The spin-only magnetic moment of Cu2+ is 1.73 BM. The number of unpaired electrons in Cu2+ is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1
The spin-only formula links the magnetic moment μ (in BM) with the number of unpaired electrons n:
\( \displaystyle \mu=\sqrt{n(n+2)} \)
\( \displaystyle 1.73=\sqrt{n(n+2)}\quad\Rightarrow\quad n(n+2)=3 \)
\( \displaystyle n=1 \)
The configuration agrees:
Configuration: Cu (Z = 29) is [Ar]3d104s1.
Electrons removed: \( \ce{Cu^2+} \) loses the 4s electron and one 3d electron: 3d9.
Unpaired electrons: four 3d orbitals hold pairs and one holds a single electron, so there is 1 unpaired electron.
NCERT (2026-27 edition):
4.3.9 Magnetic Properties – “Paramagnetism arises from the presence of unpaired electrons, each such electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum. … For these, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the ‘spin-only’ formula, i.e., \( \mu=\sqrt{n(n+2)} \) where n is the number of unpaired electrons and μ is the magnetic moment in units of Bohr magneton (BM). A single unpaired electron has a magnetic moment of 1.73 Bohr magnetons (BM).”