Correct order of ionic character is _____

$N a F$ > $N a_{2} O$ > $N a_{3} N$

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$N a_{3} N$ > $N a F$ > $N a_{2} O$

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$N a F$ > $N a_{3} N$ > $N a_{2} O$

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$N a_{3} N$ < $N a F$ > $N a_{2} O$

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Solution

The correct option is A
$N a F$ > $N a_{2} O$ > $N a_{3} N$

All the compounds have same cation.

Among the anions, $F^{-}$ is smallest followed by $O^{2 -}$ and $N^{3 -}$

$∴$ $N^{3 -}$ is easily polarized $\Rightarrow$ $N a_{3} N$ is most covalent.

$∴$ The order is NaF > $N a_{2} O$ > $N a_{3} N$

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