= Solution
For $n$ <unpaired electrons>, the <spin-only magnetic moment> is
$$
\boxed{\mu_{\rm eff}=\sqrt{n(n+2)}\,\mu_{\rm B}
=2\sqrt{S(S+1)}\,\mu_{\rm B},}
$$
where $\mu_{\rm B}$ is the <Bohr magneton> and the second expression uses $g\simeq2$.
For the square, a large separation of the highest $B_{1g}$ level favours pairing in the lower four <molecular orbitals>: \b[$S=0$, $n=0$, $\mu_{\rm eff}=0$]. If the cost of occupying $B_{1g}$ were smaller than the relevant <electron pairing energy>, a <high-spin complex> with two <unpaired electrons> and $S=1$ could instead occur, giving \b[$\sqrt8\,\mu_{\rm B}\simeq2.83\,\mu_{\rm B}$].
For ideal tetrahedral $d^8$, the occupation is $e^4t_2^4$. <Hund's rules> distribute four <electrons> over the three degenerate $t_2$ <molecular orbitals> as one pair and two single occupations. Thus \b[$S=1$ and $\mu_{\rm eff}\simeq2.83\,\mu_{\rm B}$]. Merely increasing the $e$–$t_2$ splitting does not create a conventional paired tetrahedral $d^8$ alternative: four <electrons> must still occupy $t_2$.
\b[The absence of <paramagnetism> therefore favours a square-planar, low-spin $d^8$ complex.]
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