Question

The colour of the coordination compounds depends on the crystal field splitting. What will be the correct order of absorption of wavelength of light in the visible region, for the complexes, $${\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }},{\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }},$$       $${\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}?$$

A. $${\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }} > {\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$$       $$ > {\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}$$
B. $${\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }} > {\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}$$       $$ > {\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }}$$
C. $${\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }} > {\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$$       $$ > {\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }}$$  
D. $${\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }} > {\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }}$$       $$ > {\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}$$
Answer :   $${\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }} > {\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$$       $$ > {\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }}$$
Solution :
The $$CFSE$$   of the ligands is in the order : $${H_2}O < N{H_3} < C{N^ - }$$
Hence, excitation energies are in the order : $${\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }} < {\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$$       $$ < {\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }}$$
From the relation $$E = \frac{{hc}}{\lambda } \Rightarrow E \propto \frac{1}{\lambda }$$
The order of absorption of wavelength of light in the visible region is : $${\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }} > {\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$$       $$ > {\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }}$$

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. $${\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}$$
View Answer
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. $$X = {\left[ {Au{{\left( {CN} \right)}_4}} \right]^ - },Y = {\left[ {Zn{{\left( {CN} \right)}_4}} \right]^{2 - }}$$
View Answer

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Co - ordination Compounds

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