\left( {CO} \right)_4}$$   is paramagnetic
C. $$Ni{\left( {CO} \right)_4}\,{\text{and}}\,{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$     are diamagnetic and $$NiCl_4^{2 - }$$  is paramagnetic
D. $$Ni{\left( {CO} \right)_4}$$  is diamagnetic and $$NiCl_4^{2 - }\,{\text{and}}\,{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$     are paramagnetic
View Answer
Releted Question 2

The geometry of $$Ni{\left( {CO} \right)_4}\,{\text{and}}\,Ni{\left( {PP{h_3}} \right)_2}C{l_2}\,{\text{are}}$$

A. both square planar
B. tetrahedral and square planar, respectively
C. both tetrahedral
D. square planar and tetrahedral, respectively
View Answer
Releted Question 3

The complex ion which has no $$'d'$$ electron in the central metal atom is

A. $${\left[ {Mn{O_4}} \right]^ - }$$
B. $${\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$$
C. $${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}$$
D. $${\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}$$
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Releted Question 4

In the process of extraction of gold,
Roasted gold ore $$ + C{N^ - } + {H_2}O\mathop \to \limits^{{O_2}} \left[ X \right] + O{H^ - }$$
$$\left[ X \right] + Zn \to \left[ Y \right] + Au$$
Identify the complexes $$\left[ X \right]\,\,{\text{and}}\,\,\left[ Y \right]$$

A. $$X = {\left[ {Au{{\left( {CN} \right)}_2}} \right]^ - },Y = {\left[ {Zn{{\left( {CN} \right)}_4}} \right]^{2 - }}$$
B. $$X = {\left[ {Au{{\left( {CN} \right)}_4}} \right]^{3 - }},Y = {\left[ {Zn{{\left( {CN} \right)}_4}} \right]^{2 - }}$$
C. $$X = {\left[ {Au{{\left( {CN} \right)}_2}} \right]^ - },Y = {\left[ {Zn{{\left( {CN} \right)}_6}} \right]^{4 - }}$$
D. $$X = {\left[ {Au{{\left( {CN} \right)}_4}} \right]^ - },Y = {\left[ {Zn{{\left( {CN} \right)}_4}} \right]^{2 - }}$$
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Co - ordination Compounds

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