Magnetic moment, $\mu=\sqrt{n(n+2)}$
Where, $n=$ number of unpaired electrons
The electronic configuration for the given ions is as follow:
For $M n^{2+}[Z=25]=1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2} 3 p^{6}, 3 d^{2} 4 s^{0}$
$\mathop {\boxed \uparrow \boxed \uparrow \boxed \uparrow \boxed \uparrow \boxed \uparrow }\limits^{3{d^2}} $
$\mu=\sqrt{5(5+2)}=\sqrt{35}$
For $C r^{2+}[Z=24]=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6}, 3 d^{4}, 4 s^{0}$
$\mathop {\boxed \uparrow \boxed \uparrow \boxed \uparrow \boxed \uparrow }\limits^{3{d^4}} $
$\mu=\sqrt{4(4+2)}=\sqrt{24}$
For $T i^{2+}[Z=22]=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6}, 3 d^{2}, 4 s^{0}$
$\mathop {\boxed \uparrow \boxed \uparrow \boxed \uparrow }\limits^{3{d^2}} $
$\mu=\sqrt{2(2+2)}=\sqrt{8}$
$\therefore$ correct order of spin only magnetic moment is
$M n^{2}+C r^{2+}>T i^{2+}$
$(I)\, K_4 [Fe(CN)_6]$ $(II)\, K_3[Cr(CN)_6]$ $(III)\, K_3 [Co(CN)_6]$ $(IV)\, K_2[Ni(CN)_4]$
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