Question

The atomic number of an atom $\mathrm{A}$ is $\mathrm{x}$. Valence electronic configuration of another atom $\mathrm{B}$ is $6{\mathrm{s}}^{2}5{\mathrm{d}}^3$ and its atomic number is $(\mathrm{x}+5)$. Write the electronic configuration of $\mathrm{A}$.

Answer 1

1. The atomic number of $B$ is $73$ because its electronic configuration is $\left[\mathrm{Xe}\right]6{\mathrm{s}}^{2}4{\mathrm{f}}^{14}5{\mathrm{d}}^3$.
2. Thus, The atomic number of $\mathrm{A}$ is $\mathrm{x}$, that is $73-5=68$.
3. So, its electronic configuration is $\left[\mathrm{Xe}\right]6{\mathrm{s}}^2{5\mathrm{d}}^{1}4{\mathrm{f}}^{11}$