Question

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

• A. \(\mathrm{H_2O}\)
• B. trans-planar conformer of \(\mathrm{H_2O_2}\)
• C. eclipsed conformer of ethane
• D. \(\mathrm{CH_4}\)

Hint:

The rule of mutual exclusion (IR active modes are Raman silent and vice versa) works only for molecules with a center of inversion; check which option has an $i$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept.
The behavior described, every IR active mode being Raman inactive and every Raman active mode being IR inactive, is called the rule of mutual exclusion. This rule applies only to molecules that have a center of inversion (a center of symmetry, \(i\)). In a centrosymmetric molecule, a vibration either keeps its sign under inversion (symmetric, \(g\)) or changes sign (antisymmetric, \(u\)); IR activity needs a mode that transforms like \(x, y, z\) (always \(u\)), while Raman activity needs a mode that transforms like a quadratic function such as \(x^2, xy\) (always \(g\)), so no single mode can ever be both.

Step 2: Check option (A), \(\mathrm{H_2O}\).
Water is bent, point group \(C_{2v}\), which has no center of inversion. All three vibrational modes of water (symmetric stretch, bend, asymmetric stretch) are both IR and Raman active, so it does not obey mutual exclusion.

Step 3: Check option (B), trans-planar \(\mathrm{H_2O_2}\).
If \(H_2O_2\) is forced into the fully planar, trans (anti) arrangement, with both \(O-H\) bonds and the \(O-O\) bond all in one plane and the two \(H\) atoms on opposite sides of the \(O-O\) bond, the molecule has a \(C_2\) axis, a horizontal mirror plane (the molecular plane itself), and a center of inversion at the midpoint of the \(O-O\) bond. This combination of elements is the point group \(C_{2h}\), which does have a center of symmetry \(i\).
Because it is centrosymmetric, every vibration of trans-planar \(H_2O_2\) splits cleanly into \(A_g\)/\(B_g\) modes (Raman active, IR inactive) and \(A_u\)/\(B_u\) modes (IR active, Raman inactive), so it obeys mutual exclusion exactly.

Step 4: Check options (C) and (D).
The eclipsed conformer of ethane has point group \(D_{3h}\); this has a \(\sigma_h\) but no center of inversion (it is the staggered conformer, \(D_{3d}\), that has \(i\), not the eclipsed one), so mutual exclusion does not apply to eclipsed ethane. \(CH_4\) is tetrahedral, point group \(T_d\), which also has no center of inversion, so it too does not follow mutual exclusion (its IR and Raman active modes overlap for some vibrations).

Final Answer:
Only the trans-planar conformer of \(H_2O_2\) has a center of symmetry, so it is the one that shows perfect mutual exclusion between IR and Raman activity. \[ \boxed{\text{(B) trans-planar } \mathrm{H_2O_2}} \]