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(B) Hydrogen atoms do not undergo hybridization. The number of hybrid orbitals is calculated as follows:$(A)$ $C_6H_6$: All six $C$-atoms are $sp^2$-h

Vedclass · Accepted Answer

$(CH_3)_4C$

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(B) Hydrogen atoms do not undergo hybridization. The number of hybrid orbitals is calculated as follows:$(A)$ $C_6H_6$: All six $C$-atoms are $sp^2$-hybridized (each has $3$ hybrid orbitals). Total $= 6 	imes 3 = 18$.$(B)$ $(CH_3)_4C$: All five $C$-atoms are $sp^3$-hybridized (each has $4$ hybrid orbitals). Total $= 5 	imes 4 = 20$.$(C)$ $(CH_3)_2C=O$: Two $C$-atoms in $CH_3$ groups are $sp^3$-hybridized ($2 	imes 4 = 8$ orbitals). The carbonyl $C$-atom is $sp^2$-hybridized ($3$ orbitals). The $O$-atom is $sp^2$-hybridized ($3$ orbitals). Total $= 8 + 3 + 3 = 14$.$(D)$ $CH_3-CH=CH-CN$: $C_4$ is $sp^3$ ($4$ orbitals),$C_3$ and $C_2$ are $sp^2$ ($2 	imes 3 = 6$ orbitals),$C_1$ is $sp$ ($2$ orbitals),and $N$ is $sp$ ($2$ orbitals). Total $= 4 + 6 + 2 + 2 = 14$.Thus,$(CH_3)_4C$ has the maximum number of hybrid orbitals. The correct option is $(B)$.

List-$I$ (molecule)	List-$II$ (hybridization,shape)
$A. XeO_3$	$I. sp^3$; pyramidal
$B. XeF_2$	$II. sp^3d$; linear
$C. XeOF_4$	$III. sp^3d^2$; square pyramidal
$D. XeF_6$	$IV. sp^3d^3$; distorted octahedral

Which one of the following has the maximum number of hybrid orbitals?

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Match List-$I$ with List-$II$.

List-$I$ (molecule) List-$II$ (hybridization,shape)

$A. XeO_3$ $I. sp^3$; pyramidal

$B. XeF_2$ $II. sp^3d$; linear

$C. XeOF_4$ $III. sp^3d^2$; square pyramidal

$D. XeF_6$ $IV. sp^3d^3$; distorted octahedral

Choose the correct answer from the options given below.

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