Chemical Bonding JEE 2026: 6 Paramagnetic Complexes Numerical

How many paramagnetic complexes are there in the JEE Main 8 April 2026 chemical bonding question?

Six of the ten coordination complexes are paramagnetic. The numerical answer is 6.

The question from the 8 April 2026 evening slot JEE Main paper, a hard-tagged question worth 4 marks, requires classification of each species: [MnBr4]2-, [NiCl4]2-, [Ni(CN)4]2-, Ni(CO)4, [CoF6]3-, [Fe(CN)6]4-, [Mn(CN)6]3-, [Ti(CN)6]3-, [Cu(H2O)6]2+, [Co(C2O4)3]3-. The task is to report the number that are paramagnetic as a numerical answer.

What is the official line-by-line solution for each complex in this chemical bonding jee 2026 numerical?

The official method gives a total of six paramagnetic complexes. It follows one rigid sequence for every complex: oxidation state, d-electron count, geometry, ligand field strength, spin state, and unpaired electron count.

• [MnBr4]2-: Mn is +2 (d5), tetrahedral, Br- weak field, high-spin d5, five unpaired electrons, paramagnetic.
• [NiCl4]2-: Ni is +2 (d8), tetrahedral, Cl- weak field, high-spin d8, two unpaired electrons, paramagnetic.
• [Ni(CN)4]2-: Ni is +2 (d8), square planar, CN- strong field, low-spin d8, zero unpaired electrons, diamagnetic.
• Ni(CO)4: Ni is 0 (d10), tetrahedral, CO strong field, all electrons paired, diamagnetic.
• [CoF6]3-: Co is +3 (d6), octahedral, F- weak field, high-spin d6, four unpaired electrons, paramagnetic.
• [Fe(CN)6]4-: Fe is +2 (d6), octahedral, CN- strong field, low-spin t2g6, zero unpaired electrons, diamagnetic.
• [Mn(CN)6]3-: Mn is +3 (d4), octahedral, CN- strong field, low-spin t2g4, two unpaired electrons, paramagnetic.
• [Ti(CN)6]3-: Ti is +3 (d1), octahedral, one unpaired electron, paramagnetic.
• [Cu(H2O)6]2+: Cu is +2 (d9), octahedral, one unpaired electron, paramagnetic.
• [Co(C2O4)3]3-: Co is +3 (d6), octahedral, oxalate strong for Co(III), low-spin t2g6, zero unpaired electrons, diamagnetic.

The six paramagnetic complexes are [MnBr4]2- (high-spin d5), [NiCl4]2- (high-spin d8), [CoF6]3- (high-spin d6), [Mn(CN)6]3- (low-spin d4 with two unpaired), [Ti(CN)6]3- (d1), [Cu(H2O)6]2+ (d9). The four diamagnetic species are [Ni(CN)4]2-, Ni(CO)4, [Fe(CN)6]4- and [Co(C2O4)3]3-. Final total is 6.

What single method mistake turns the correct count of 6 into 5 or 7?

One precise procedural error shifts the count: breaking the fixed decision order or misplacing borderline ligands on the spectrochemical series.

Assigning strong-field character to F- makes [CoF6]3- low-spin d6 with zero unpaired electrons and drops the total to 5. Assigning weak-field character to oxalate makes [Co(C2O4)3]3- high-spin d6 with four unpaired electrons and raises the total to 7. Skipping the geometry check for Ni(II) species and treating [NiCl4]2- as square planar instead of tetrahedral also yields 5 by wrongly pairing its electrons.

These errors share one root: the decision sequence was not applied uniformly to all ten species. Stick to the order and the count stays 6.

What four-step decision tree eliminates errors in chemical bonding jee 2026 coordination numericals?

The four-step decision tree is oxidation state first, standard coordination geometry second, ligand position in the spectrochemical series third, and high-spin versus low-spin configuration fourth. Run every complex through this sequence in under two minutes and the count matches the official total of 6.

1. Calculate oxidation state of the central metal from the overall charge and ligand charges.
2. Assign standard geometry: tetrahedral for four-coordinate halides, square planar for Ni2+ with strong ligands, octahedral for six-coordinate.
3. Place the ligand in the spectrochemical series: Br-, Cl-, F- weak; oxalate, H2O intermediate but strong for Co(III); CN-, CO strong.
4. For d4–d8 octahedral and tetrahedral cases, compare pairing energy versus Δ. Weak field (small Δ) gives high-spin with maximum unpaired electrons. Strong field (large Δ) gives low-spin with maximum pairing. Tetrahedral cases are almost always high-spin.

Ligands in this question sort cleanly once geometry is fixed:

• Br- (weak, tet)
• CN- (strong)
• CO (strong)
• oxalate (strong for Co3+)

Which two related hard questions from chemical bonding test the same oxidation state and spin logic?

Question 1: Find the number of unpaired electrons in [CrF6]3- and [Cr(CN)6]3-.

Both are octahedral Cr3+ (d3). Field strength does not matter for d3: electrons occupy t2g orbitals singly in either case. Each complex has three unpaired electrons.

Question 2: Predict whether [Fe(H2O)6]2+ and [Fe(CN)6]4- are paramagnetic or diamagnetic and give their spin-only magnetic moments.

[Fe(H2O)6]2+ is Fe2+ (d6), H2O weak field, high-spin t2g4 eg2, four unpaired electrons, paramagnetic.

$\sqrt{4(4+2)}=\sqrt{24}\approx 4.9$ BM.

[Fe(CN)6]4- is Fe2+ (d6), CN- strong field, low-spin t2g6, zero unpaired electrons, diamagnetic. Spin-only magnetic moment is 0 BM.

These items reinforce the same four-step sequence.

How should you prepare this lesson for JEE Main Session 2 2026?

You should expect to solve this difficulty in 120 seconds once the decision tree is internalised. Session 2 2026 is likely to open from 7 April 2026 onward.

Practise the four-step decision tree on every coordination complex until classification becomes automatic. Search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free). Start with the adjacent chapter practice at redox reactions and electrochemistry JEE 2013 full solution and run a mock test error audit at [/blog/jee-main-mock-test-analysis-a-7-step-error-audit/].