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

$$Cr - C\pi $$   bond is formed by the donation of a pair of electrons from a filled $$d$$ - orbital of $$Cr$$  into the vacant antibonding $${\pi ^ * }$$  orbital of $$CO.$$  This is synergic bonding due to which both $$\sigma $$  and $$\pi $$  character is shown in $$Cr-C$$  bond.

The complex is square planar and is of the type $$\left[ {M\left( {abcd} \right)} \right].$$   It has three geometrical isomers.

Octahedral coordination entities of the type$$M{a_3}{b_3}$$  exhibit Geometrical isomerism. The compound exists both as facial and meridional isomers.

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In silver plating $$K\left[ {Ag{{\left( {CN} \right)}_2}} \right]$$   is used which provides constant and required supply of $$A{g^ + }\,ions$$   as $$Ag\left( {CN} \right)_2^ - $$   is very stable. But if $$AgN{O_3}$$  is used concentration of $$\left[ {A{g^ + }} \right]$$  in solution will be very large. In that case $$Ag$$  will be deposited at faster rate without any uniformity.

$$\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]C{l_3}$$   is an inner orbital complex, because in this complex $$d{\text{ - orbitals}}$$   used is of lower quantum number ie $$\left( {n - 1} \right).$$  It results from $${d^2}s{p^3}$$  (inner orbital) hybridization.

$$\left[ {Co{{\left( {py} \right)}_2}{{\left( {{H_2}O} \right)}_2}C{l_2}} \right]Cl$$      and $$\left[ {Co{{\left( {py} \right)}_2}\left( {{H_2}O} \right)C{l_3}} \right]{H_2}O$$      show hydrate isomerism.

For a substance to be optical isomer following conditions should be fulfiled
(A) A coordination compound which can rotate the plane of polarised light is said to be optically active.
(B) When the coordination compounds have same formula but differ in their abilities to rotate directions of the plane of polarised light are said to exhibit optical isomerism and the molecules are optical isomers. The optical isomers are pair of molecules which are nonsuperimposable mirror images of each other.
(C) This is due to the absence of elements of symmetry in the complex.
(D) Optical isomerism is expected in tetrahedral complexes of the type $$Mabcd.$$
Based on this only option (B) shows optical isomerism $${\left[ {Co{{\left( {en} \right)}_3}} \right]^{3 + }}$$

Complexes of $$Z{n^{ + + }}$$ cannot show optical isomerism as they are tetrahedral complexes with plane of symmetry.
$${\left[ {Co{{\left( {{H_2}O} \right)}_4}\left( {en} \right)} \right]^{3 + }}$$    have two planes of symmetry hence it is also optically inactive.
Hence the formula of the complex is $$\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]C{l_3}$$

$$Al{\left( {O{C_2}{H_5}} \right)_3}$$   does not have metal-carbon bond, ( i.e. it is not an example of organometallic compound )
Structure

Since the hybridisation of central metal tin $${\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$$   complex ion is $$s{p^3}{d^2}$$  and coordination number of $$C{o^{3 + }}$$  is 6. So, its geometry is octahedral.