- A$Ar _{( g )}+ e ^{-} \rightarrow Ar _{( g )}^{-}$
- ✓$H _{( g )}+ e ^{-} \rightarrow H _{( g )}^{-}$
- C$Na _{( g )} \rightarrow Na _{( g )}^{+}+ e ^{-}$
- D$O _{( g )}^{-}+ e ^{-} \rightarrow O _{( g )}^{2-}$
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$2N{O_2}\left( g \right) \rightleftharpoons {N_2}{O_4}\left( g \right)$ is
$\left( {R = \frac{{25}}{3}J/K.mol,\ln 2 = 0.7,\ln 3 = 1.1} \right)$
Assertion $(A)$ : $\mathrm{NH}_3$ and $\mathrm{NF}_3$ molecule have pyramidal shape with a lone pair of electrons on nitrogen atom. The resultant dipole moment of $\mathrm{NH}_3$ is greater than that of $\mathrm{NF}_3$.
Reason $(R)$ : In $\mathrm{NH}_3$, the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the $\mathrm{N}-\mathrm{H}$ bonds. $\mathrm{F}$ is the most electronegative element.
In the light of the above statements, choose the correct answer from the options given below:


$Ag(s)\,\,|\,\,A{g_2}{C_2}{O_4}(s)\,\,|\,\,{C_2}O_4^{2 - }(0.02\,M)\,\,||\,\,A{g^ + }(0.5\,M)\,\,|\,\,Ag(s)$
$EMF$ of cell is $0.264\ V$ at $25\,^oC$ . $K_{sp}$ of $Ag_2C_2O_4(s)$ is
$\left[ {Given = \frac{{2.303 \times 8.314 \times 298}}{{96500}} = \frac{{2.303RT}}{F} = 0.06,\,\log 2 = 0.3} \right]$

$HCl{O_4} + {H_2}O$ $ \rightleftharpoons $ ${H_3}{O^ + } + ClO_4^ - $