Question

How many unpaired electrons are there in the following complex ions?

$$\begin{gathered} \mathbf{a}\mathbf{)}\mathbf{}\mathbf{Ru}{{\mathbf{\left(}{\mathbf{NH}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{6}}}^{\mathbf{2}\mathbf{+}}\mathbf{(}\mathbf{Low}\mathbf{}\mathbf{spin}\mathbf{}\mathbf{case}\mathbf{)} \\ \mathbf{b}\mathbf{)}\mathbf{}\mathbf{Ni}\mathbf{}{{\mathbf{\left(}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\right)}}_{\mathbf{6}}}^{\mathbf{2}\mathbf{+}} \\ \mathbf{c}\mathbf{)}\mathbf{}\mathbf{V}{{\mathbf{\left(}\mathbf{en}\mathbf{\right)}}_{\mathbf{3}}}^{\mathbf{3}\mathbf{+}}\mathbf{} \end{gathered}$$

Short answer

1. 0

2. 2

3. 2

Step-by-step solution

All the 6 electrons are paired in the ruthenium complex ion.

As ${\mathrm{NH}}_3$is a strong-field ligand for low-spin case, it provides greater splitting energy data-custom-editor="chemistry" $∆$, therefore all the 6 electrons of 4d orbitals of free ruthenium ion remain paired in the ${\mathrm{t}}_{2\mathrm{g}}$ orbitals of the complex ion. Hence zero.

Four electrons are unpaired in nickel complex ion.

As ${\mathrm{H}}_2\mathrm{O}$is a strong-field ligand, it gives high splitting energy gap data-custom-editor="chemistry" $∆$, therefore 8 electrons in 3d orbitals of free nickel ion exist as six pairs of electrons in ${\mathrm{t}}_{2\mathrm{g}}$ and two unpaired electrons in e_gorbitals respectively of the nickel complex ion. Hence 2.

Two electrons are unpaired in vanadium complex ion.

As ‘en’ is a strong-field ligand, it gives greater splitting energy gap $∆$, therefore 2 electrons in 3d orbital of free vanadium ion are present as unpaired in t_2gorbitals of the vanadium complex ion. Hence 2.