both ni co 4 and ni cn 4 pow 2 are diamagnetic the

Both $$\left[ {Ni{{\left( {CO} \right)}_4}} \right]\,\,{\text{and}}\,\,{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ are diamagnetic. The hybridisation of nickel in these complexes,respectively,

A. $$s{p^3},s{p^3}$$

B. $$s{p^3},ds{p^2}$$

C. $$ds{p^2},s{p^3}$$

D. $$ds{p^2},s{p^2}$$

Answer : $$s{p^3},ds{p^2}$$

Solution :
NOTE: : In carbonyls $$O.S.$$ of metal is zero
In $$\left[ {Ni{{\left( {CO} \right)}_4}} \right]$$ , the oxidation state of nickel is zero. Its configuration in $${Ni{{\left( {CO} \right)}_4}}$$ is
Co-ordination Compounds mcq solution image

In $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ the oxidation state of $${Ni}$$ is $$2 + $$ and its configuration is
Co-ordination Compounds mcq solution image

Thus the hybridisation of nickel in these compounds are $$s{p^3}$$ and $$ds{p^2}$$ respectively.
Hence (B) is the correct answer.

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

Both $$\left[ {Ni{{\left( {CO} \right)}_4}} \right]\,\,{\text{and}}\,\,{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ are diamagnetic. The hybridisation of nickel in these complexes,respectively,

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

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

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

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

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 $$\left[ X \right]\,\,{\text{and}}\,\,\left[ Y \right]$$

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Co - ordination Compounds

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Both $$\left[ {Ni{{\left( {CO} \right)}_4}} \right]\,\,{\text{and}}\,\,{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$ are diamagnetic. The hybridisation of nickel in these complexes,respectively,

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

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

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

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

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 $$\left[ X \right]\,\,{\text{and}}\,\,\left[ Y \right]$$

Practice More Releted MCQ Question on Co - ordination Compounds

Practice More MCQ Question on Chemistry Section

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

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 $$\left[ X \right]\,\,{\text{and}}\,\,\left[ Y \right]$$

Practice More Releted MCQ Question on
Co - ordination Compounds