Sodium forms ${\rm{N}}{{\rm{a}}^ + }$ and not ${\rm{N}}{{\rm{a}}^{2 + }}$ because:

First ionization potential is small and the difference in first and second ionization potentials is very large

Radius of ${\rm{N}}{{\rm{a}}^ + }$ is much smaller than of ${\rm{N}}{{\rm{a}}^ + }$

NEET UG CHEMISTRY MCQ: S-BLOCK ELEMENTS

1 Easy

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Understanding +4 | -1

Sodium forms

\mathrm{N}{\mathrm{a}^ + }

and not

\mathrm{N}{\mathrm{a}^{2 + }}

because:

View answer and explanation

Correct option: A

Sodium contains only one electron in outermost shell

Option A is correct because sodium readily loses its single valence electron to achieve a stable octet configuration, making the formation of

\mathrm{Na}^{2+}

energetically unfavorable.