Question

What is the electron configuration of ${\mathrm{Pt}}^{2+}$, according to the Madelung rule?

Answer 1

Electronic Configuration:

• Arrangement of electrons in different orbits and orbitals around an atomic nucleus is defined as electronic configuration
• There are four different kinds of orbitals $s\mathit{,}\mathit{}p\mathit{,}\mathit{}d\mathit{}\mathit{\&}\mathit{}f$

Madelung Rule:

• The energy of an atomic orbital increases with $n+1$ increase
• For identical values of $n+1$, energy increases with increasing n

For ${\mathrm{Pt}}^{2+}$:

• The electronic configuration of ${\mathrm{Pt}}^{2+}$is $\left[\mathrm{Xe}\right]\mathit{4}{f}^{\mathit{14}}\mathit{5}{d}^{\mathit{8}}$
• Now by applying the Madelung's Rule: $\left[\mathrm{Xe}\right]\mathit{6}{s}^{\mathit{2}}\mathit{4}{f}^{\mathit{14}}\mathit{5}{d}^{\mathit{8}}$

For $6s$ we get, $(n+1)=6+0=6$

For $4f$ we get, $(n+1)=4+3=7$

For $5d$we get, $(n+1)=5+2=7$

When two sub shells have the same value i.e $7$ the one with the higher value of $n$is said to be in highest energy state. Hence the electron configuration of ${\mathrm{Pt}}^{2+}$is $\left[\mathrm{Xe}\right]\mathit{4}{f}^{\mathit{14}}\mathit{5}{d}^{\mathit{8}}$.