Question

On the basis of the spectro-chemical series, determine whether each of the following complexes is inner orbital or outer orbital and diamagnetic or paramagnetic.
(a) ${[Co{(N{H_3})_6}]^{3 + }}$
(b) ${[Mn{F_6}]^{3 - }}$
(c) ${[Co{(CN)_6}]^{3 - }}$
(d) ${[Cr{(N{H_3})_6}]^{3 + }}$
A.Inner, diamagnetic; (b) outer, paramagnetic; (c) inner, diamagnetic; (d) inner, paramagnetic.
B.Outer, paramagnetic; (b) inner, paramagnetic; (c) inner, diamagnetic; (d) inner, paramagnetic.
C.Inner, paramagnetic; (b) outer, paramagnetic; (c) outer, diamagnetic; (d) inner, diamagnetic.
D.Inner, paramagnetic; (b) inner, diamagnetic; (c) inner, diamagnetic; (d) inner, paramagnetic.

Answer 1

Hint:
Crystal field theory depicts the net change in gem energy coming about because of the direction of d orbitals of a progress metal cation inside a planning gathering of anions likewise called ligands.

Complete step by step answer:
1. ${[Cr{(N{H_3})_6}]^{3 + }}$
$Co$ is in +3 oxidation state.
$C{o^{3 + }} - [Ar]3{d^6}$
It has no unpaired electron in 3d orbital in light of the fact that $N{H_3}$ is a solid field ligand and it matches the electrons. Subsequently it is diamagnetic.
Hybridization included is ${d^2}s{p^3}$ . Thus it is an inward orbital complex.
2. ${[Mn{F_6}]^{3 - }}$
$Mn$ is in +3 oxidation state.
$M{n^{3 + }} - [Ar]3{d^4}$
It has four unpaired electrons in 3d orbital henceforth it is paramagnetic.
Since F is a feeble field ligand that it won't compel the 3d electrons to 4s orbital the approaching F iotas will have their spot in 4s, 4p orbital. Hybridization included is $s{p^3}{d^2}$ . Thus it is an external orbital complex.
3. ${[Co{(CN)_6}]^{3 - }}$
$Co$ is in +3 oxidation state.
$C{o^{3 + }} - [Ar]3{d^6}$
It has no unpaired electron in 3d orbital on the grounds that $CN$ is solid field ligand and thus it is diamagnetic.
Since $CN$ is a solid field ligand that it will compel the 3d electrons to match the approaching $CN$ ligand will have their spot in 4s, 3d and 4p orbital. Hybridization included is ${d^2}s{p^3}$ . Henceforth it is an internal orbital complex.
4. ${[Co{(N{H_3})_6}]^{3 + }}$
$Cr$ is in +3 oxidation state.
$C{r^{3 + }} - [Ar]3{d^3}$
It has three unpaired electrons in 3d orbital; subsequently it is paramagnetic.
Since $N{H_3}$ is a solid field ligand the hybridization will be ${d^2}s{p^3}$ . It is an inward orbital complex.
Hence, option A is the correct answer.

Note: Weak field ligands prompt less parting of the precious stone fields. They structure buildings with high spins. Strong field ligands bring about more prominent parting of the gem field. They structure edifices with low twists.