Question

$K:{K_4}\,\left[ {Mn{{(CN)}_6}} \right],$
$L:{K_3}\,\left[ {Co{{(N{O_2})}_6}} \right],$

M:\,\,\left[ {Co{{({H_2}O)}_6}} \right]C{l_3},\\\ N:\,\,\left[ {Cr{{(CO)}_6}} \right],\\\ O:\,\,\left[ {Ni{{(en)}_3}} \right]{\left( {N{O_3}} \right)_3}\\\ P:\,\,\left[ {Pt{{(N{H_3})}_2}C{l_2}} \right] \end{array}$$ The diamagnetic complexes are: (A) $$L,M,N,P$$ (B) $$K,L,N,O$$ (C) $$L,M,O,P$$ (D) $$K,N,O,P$$

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 paired in ground state and the weak field ligands split d- orbitals of metals in small energy gaps, so the electrons of d- orbitals get excited to excited level.
The number of electrons of central atom can be calculated by calculating its oxidation state as-
In $K:{K_4}\,\left[ {Mn{{(CN)}_6}} \right]$complex
Suppose $x$ is the oxidation state of manganese

{K_4}\,\left[ {Mn{{(CN)}_6}} \right],\\\ 4( + 1) + x + 6( - 1) = 0\\\ x = + 2 \end{array}$$ Where the formal charge of$$K$$is$$ + 1$$ and of $$CN$$ is$$ - 1$$. Now, we calculate the number of electrons of manganese as: $$\begin{array}{c} Mn\,(in\,free\,state) = 3{d^5}4{s^2}\\\ M{n^{ + 2}} = \,3{d^5} \end{array}$$ $$CN$$ is a strong field ligand and has two paired electrons and one unpaired electron so the complex is paramagnetic. In $$L:{K_3}\,\left[ {Co{{(N{O_2})}_6}} \right]$$complex Suppose $$x$$ is the oxidation state of cobalt. $$\begin{array}{c} {K_3}\,\left[ {Co{{(N{O_2})}_6}} \right]\\\ 3 + x + 6( - 1) = 0\\\ x = + 3 \end{array}$$ Where the formal charge of$$K$$is$$ + 1$$ and of $$N{O_2}$$is$$ - 1$$. Now we calculate the number of electrons of cobalt as: $$\begin{array}{c} Co\,(in\,free\,state) = 3{d^7}4{s^2}\\\ C{o^{ + 3}} = \,3{d^6} \end{array}$$ $$N{O_2}$$ is a strong field ligand and has three paired electrons, so the complex is diamagnetic. In $$M:\,\left[ {Co{{({H_2}O)}_6}} \right]C{l_3}$$complex Suppose $$x$$is the oxidation state of cobalt. $$\begin{array}{c} \,\left[ {Co{{({H_2}O)}_6}} \right]C{l_3}\\\ x + 6 \times 0 = + 3\\\ x = + 3 \end{array}$$ Where the formal charge of$${H_2}O$$ is$$0$$ and of $$Cl$$is$$ - 1$$. Now we calculate the number of electrons of cobalt as: $$\begin{array}{c} Co\,(in\,free\,state) = 3{d^7}4{s^2}\\\ C{o^{ + 3}} = \,3{d^6} \end{array}$$ $${H_2}O$$ is a weak field ligand but the cobalt is having $$ + 3$$ oxidation state so the electrons will be paired up in ground state and the complex will be diamagnetic. In $$N:\,\,\left[ {Cr{{(CO)}_6}} \right]$$complex Suppose $$x$$is the oxidation state of chromium. $$\begin{array}{c} \,\left[ {Cr{{(CO)}_6}} \right]\\\ x + 6 \times 0 = 0\\\ x = 0 \end{array}$$ Where the formal charge of$$CO$$ is$$0$$. Now, we calculate the number of electrons of chromium as: $$\begin{array}{c} Cr\,(in\,free\,state) = 3{d^5}4{s^1}\\\ C{r^0} = \,3{d^5}4{s^1} \end{array}$$ $$CO$$is a very strong field ligand and chromium having three paired electrons, so the complex is diamagnetic. In $$O:\,\,\left[ {Ni{{(en)}_3}} \right]{\left( {N{O_3}} \right)_3}$$complex Suppose $$x$$is the oxidation state of nickel. $$\begin{array}{c} \,\left[ {Ni{{(en)}_3}} \right]{\left( {N{O_3}} \right)_3}\\\ x + 3 \times 0 = + 3\\\ x = + 3 \end{array}$$ Where the formal charge of$$en$$ is $$0$$and of $$N{O_3}$$is $$ - 1$$. Now, we calculate the number of electrons of nickel as: $$\begin{array}{c} Ni\,(in\,free\,state) = 3{d^8}4{s^2}\\\ N{i^{ + 3}} = \,3{d^7} \end{array}$$ $$en$$is a strong field ligand and thus, nickel having three paired electrons and one unpaired electrons so, the complex is paramagnetic. In $$P:\,\,\left[ {Pt{{(N{H_3})}_2}C{l_2}} \right]$$complex Suppose $$x$$is the oxidation state of platinum. $$\begin{array}{c} \,\left[ {Pt{{(N{H_3})}_2}C{l_2}} \right]\\\ x + 2 \times 0 + 2\left( { - 1} \right) = 0\\\ x = + 2 \end{array}$$ Where the formal charge of$$N{H_3}$$ is $$0$$and of $$Cl$$is $$ - 1$$. Now, we calculate the number of electrons of nickel as: $$\begin{array}{c} Pt\,(in\,free\,state) = 5{d^9}4{s^1}\\\ P{t^{ + 2}} = 5{d^8} \end{array}$$ Platinum is a $${3^{rd}}$$ row transition metal and the metals belong to second and third row transition metals always form low spin complexes. Hence this complex is also diamagnetic. **So, the correct answer is “Option A”.** **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.