[Ni(CN)4]2- hybridisation

Key Ideas to Understand

1. Oxidation State of Ni
   • In [Ni(CN)₄]²⁻, each CN⁻ carries −1 charge.
   • Let oxidation state of Ni = x.
     $x + 4(-1) = -2 \Rightarrow x = +2$
2. Electronic Configuration
   • Ground state Ni: $[Ar]\,3d^8\,4s^2$.
   • Ni²⁺ loses the two 4s electrons: $[Ar]\,3d^8$.
3. Strong-Field Ligand Effect
   • CN⁻ is high in the spectrochemical series.
   • It causes pairing of the 3d electrons to minimise repulsion.
4. Availability of Orbitals
   • After pairing, one 3d orbital ($d_{x^2-y^2}$) becomes empty, while the remaining 3d orbitals are doubly occupied.
5. Hybridisation Choice
   • For a coordination number (CN) = 4, two possibilities exist:
     • sp³ → tetrahedral (weak-field ligands).
     • dsp² → square planar (strong-field ligands).
   • Because CN⁻ is strong, energy cost of pairing < crystal field stabilisation, so Ni adopts dsp² hybridisation.
6. Resulting Shape
   • Hybrid orbitals (d, s, p_x, p_y) point to the corners of a square → square planar geometry.

Hence, Nickel(II) in [Ni(CN)₄]²⁻ is dsp² hybridised and the complex is square planar.

Imagine Lego Blocks and Parking Slots

• Nickel (Ni) ion is like a parking garage with 10 special parking slots called orbitals.
• In the garage, 8 cars (electrons) are already parked in 3 rows called 3d.
• Four friends (the cyanide CN⁻ ligands) want to park around Nickel.
• Because CN⁻ is a bossy (strong‐field) friend, it tells some cars to double-park so that an empty VIP slot appears.
• Once a free slot shows up, Nickel rearranges 4 slots into one single VIP floor: d, s, p_x, p_y. These 4 mix to form a square-shaped parking area (called dsp² hybridisation).
• The four CN⁻ friends park at the corners of this square. That’s why the whole structure is square planar.

So, the hybridisation of Ni in [Ni(CN)₄]²⁻ is dsp².

Step-by-Step Solution

1. Find Oxidation State of Ni
   $x + 4(-1) = -2 \implies x = +2$
2. Write Electronic Configuration of Ni²⁺
   $\text{Ni: }[Ar]\,3d^8\,4s^2 \rightarrow \text{Ni}^{2+}: [Ar]\,3d^8$
3. Account for Strong Field Ligand (CN⁻)
   • CN⁻ causes pairing of the 3d electrons, giving:
     $\text{Paired configuration: } (3d)^8 = t_{{2g}}^6 e_g^2$
   • One $d_{x^2-y^2}$ orbital becomes vacant.
4. Choose Hybridisation for CN = 4
   • Vacant $d_{x^2-y^2}$, one 4s, and two 4p (p_x, p_y) orbitals mix: $d_{x^2-y^2} + s + p_x + p_y \longrightarrow \text{dsp}^2$
5. Describe Geometry
   • The four dsp² hybrid orbitals orient at 90° in a single plane → square planar.

[ \boxed{\text{Hybridisation of Ni in }[Ni(CN)_4]^{2-} = \text{dsp}^2\text{ (square planar)}} ]