Question

Which of the following pairs represent linkage isomers?

A. $$\left[ {Pd{{\left( {P\,P{h_3}} \right)}_2}{{\left( {NCS} \right)}_2}} \right]\,{\text{and}}\,\left[ {Pd{{\left( {P\,P{h_3}} \right)}_2}{{\left( {SCN} \right)}_2}} \right]$$  
B. $$\left[ {Co{{\left( {N{H_3}} \right)}_5}N{O_3}} \right]S{O_4}\,{\text{and}}\,\left[ {Co{{\left( {N{H_3}} \right)}_5}S{O_4}} \right]N{O_3}$$
C. $$\left[ {PtC{l_2}{{\left( {N{H_3}} \right)}_4}} \right]B{r_2}\,{\text{and}}\,\left[ {Pt\,B{r_2}{{\left( {N{H_3}} \right)}_4}} \right]C{l_2}$$
D. $$\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]\left[ {Pt\,C{l_4}} \right]\,{\text{and}}\,\left[ {Pt{{\left( {N{H_3}} \right)}_4}} \right]\left[ {CuC{l_4}} \right]$$
Answer :   $$\left[ {Pd{{\left( {P\,P{h_3}} \right)}_2}{{\left( {NCS} \right)}_2}} \right]\,{\text{and}}\,\left[ {Pd{{\left( {P\,P{h_3}} \right)}_2}{{\left( {SCN} \right)}_2}} \right]$$
Solution :
The $$SC{N^ - }$$  ion can coordinate through $$S$$ or $$N$$ atom giving rise to linkage isomerism
$$M \leftarrow SCN$$   thiocyanato
$$M \leftarrow NCS$$   isothiocyanato.

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

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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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