Co - ordination Compounds MCQ Questions & Answers in Inorganic Chemistry | Chemistry

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171. An octahedral complex with molecular composition $$M.5{\text{ }}N{H_3}.Cl.S{O_4}$$    has two isomers, $$A$$ and $$B.$$ The solution of $$A$$ gives a white precipitate with $$AgN{O_3}$$  solution and the solution of $$B$$  gives white precipitate with $$BaC{l_2}$$  solution. The type of isomerism exhibited by the complex is :

A Linkage isomerism
B Ionisation isomerism
C Coordinate isomerism
D Geometrical isomerism
Answer :   Ionisation isomerism

172. The correct order of ligands in the trans-directing series is

A $$C{N^ - } > CH_3^ - > NO_2^ - > B{r^ - }$$
B $$C{N^ - } > B{r^ - } > NO_2^ - > CH_3^ - $$
C $$B{r^ - } > NO_2^ - > C{N^ - } > CH_3^ - $$
D $$CH_3^ - > C{N^ - } > NO_2^ - > B{r^ - }$$
Answer :   $$C{N^ - } > CH_3^ - > NO_2^ - > B{r^ - }$$

173. In which of the following complex hybridization of central metal is not same as that of donor atom of ligand

A $$\left[ {Ni{{\left( {P{F_3}} \right)}_4}} \right]$$
B $$\left[ {Fe{{\left( {dmg} \right)}_2}} \right]$$
C $${\left[ {Zn{{\left( {en} \right)}_2}} \right]^{2 + }}$$
D $${\left[ {Ni{{\left( {PM{e_3}} \right)}_4}} \right]^{2 + }}$$
Answer :   $$\left[ {Fe{{\left( {dmg} \right)}_2}} \right]$$

174. In $$\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]C{l_3},$$    the number of covalent bonds is

A 3
B 6
C 9
D 18
Answer :   18

175. Relative to the average energy in the spherical crystal field, the $${t_{2g}}$$  orbitals is tetrahedral field is

A $${\text{Raised by}}\left( {\frac{2}{5}} \right){\Delta _t}$$
B $${\text{Lowered by }}\left( {\frac{2}{5}} \right){\Delta _t}$$
C $${\text{Raised by}}\,\left( {\frac{3}{5}} \right){\Delta _t}$$
D $${\text{Lowered by}}\,\left( {\frac{1}{5}} \right){\Delta _t}$$
Answer :   $${\text{Raised by}}\left( {\frac{2}{5}} \right){\Delta _t}$$

176. The correct IUPAC name of the coordination compound $${K_3}\left[ {Fe{{\left( {CN} \right)}_5}NO} \right]$$    is

A potassium pentacyanonitrosylferrate(II)
B potassium pentacyanonitroferrate(III)
C potassium nitritopentacyanoferrate(IV)
D potassium nitritepentacyanoiron(II)
Answer :   potassium pentacyanonitrosylferrate(II)

177. The $$d$$  electron configurations of $$C{r^{2 + }},M{n^{2 + }},F{e^{2 + }}$$    and $$N{i^{2 + }}$$  are $$3{d^4},3{d^5},3{d^6}$$    and $$3{d^8}$$  respectively. Which one of the following aqua complexes will exhibit the minimum paramagnetic behaviour ?
$$\left( {{\text{At}}{\text{. no}}{\text{. of}}\,\,Cr = 24,Mn = 25,} \right.$$      $$\left. {Fe = 26,Ni = 28} \right)$$

A $${\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }}$$
B $${\left[ {Ni{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }}$$
C $${\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }}$$
D $${\left[ {Mn{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }}$$
Answer :   $${\left[ {Ni{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }}$$

178. The geometry possessed by $$\left[ {Ni{{\left( {CO} \right)}_4}} \right]$$   is

A tetrahedral
B square planar
C linear
D octahedral
Answer :   tetrahedral

179. An octahedral complex of $$C{o^{3 + }}$$  is diamagnetic. The hybridisation involved in the formation of the complex is :

A $$s{p^3}{d^2}$$
B $$ds{p^2}$$
C $${d^2}s{p^3}$$
D $$s{p^3}d$$
Answer :   $${d^2}s{p^3}$$

180. 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 :   $$X = \left[ {Co{{\left( {N{H_3}} \right)}_5}S{O_4}} \right]Br,$$     $$Y = \left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]S{O_4}$$