Question

A coordination compound $$X$$ gives pale yellow colour with $$AgN{O_3}$$  solution while its isomer $$Y$$ gives white precipitate with $$BaC{l_2}.$$  Two compounds are isomers of $$CoBrS{O_4} \cdot 5N{H_3}.$$    What could be the possible formula of $$X$$ and $$Y?$$

A. $$X = \left[ {Co{{\left( {N{H_3}} \right)}_5}S{O_4}} \right]Br,$$     $$Y = \left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]S{O_4}$$  
B. $$X = \left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]S{O_4},$$     $$Y = \left[ {Co{{\left( {N{H_3}} \right)}_5}S{O_4}} \right]Br$$
C. $$X = \left[ {Co{{\left( {N{H_3}} \right)}_5}Br\left( {S{O_4}} \right)} \right],$$     $$Y = \left[ {CoBr\left( {S{O_4}} \right){{\left( {N{H_3}} \right)}_5}} \right]$$
D. $$X = \left[ {Co{{\left( {Br} \right)}_5}N{H_3}} \right]S{O_4},$$     $$Y = \left[ {CoBr\left( {S{O_4}} \right)} \right]N{H_3}$$
Answer :   $$X = \left[ {Co{{\left( {N{H_3}} \right)}_5}S{O_4}} \right]Br,$$     $$Y = \left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]S{O_4}$$
Solution :
$$\mathop {\left[ {Co{{\left( {N{H_3}} \right)}_5}S{O_4}} \right]}\limits_X Br \to $$     $${\left[ {Co{{\left( {N{H_3}} \right)}_5}S{O_4}} \right]^ + } + B{r^ - }$$
$$AgN{O_3} + B{r^ - } \to \mathop {AgBr}\limits_{{\text{pale yellow}}} + NO_3^ - $$
$$\mathop {\left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]S{O_4}}\limits_Y \to $$      $${\left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]^{2 + }} + SO_4^{2 - }$$
$$BaC{l_2} + SO_4^{2 - } \to \mathop {BaS{O_4}}\limits_{{\text{white}}\,{\text{ppt}}{\text{.}}} + 2C{l^ - }$$

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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