Thus is does have plane of symmetry.
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$A.$ has completely filled $\pi$ antibonding molecular orbitals
$B.$ is diamagnetic
$C.$ has bond order one
$D.$ is isoelectronic with Neon
$(a)$ Diborane is prepared by the oxidation of $\mathrm{NaBH}_{4}$ with $\mathrm{I}_{2}$.
$(b)$ Each boron atom is in sp $^{2}$ hybridized state.
$(c)$ Diborane has one bridged $3$ centre$-2-$electron bond.
$(d)$ Diborane is a planar molecule.
The option with correct statement(s) is :
$(I)$ $\begin{array}{*{20}{c}}
{C{H_3} - C - {O^ - }} \\
{\,\,\,\,\,\,||\,\,} \\
{\,\,\,\,\,\,O\,\,}
\end{array}$
$(II)$ $CH_3O^-$
$(III)$ ${C_6}{H_5}{O^\Theta }$
$(IV)$ (image)