Co - ordination Compounds MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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181.
Coordination number of $$Ni$$ in $${\left[ {Ni{{\left( {{C_2}{O_4}} \right)}_3}} \right]^{4 - }}$$ is
A
3
B
6
C
4
D
2
Answer :
6
Coordination number of nickel in $${\left[ {Ni{{\left( {{C_2}{O_4}} \right)}_3}} \right]^{4 - }}$$ is 6 because $${C_2}O_4^{2 - }$$ is a bidentate ligand.
182.
Which of the following does not have a metalcarbon bond ?
A
$$Al{\left( {O{C_2}{H_5}} \right)_3}$$
B
$${C_2}{H_5}MgBr$$
C
$$K\left[ {Pt\left( {{C_2}{H_4}} \right)C{l_3}} \right]$$
D
$$Ni{\left( {CO} \right)_4}$$
Answer :
$$Al{\left( {O{C_2}{H_5}} \right)_3}$$
Triethoxyaluminium has no $$Al – C$$ linkage
183.
Which of the following ligands form a chelate?
A
Acetate
B
Oxalate
C
Cyanide
D
Ammonia
Answer :
Oxalate
Oxalate is a bidentate ligand hence forms a chelate.
184.
Complexes $$\left[ {Co{{\left( {N{H_3}} \right)}_5}S{O_4}} \right]Br$$ and $$\left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]S{O_4}$$ can be distinguished by
A
conductance measurement
B
using $$BaC{l_2}$$
C
using $$AgN{O_3}$$
D
Both (B) and (C)
Answer :
Both (B) and (C)
$$\left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]S{O_4}$$ gives white precipitate of $$BaS{O_4}$$ with $$BaC{l_2}\left( {aq} \right)$$ whereas $$\left[ {Co{{\left( {N{H_3}} \right)}_5}S{O_4}} \right]Br$$ gives yellow precipitate $$(AgBr)$$ with $$AgN{O_3}\left( {aq} \right).$$
185.
Which one of the following statements is not correct ?
A
Mercury (II) iodide dissolves in excess of potassium iodide solution
B
Tin (IV) chloride is made by dissolving tin solution in concentrated hydrochloric acid
C
Zinc dissolves in sodium hydroxide solution
D
Carbon monoxide reduces iron (III) oxide to iron
Answer :
Tin (IV) chloride is made by dissolving tin solution in concentrated hydrochloric acid
$$SnC{l_4}$$ is obtained by passing chlorine over tin. So $$Sn$$ (IV) chloride is made by dissolving tin solution in concentrated solution statement is incorrect and answer is (B).
186.
Which kind of isomerism is exhibited by octahedral $$Co{\left( {N{H_3}} \right)_4}B{r_2}Cl$$ ?
A
Geometrical and Ionization
B
Geometrical and Optical
C
Optical and Ionization
D
Geometrical only
Answer :
Geometrical and Ionization
$$Co{\left( {N{H_3}} \right)_4}B{r_2}Cl\,{\text{will show both geometrical and ionization isomerism}}{\text{. }}$$.
$$\left[ {Co{{\left( {N{H_3}} \right)}_4}B{r_2}} \right]Cl\,\,{\text{and}}\,\,\left[ {Co{{\left( {N{H_3}} \right)}_4}B{r_2}Cl} \right]Br\,\,{\text{are}}\,{\text{ionization isomers and geometrical isomers are }}$$
187.
The colour of the coordination compounds depends on the crystal field splitting. What will be the correct order of absorption of wavelength of light in the visible region, for the complexes, $${\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }},{\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }},$$ $${\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}?$$
The $$CFSE$$ of the ligands is in the order : $${H_2}O < N{H_3} < C{N^ - }$$
Hence, excitation energies are in the order : $${\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }} < {\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$$ $$ < {\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }}$$
From the relation $$E = \frac{{hc}}{\lambda } \Rightarrow E \propto \frac{1}{\lambda }$$
The order of absorption of wavelength of light in the visible region is : $${\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }} > {\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$$ $$ > {\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }}$$
188.
The formula of dichlorobis (urea) copper (II) is
A
$$\left[ {Cu\left\{ {O = C{{\left( {N{H_2}} \right)}_2}Cl} \right\}} \right]Cl$$
B
$$\left[ {CuC{l_2}{{\left\{ {O = C{{\left( {N{H_2}} \right)}_2}} \right\}}_2}} \right]$$
C
$$\left[ {Cu\left\{ {O = C{{\left( {N{H_2}} \right)}_2}} \right\}} \right]C{l_2}$$
D
$$\left[ {CuC{l_2}\left\{ {O = C{{\left( {N{H_2}} \right)}_2}{H_2}} \right\}} \right]$$
In $$\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]S{O_4}$$ primary valency is 2 and secondary valency is 4.
190.
A compound contains $$1.08\,mole$$ of $$Na,0.539\,mole$$ of $$Cu$$ and $$2.16\,mole$$ of $$F.$$ Its aqueous solution shows osmotic pressure which is three times that of urea having same molar concentration. The formula of the compound is :
A
$$N{a_4}\left[ {Cu{F_6}} \right]$$
B
$$Na\left[ {Cu{F_4}} \right]$$
C
$$N{a_2}\left[ {Cu{F_4}} \right]$$
D
$$N{a_2}\left[ {Cu{F_3}} \right]$$
Answer :
$$N{a_2}\left[ {Cu{F_4}} \right]$$
Mole ratio of $$Na, Cu$$ and
$$F = \frac{{1.08}}{{0.539}}:\frac{{0.539}}{{0.539}}:\frac{{2.16}}{{0.539}} = 2:1:4$$
Empirical formula $$ = N{a_2}Cu{F_4}:{\text{Van't}}\,{\text{Hoff}}$$
factor $$\left( i \right) = 3$$ (given)
Hence, formula of the compound :
$$N{a_2}\left[ {Cu{F_4}} \right] \to 2N{a^ + } + {\left[ {Cu{F_4}} \right]^{2 - }}$$