Question

Which of the following will give a pair of enantiomers ? $$\left( {en = N{H_2}C{H_2}C{H_2}N{H_2}} \right)$$

A. $$\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]\left[ {Co{{\left( {CN} \right)}_6}} \right]$$
B. $$\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]Cl$$  
C. $$\left[ {Pt{{\left( {N{H_3}} \right)}_4}} \right]\left[ {PtC{l_6}} \right]$$
D. $$\left[ {Co{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]N{O_2}$$
Answer :   $$\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]Cl$$
Solution :
Enantiomorphs or Enantiomers A pair of molecules related to each other as an object and its mirror images are known as enantiomorphs or enantiomers. These are not superimposable on its mirror image.
The example is $${\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]^ + }$$
Dichlorobis ( ethylene diamine ) cobalt (III)
Co-ordination Compounds mcq solution image

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
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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
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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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Physical ChemistrySome Basic Concepts in ChemistryStoichiometryStates of Matter Solid, Liquid and GasAtomic StructureChemical ThermodynamicsSolutionsSolid StateSurface ChemistryChemical EquilibriumIonic EquilibriumChemical KineticsRedox ReactionElectrochemistryNuclear Chemistry
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