Using valence bond theory,explain the following in relation to the complexes given below: $[Mn(CN)_{6}]^{3-}, [Co(NH_{3})_{6}]^{3+}, [Cr(H_{2}O)_{6}]^{3+}, [FeCl_{6}]^{4-}$
$(i)$ Type of hybridisation.
$(ii)$ Inner or outer orbital complex.
$(iii)$ Magnetic behaviour.
$(iv)$ Spin only magnetic moment value.

(N/A) $(i)$ $[Mn(CN)_{6}]^{3-}$: $Mn^{3+}$ is $3d^{4}$. $CN^-$ is a strong field ligand,causing pairing. Hybridization: $d^{2}sp^{3}$. Inner orbital complex. Paramagnetic (two unpaired electrons). Magnetic moment,$\mu = \sqrt{2(2+2)} = 2.83 \ B.M.$
$(ii)$ $[Co(NH_{3})_{6}]^{3+}$: $Co^{3+}$ is $3d^{6}$. $NH_{3}$ is a strong field ligand,causing pairing. Hybridization: $d^{2}sp^{3}$. Inner orbital complex. Diamagnetic. Magnetic moment,$\mu = 0 \ B.M.$
$(iii)$ $[Cr(H_{2}O)_{6}]^{3+}$: $Cr^{3+}$ is $3d^{3}$. Hybridization: $d^{2}sp^{3}$. Inner orbital complex. Paramagnetic (three unpaired electrons). Magnetic moment,$\mu = \sqrt{3(3+2)} = 3.87 \ B.M.$
$(iv)$ $[FeCl_{6}]^{4-}$: $Fe^{2+}$ is $3d^{6}$. $Cl^-$ is a weak field ligand,no pairing. Hybridization: $sp^{3}d^{2}$. Outer orbital complex. Paramagnetic (four unpaired electrons). Magnetic moment,$\mu = \sqrt{4(4+2)} = 4.90 \ B.M.$