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The gross requirement is an anisotropic polarizability tensor, so that changing molecular orientation changes the induced response. A permanent electric dipole moment is not required; a homonuclear diatomic can therefore have a rotational Raman scattering spectrum.

For the inelastic lines of a linear rigid rotor, the detailed rule is $\Delta J=\pm 2$, with $J\ge 0$. The $\Delta J=0$ contribution is unshifted Rayleigh scattering, not a rotational Stokes Raman scattering line. Stokes Raman scattering has $\Delta J=+2$ because the molecule gains rotational energy.

Because $\Delta K=0$, the $(\tilde{C}-\tilde{B})K^2$ terms cancel between the initial and final states. The two positive-shift branches are consequentlyThe $+2$ branch gives shifts $6\tilde{B},10\tilde{B},14\tilde{B},\dots$, as for the diatomic. The $+1$ branch fills in a series spaced by $2\tilde{B}$, so the combined nonzero series is $4\tilde{B},6\tilde{B},8\tilde{B},10\tilde{B},\dots$.

There is a boundary qualification to the abbreviated rules printed in the question: for an ordinary symmetric polarizability tensor, $\Delta J=\pm 1$ has zero strength when $K=0$. Thus $J=0,K=0\to J=1,K=0$ does not produce a $2\tilde{B}$ line; the first $+1$ transition can start at $J=1,|K|=1$ and has shift $4\tilde{B}$. This follows from the rank-two angular-momentum coupling: the all-zero-projection coupling for $J$ and $J+1$ vanishes because $J+2+(J+1)$ is odd. If one applies only the printed inequalities without this matrix-element restriction, $2\tilde{B}$ is a formal position, but its intensity is zero.

The lower panel of the figure in part (a)(ii) marks the two branches and their overlaps. The useful spacing is $2\tilde{B}$, not the diatomic value $4\tilde{B}$.

The D4h molecular point group contains an inversion centre, so infrared and Raman mutual exclusion applies. A normal mode of $u$ parity can have vibrational infrared activity, whereas the quadratic polarizability tensor components responsible for Raman vibrational activity have $g$ parity. No individual fundamental can be active in both spectra.

The Td molecular point group lacks an inversion centre. In particular, $T_2$ normal modes transform both as electric-dipole components and as off-diagonal polarizability tensor components, so they occur in both spectra. Coincident IR and Raman fundamentals therefore favour the tetrahedral structure; symmetry requires no such coincidences for the square. Weak or unresolved features can conceal coincidences experimentally, while accidental equality of different frequencies is not a violation of mutual exclusion.

In this C2v point group convention, $z\sim A_1$, $x\sim B_1$, and $y\sim B_2$. Therefore all six normal modes have symmetry-allowed infrared fundamentals: $\boxed{6\text{ IR features}}$, absent accidental coincidences or vanishing intensities.

For Raman spectroscopy, examine the polarizability tensor: $x^2,y^2,z^2\sim A_1$, $xz\sim B_1$, and $yz\sim B_2$; $xy\sim A_2$ is also Raman-allowed but no $A_2$ vibration occurs here. All six vibrations are therefore Raman-active too, giving $\boxed{6\text{ Raman features and }6\text{ IR/Raman coincidences}}$. The centrosymmetric mutual-exclusion rule does not apply to $C_{2v}$ formaldehyde.