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

$$\left[ {Co{{\left( {en} \right)}_2}\left( {N{O_2}} \right)Cl} \right]Br$$     and $$\left[ {Co{{\left( {en} \right)}_2}\left( {ONO} \right)Cl} \right]Br$$     are linkage isomers.

The complex $$ion{\left[ {Cr{{\left( {SCN} \right)}_2}{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}$$     can exhibit geometrical and linkage isomerism

Atoms, ions or molecules having unpaired electrons are paramagnetic. In $${\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }},Cr$$     is present as $$Cr\left( {{\text{III}}} \right).$$
$$C{r^{3 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^3}$$
In excited state

Number of unpaired electrons =3, so it is paramagnetic while rest of the species are diamagnetic.

The correct structure of EDTA is

No explanation is given for this question. Let's discuss the answer together.

$${\left[ {NiC{l_4}} \right]^{2 - }},O.S.$$    of $$Ni = + 2$$
$$Ni\left( {28} \right) = 3{d^8}4{s^2}$$

$$C{l^ - }$$  being weak ligand it cannot pair up the two electrons present in $$3d$$  orbital

No. of unpaired electrons $$= 2$$
Magnetic moment, $$\mu = 2.82\,BM.$$

$$CuS{O_4} + 2KCN \to Cu{\left( {CN} \right)_2} + {K_2}S{O_4}$$
$$\mathop {Cu{{\left( {CN} \right)}_2}}\limits_{{\text{unstable}}} \rightleftharpoons CuCN + \frac{1}{2}{\left( {CN} \right)_2}$$
$$CuCN + 3KCN \rightleftharpoons \mathop {{K_3}{{\left[ {Cu{{\left( {CN} \right)}_4}} \right]}^{3 - }}}\limits_{{\text{colourless}}} $$

Number of donor atoms in $$N{\left( {C{H_2}C{H_2}N{H_2}} \right)_3}$$     is four hence it is a tetradentate ligand.

In both states ( paramagnetic and diamagnetic ) of the given complex, $$Ni$$  exists as $$N{i^{2 + }}$$  whose electronic configuration is $$\left[ {Ar} \right]3{d^8}4{s^0}.$$

In the above paramagnetic state, geometry of the complex is $$s{p^3}$$  giving tetrahedral geometry. The diamagnetic state is achieved by pairing of electrons in $$3d$$  orbital.

Thus the geometry of the complex will be $$ds{p^2}$$  giving square planar geometry.

$$\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]C{l_3} \to $$     $${\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }} + 3C{l^ - }$$
$$\left[ {Co{{\left( {N{H_3}} \right)}_3}C{l_3}} \right] \to \left[ {Co{{\left( {N{H_3}} \right)}_3}C{l_3}} \right]$$
$$\left[ {Co{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]Cl \to $$     $${\left[ {Co{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]^ + } + C{l^ - }$$
$$\left[ {Co{{\left( {N{H_3}} \right)}_5}Cl} \right]C{l_2} \to $$     $${\left[ {Co{{\left( {N{H_3}} \right)}_5}Cl} \right]^{2 + }} + 2C{l^ - }$$
So, $$\left[ {Co{{\left( {N{H_3}} \right)}_3}C{l_3}} \right]$$    does not ionise so does not give test for chloride ions.