Question

Which ion with a +2 charge has the electron configuration\({\bf{1}}{{\bf{s}}^{\bf{2}}}{\bf{2}}{{\bf{s}}^{\bf{2}}}{\bf{2}}{{\bf{p}}^{\bf{6}}}{\bf{3}}{{\bf{s}}^{\bf{2}}}{\bf{3}}{{\bf{p}}^{\bf{6}}}{\bf{3}}{{\bf{d}}^{{\bf{10}}}}{\bf{4}}{{\bf{s}}^{\bf{2}}}{\bf{4}}{{\bf{p}}^{\bf{6}}}{\bf{4}}{{\bf{d}}^{\bf{5}}}\)? Which ion with a +3 charge has this configuration?

Short answer

\(T{c^{2 + }}and{\rm{ }}R{u^{3 + }}\)

Step-by-step solution

Cations

When an atom loses an electron from its valence shell then a positively charged species is formed and is called cation.

The charge on the cation represents the number of electrons lost by it.

Cation with the given electronic configuration and +2 charge

The given electronic configuration is\({\bf{1}}{{\bf{s}}^{\bf{2}}}{\bf{2}}{{\bf{s}}^{\bf{2}}}{\bf{2}}{{\bf{p}}^{\bf{6}}}{\bf{3}}{{\bf{s}}^{\bf{2}}}{\bf{3}}{{\bf{p}}^{\bf{6}}}{\bf{3}}{{\bf{d}}^{{\bf{10}}}}{\bf{4}}{{\bf{s}}^{\bf{2}}}{\bf{4}}{{\bf{p}}^{\bf{6}}}{\bf{4}}{{\bf{d}}^{\bf{5}}}\).

If it has +2 charge, then the atom’s original electronic configuration becomes

\({\bf{1}}{{\bf{s}}^{\bf{2}}}{\bf{2}}{{\bf{s}}^{\bf{2}}}{\bf{2}}{{\bf{p}}^{\bf{6}}}{\bf{3}}{{\bf{s}}^{\bf{2}}}{\bf{3}}{{\bf{p}}^{\bf{6}}}{\bf{3}}{{\bf{d}}^{{\bf{10}}}}{\bf{4}}{{\bf{s}}^{\bf{2}}}{\bf{4}}{{\bf{p}}^{\bf{6}}}{\bf{4}}{{\bf{d}}^{\bf{7}}}\).

The atom with this electronic configuration has the atomic number 43.

The element is Tc.

The symbol is\(T{c^{2 + }}\).

Cation with the given electronic configuration and +2 charge

The given electronic configuration is\({\bf{1}}{{\bf{s}}^{\bf{2}}}{\bf{2}}{{\bf{s}}^{\bf{2}}}{\bf{2}}{{\bf{p}}^{\bf{6}}}{\bf{3}}{{\bf{s}}^{\bf{2}}}{\bf{3}}{{\bf{p}}^{\bf{6}}}{\bf{3}}{{\bf{d}}^{{\bf{10}}}}{\bf{4}}{{\bf{s}}^{\bf{2}}}{\bf{4}}{{\bf{p}}^{\bf{6}}}{\bf{4}}{{\bf{d}}^{\bf{5}}}\).

If it has +2 charge, then the atom’s original electronic configuration becomes

\({\bf{1}}{{\bf{s}}^{\bf{2}}}{\bf{2}}{{\bf{s}}^{\bf{2}}}{\bf{2}}{{\bf{p}}^{\bf{6}}}{\bf{3}}{{\bf{s}}^{\bf{2}}}{\bf{3}}{{\bf{p}}^{\bf{6}}}{\bf{3}}{{\bf{d}}^{{\bf{10}}}}{\bf{4}}{{\bf{s}}^{\bf{2}}}{\bf{4}}{{\bf{p}}^{\bf{6}}}{\bf{4}}{{\bf{d}}^8}\).

The atom with this electronic configuration has the atomic number 44.

The element is Ru.

The symbol is\(R{u^{3 + }}\).