$\left[P t\left(N H_{3}\right)_{5} C l\right] C l_{3}$ on ionization gives 4 ions in solution i.e $3 C l^{-}+\left[P t\left(N H_{3}\right)_{5} C l\right]^{+}$
$(I)\, [(Ph_3P)_2PdCl_2PdCl_2]$ $(II)\,[NiBrCl(en)]$
$(III)\, Na_4 [Fe(CN)_5 NOS]$ $(IV)\, Cr(CO)_3(NO)_2$
$(I)\,-\,(II)\,-\,,(III)\,-\,(IV)$