= Solution
There are four internal C–N stretching coordinates. A stretch length is unchanged in sign when a <molecular symmetry operation> carries a bond to an equivalent bond, so these coordinates have the same permutation <group representation> as the four ligand sigma functions.
For the square,
$$
\Gamma_{\rm CN}=A_{1g}+B_{1g}+E_u.
$$
Only $E_u\sim(x,y)$ has <vibrational infrared activity>. The $A_{1g}$ and $B_{1g}$ stretches have <Raman vibrational activity>. Thus the ideal square gives \b[one IR feature and two Raman features, with no common fundamental]. The two components of $E_u$ have one frequency, not two different lines.
For the tetrahedron,
$$
\Gamma_{\rm CN}=A_1+T_2.
$$
The $T_2$ stretch is IR active, while both $A_1$ and $T_2$ are Raman active. Thus the tetrahedron also gives \b[one IR feature and two Raman features], but \b[one feature, $T_2$, is shared between the spectra]. Its three components likewise describe three independent displacements at one frequency.
\b[The separate line counts alone do not distinguish the geometries. Comparing their frequencies does:] the tetrahedral IR C–N feature coincides with one Raman C–N feature, whereas the square has no symmetry-enforced coincidence. This assumes resolved fundamentals and excludes accidental equality of unrelated frequencies.
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