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

$${\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]^ + }$$   shows geometrical as well as optical isomerism.

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Since (A) and (B), each has 4 unpaired electrons, they will have same magnetic moment.

For complexes $$L$$ and $$M,$$  i.e.,$${\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$$   and $${\left[ {Co{{\left( {ox} \right)}_3}} \right]^{3 - }},$$    $$C{o^{3 + }}\left( {3{d^6}} \right)$$   is $${d^2}s{p^3}$$  hybridised.

For complex $$K,$$  i.e., $${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }},F{e^{3 + }}\left( {3{d^5}} \right)$$     is $${d^2}s{p^3}$$  hybridised.

For complex $$O$$  i.e., $${\left[ {Pt{{\left( {CN} \right)}_4}} \right]^{2 - }},$$   $$P{t^{2 + }}\left( {5{d^8}} \right)$$   is $$ds{p^2}$$  hybridised.

For complex $$P,$$ i.e., $${\left[ {Zn{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }},Z{n^{2 + }}\left( {3{d^{10}}} \right)$$       is $$s{p^3}{d^2}$$  hybridised.

For complex $$N,$$ i.e., $${\left[ {Ni{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }},N{i^{2 + }}\left( {3{d^8}} \right)$$       is $$s{p^3}{d^2}$$  hybridised.

Hence, complexes $$L, M, O$$   and $$P$$ are diamagnetic.

Square planar complexes of type $${\text{M}}\left[ {{\text{ABCD}}} \right]$$   form three isomers. Their position may be obtained by fixing the position of one ligand and placing at the $$trans$$  position any one of the remaining three ligands one by one.

A substance absorbs light at specific wavelength in the visible part of the spectrum and reflects the rest of the wavelengths. Each wavelength represents a different colour hence corresponding colour is observed.

$$\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]Cl$$
Possible isomers are

Hence, total number of stereoisomers = 2 + 1= 3

In $${\left[ {Mn{O_4}} \right]^ - },Mn\,{\text{is}}\,{\text{in}}\, + 7$$     oxidation state.
Electronic configuration of $$\left( {Z = 25} \right):\left[ {Ar} \right]3{d^5}4{s^2}$$
Electronic configuration of $$M{n^{7 + }}:\left[ {Ar} \right]3{d^0}4{s^0}$$
Central atom in other ions have definite number of $$d$$ electrons
No. of electrons
$$\mathop {{{\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]}^{3 + }}}\limits_{{\text{in}}\,C{o^{3 + }} = {\text{Six}}} \,\mathop {{{\left[ {Fe\left( {CN} \right)6} \right]}^{3 - }}}\limits_{{\text{in}}\,F{e^{3 + }} = {\text{Five}}} \,\mathop {{{\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]}^{3 + }}}\limits_{{\text{in}}\,C{r^{3 + }} = {\text{three}}} $$

More the number of unpaired electrons, higher is its paramagnetism.
$$C{r^{3 + }}:3{d^3},F{e^{2 + }}:3{d^6},$$     $$C{u^{2 + }}:3{d^9},Z{n^{2 + }}:3{d^{10}}$$
$$F{e^{2 + }}$$  has four unpaired electrons hence it shows highest paramagnetism.

The compound $$\left[ {Co{{\left( {N{H_3}} \right)}_4}{{\left( {N{O_2}} \right)}_2}} \right]Cl$$      exhibits linkage, ionisation and geometrical isomerism.
Hence,
(i) its linkage isomers are $$\left[ {Co{{\left( {N{H_3}} \right)}_4}{{\left( {N{O_2}} \right)}_2}} \right]Cl$$      and $$\left[ {Co{{\left( {N{H_3}} \right)}_4}{{\left( {ON{O_2}} \right)}_2}} \right]Cl$$
(ii) its ionisation isomers are $$\left[ {Co{{\left( {N{H_3}} \right)}_4}\left( {N{O_2}} \right)Cl} \right]N{O_2}$$      and $$\left[ {Co{{\left( {N{H_3}} \right)}_4}{{\left( {N{O_2}} \right)}_2}} \right]Cl$$
(iii) its geometrical isomers are