Select the molecule in which all the atoms ma | vedclass

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(A) For all atoms to lie in a single plane, all carbon atoms and the atoms attached to the benzene ring must be $sp^2$ hybridized to maintain planarit

Vedclass · Accepted Answer

$4-$Nitrobenzaldehyde

Vedclass · Answer

(A) For all atoms to lie in a single plane, all carbon atoms and the atoms attached to the benzene ring must be $sp^2$ hybridized to maintain planarity through conjugation.In $4-$Nitrobenzaldehyde $(O_2N-C_6H_4-CHO)$, the benzene ring, the nitro group $(-NO_2)$, and the aldehyde group $(-CHO)$ are all $sp^2$ hybridized.The conjugation extends throughout the molecule, allowing all atoms (including the hydrogen of the aldehyde group) to lie in the same plane.In other options, groups like $-OCH_3$ (in $4-$Methoxybenzaldehyde), $-CH_3$ (in $4-$Methylnitrobenzene), and $-COCH_3$ (in $4-$Nitroacetophenone) contain $sp^3$ hybridized carbon atoms, which possess tetrahedral geometry and prevent all atoms from lying in a single plane.

Column-$I$	Column-$II$
$A$. Boron in $[B(OH)_4]^-$	$1$. $sp^2$
$B$. Aluminum in $[Al(H_2O)_6]^{3+}$	$2$. $sp^3$
$C$. Boron in $B_2H_6$	$3$. $sp^3d^2$
$D$. Carbon in Buckminsterfullerene
$E$. $Si$ in $SiO_4^{4-}$
$F$. $Ge$ in $[GeCl_6]^{2-}$

Select the molecule in which all the atoms may lie on a single plane.

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