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)$\left[Co\right(N{H}_3{)}_6{]}^{3+}$
(b)$[Mn{F}_6{]}^{3-}$
(c)$\left[Co\right(CN{)}_6{]}^{3-}$
(d)$\left[Cr\right(N{H}_3{)}_6{]}^{3+}$

• A. (a) inner, diamagnetic; (b) outer, paramagnetic; (c) inner, diamagnetic; (d) inner, paramagnetic.
• B. (a) outer, paramagnetic; (b) inner, paramagnetic; (c) inner, diamagnetic; (d) inner, paramagnetic.
• C. (a) inner, paramagnetic; (b) outer, paramagnetic; (c) outer, diamagnetic; (d) inner, diamagnetic.
• D. (a) inner, paramagnetic; (b) inner, diamagnetic; (c) inner, diamagnetic; (d) inner, paramagnetic.

Answer 1

The correct option is A (a) inner, diamagnetic; (b) outer, paramagnetic; (c) inner, diamagnetic; (d) inner, paramagnetic.

1. $\left[Cr\right(N{H}_3{)}_6{]}^{3+}$

Co is in +3 oxidation state.
$C{o}^{3+}-\left[Ar\right]3d^6$
It has no unpaired electron in 3d orbital because $N{H}_3$ is a strong field ligand and it pairs the electrons. Hence it is diamagnetic.
Hybridization involved is $d^{2}s{p}^3$. Hence it is inner orbital complex.

2. $[Mn{F}_6{]}^{3-}$

Mn is in +3 oxidation state.
$M{n}^{3+}-\left[Ar\right]3d^4$
It has four unpaired electron in 3d orbital hence it is paramagnetic.
Since $F$ is a weak field ligand that it will not force the 3d electrons to 4s orbital the incoming $F$ atoms will take their place in 4s, 4p orbital. Hybridization involved is $s{p}^3{d}^2$. Hence it is outer orbital complex.

3. $\left[Co\right(C{N}_6{]}^{3-}$

Co is in +3 oxidation state.
$C{o}^{3+}-\left[Ar\right]3d^6$
It has no unpaired electron in 3d orbital because $CN$ is strong field ligand and hence it is diamagnetic.
Since $CN$ is a strong field ligand that it will force the 3d electrons to pair the incoming $CN$ ligand will take their place in 4s, 3d and 4p orbital. Hybridization involved is $d^{2}s{p}^3$. Hence it is inner orbital complex.

4. $\left[Cr\right(N{H}_3{)}_6{]}^{3+}$

Cr is in +3 oxidation state.
$C{r}^{3+}-\left[Ar\right]3d^3$
It has three unpaired electron in 3d orbital hence it is paramagnetic.
Since $N{H}_3$ is a strong field ligand and the hybridization will be $d^{2}s{p}^3$. It is inner orbital complex