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

The donor atoms, molecules or anions which donate a pair of electrons to the metal atom or ion and form a coordinate bond with it are called ligands. In methane there is no electrons for donation to central metal atom/ion it is stable with complete octet configuration.

In $${d^6}$$  (low spin), electrons get paired up to make two $$d$$ - orbitals empty. Hybridisation is $${d^2}s{p^3}$$  (octahedral) and the complex is low spin.

$$F{e^{3 + }}{\text{ in}}{\left[ {Fe\left( {C{N_6}} \right)} \right]^{3 - }}$$

$$\eqalign{ & \Rightarrow {\text{low spin complex }}\mu = 1.732BM \cr & Ni{\text{ in}}\left[ {Ni{{\left( {CO} \right)}_4}} \right] \cr} $$

$$\eqalign{ & \Rightarrow {\text{low and high spin complex is applicable}} \cr & {\text{for }}{d^4}{\text{ to }}{d^7}{\text{ configuration}} \cr & M{n^{2 + }}{\text{ in}}{\left[ {Mn{{\left( {CN} \right)}_6}} \right]^{4 - }} \cr} $$

$$ \Rightarrow {\text{low spin complex }}\mu = 1.732BM$$

Coordination number is the number of ligands that are directly bound to the central metal atom by coordinate bonds. Oxidation state is the residual charge on the central metal atom left after removing all the ligands along with the electron pairs that are shared with the central atom.

Complex $$X$$  is $${\left[ {Fe{{\left( {en} \right)}_3}} \right]^{2 + }};$$   as $$'en'$$  is a strong field ligand pairing of electrons will take place.

Hence, hybridisation is $${d^2}s{p^3}$$  and complex is diamagnetic. As it has 3 bidentate symmetrical $$'en'$$  ligands so it will not show geometrical isomerism.

∴ Since $$C{o^{2 + }}$$  has lowest no. of unpaired electrons hence lowest paramagnetic behaviour is shown by $${\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }}$$

$$NH_4^ + $$  has no lone pair of electrons.

$$0.2\,mol$$   of $$AgCl$$  are obtained when $$0.1\,mol$$   $$CoC{l_3}{\left( {N{H_3}} \right)_5}$$    is treated with excess of $$AgN{O_3}$$  which shows that one molecule of the complex gives two $$C{l^ - }$$ ions in solution. Thus, the formula of the complex is $$\left[ {Co{{\left( {N{H_3}} \right)}_5}Cl} \right]$$   $$C{l_2}$$  i.e., 1 : 2 electrolyte.

NOTE: Colour is due to $$d - d$$  transitions. Coloured compounds contain partly filled $$d - {\text{orbital}}.$$
The oxidation state of copper in various compounds is $$ + 1\,\,{\text{and}}\, + 2.$$   In $$Cu{F_2}$$   it is in $$ + 2$$   oxidation state. In $$ + 2$$  state its configuration is
$$C{u^{2 + }} = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^9}$$
i.e.
It has one unpaired electron due to this it is coloured. The colour is due to $$d - d$$  transitions.
(NOTE : $$Cu{F_2}$$  possesses blue colour in crystalline form)

In $${\left[ {FeC{l_6}} \right]^{4 - }}$$  oxidation state of $$Fe = + 2,F{e^{2 + }} = 3{d^6}$$

Paramagnetic due to presence of four unpaired electrons.