For high spin $d^4$ octahedral complex,
therefore, Crystal field stabilisation energy
$=(-3 \times 0.4+1 \times 0.6) \Delta_{0}$
$=-(-1.2 \times 0.6) \Delta_{0}$
$=-0.6\Delta_{0}$
$(I)\, [CrCl_2(NO_2)_2(NH_3)_2]^-$ $(II)\, [Co(NO_2)_3(NH_3)_3]$
$(III)\, [PtCl(NO_2)(NH_3 )(py)]$ $(IV)\, [PtBrCl(en)]$
$(I)\,\,\,-\,\,\,(II)\,\,\,-\,\,\,(III)\,\,\,-\,\,\,(IV)$
(પ.ક્ર.: $Mn\, = 25, Co\, = 27, Ni\, = 28, Zn\, = 30$)