Question

Correct increasing order for the wavelengths of absorption in the visible region for the complexes of $$C{O^{3 + }}$$  is

A. $${\left[ {Co{{\left( {en} \right)}_3}} \right]^{3 + }},{\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }},{\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}$$  
B. $${\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }},{\left[ {Co{{\left( {en} \right)}_3}} \right]^{3 + }},{\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$$
C. $${\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }},{\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }},{\left[ {Co{{\left( {en} \right)}_3}} \right]^{3 + }}$$
D. $${\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }},{\left[ {Co{{\left( {en} \right)}_3}} \right]^{3 + }},{\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}$$
Answer :   $${\left[ {Co{{\left( {en} \right)}_3}} \right]^{3 + }},{\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }},{\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}$$
Solution :
Key concept Wavelength $$\left( \lambda \right)$$  of absorption is inversely proportional to $$CFSE$$  $$\left( {{\vartriangle _o}\,{\text{value}}} \right)$$   of ligands attached with the central metal ion
$${\text{i}}{\text{.e}}{\text{.}}\,\,\lambda \propto \frac{1}{{{\Delta _O}}}$$
According to spectrochemical series.
$${I^ - } < B{r^ - } < {S^{2 - }} < SC{N^ - }$$     $$ < C{l^ - } < {F^ - } < O{H^ - } < {C_2}O_4^{2 - }$$     $$ < {O^{2 - }} < {H_2}O < NS{S^ - } < N{H_3}$$       $$ < en < NO_2^ - < C{N^ - }$$
Co-ordination Compounds mcq solution image
The $$CFSE$$   of ligands attached with $$C{O^{3 + }}$$  $$ion$$  is in the order $$en > N{H_3} > {H_2}O$$     ( From spectrochemical series )
$$\because $$  Wavelength of absorbed light $$\left( \lambda \right) \propto \frac{1}{{{\vartriangle _o}}}$$
$$\therefore $$  For ligand the order of wavelength of absorption in the visible region will be : $$en < N{H_3} < {H_2}O$$
$${\text{or,}}{\left[ {Co{{\left( {en} \right)}_3}} \right]^{3 + }} < {\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$$       $$ < {\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}$$

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. $$X = {\left[ {Au{{\left( {CN} \right)}_4}} \right]^ - },Y = {\left[ {Zn{{\left( {CN} \right)}_4}} \right]^{2 - }}$$
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Co - ordination Compounds

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