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Why is [Ni(CN)₄]²⁻ square planar and diamagnetic while [NiCl₄]²⁻ is tetrahedral and paramagnetic, when both contain Ni²⁺? Valence bond theory (VBT) answers this with a short procedure. This post gives that procedure, works it for six complexes from NCERT, and shows where the theory stops working.
Where does VBT sit in the syllabus?
The rationalised NCERT Class 12 chapter on Coordination Compounds (Reprint 2026-27) covers VBT in Section 5.5.1, magnetic properties in 5.5.2 and its limitations in 5.5.3. The JEE (Main) 2026 syllabus lists the "valence bond approach" in Unit 12, and the JEE (Advanced) 2026 brochure asks for hybridisation and geometries (linear, tetrahedral, square planar and octahedral), VBT and spin-only magnetic moments. Check the syllabus for your exam year.
The approach goes back to Linus Pauling, whose 1931 paper on the nature of the chemical bond applied quantum mechanics and a theory of paramagnetic susceptibility to the structure of molecules. He received the 1954 Nobel Prize in Chemistry "for his research into the nature of the chemical bond".
What is the five-step method?
Find the oxidation state of the metal from the charges on the ligands and the complex.
Write the d-electron count of the metal ion, for example Co³⁺ is 3d⁶ and Ni²⁺ is 3d⁸.
Use the magnetic data to place the electrons. If the complex has fewer unpaired electrons than the free ion, electrons have paired up to empty inner d orbitals.
Choose the hybridisation that matches the coordination number and the empty orbitals. NCERT's Table 5.2 gives sp³ (tetrahedral), dsp² (square planar), sp³d (trigonal bipyramidal), and sp³d² or d²sp³ (octahedral).
State the magnetic moment with the spin-only formula μ = √[n(n + 2)] BM, where n is the number of unpaired electrons. The formula and values are in the NCERT d- and f-block chapter.
Worked examples: six complexes from NCERT
[Co(NH₃)₆]³⁺. NH₃ is neutral, so cobalt is +3: 3d⁶. The complex is diamagnetic, so all six d electrons are paired in three 3d orbitals, leaving two 3d orbitals empty. Hybridisation d²sp³, octahedral, inner orbital, μ = 0.
[CoF₆]³⁻. Six F⁻ give −6, so cobalt is +3 again: 3d⁶. This complex has four unpaired electrons, so the 3d orbitals are not emptied and the outer 4d orbitals are used. Hybridisation sp³d², octahedral, outer orbital. μ = √(4 × 6) = √24 ≈ 4.90 BM.
[NiCl₄]²⁻. Four Cl⁻ give −4, so nickel is +2: 3d⁸. It has two unpaired electrons, so the 3d set is left alone and 4s and 4p are used. Hybridisation sp³, tetrahedral. μ = √(2 × 4) = √8 ≈ 2.83 BM.
[Ni(CN)₄]²⁻. Nickel is +2: 3d⁸. The complex is diamagnetic, so the eight electrons pair up in four 3d orbitals, freeing one 3d orbital. Hybridisation dsp², square planar, μ = 0. This answers the opening question.
[Fe(CN)₆]³⁻ and [FeF₆]³⁻. Iron is +3 in both: 3d⁵. NCERT reports one unpaired electron for the cyanide complex (d²sp³, μ = √3 ≈ 1.73 BM) and five for the fluoride complex (sp³d², μ = √35 ≈ 5.92 BM).
Running it backwards. NCERT's Example 5.7 gives [MnBr₄]²⁻ a spin-only moment of 5.9 BM. Mn²⁺ is 3d⁵; 5.9 BM means five unpaired electrons, so no 3d orbital was emptied and dsp² is impossible. The complex is tetrahedral, sp³.
What can VBT not do?
NCERT lists its limitations. VBT involves a number of assumptions, gives no quantitative interpretation of magnetic data, does not explain the colour of complexes, does not explain thermodynamic or kinetic stability quantitatively, does not make exact predictions about tetrahedral versus square planar four-coordinate complexes, and does not distinguish between weak and strong ligands.
That last point matters in practice. In the examples above, VBT did not tell us that CN⁻ forces pairing and F⁻ does not; we used the measured magnetism. Crystal field theory, the next section of the chapter, explains that through ligand field strength. Our post on coordination compound naming and crystal field basics covers that side.
Which mistakes cost marks?
Wrong oxidation state. Count ligand charges first; CN⁻, Cl⁻ and F⁻ are −1, NH₃ and H₂O are 0.
Writing sp³d² and d²sp³ as the same thing. Both are octahedral, but the order tells you outer versus inner orbitals.
Assuming every four-coordinate complex is tetrahedral. Check the magnetic data or the d count.
Using μ = n(n + 2) without the square root.
A practice routine
Day 1: Draw orbital boxes for the six complexes above from memory.
Day 2: Calculate μ for n = 1 to 5 and keep the list: 1.73, 2.83, 3.87, 4.90 and 5.92 BM.
Day 3: Solve the NCERT exercises on [Ni(CN)₄]²⁻ and on [Co(NH₃)₆]³⁺ versus [Ni(NH₃)₆]²⁺.
Day 4: Five "given μ, find the geometry" questions, like Example 5.7.
If orbital boxes still feel abstract, Fermi's AI tutor works out what you already know and builds a plan for the session, then draws the figure as it teaches.
Tonight, predict the hybridisation and magnetic moment of [Ni(CO)₄] and check your answer against NCERT: tetrahedral and diamagnetic.
