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

The compound shows linkage isomerism because the ligand in the compound is an ambidenate ligand that can bond at more than one atomic site.
$$i.e.,:NC{S^ - }{\text{ and : }}SC{N^ - }$$

Geometrical isomers of the complex $$\left[ {Co{{\left( {N{O_2}} \right)}_3}{{\left( {N{H_3}} \right)}_3}} \right]$$     are two. These are

Optical isomerism is exhibited only by those complexes in which plane of symmetry are absent. Octahedral complexes of the types $$\left[ {M{{\left( {aa} \right)}_3}} \right],\left[ {M\left( {aa} \right){x_2},{y_2}} \right]$$     and $$\left[ {M{{\left( {aa} \right)}_2}{x_2}} \right]$$   have absence of plane of symmetry, thus exhibit optical isomerism. Here, $$aa$$  represents bidentate ligand, $$x$$ or $$y$$ represents monodentate ligand and $$M$$  represents central metal ion.
Hence, $${\left[ {Co{{\left( {N{H_3}} \right)}_3}C{l_3}} \right]^0}$$   due to presence of symmetry elements does not exhibit optical isomerism.
or
Octahedral complexes of $$\left[ {M{{\left( {AA} \right)}_2}{B_2}} \right]$$    type, e.g. $${\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]^ + },\left[ {M\left( {AA} \right){B_2}{C_2}} \right]$$       type, e.g. $$\left[ {Co\left( {en} \right)C{l_2}{{\left( {N{H_3}} \right)}_2}} \right]$$     and $$\left[ {M{{\left( {AA} \right)}_3}} \right]$$   type, e.g. $${\left[ {Co{{\left( {en} \right)}_3}} \right]^{3 + }}$$   show optical isomensim, whereas complexes of $$\left[ {M{A_3}{B_3}} \right]$$   type, e.g. $${\left[ {Co\left( {N{H_3}} \right)C{l_2}} \right]^0}$$    do not show optical isomerism.

Crystal field splitting depends upon the nature of ligand. The nature of ligand $$\Delta $$  decreases as shown below
$${C_2}O_4^{2 - } < {H_2}O < N{H_3} < C{N^ - }$$
hence the crystal field splitting will be maximum for $${\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }}$$

In $${\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }},Cr$$     is present as $$C{r^{3 + }}.$$
$$C{r^{3 + }} = \left[ {Ar} \right]3{d^3},4{s^0}$$

$${\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }} = \left[ {Ar} \right]3{d^3}$$

Since, this complex has three unpaired electrons, excitation of electrons is possible and thus, it is expected that this complex will absorb visible light.

Coordination number of $$Cr$$  is 6 ( oxalate is bidentate ligand ) and oxidation state of $$Cr$$  in $${K_3}\left[ {Cr{{\left( {{C_2}{O_4}} \right)}_3}} \right]$$    is calculated below.
$$\eqalign{ & 3\left( 1 \right) + x + 3\left( { - 2} \right) = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,3 + x + \left( { - 6} \right) = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 6 - 3 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = + 3 \cr} $$

The chemical formula of nitropentammine chromium (III) chloride is
$$\left[ {Cr{{\left( {N{H_3}} \right)}_5}N{O_2}} \right]C{l_2}$$
It can exist in following two structures
$$\left[ {Cr{{\left( {N{H_3}} \right)}_5}N{O_2}} \right]C{l_2}$$     and
nitropentammine chromium (III) chloride
$$\left[ {Cr{{\left( {N{H_3}} \right)}_5}ONO} \right]C{l_2}$$
Nitropentammine chromium (III) chloride
Therefore the type of isomerism found in this compound is linkage isomerism as nitro group is linked through $$N$$ as $$-N{O_2}$$  or through $$O$$ as $$-ONO.$$

$$\left[ {Cr{{\left( {en} \right)}_2}B{r_2}} \right]Br$$
dibromidobis (ethylenediamine) chromium (III) Bromide.

$$C{o^{3 + }} + 2C{l^ - } + 2\left( {en} \right)$$
net charge = + 1
Thus, complex is $${\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]^ + }C{l^ - } \to $$     $${\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]^ + } + C{l^ - }$$
Only one $$C{l^ - }$$ which is precipitated as $$AgCl$$
100$$\,mL$$  of 0.024$$\,M$$  complex = 2.4 millimol = 0.0024 $$mol$$