With cold and dilute sodium sodium hydroxide fluorine reacts to give :

N a F

and

O F_{2}

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N a F + O_{3}

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O_{2}

and

O_{3}

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N a F + O_{2}

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Solution

The correct option is A $N a F$ and $O F_{2}$
$F_{2}$ on reaction with NaOH gives different products under different condition
(i) $F_{2} + d i l, c o l d N a O H$
$2 F_{2} + 2 N a O H_{(c o l d) (d i l)} ⟶ 2 N a F + H_{2} O + O F_{2} o x y g e n d i f l u o r i d e$
(ii) $F_{2} + h o t, c o n c N a O H$
$4 F_{2} + N a O H_{(h o t) (d i l)} ⟶ 4 N a F + 2 H_{2} O + 2 H_{2} O + O_{2}$

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