Step 1: Electronic configuration of Ni and $Ni^{2+}$.
Ni (Z=28): $[Ar]3d^84s^2$. $Ni^{2+}$ (remove two $4s$ electrons): $[Ar]3d^8$. The $3d^8$ configuration has 8 electrons in 5 $d$-orbitals; by Hund's rule, 2 remain unpaired in the absence of strong ligand field.
Step 2: Part (i): Hybridisation of $[NiCl_4]^{2-}$ (weak field ligand $Cl^-$).
$Cl^-$ is a weak field ligand and does NOT force electron pairing. $Ni^{2+}$ retains $3d^8$ with 2 unpaired electrons. The complex uses outer orbitals: one $4s$ + three $4p$ = $sp^3$ hybridisation, giving a tetrahedral geometry.
Step 3: Part (i): Hybridisation of $[Ni(CN)_4]^{2-}$ (strong field ligand $CN^-$).
$CN^-$ is a strong field ligand and forces electron pairing. In $3d^8$, the two unpaired electrons pair up in the lower $d$-orbitals, freeing one $3d$ orbital. Ni now uses: one inner $3d$ + one $4s$ + two $4p$ = $dsp^2$ hybridisation, giving a square planar geometry.
Step 4: Part (ii): Inner vs. outer orbital complex.
$[NiCl_4]^{2-}$: Uses outer $4s$ and $4p$ orbitals for bonding ($sp^3$). This is an outer orbital complex. $[Ni(CN)_4]^{2-}$: Uses an inner $3d$ orbital for bonding ($dsp^2$). This is an inner orbital complex.
Step 5: Part (iii): Compare magnetic properties.
$[NiCl_4]^{2-}$: 2 unpaired electrons (weak field, no pairing). Magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{2 \times 4} = \sqrt{8} \approx 2.83$ BM. This complex is PARAMAGNETIC (attracted to magnetic field). $[Ni(CN)_4]^{2-}$: 0 unpaired electrons (strong field forces all 8 $3d$ electrons to pair). This complex is DIAMAGNETIC (all electrons paired, weakly repelled by magnetic field).
Step 6: Summarise all three answers.
(i) $[NiCl_4]^{2-}$: $sp^3$; $[Ni(CN)_4]^{2-}$: $dsp^2$. (ii) $[NiCl_4]^{2-}$: outer orbital; $[Ni(CN)_4]^{2-}$: inner orbital. (iii) $[NiCl_4]^{2-}$: paramagnetic ($\mu \approx 2.83$ BM); $[Ni(CN)_4]^{2-}$: diamagnetic ($\mu = 0$). \[ \boxed{[NiCl_4]^{2-}: sp^3,\;\text{outer, paramagnetic};\;[Ni(CN)_4]^{2-}: dsp^2,\;\text{inner, diamagnetic}} \]