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

$$\mathop {CoC{l_3}.6N{H_3}}\limits_{2.675g} \to xC{l^ - }$$
$$xC{l^ - } + AgN{O_3} \to \mathop {x\,AgCl \downarrow }\limits_{4.78g} $$
Number of moles of the complex $$ = \frac{{2.675}}{{267.5}} = 0.01\,{\text{moles}}$$
Number of moles of $$AgCl$$  obtained $$ = \frac{{4.78}}{{143.5}} = 0.03\,{\text{moles}}$$
$$\therefore {\text{No}}{\text{. of moles of }}AgCl{\text{ obtained}} = 3 \times {\text{No}}{\text{. of moles of complex }}$$
$$\therefore n = \frac{{0.03}}{{0.01}} = 3$$

The absorption of visible light and hence coloured nature of the transition metal cation is due to the promotion of one or more unpaired $$- d -$$  electron from a lower to higher level withing same $$d$$ - subshell. Hence higher will be the number of unpaired electron higher will be the absorpion in visible light.
The electronic configuration of the given elements is
$$\eqalign{ & S{c^{3 + }}\left( {18} \right) = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^0}4{s^0} - {\text{no}}\,{\text{unpaired}}\,{e^ - }. \cr & T{i^{4 + }}\left( {18} \right) = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^0}4{s^0} - {\text{no}}\,{\text{unpaired}}\,{e^ - }. \cr & {V^{3 + }}\left( {20} \right) = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^2}4{s^0} - {\text{Two}}\,{\text{unpaired}}\,{e^ - }. \cr & Z{n^{2 + }}\left( {28} \right) = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}4{s^0} - {\text{no}}\,{\text{unpaired}}\,{e^ - }. \cr} $$
hence $${\left[ {V{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$$   will most likely absorb visible light.

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$$\eqalign{ & C{r^{3 + }},{C_2}O_4^{2 - },{\left( {en} \right)^0},{O_2} \cr & {C_2}O_4^{2 - }{\text{ and }}{\left( {en} \right)^0}{\text{are bidenate ligand}}{\text{.}} \cr & CN{\text{ of }}C{r^{3 + }} = 6, \cr} $$

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Structure of $$\left[ {C{o_2}{{\left( {CO} \right)}_8}} \right]$$

Total $$M – C$$  bonds $$= 10,$$
Total $$M – M$$  bonds $$= 1$$

$$\eqalign{ & {\text{Hybridisation}} \cr & \mathop {{{\left[ {Fe{{\left( {CN} \right)}_6}} \right]}^{4 - }}}\limits_{{d^2}s{p^3}} ,\mathop {{{\left[ {Mn{{\left( {CN} \right)}_6}} \right]}^{4 - }},}\limits_{{d^2}s{p^3}} \cr & \mathop {{{\left[ {Co(N{H_3}} \right]}^{3 + }}}\limits_{{d^2}s{p^3}} ,\mathop {{{\left[ {Ni{{\left( {N{H_3}} \right)}_6}} \right]}^{2 + }}}\limits_{s{p^3}{d^2}} \cr} $$
Hence $${\left[ {Ni{{\left( {N{H_3}} \right)}_6}} \right]^{2 + }}$$   is outer orbital complex.

In it the $$E.A.N.$$  of $$V$$  is $$35\left( {23 + 12 = 35} \right)$$    so it can be reduced easily.
In all other cases the $$E.A.N.$$  is $$36.$$

Tetraethyl lead $$Pb{\left( {{C_2}{H_5}} \right)_4}$$   is not $$\pi $$  bonded complex. It is $$\sigma $$  bonded organometallic compound.

$${\left[ {Co{{\left( {CO} \right)}_5}N{H_3}} \right]^{2 + }}.$$     In this complex, $$Co$$ -atom attached with $$N{H_3}$$  through $$\sigma $$ - bonding and with $$CO$$  through dative $$\pi $$ - bond.