Co - ordination Compounds MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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101.
Out of $$TiF_6^{2 - },CoF_6^{3 - },C{u_2}C{l_2}$$ and $$NiCl_4^{2 - }$$ $$\left( {{\text{At}}.\,{\text{no}}{\text{.}}\,\,Z\,\,{\text{of}}\,\,Ti = 22,Co = 27,} \right.$$ $$\left. {Cu = 29,Ni = 28} \right),$$ the colourless species are
In $$TiF_6^{2 - },Ti$$ is present as $$T{i^{4 + }}.$$
$$T{i^{4 + }} = \left[ {Ar} \right]3{d^0}4{s^0}$$
Hence, $$TiF_6^{2 - }$$ is colourless due to the absence of unpaired electrons.
In $$C{u_2}C{l_2},Cu$$ is present as $$C{u^ + }.$$
$$C{u^ + } = \left[ {Ar} \right]$$
Due to absence of unpaired electrons, $$C{u_2}C{l_2}$$ is colourless.
102.
$$\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]C{l_3}$$ ( at. no. of $$Cr =$$ 24 ) has a magnetic moment of $$3.83\,B.M.$$ The correct distribution of $$3d$$ electrons in the chromium present in the complex is
A
$$3{d_{xy}},3{d_{yz}},3{d_{zx}}$$
B
$$3{d_{xy}},3{d_{yz}},3{d_{{z^2}}}$$
C
$$3{d_{\left( {{x^2} - {y^2}} \right)}},3{d_{{z^2}}},3{d_{xz}}$$
D
$$3{d_{xy}},3{d_{\left( {{x^2} - {y^2}} \right)}},3{d_{yz}}$$
Answer :
$$3{d_{xy}},3{d_{yz}},3{d_{zx}}$$
Magnetic moment indicates that there are three unpaired electrons present in chromium. These must be present in lower energy orbitals which are $$3{d_{xy}},3{d_{yz}}$$ and $$3{d_{zx}}.$$
103.
Mark the incorrect match.
A
Insulin - Zinc
B
Haemoglobin - Iron
C
Vitamin $${B_{12}}$$ - Cobalt
D
Chlorophyll - Chromium
Answer :
Chlorophyll - Chromium
Chlorophyll contains magnesium.
104.
The terminal and bridged $$CO$$ ligands in the compound $$\left[ {C{o_2}{{\left( {CO} \right)}_8}} \right]$$ are respectively
A
0, 2
B
6, 1
C
5, 2
D
6, 2
Answer :
6, 2
The structure of $$\left[ {C{o_2}{{\left( {CO} \right)}_8}} \right]$$ is
Terminal $$CO = 6$$
Bridged $$CO = 2$$
105.
Nickel $$(Z = 28)$$ combines with a uninegative monodentate ligand to form a diamagnetic complex $${\left[ {Ni{L_4}} \right]^{2 - }}.$$ The hybridisation involved and the number of unpaired electrons present in the complex are respectively :
A
$$s{p^3},{\text{two}}$$
B
$$ds{p^2},{\text{zero}}$$
C
$$ds{p^2},{\text{one}}$$
D
$$s{p^3},{\text{zero}}$$
Answer :
$$s{p^3},{\text{two}}$$
$${\left[ {Ni{L_4}} \right]^{2 - }}$$
i.e, no. of unpaired electron $$= 2$$
hybridization $$ - s{p^3}.$$
106.
Among $$\left[ {Ni{{\left( {CO} \right)}_4}} \right],{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }},{\left[ {NiC{l_4}} \right]^{2 - }}$$ species, the hybridisation states of the $$Ni$$ atom are, respectively ( At. no. of $$Ni= 28$$ )
A
$$s{p^3},ds{p^2},ds{p^2}$$
B
$$s{p^3},ds{p^2},s{p^3}$$
C
$$s{p^3},s{p^3},ds{p^2}$$
D
$$ds{p^2},s{p^3},s{p^3}$$
Answer :
$$s{p^3},ds{p^2},s{p^3}$$
In $$Ni{\left( {CO} \right)_4},$$ nickel is $$s{p^3}$$ hybridised because in it oxidation state of $$Ni$$ is zero. So, configuration of $$_{28}Ni = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^8},4{s^2}$$
( $$CO$$ is a strong field ligand, hence pairing of electrons will occur )
In $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }},$$ nickel is present as $$N{i^{2 + }},$$ so its configuration$$ = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^8}$$
$$C{N^ - }$$ is strong field ligand, hence it makes $$N{i^{2 + }}$$ electrons to be paired up.
In $${\left[ {NiC{l_4}} \right]^{2 - }}.$$ nickel is present as $$N{i^{2 + }},$$ so its configuration $$ = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6},3{d^8}$$
$$C{l^ - }$$ is a weak field ligand, hence in $$N{i^{2 + }}$$ electrons are not paired.
107.
Which one of the following complexes will have four isomers?
A
$$\left[ {Co{{\left( {en} \right)}_3}} \right]C{l_3}$$
B
$$\left[ {Co{{\left( {PP{h_3}} \right)}_2}{{\left( {N{H_3}} \right)}_2}C{l_2}} \right]$$
C
$$\left[ {Co\left( {en} \right){{\left( {N{H_3}} \right)}_2}C{l_2}} \right]Cl$$
D
$$\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]Br$$
The four isomers are $$trans$$ - isomer, $$cis$$ - isomer, and
ionisation isomer.
108.
The complex ion which has no $$d$$ - electrons 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
$${\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }}$$
Answer :
$${\left[ {Mn{O_4}} \right]^ - }$$
Oxidation state of $$Mn$$ in $${\left[ {Mn{O_4}} \right]^ - } = + 7,$$ hence it does not have any $$d$$ - electron present in it.
$$M{n^{7 + }}\left( {Z = 25} \right):\left[ {Ar} \right]4{s^0}3{d^0}$$
109.
Which of the following complexes formed by $$C{u^{2 + }}$$ ions is most stable?
The cyano and hydroxo complexes are far more stable than those formed by halide ion. This is due to the fact that $$C{N^ - }\,{\text{and}}\,O{H^ - }$$. are strong lewis bases (nucleophiles).
Further $${\left[ {Fe{{\left( {OH} \right)}_5}} \right]^{3 - }}$$ is not formed. hence most stable ion is $${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}$$