A critical study of the magnetic data of coordination compounds of metals of the first transition series reveals some complications. For metal ions with up to three electrons in the d orbitals, like Ti3+(d1); V3+(d2); Cr3+(d3); two vacant d orbitals are available for octahedral hybridization with 4s and 4p orbitals. The magnetic behaviour of these free ions and their coordination entities is similar. When more than three 3d electrons are present, the required pair of 3d orbitals for octahedral hybridization is not directly available (as a consequence of Hund’s rule). Thus, for d4 (Cr2+, Mn3+), d5 (Mn2+, Fe3+), d6 (Fe2+, Co3+) cases, a vacant pair of d orbitals results only by pairing of 3d electrons which leaves two, one and zero unpaired electrons, respectively. The magnetic data agree with maximum spin pairing in many cases, especially with coordination compounds containing d6 ions. However, with species containing d4 and d5 ions there are complications.
What is the spin-only magnetic moment of the complex [Cr(NH3)4Cl2]Br?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 3.87 BM
The spin-only magnetic moment depends only on the number of unpaired electrons, n, in the metal ion.
Charge on the complex ion: one \( \ce{Br^-} \) ion lies outside the bracket, so the complex ion is \( \ce{[Cr(NH3)4Cl2]^+} \).
Oxidation state of Cr: \( \ce{NH3} \) is neutral and each \( \ce{Cl^-} \) is −1, so x + 4(0) + 2(−1) = +1 and x = +3.
Configuration: Cr (Z = 24) is [Ar]3d54s1; losing the 4s electron and two 3d electrons gives \( \ce{Cr^3+} \): 3d3.
Unpaired electrons: the three d electrons occupy three d orbitals singly, so n = 3. With only three d electrons no pairing is needed, whatever the ligands.
Magnetic moment:
\( \displaystyle \mu=\sqrt{n(n+2)} \)
\( \displaystyle \mu=\sqrt{3(3+2)}=\sqrt{15} \)
\( \displaystyle \mu=3.87\,\mathrm{BM} \)
NCERT (2026-27 edition):
4.3.9 Magnetic Properties – “… For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and hence is of no significance. 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).”