- A$(CH_3COO)_2Pb$
- B$Na_2[Fe(CN)_5NO]$
- ✓both $(A)$ and $(B)$
- DNone of these

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Reason : Iodine is a polar compound.
$(i)\,\,C\,({\rm{graphite}})\, + \,{O_2}{\kern 1pt} (g)\, \to \,C{O_2}\,(g);\,\Delta r{H^\circleddash} = x\,\,kJ\,mo{l^{ - 1}}$
$(ii)\,\,C\,({\rm{graphite}})\, + \,\frac{1}{2}{O_2}{\kern 1pt} (g)\, \to \,CO\,(g);\,\Delta r{H^\circleddash} = y\,\,kJ\,mo{l^{ - 1}}$
$(iii)\,\,CO\,(g)\, + \,\frac{1}{2}{O_2}{\kern 1pt} (g)\, \to \,C{O_2}\,(g);\,\Delta r{H^\circleddash} = z\,\,kJ\,mo{l^{ - 1}}$
Based on the above thermochemical equations, find out which one of the following algebraic relationships is correct?
|
List-$I$ Reaction |
List-$II$ Type of redox reaction |
| $(A)$ $\mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{(\mathrm{g})}$ | $(I)$ Decomposition |
| $(B)$ $\begin{aligned} & 2 \mathrm{~Pb}\left(\mathrm{NO}_3\right)_{2(\mathrm{~s})} \rightarrow 2 \mathrm{PbO}_{(\mathrm{s})}+4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}\end{aligned}$ | $(II)$ Displacement |
| $(C)$ $\begin{aligned} 2 \mathrm{Na}_{(\mathrm{s})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \rightarrow 2 \mathrm{NaOH}_{(\mathrm{aq} .)}+\mathrm{H}_{2(\mathrm{~g})}\end{aligned}$ | $(III)$ Disproportionation |
| $(D)$ $\begin{aligned} 2 \mathrm{NO}_{2(\mathrm{~g})}+2-\mathrm{OH}_{(\mathrm{aq})} \rightarrow \mathrm{NO}_{2(\mathrm{aq} .)}^{-}+\mathrm{NO}_{3(\mathrm{qq} .)}^{-}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}\end{aligned}$ | $(IV)$ Combination |
Choose the correct answer from the options given below:
$\begin{array}{*{20}{c}}
{HO - C = O\,\,\,\,\,\,\,C{H_3}} \\
{\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{C{H_3} - C = C-C - H} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,N{H_2}\,\,\,\,\,Cl}
\end{array}$
is :