- A$BF_3 , BrF_3$
- B$ICl_2^{\ominus}, BeCl_2$
- C$BCl_3 , PCl_3$
- D$PCl_3,NCl_3$
$BrF_3$ $sp^3d$ Bent $'T'$ shape
$IC^{\ominus}_2$ $sp^3d$ Linear
$BeCl_2$ $sp$ Linear
$BCl_3$ $sp^2$ Trigonal planar
$PCl_3$ $sp^3$ Pyramidal
$NCl_3$ $sp^3$ Pyramidal
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$C{H_3} - CH = C{H_2} + HBr\xrightarrow{{{{({C_6}{H_5}CO)}_2}{O_2}}}$
$\begin{array}{*{20}{c}}
{BrC{H_2} - CH - CO - C{H_2} - C{H_2}C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{CON{H_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
is :