- A${F_2}$
- B$NaF$
- C$Ca{F_2}$
- ✓$B{F_3}$
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$(I)$ ${P_4} + S{O_2}C{l_2} \to $
$(II)$ ${P_4} + SOC{l_2} \to $
$(III)$ $PC{l_5} + S{O_2} \to $
$(IV)$ $PC{l_5} + {C_2}{H_5}OH \to $
$(i)\,\mathop {\begin{array}{*{20}{c}}
{\,\,\,\,\,C{H_3}} \\
| \\
{C{H_3} - C - C{H_3}} \\
| \\
{\,\,\,\,\,C{H_3}}
\end{array}}\limits_{(Neo - pen\tan e)\,(i)} $
$(ii\mathop {)\,\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_2} - C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,\,\,\,C{H_{3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}}
\end{array}}\limits_{(Iso - pen\tan e)\,\,(ii)} $
$(iii)\,\mathop {C{H_3} - C{H_2} - C{H_2} - C{H_2} - C{H_3}}\limits_{(n - pen\tan e)} $

$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g})$
the correct option is
In this reaction, the end product $C$ is :
