Chemistry · Structure Of Atom · NEET
'n' is the number of UNPAIRED electrons only. It is NOT the total number of electrons and NOT the atomic number. First write the electron configuration, then count how many boxes (orbitals) have a single electron sitting alone. That count is n.
Step 1: Write the electronic configuration of the atom or ion. Step 2: For ions, remove electrons from the outermost shell first (for transition metals, remove 4s before 3d). Step 3: Fill the last subshell using Hund's rule (one electron in each orbital first, before pairing). Step 4: Count the orbitals that have just one electron. Example: Fe²⁺ is 3d⁶ → the d orbitals hold 6 electrons in 5 boxes, so 4 boxes are alone and 1 box is paired → n = 4.
Always remove the electrons from the shell with the highest principal quantum number (n) first. For 3d transition metals, this means you remove the 4s electrons BEFORE the 3d electrons. So Ti (3d² 4s²) becomes Ti²⁺ = 3d² (the two 4s electrons leave first). This is a very common NEET trap.
Each unpaired electron spins and acts like a tiny magnet. When electrons are paired, their spins point in opposite directions and cancel out, giving zero magnetism. Only unpaired (lonely) electrons add up, so more unpaired electrons = larger μ = more paramagnetic.
Paramagnetic means the species HAS unpaired electrons (n ≥ 1), so μ is greater than zero and it is attracted by a magnet. Diamagnetic means ALL electrons are paired (n = 0), so μ = 0 and it is weakly pushed away. If a question asks for a diamagnetic ion, the answer is the one with zero unpaired electrons.
BM stands for Bohr Magneton. It is just the unit for magnetic moment, like 'metre' is a unit for length. You do not need to calculate its value in NEET; you only report your final answer in BM. One BM equals eh/4πm, but that number is not asked.
The calculated 'spin-only' magnetic moment of Ti²⁺ (3d²) is:
Two electrons occupying the same orbital are distinguished by:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
μ = √n(n+2) Bohr Magnetons (BM), where n is the number of unpaired electrons. It ignores the orbital contribution and uses only electron spin, which is a good approximation for most first-row transition metal ions in NEET.
n=1 → √3 = 1.73 BM; n=2 → √8 = 2.84 BM; n=3 → √15 = 3.87 BM; n=4 → √24 = 4.90 BM; n=5 → √35 = 5.92 BM. Memorising these five values saves time in the exam.
Yes. Reverse the formula. If μ = 3.87 BM, then √n(n+2) = 3.87, so n(n+2) = 15, which gives n = 3 unpaired electrons. NEET sometimes gives μ and asks for n.
Yes. A diamagnetic ion has all electrons paired, so n = 0 and μ = √0 = 0 BM. Examples include Zn²⁺ (3d¹⁰) and Sc³⁺ (3d⁰).
For 3d transition metal ions the orbital angular momentum is largely 'quenched' (cancelled) by the surroundings, so only spin matters. This makes the spin-only value very close to experiment, which is why NEET uses it.
