The elements with atomic number $8, 7, 11, 12, 13,$ and $9$ are oxygen nitrogen sodium magnesium aluminium and fluorine respectively. Generally these elements form $O^{- 2}$ , $N^{- 3}$ , $N a^{+}$ , $M g^{+ 2}$ , $A I^{3 +}$ and $F^{-}$ ions. The decreasing order of ionic size is?

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O^{- 2} > N^{- 3} > N a^{+} > M g^{+ 2} > A I^{3 +} > F^{-}

a^{+} > M g^{+ 2} > A I^{3 +} > N^{- 3} > O^{- 2} > F^{-}

N^{- 3} > N a^{+} > M g^{+ 2} > O^{- 2} > A I^{3 +}

> F^{-}

N^{- 3} > O^{- 2} > F^{-} > N a^{+}

M g^{+ 2} > A I^{3 +}

answer is C.

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Detailed Solution

The elements with atomic number $8, 7, 11, 12, 13,$ and $9$ are oxygen nitrogen sodium magnesium aluminium and fluorine respectively. Generally these elements form
$O^{- 2}$ , $N^{- 3}$ , $N a^{+}$ , $M g^{+ 2}$ , $A I^{3 +}$ and $F^{-}$ ions.

Oxygen, nitrogen, sodium, magnesium, aluminium, and fluorine are the elements having atomic numbers $8, 7, 11, 12, 13$ and $9$ correspondingly.
A positive sign indicates a lack of electrons, while a negative sign indicates more electrons.

⇒Here, $N^{- 3}$ with atomic no. $7$ contains $7 + 3 = 10$ electrons.
⇒ $O^{- 2}$ with an atomic number of $8$ , $O^{- 2}$ possesses $8 + 2 = 10$ electrons.
⇒ $F^{-}$ with an atomic number of $9$ , $F^{-}$ contains $9 + 1 = 10$ electrons.
⇒ $N a^{+}$ with an atomic number of 11, Na+ contains $11 - 1 = 10$ electrons.
⇒ $M g^{+ 2}$ with an atomic number $12$ of $M g^{+ 2}$ has $12 - 2 = 10$ electrons.
⇒ $A I^{3 +}$ , atomic number $13$ , contains $13 - 3 = 10$ electrons.

Hence, The decreasing order of ionic size is
$N^{- 3} > O^{- 2} > F^{-} > N a^{+}$ > $M g^{+ 2} > A I^{3 +}$ .