- A$Pb$
- ✓$Sn$
- C$Si$
- D$C$
It is due to the poor shielding of $d$- and $f-$electron in $Pb$ due to which it feels greater attraction from nucleus.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

$\begin{array}{*{20}{c}}
{C{H_3} - CH - CH = C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}\xrightarrow{{HBr}}A$
$\begin{array}{*{20}{c}}
{C{H_3} - CH - CH = C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}\xrightarrow{{HBr + ROOR}}B$
$\begin{array}{*{20}{c}}
{C{H_3} - CH - CH = C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}\xrightarrow{{NBS}}C$