Question

The oxidation states of $$Cr$$  in $$\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]C{l_3},\left[ {Cr{{\left( {{C_6}{H_6}} \right)}_2}} \right]$$       and $${K_2}\left[ {Cr{{\left( {CN} \right)}_2}{{\left( O \right)}_2}{{\left( O \right)}_2}\left( {N{H_3}} \right)} \right]$$       respectively are :

A. $$ + 3, + 4,\,{\text{and}}\, + 6$$
B. $$ + 3, + 2,\,{\text{and}}\, + 4$$
C. $$ + 3,0,\,{\text{and}}\, + 6$$  
D. $$ + 3,0,\,{\text{and}}\, + 4$$
Answer :   $$ + 3,0,\,{\text{and}}\, + 6$$
Solution :
$$\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]C{l_3}$$
$$ \Rightarrow x + 6 \times 0 + \left( { - 1} \right) \times 3 = 0$$
$$ \Rightarrow x = + 3$$
$$\left[ {Cr{{\left( {{C_6}{H_6}} \right)}_2}} \right]$$
$$x + 2 \times 0 = 0;x = 0$$
$${K_2}\left[ {Cr{{\left( {CN} \right)}_2}{{\left( O \right)}_2}\left( {N{H_3}} \right)} \right]$$
$$2 \times 1 + x + 2 \times \left( { - 1} \right) + 2 \times \left( { - 2} \right) + \left( { - 2} \right) + 0 = 0$$
$$x = + 6$$

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
Inorganic ChemistryChemical Bonding and Molecular StructureOres and MetallurgyClassification of Elements and Periodicity in PropertiesHydrogenS - Block ElementsP - Block ElementsD and F Block ElementsCo - ordination CompoundsQualitative AnalysisPreparation and Properties of CompoundsEnvironmental Chemistry
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