Simulation · Chemistry · Class 12

Counting unpaired electrons with a magnet

From the lesson Magnetic properties in The d- and f-Block Elements. Change the values and watch what happens.

The idea behind it

NCERT §4.3.9

A diamagnetic substance is pushed out of a magnetic field; a paramagnetic one is pulled in. A ferromagnetic substance is pulled in very strongly, and ferromagnetism is an extreme form of paramagnetism.
Paramagnetism comes from unpaired electrons: each has a magnetic moment from its spin and its orbital motion. In first-row compounds the orbital part is effectively quenched, so only spin counts.
Spin-only formula: μ = √[n(n + 2)] BM, where n is the number of unpaired electrons and BM is the Bohr magneton. One unpaired electron gives 1.73 BM.
The moment rises with n: n = 1, 2, 3, 4, 5 give 1.73, 2.83, 3.87, 4.90 and 5.92 BM (√8 = 2.83; NCERT's Table 4.7 prints 2.84 for the n = 2 ions Ti²⁺ and Ni²⁺). A measured moment therefore tells you how many electrons are unpaired.
Worked case: a divalent ion of Z = 25 is Mn²⁺, 3d⁵, with five unpaired electrons, so μ = √35 = 5.92 BM. For Z = 27, Co²⁺ is 3d⁷ with three unpaired electrons: μ = √15 = 3.87 BM.
d⁰ and d¹⁰ ions (Sc³⁺, Ti⁴⁺, Zn²⁺) have no unpaired electrons and are diamagnetic.
Measured moments of hydrated ions match the spin-only values closely for Ti³⁺ to Mn²⁺ (Mn²⁺: 5.92 calculated, 5.96 observed) but run higher for Fe²⁺ (5.3-5.5 against 4.90), Co²⁺ (4.4-5.2 against 3.87) and Cu²⁺ (1.8-2.2 against 1.73).

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