Select the molecule in which all the atoms ma | vedclass

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Question

(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

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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 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.

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

Similar Questions

The hybridization of atomic orbitals of nitrogen in $NO_2^+$,$NO_3^-$,and $NH_4^+$ are:

In the series ethane,ethylene and acetylene,the $C-H$ bond energy is

The percentage $s$-character of the hybrid orbitals in methane,ethene,and ethyne are respectively:

What is $sp^{2}$ hybridization? Give its characteristics.

Identify the molecules which have $sp$-hybridised carbon$(s)$ from the list given below:
$I$. Ethane nitrile $(CH_3CN)$
$II$. Buta$-1,3-$diene $(CH_2=CH-CH=CH_2)$
$III$. Propan$-1,2-$diene $(CH_2=C=CH_2)$
$IV$. Ethyne $(HC \equiv CH)$
$V$. Benzene $(C_6H_6)$

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