Question

Which of the following is diamagnetic in nature?
(A) $\,{\left[ {{\rm{Fe(CN}}{{\rm{)}}_{\rm{6}}}} \right]^{3 - }}$
(B) ${\rm{NiC}}{{\rm{l}}_{\rm{4}}}^{2 - }$
(C) $\,\left[ {{\rm{Ni(CO}}{{\rm{)}}_{\rm{4}}}} \right]$
(D) $\,{\left[ {{\rm{Mn(Cl}}{{\rm{)}}_4}} \right]^{2 - }}$

Answer 1

As we know that the complexes which are given in the question are known as coordination complexes. The central atom is a transition metal in the given complexes which has d electrons in its outermost shell.

Complete step by step answer:
The diamagnetic term is used for those complexes in which the central atom has paired electrons in its d-orbital. So, more the number of paired electrons more will be the diamagnetism of complexes.
The strong field ligands form low spin complexes and weak field ligands form high spin complexes. The strong field ligand causes splitting of d- orbitals of metals in large energy gaps so that electrons pair up in ground state and the weak field ligands split d- orbitals of metals in small energy gaps, so the electrons of d- orbitals are excited to an excited level.
The number of electrons of central atom can be calculated by calculating its oxidation state as-
In $\,{\left[ {{\rm{Fe(CN}}{{\rm{)}}_6}} \right]^{3 - }}$complex
Suppose $x$ is the oxidation state of Iron

\,{\left[ {{\rm{Fe(CN}}{{\rm{)}}_{\rm{6}}}} \right]^{3 - }}\\\ x + 6( - 1) = - 3\\\ x = + 3 \end{array}$$ Where the formal charge $${\rm{CN}}$$ is$$ - 1$$. Now, we calculate the number of electrons of iron as: $$\begin{array}{c} {\rm{Fe(in}}\,{\rm{free}}\,{\rm{state)}} = 3{d^6}4{s^2}\\\ {\rm{F}}{{\rm{e}}^{ + 3}} = \,3{d^5} \end{array}$$ $${\rm{CN}}$$ is a strong field ligand, therefore the pairing energy is less than excitation energy and pairing occurs between the d electrons. There are two paired electrons and one unpaired electron. Hence, the complex is paramagnetic. In $${\rm{NiC}}{{\rm{l}}_4}^{2 - }$$ complex Suppose $$x$$ is the oxidation state of nickel $$\begin{array}{c} {\rm{NiC}}{{\rm{l}}_4}^{2 - }\\\ x + 4( - 1) = - 2\\\ x = + 2 \end{array}$$ Where the formal charge $${\rm{Cl}}$$ is$$ - 1$$. Now, we calculate the number of electrons of nickel as: $$\begin{array}{c} {\rm{Ni(in}}\,{\rm{free}}\,{\rm{state)}} = 3{d^8}4{s^2}\\\ {\rm{N}}{{\rm{i}}^{ + 2}} = \,3{d^8} \end{array}$$ $${\rm{Cl}}$$is a weak field ligand, therefore the pairing energy is larger than excitation energy and excitation occurs between the d orbital electrons. There are three paired electrons and two unpaired electrons. Hence the complex is paramagnetic. In $$\,\left[ {{\rm{Ni(CO}}{{\rm{)}}_{\rm{4}}}} \right]$$complex Suppose $$x$$ is the oxidation state of nickel $$\begin{array}{c} \,\left[ {{\rm{Ni(CO}}{{\rm{)}}_{\rm{4}}}} \right]\\\ x + 4(0) = 0\\\ x = 0 \end{array}$$ Where the formal charge $${\rm{CO}}$$ is$$0$$. Now, we calculate the number of electrons of nickel as: $${\rm{Ni(in}}\,{\rm{free}}\,{\rm{state}}) = 3{d^8}4{s^2}$$ $${\rm{CO}}$$is a strong field ligand, therefore the pairing energy is less than excitation energy and pairing occurs between the d orbital electrons. There are five paired electrons. Hence, the complex is diamagnetic. In $$\,{\left[ {{\rm{Mn(Cl}}{{\rm{)}}_{\rm{4}}}} \right]^{2 - }}$$complex Suppose $$x$$ is the oxidation state of manganese $$\begin{array}{c} {\left[ {{\rm{Mn(Cl}}{{\rm{)}}_{\rm{4}}}} \right]^{2 - }}\\\ x + 4( - 1) = - 2\\\ x = + 2 \end{array}$$ Now, we calculate the number of electrons of manganese as: $$\begin{array}{c} {\rm{Mn(in}}\,{\rm{free}}\,{\rm{state}}) = 3{d^5}4{s^2}\\\ {\rm{M}}{{\rm{n}}^{ + 2}} = 3{d^5} \end{array}$$ $${\rm{Cl}}$$is a weak field ligand, therefore the pairing energy is larger than excitation energy and excitation occurs between the d orbital electrons. There are three paired electrons and two unpaired electrons. Hence, the complex is paramagnetic. **Thus, the correct option is C** **Note:** The coordination complexes are the complexes which have a transition metal as the central atom and the ligands are attached with the central atom by donating lone pairs of electrons to the central atom.