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

$${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}$$   has magnetic moment of a single unpaired electron whereas $${\left[ {Fe{F_6}} \right]^{3 - }}$$  has a magnetic moment of five unpaired electrons. $${\left[ {Co{F_6}} \right]^{3 - }}$$  is paramagnetic with four unpaired electrons while $${\left[ {Co{{\left( {{C_2}{O_4}} \right)}_3}} \right]^3}$$   is diamagnetic. This anomaly is explained by valence bond theory in terms of formation of inner and outer orbital coordination entities. $${\left[ {Co{{\left( {{C_2}{O_4}} \right)}_3}} \right]^{3 - }}$$    is an inner orbital complexes having $${d^2}s{p^3}$$  hybridization.

The overlap of a lone pair on the $$C$$ atom with the empty hybrid metal orbital forms a metal-to-carbon $$\sigma $$  - bond. The transition metal atom in a metal carbonyl has filled non-bonding $$d$$  - orbitals which are of proper symmetry to overlap with the anti-bonding orbitals of $$CO.$$  The electronic charge is transferred from the filled non-bonding orbitals of the metals to $${\pi ^ * }$$ orbitals of the ligand $$CO.$$  This reduces the bond order of $$CO.$$  The $$\pi $$  back-bonding strengthens the $$M- C$$   bond order, it weakens the $$C - O$$  bond order.

$${\left[ {Co{{\left( {{C_2}{O_4}} \right)}_3}} \right]^{3 - }}$$    is diamagnetic as oxalate is a strong ligand causing pairing of $$3d$$  electrons in $$C{o^{3 + }}$$  there by leading to $${d^2}s{p^3}$$  hybridisation.

$$CuS{O_4} + 4N{H_3} \to $$    $$\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]S{O_4}$$

$${\left[ {NiC{l_4}} \right]^{2 - }}\left( {{d^8}} \right)$$

i.e the number of unpaired electrons in $${\left[ {NiC{l_4}} \right]^{2 - }}\,{\text{is}}\,2.$$
$$n = \sqrt {n\left( {n + 2} \right)} = \sqrt {2\left( 4 \right)} = 2\sqrt 2 = 2 \times 1.41 = 2.82\,{\text{BM}}$$

$$Trans - {\left[ {Co\left( {N{H_3}} \right)Cl{{\left( {en} \right)}_2}} \right]^{2 + }}$$      has a plane of symmetry and hence it is optically inactive.

As positive charge on the central metal atom increases, the less readily the metal can donate electron density into the $${\pi ^ * }$$ orbitals of $$CO$$  ligand ( donation of electron density into $${\pi ^ * }$$ orbitals of $$CO$$  result in weakening of $$C – O$$  bond ). Hence, the $$C – O$$  bond would be strongest in $${\left[ {Mn{{\left( {CO} \right)}_6}} \right]^ + }.$$

Chlorodiaquatriammine cobalt (III) chloride is $$\left[ {CoCl{{\left( {N{H_3}} \right)}_3}{{\left( {{H_2}O} \right)}_2}} \right]C{l_2}.$$

Octahedral complex of type $$\left[ {M{A_5}B} \right]$$  cannot show geometrical isomerism.