Among the following, the one for which the infrared active vibrational modes are Raman inactive and vice versa is

Question

The UV-visible absorption bands in the spectra of lanthanoid ions are ‘X’, probably because of the excitation of electrons involving ‘Y’. The ‘X’ and ‘Y’, respectively, are :

Answer 1

Instead of deriving the rule from scratch, use the direct test: mutual exclusion between IR and Raman only ever happens in a molecule that possesses a center of symmetry, so this question reduces to spotting which one option is centrosymmetric.

$\mathrm{H_2O}$: bent shape, point group $C_{2v}$. A $C_{2v}$ molecule never has an inversion center because an inversion would need atoms directly opposite the center in every direction, which a bent triatomic simply cannot provide. Not centrosymmetric.
trans-planar $\mathrm{H_2O_2}$: laying the whole molecule flat with the two $O-H$ bonds pointing to opposite sides of the $O-O$ bond puts an inversion center exactly at the midpoint of the $O-O$ bond, since every atom (each $H$, each $O$) has an identical atom diagonally opposite through that point. Point group $C_{2h}$, which by definition contains $i$. Centrosymmetric.
eclipsed ethane: point group $D_{3h}$. $D_{3h}$ has a $\sigma_h$ and a $C_3$ axis but, because the two $CH_3$ groups are eclipsed rather than staggered, there is no point exactly opposite every atom through a single center. Not centrosymmetric (only the staggered form, $D_{3d}$, would qualify).
$\mathrm{CH_4}$: tetrahedral, point group $T_d$. A regular tetrahedron has no inversion center (inverting one vertex through the middle does not land on another vertex). Not centrosymmetric.

Since trans-planar $\mathrm{H_2O_2}$ is the only centrosymmetric species in the list, it is the only one where the IR active modes are guaranteed to be completely Raman silent and the Raman active modes completely IR silent.

Let's summarize:

Mutual exclusion is a shortcut test for a center of inversion, nothing more.
Bent, eclipsed, and tetrahedral shapes here all fail to have that center; only the flat, trans arrangement of $H_2O_2$ has it.

The correct option is (B), the trans-planar conformer of $\mathrm{H_2O_2}$.

Answer 2

Instead of deriving the rule from scratch, use the direct test: mutual exclusion between IR and Raman only ever happens in a molecule that possesses a center of symmetry, so this question reduces to spotting which one option is centrosymmetric.

$\mathrm{H_2O}$: bent shape, point group $C_{2v}$. A $C_{2v}$ molecule never has an inversion center because an inversion would need atoms directly opposite the center in every direction, which a bent triatomic simply cannot provide. Not centrosymmetric.
trans-planar $\mathrm{H_2O_2}$: laying the whole molecule flat with the two $O-H$ bonds pointing to opposite sides of the $O-O$ bond puts an inversion center exactly at the midpoint of the $O-O$ bond, since every atom (each $H$, each $O$) has an identical atom diagonally opposite through that point. Point group $C_{2h}$, which by definition contains $i$. Centrosymmetric.
eclipsed ethane: point group $D_{3h}$. $D_{3h}$ has a $\sigma_h$ and a $C_3$ axis but, because the two $CH_3$ groups are eclipsed rather than staggered, there is no point exactly opposite every atom through a single center. Not centrosymmetric (only the staggered form, $D_{3d}$, would qualify).
$\mathrm{CH_4}$: tetrahedral, point group $T_d$. A regular tetrahedron has no inversion center (inverting one vertex through the middle does not land on another vertex). Not centrosymmetric.

Since trans-planar $\mathrm{H_2O_2}$ is the only centrosymmetric species in the list, it is the only one where the IR active modes are guaranteed to be completely Raman silent and the Raman active modes completely IR silent.

Let's summarize:

Mutual exclusion is a shortcut test for a center of inversion, nothing more.
Bent, eclipsed, and tetrahedral shapes here all fail to have that center; only the flat, trans arrangement of $H_2O_2$ has it.

The correct option is (B), the trans-planar conformer of $\mathrm{H_2O_2}$.

Among the following, the one for which the infrared active vibrational modes are Raman inactive and vice versa is

Show Hint

The Correct Option is B

Solution and Explanation

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