Question

Which of the following complex compounds will exhibit the highest paramagnetic behaviour?

$[At.no.Ti=22,Cr=24,Co=27,Zn=30]$

• A. $\left[Ti\right(N{H}_3{)}_6{]}^{3+}$
• B. $\left[Cr\right(N{H}_3{)}_6{]}^{3+}$
• C. $\left[Co\right(N{H}_3{)}_6{]}^{3+}$
• D. $\left[Zn\right(N{H}_3{)}_6{]}^{2+}$

Answer 1

The correct option is B $\left[Cr\right(N{H}_3{)}_6{]}^{3+}$

(a) Electronic configuration of $T{i}^{3+}$ in $\left[Ti\right(N{H}_3{)}_6{]}^{3+}$
$T{i}^{3+}=3d^1$ [ref. image (a)]

(b) Electronic configuration of $C{r}^{3+}$ in $\left[Cr\right(N{H}_3{)}_6{]}^{3+}$
$C{r}^{3+}=3d^3$; [ref. image (b)]

(c) Electronic configuration of $C{o}^{3+}$ in $\left[Co\right(N{H}_3{)}_6{]}^{3+}$;
$C{o}^{3+}=3d^6$.
In the presence of strong field ligand $N{H}_3$, pairing of electrons takes place and hence, octahedral complex, $\left[Co\right(N{H}_3{)}_6{]}^{3+}$ is diamagnetic.
$\left[Co\right(N{H}_3{)}_6{]}^{3+}\left[\begin{array}{c}\text{inner orbital or}\\ \text{low spin complex}\end{array}\right]\left(6N{H}_{3}molecles\right)$ [ref. image (c)]

(d) Electronic configuration of $Z{n}^{2+}$ in $\left[Zn\right(N{H}_3{)}_6{]}^{2+}$
$\therefore \left[Zn\right(N{H}_3{)}_6{]}^{2+}$ is an outer orbital complex and is diamagnetic. [ref. image (d)]
$\left[Cr\right(N{H}_3{)}_6{]}^{3+}$ complex has the highest unpaired electron. Hence it will be highly paramagnetic.
Option B is correct.