Solution

(I) $$[Cr(H_2O)_6]Br_2$$ $$[Cr(H_2O]_6]^{2+} + 2Br^-$$ So, here we know $$H_2O$$ is neutral ligand So, $$ x + 6\times H_2O = +2$$ (zero is charge on $$H_2O$$) $$x + 6\times O = +2$$ $$x = +2$$ $$Cr^{+2} [Ar]3d^4$$ [Ref. image 1] Formula for spin only magnetic moment is, $$\mu= \sqrt{n(n+2)}$$ $$\mu = \sqrt{4(4+2)} = \sqrt{24}$$ (II) $$Na_4[Fe(CN)_6]$$ $$[Fe(CN)_6]^{4-} + 4Na^+$$ (on dissociation) $$CN$$ has $$-1$$ charge so, $$x+6\times CN = -4$$ $$x+ 6\times (-1) = -4$$ $$x= -4 + 6 = +2$$ $$Fe^{2+}[Ar]3d^6$$ [Ref. image 2] But here in this case $$CN$$ is there which is strong field so pairing of $$e^-$$ takes place as. [Ref. image 3] $$\mu= \sqrt{n(n+2)} = 0$$ No unpaired $$e^-$$ so, dimagnetic (III) $$Na_3[Fe(C_2O_4)_3] (\Delta_0 > P)$$ On dissociation it gives $$[Fe(C_2O_4)_3]^{3-}+3Na^+$$ $$C_2O_4$$ that is oxaiate has $$-2$$ chnage $$(^{\circleddash }O-\overset{O}{\overset{||}{C}}-\overset{O}{\overset{||}{C}}-O^{\circleddash })$$ $$Fe + 3(C_2O_4) = -3$$ $$Fe+3(-2) = -3$$ $$Fe = -3 + 6 = +3$$ Here we have given $$\Delta_0 > P $$ means distance between $$t_2g$$ & $$e_g$$ is more than pairing energy. $$\mu= \sqrt {1(1+2)} = \sqrt {3}$$ So, here pairing occur because less energy required for pairing. [Ref. image 4] (IV) $$(Et_4N)_2[CoCl_4]$$ $$(NEt_4) $$ has +1 charge so, $$[CoCl_4]^{2-}$$ $$Cl$$ has $$-1$$ charge $$x + 4\times Cl = -2$$ $$x = -2+ 4 = +2$$ $$x + 4\times (-1) = -2$$ $$Co^{2+} [Ar]3d^7$$ unpaired $$e^-$$= $$3$$ $$\mu = \sqrt{n(n+1)} = \sqrt{3(3+2)} = \sqrt{15}$$ [Ref. image 5]