- A

- ✓

- C

- D






Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(I)$ $\begin{array}{*{20}{c}}
{{{\left( {C{H_3}} \right)}_3}Si - NB{H_2}}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Si{{(C{H_3})}_3}}
\end{array}$ $(II)$ $\begin{array}{*{20}{c}}
{{{\left( {C{H_3}} \right)}_3}C - NB{H_2}}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{{(C{H_3})}_3}}
\end{array}$
$(A)$Covalent radius decreases down the group from $\mathrm{C}$ to $\mathrm{Pb}$ in a regular manner.
$(B)$ Electronegativity decreases from $\mathrm{C}$ to $\mathrm{Pb}$ down the group gradually.
$(C)$ Maximum covalence of $\mathrm{C}$ is $4$ whereas other elements can expand their covalence due to presence of $d$ orbitals.
$(D)$ Heavier elements do not form $\mathrm{p} \pi-p \pi$ bonds.
$(E)$ Carbon can exhibit negative oxidation states.
Choose the correct answer from the options given below:
