$\mathrm{Pt}^{2+} \rightarrow \mathrm{dsp}^2$ hybridization and have no unpaired e-s.
$\therefore$ Magnetic moment $\mathrm{o} 0$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$M(s) \to M(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\, ........(1)$
$M(s) \to M^{2+} (g) + 2e^-\,\,\,\,\,\,\,\,.......(2)$
$M(g) \to M^+(g) + e^-\,\,\,\,\,\,\,\,\,\,\,.........(3)$
$M^+ (g) \to M^{2+} (g) + e^-\,\,\,\,\,\,\,\,\,.........(4)$
$M(g) \to M^{2+} (g) +2e^-\,\,\,\,\,\,\,\,\,\,\,..........(5)$
The second ionization energy of $M$ could be calculated from the energy values assoclated with
$SO_3(g) \rightleftharpoons SO_2(g)+ \frac{1}{2} O_2(g)$
is $K_c= 4.9 \times 10^{-2}.$ The value of $K_c$ for the reaction
$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$
will be