$Co = 6$;
$O. N. = x + 5 \times (0) + 1 \times (-2) + 1\times (-1) = 0$
$\therefore $ $x = + 3$ ; electronic configuration of
$Co^{3+}[Ar]\, 3d^64s^0$ hence number of $d$ electrons is $6$. All $d-$ electrons are paired due to strong ligand hence unpaired electron is zero.
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$2N{H_3}(g)\xrightarrow[{Pt\,catalyst}]{{1130\,K}}{N_2}(g) + 3{H_2}(g)$
Choose the option which is not true with respect to the reaction, from the choices given below :-