Arrange Na, Na^+, F, F^- in the decreasing or | vedclass

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(A) The atomic radius of a neutral atom is larger than its cation because the removal of an electron decreases the electron-electron repulsion and inc

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$Na > F^- > Na^+ > F$

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(A) The atomic radius of a neutral atom is larger than its cation because the removal of an electron decreases the electron-electron repulsion and increases the effective nuclear charge. Thus, $Na > Na^+$.The ionic radius of an anion is larger than its neutral atom because the addition of an electron increases the electron-electron repulsion, leading to a decrease in the effective nuclear charge per electron. Thus, $F^- > F$.Comparing the sizes based on atomic/ionic radii values (in $pm$):$Na$ $(186 \ pm)$, $F^-$ $(136 \ pm)$, $Na^+$ $(86 \ pm)$, $F$ $(64 \ pm)$.Therefore, the decreasing order of size is $Na > F^- > Na^+ > F$.

Arrange $Na, Na^+, F, F^-$ in the decreasing order of their radii (size).

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