= Solution
The angular factors of the nickel $3d$ <atomic orbitals> transform like quadratic Cartesian functions. In the <D4h molecular point group>, the assignments are
$$
\boxed{\begin{array}{c|c}
\text{orbital}&D_{4h}\\\hline
d_{z^2}&A_{1g}\\
d_{x^2-y^2}&B_{1g}\\
d_{xy}&B_{2g}\\
(d_{xz},d_{yz})&E_g
\end{array}}
$$
All five are even under <inversion symmetry>, hence the $g$ suffixes. The pair $(d_{xz},d_{yz})$ is mixed by a quarter-turn and spans a two-dimensional <irreducible representation>.
In the <Td molecular point group>, the corresponding result is
$$
\boxed{(d_{z^2},d_{x^2-y^2})\sim E,\qquad
(d_{xy},d_{xz},d_{yz})\sim T_2.}
$$
There is no inversion label because $T_d$ has no <inversion centre>. In both cases the dimensions add to five.
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