Question

A coordination compound $$CrC{l_3} \cdot 4{H_2}O$$    gives white precipitate of $$AgCl$$  with $$AgN{O_3}.$$  The molar conductance of the compound corresponds to two ions. The structural formula of the compound is

A. $$\left[ {Cr{{\left( {{H_2}O} \right)}_4}C{l_3}} \right]$$
B. $$\left[ {Cr{{\left( {{H_2}O} \right)}_3}C{l_3}} \right]{H_2}O$$
C. $$\left[ {Cr{{\left( {{H_2}O} \right)}_4}C{l_2}} \right]Cl$$  
D. $$\left[ {Cr{{\left( {{H_2}O} \right)}_4}Cl} \right]C{l_2}$$
Answer :   $$\left[ {Cr{{\left( {{H_2}O} \right)}_4}C{l_2}} \right]Cl$$
Solution :
It gives precipitate with $$AgN{O_3}$$  it means it gives $$C{l^ - }$$  ions in the solution. Since conductivity corresponds to two ions, it shows one $$C{l^ - }$$  is outside the coordination sphere. The structure will be
$$\left[ {Cr{{\left( {{H_2}O} \right)}_4}C{l_2}} \right]Cl \to $$     $${\left[ {Cr{{\left( {{H_2}O} \right)}_4}C{l_2}} \right]^ + } + C{l^ - }$$
$$AgN{O_3} + C{l^ - } \to \mathop {AgCl}\limits_{{\text{white ppt}}{\text{.}}} + NO_3^ - $$

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
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
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. $${\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}$$
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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