VBT: [NiCl₄]²⁻ (sp³), [Ni(CN)₄]²⁻ (dsp²), [Ni(CO)₄] (sp³)VBT: [NiCl₄]²⁻ (sp³), [Ni(CN)₄]²⁻ (dsp²), [Ni(CO)₄] (sp³)VBT: [NiCl₄]²⁻ (sp³), [Ni(CN)₄]²⁻ (dsp²), [Ni(CO)₄] (sp³)

NCERT §5.5.1 · Class 12 Chemistry · Chapter 5 · interactive diagram

Parts and what they mean

• Free Ni²⁺ ion: 3d⁸ (मुक्त Ni²⁺ आयन: 3d⁸) — Ni = [Ar]3d⁸4s²; Ni²⁺ = 3d⁸ with two unpaired electrons.
• [NiCl₄]²⁻: sp³, tetrahedral ([NiCl₄]²⁻: sp³, चतुष्फलकीय) — Cl⁻ is weak, so the two unpaired 3d electrons stay; 4s + three 4p → four sp³ hybrids for four Cl⁻ pairs.
• Paramagnetic (2 unpaired) (अनुचुंबकीय (2 अयुग्मित)) — μ = √(2×4) = √8 ≈ 2.83 BM.
• [Ni(CN)₄]²⁻: dsp², square planar ([Ni(CN)₄]²⁻: dsp², वर्ग समतली) — CN⁻ forces the two odd electrons to pair, freeing one 3d orbital; 3d + 4s + two 4p → four dsp² hybrids.
• Diamagnetic (प्रतिचुंबकीय) — No unpaired electron.
• [Ni(CO)₄]: sp³, tetrahedral ([Ni(CO)₄]: sp³, चतुष्फलकीय) — Ni is in the zero oxidation state; with CO the 4s electrons pair into 3d (3d¹⁰), so 4s + three 4p give four sp³ hybrids.
• Diamagnetic although tetrahedral (चतुष्फलकीय फिर भी प्रतिचुंबकीय) — NCERT intext 5.6: [NiCl₄]²⁻ is paramagnetic but [Ni(CO)₄] is diamagnetic, though both are tetrahedral.

Original diagram drawn by Simple Learning Hub, based on NCERT §5.5.1. Report a mistake