Question 26 of 34advanced🔍 AnalyzeLong Answer5 marks

Explain on the basis of valence bond theory that [Ni(CN)4]2- ion with square planar structure is diamagnetic and the [NiCl4]2- ion with tetrahedral geometry is paramagnetic.

Correct Answer

[Ni(CN)₄]²⁻ (Square Planar and Diamagnetic):

In [Ni(CN)₄]²⁻, nickel is in +2 oxidation state with electronic configuration 3d⁸. CN⁻ is a strong ligand which causes the 3d electrons to pair up, leaving one d orbital empty. The hybridisation involved is dsp², forming a square planar structure. Each of the hybridised orbitals receives a pair of electrons from a cyanide ion. Since all electrons are paired with no unpaired electron, the complex is diamagnetic.

[NiCl₄]²⁻ (Tetrahedral and Paramagnetic):

In [NiCl₄]²⁻, nickel is in +2 oxidation state with electronic configuration 3d⁸. Cl⁻ is a weak ligand which does not cause pairing of 3d electrons. The hybridisation involved is sp³, forming a tetrahedral geometry. Each Cl⁻ ion donates a pair of electrons to the hybrid orbitals. The compound is paramagnetic since it contains two unpaired electrons in the 3d orbitals.

Key Difference:

The difference arises due to the nature of ligands. CN⁻ being a strong ligand causes electron pairing and uses inner d orbital (inner orbital complex), while Cl⁻ being a weak ligand does not pair electrons and uses outer orbitals (outer orbital complex).

Exercise: Intext Questions | Q: 5.5 | (Chapter: Page 18)
For More Understanding

Explanation

The answer explains both complexes using valence bond theory as per the textbook context. For [Ni(CN)₄]²⁻, the strong CN⁻ ligand causes pairing of 3d⁸ electrons, enabling dsp² hybridisation (square planar) with no unpaired electrons (diamagnetic). For [NiCl₄]²⁻, the weak Cl⁻ ligand does not pair the 3d⁸ electrons, resulting in sp³ hybridisation (tetrahedral) with two unpaired electrons (paramagnetic). The context clearly distinguishes between inner orbital complexes (strong ligands) and outer orbital complexes (weak ligands).

Solution Steps

  1. Step 1: Identify oxidation state of Ni in both complexes as +2 with electronic configuration 3d⁸

  2. Step 2: For [Ni(CN)₄]²⁻ - CN⁻ is strong ligand, causes electron pairing, dsp² hybridisation, square planar, no unpaired electrons → diamagnetic

  3. Step 3: For [NiCl₄]²⁻ - Cl⁻ is weak ligand, no electron pairing, sp³ hybridisation, tetrahedral, two unpaired electrons → paramagnetic

  4. Step 4: Conclude that ligand strength determines hybridisation and magnetic properties