= Solution
{c}
Use the indicated out-of-plane displacements and the equivalence of the four <oxygen> sites and of the two <carbon> sites in the isolated planar <oxalate> ion. In $V_1$, the two <oxygen> atoms on one side move oppositely to those on the other side, and the two <carbon> atoms also move oppositely. Their dipole changes cancel. The mode is even under molecular <inversion symmetry>, which independently forbids an electric-dipole transition.
In $V_2$, all four <oxygen> atoms move together out of the plane while both <carbon> atoms move in the opposite direction. The positive and negative charge distributions acquire a relative out-of-plane displacement, producing a nonzero derivative of the molecular <electric dipole moment>.
In $V_3$, two <oxygen> atoms move out and two move in, in a checkerboard arrangement, while the <carbon> atoms are stationary. The equivalent oxygen contributions cancel. This mode illustrates that being odd under <inversion symmetry> is necessary but not sufficient for <vibrational infrared activity>: its full symmetry does not transform like any component of an <electric dipole moment>.
$$
\boxed{V_2\text{ is IR active};\qquad V_1,V_3\text{ are IR inactive}.}
$$
These conclusions concern the isolated-molecule displacements drawn in the source; a different crystal environment can alter selection rules through additional symmetry breaking.
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