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

Ionic conductivity depends upon the number of ions produced in aqueous solution. $${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$$   produces maximum number of ions, i.e. 5.
$$\underbrace {4{K^ + } + {{\left[ {Fe{{\left( {CN} \right)}_6}} \right]}^{4 - }}}_{{\text{Total 5 ions}}}$$
$$\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]C{l_3}$$    produces $$3,\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]C{l_2}$$     produces 3 and $$\left[ {Ni{{\left( {Co} \right)}_4}} \right]$$   gives zero ions.

$$Cis$$  - platin is known as anticancer agent. The formula of $$cis$$  - platin is $$cis - \left[ {PtC{l_2}{{\left( {N{H_3}} \right)}_2}} \right].$$     Here, the word $$cis$$  refers to $$cis$$  geometrical isomer of $$\left[ {PtC{l_2}{{\left( {N{H_3}} \right)}_2}} \right].$$    It is used as an antitumour agent.

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

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

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

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

It gives precipitate with $$AgN{O_3}$$  it means it gives $$C{l^ - }$$  ions in the solution. Since conductivity corresponds to two ions, it shows one $$C{l^ - }$$  is outside the coordination sphere. The structure will be
$$\left[ {Cr{{\left( {{H_2}O} \right)}_4}C{l_2}} \right]Cl \to $$     $${\left[ {Cr{{\left( {{H_2}O} \right)}_4}C{l_2}} \right]^ + } + C{l^ - }$$
$$AgN{O_3} + C{l^ - } \to \mathop {AgCl}\limits_{{\text{white ppt}}{\text{.}}} + NO_3^ - $$

$${\text{Moles of}}\,{\text{complex = }}\frac{{{\text{Molarity}}\, \times {\text{Volume}}\left( {mL} \right)}}{{1000}}{\text{ = }}\frac{{100 \times 0.1}}{{1000}} = 0.01\,{\text{mole}}$$
Moles of ions precipitated with excess of $$AgN{O_3}$$
$$ = \frac{{1.2 \times {{10}^{22}}}}{{6.02 \times {{10}^{23}}}} = 0.02\,{\text{moles}}$$
$$0.01 \times n = 0.02$$
$$\therefore n = 2$$
It means $$2C{l^ - }$$  ions present in ionization sphere
∴ Complex is $$\left[ {Co{{\left( {{H_2}O} \right)}_5}Cl} \right]C{l_2}.{H_2}O$$

As in all the given complex the central metal atom is same and contains same number of $$d$$ electrons, thus $$CFSE$$   is decided by ligands. In case of strong field ligand, $$CFSE$$   is maximum. $$C{N^ - }$$  is a strong field ligand, Hence, in $${\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }}$$   magnitude of $$CFSE$$   i.e. $${\Delta _o}$$  is maximum.

Correct order is $$:C{N^ - } > en > NC{S^ - } > C{l^ - }$$

Key Idea Complexes of $$\left[ {M{A_4}{B_2}} \right]$$  type exhibit geometrical isomerism.
The complex $${\left[ {Co{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]^ + }$$   is a $$\left[ {M{A_4}{B_2}} \right]$$  type complex and thus, fulfills the conditions that are necessary to exhibit geometrical isomerism. Hence, it has two geometrical isomers of different colours as : The structure of the geometrical isomers is as

For linkage isomerism, presence of ambidentate ligand is necessary.
For coordination isomerism, both the cation and anion of the complex must be complex ions.
For ionisation isomerism, an anion different to the ligands must be present outside the coordination sphere. All these conditions are not satisfied by this complex. Hence, it does not exhibit other given isomerisms.