Question

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

(At. No. : $Ti=22,Cr=24,Co=27,Zn=30$)

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

Answer 1

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

(A) ${\left[Ti{\left(N{H}_3\right)}_6\right]}^{3+}:Ti$ has $3d^1$ configuration and thus has one unpaired electron.

(B) ${\left[Cr{\left(N{H}_3\right)}_6\right]}^{3+}:$ The complex is inner orbital complex but $3d^3$ configuration has three unpaired electrons with weak as well as with strong field ligand.

(C) ${\left[Co{\left(N{H}_3\right)}_6\right]}^{3+}:$ The cobalt ion is in +3 oxidation state with $3d^6$ configuration and thus is diamagnetic octahedral complex, ${\left[Co{\left(N{H}_3\right)}_6\right]}^{3+}$, as the electrons paired up.

(D) ${\left[Zn{\left(N{H}_3\right)}_6\right]}^{2+}:$ Because of $3d^{10}$ configuration no $(n-1)d$ orbital is available for $d^{2}s{p}^3$ hybridisation and thus forms outer orbital complex. The complex is diamagnetic.