Question

why is \([Ni(H_2O)_6]^{2+}\) coloured ? [Atomic number of Ni = 28]

Hint:

Transition metal complexes are generally coloured because: \[ \text{Partially filled } d\text{-orbitals} \] \[ \Longrightarrow \text{Crystal field splitting} \] \[ \Longrightarrow d-d \text{ transitions} \] \[ \Longrightarrow \text{Absorption of visible light} \] \[ \Longrightarrow \text{Colour observed} \] Complexes with \(d^0\) or \(d^{10}\) configurations are generally colourless.

Answer 1

Step 1: Find the configuration of $Ni^{2+}$.
Ni ($Z = 28$): $[Ar]\,3d^8\,4s^2$. Losing 2 electrons from $4s$ gives $Ni^{2+}$: $[Ar]\,3d^8$. This is a partially filled $d$-subshell.
Step 2: Crystal field splitting by $H_2O$ ligands.
In $[Ni(H_2O)_6]^{2+}$, six water molecules in an octahedral arrangement split the five degenerate $d$-orbitals into a lower energy $t_{2g}$ set (3 orbitals) and a higher energy $e_g$ set (2 orbitals).
Step 3: $d$-$d$ transition causes colour.
The 8 electrons of $Ni^{2+}$ partially fill both $t_{2g}$ and $e_g$ sets. Electrons absorb visible light and jump from $t_{2g}$ to $e_g$ (a $d$-$d$ transition). The complementary colour of absorbed light is transmitted, making the complex appear coloured. \[ \boxed{[Ni(H_2O)_6]^{2+} \text{ is coloured because } Ni^{2+}(3d^8) \text{ undergoes } d\text{-}d \text{ transitions after crystal field splitting}} \]