Question

Using valence bond approach, predict the shape and magnetism (paramagnetic or diamagnetic) of [Ni(CN)₄]⁻.

Answer 1

Oxidation State of Nickel:

• The cyanide (CN⁻) ligand is a monodentate ligand with a charge of −1.
• The overall charge on the complex is −1, so for [Ni(CN)₄]⁻, the charge on nickel must be +2.
• Therefore, Nickel (Ni) is in the +2 oxidation state, i.e., Ni²⁺.

Electron Configuration of Ni²⁺:

• Nickel has an atomic number of 28, so the electron configuration of neutral Ni is:
  Ni : [Ar] 3d⁸ 4s²
• For Ni²⁺, it loses two electrons (the two 4s electrons):
  Ni²⁺ : [Ar] 3d⁸
• The electron configuration of Ni²⁺ is 3d⁸.

Hybridisation and Geometry:

• In [Ni(CN)₄]⁻ there are four cyanide ligands surrounding the central nickel ion.
• The coordination number of nickel is 4, and the ligands arrange themselves symmetrically around the central metal ion.
• Since cyanide is a strong field ligand (it causes pairing of electrons), the hybridisation of Ni²⁺ is sp³ (four hybrid orbitals are formed from one s-orbital and three p-orbitals).
• The geometry of the complex is therefore square planar, which is typical for d⁸ metal ions in a low-spin complex (due to strong-field ligands like cyanide).

Magnetic Behavior:

• The Ni²⁺ ion has 8 electrons in its 3d-orbitals (3d⁸).
• In a square planar geometry, the d-orbitals split into two energy levels:
  The lower-energy d_x²−y²​ orbital.
  The higher-energy d_z²​, d_xy​, d_xz​, and d_yz​ orbitals.
• The strong field ligands (cyanide) cause pairing of the electrons in the lower-energy orbitals.
• Since all the 3d-electrons are paired, there are no unpaired electrons in the complex.

Therefore, the complex [Ni(CN)₄]⁻ is diamagnetic because there are no unpaired electrons.