Animated Solution for Chemistry - Coordination Compounds: An aqueous solution of ${\text{NiCl}}_2$ was heated with excess sodium cyanide in presence of strong oxidising agent to form $[{\text{Ni(CN)}}_6{]}^{2-}$. The total change in number of unpaired electrons on metal centre is …… .

Let's embark on this fascinating journey of coordination chemistry by examining our starting material: an aqueous solution of nickel(II) chloride, ${\text{NiCl}}_2$. In this solution, nickel exists as the ${\text{Ni}}^{2+}$ ion.

To understand its magnetic properties, we must look at its electronic configuration. Nickel has an atomic number of 28, so the neutral atom has the configuration $[\text{Ar}]4s^{2}3d^8$. When it loses two electrons to form ${\text{Ni}}^{2+}$, it loses them from the outermost $4s$ orbital, leaving us with a $3d^8$ configuration.

In an octahedral aqueous environment, these eight $d$-electrons distribute themselves among the $t_{2g}$ and $e_g$ orbitals. Following Hund's rule, the first six electrons fill the lower energy $t_{2g}$ orbitals completely ($t_{2g}^6$), and the remaining two electrons occupy the higher energy $e_g$ orbitals singly. This gives us exactly 2 unpaired electrons.

The plot thickens when we introduce excess sodium cyanide ($\text{NaCN}$) and a strong oxidising agent. The cyanide ion, ${\text{CN}}^{-}$, is a notorious strong-field ligand. But wait, there's an oxidising agent present! This means our central metal ion is going to lose more electrons.

The product formed is the complex ion $[{\text{Ni(CN)}}_6{]}^{2-}$. Let's calculate the new oxidation state of nickel. Let the oxidation state be $x$. Since each cyanide ligand carries a $-1$ charge, we can set up the equation:

Solving for $x$, we find that $x=+4$. The strong oxidising agent has stripped two more electrons from the nickel ion, transforming it from ${\text{Ni}}^{2+}$ to ${\text{Ni}}^{4+}$.

Now, let's analyze the ${\text{Ni}}^{4+}$ ion. Having lost two more electrons from the $3d$ subshell, its new configuration is $3d^6$.

Here is where the nature of the ligand plays a crucial role. Cyanide is a strong-field ligand, meaning it causes a large crystal field splitting energy (${\Delta }_o$). Because ${\Delta }_o$ is greater than the pairing energy ($P$), it is energetically more favorable for the electrons to pair up in the lower energy $t_{2g}$ orbitals rather than jump to the higher $e_g$ orbitals.

Therefore, all six $d$-electrons will pair up in the $t_{2g}$ level, resulting in a $t_{2g}^6{e}_g^0$ configuration. As a result, there are 0 unpaired electrons in the final complex.

The question asks for the total change in the number of unpaired electrons on the metal center.

We started with 2 unpaired electrons in ${\text{Ni}}^{2+}$ and ended with 0 unpaired electrons in ${\text{Ni}}^{4+}$.

The total change in the number of unpaired electrons is 2.