Question

Which of the following is a low-spin (spin paired) complex?
A.${\left[ {{\text{Ni}}{{\left( {{\text{N}}{{\text{H}}_{\text{3}}}} \right)}_{\text{6}}}} \right]^{{\text{3 + }}}}$
B.${\left[ {{\text{Ti}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{6}}}} \right]^{{\text{3 + }}}}$
C.${\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$
D.${\left[ {Fe{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$

Answer 1

Hint: To solve this question, you must recall the electronic configurations of the metal cations present in the given coordination compounds. In presence of strong field ligands, low spin complexes are formed due to pairing of electrons, while complexes with weak field ligands have a high spin arrangement

Complete step by step solution:
A low spin complex is one in which the electrons are paired up to give a maximum number of doubly occupied $d$orbitals and a minimum number of unpaired electrons. Outer orbital complexes are high spin complexes and inner orbital complexes are low spin complexes.
${\text{N}}{{\text{H}}_3}$ and ${{\text{H}}_2}{\text{O}}$ are weak field ligands and do not cause pairing of electrons in the ions.

In ${\left[ {{\text{Ni}}{{\left( {{\text{N}}{{\text{H}}_{\text{3}}}} \right)}_{\text{6}}}} \right]^{{\text{3 + }}}}$, nickel ion has oxidation number $ + 3$
${\text{N}}{{\text{i}}^{3 + }}=\left[ {Ar} \right]3{d^7}4{s^0}$; Since the coordination number of nickel in the complex is $6$, its hybridisation is $s{p^3}{d^2}$
There are three unpaired electrons and thus the compound formed is an outer orbital complex or high spin complex.

In ${\left[ {{\text{Ti}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{6}}}} \right]^{{\text{3 + }}}}$, titanium has oxidation number $ + 3$
${\text{T}}{{\text{i}}^{{\text{3 + }}}} = \left[ {{\text{Ar}}} \right]3{d^1}4{s^0}$; Since there is only one electron present in the $d$orbital, it cannot undergo pairing of electrons and thus cannot form low spin complex.

In ${\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$, chromium has oxidation number $ + 3$
${\text{C}}{{\text{r}}^{3 + }} = \left[ {{\text{Ar}}} \right]3{d^3}4{s^0}$; Since there are three electrons in the $d$orbital, it cannot undergo pairing of electrons and thus cannot form low spin complex.

In ${\left[ {Fe{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$, iron has oxidation number $ + 3$
${\text{F}}{{\text{e}}^{3 + }} = \left[ {{\text{Ar}}} \right]3{d^5}4{s^0}$; the ferric ion has ${d^2}s{p^3}$ hybridisation and forms inner orbital or low spin complex having one unpaired electron.

So the correct answer is option D

Note: ${\text{CO,}} {\left( {{\text{CN}}} \right)^{\text{ - }}}{\text{,N}}{{\text{O}}_{\text{2}}}^{\text{ - }}{\text{, phen, dipy, en}}$ are strong field ligands. These ligands cause the pairing of unpaired electrons in the atom with which they form a coordination bond.
Generally, the ligands which donate through their $\pi $ orbitals are weak (known as weak field ligands) while those which donate through sigma orbitals are comparatively stronger and those which donate through sigma and accept through $\pi $orbitals are even stronger (known as strong field ligands)