[Cr(H_2O)_6]Cl_3 (atomic number of Cr = 24) has a magnetic moment of…

The correct answer is Option (4) → 3d¹_xy, 3d¹_yz, 3d¹_xz

Work out the electrons of chromium in the complex:

    Cr: Z = 24; beyond the argon core: 24 − 18 = 6 electrons
    Cr atom: [Ar] 3d⁵ 4s¹
    In \( \ce{[Cr(H2O)6]Cl3} \) chromium is +3 (six neutral \( \ce{H2O} \), three \( \ce{Cl^{-}} \) outside): remove the 4s electron and two 3d electrons
    \( \ce{Cr^{3+}} \): [Ar] 3d³

In an octahedral field the d orbitals split into the lower t_2g set (d_xy, d_yz, d_xz) and the higher e_g set (d_x²−y², d_z²). The three electrons go singly into the three t_2g orbitals (Hund’s rule):

    t_2g: [↑] [↑] [↑]     e_g: [  ] [  ]

Three unpaired electrons give μ = √[n(n + 2)] = √15 = 3.87 BM, which agrees with the measured 3.83 BM. The distribution is d_xy¹ d_yz¹ d_xz¹.

Why the other options are incorrect

• Option (1) d_xy, d_yz, d_z² — Incorrect. d_z² is a higher-energy e_g orbital, which stays empty while a t_2g orbital is still free.

• Option (2) d_x²−y², d_z², d_xz — Incorrect. Two electrons are placed in the e_g set, which would be far less stable.

• Option (3) d_xy, d_x²−y², d_yz — Incorrect. d_x²−y² is also an e_g orbital and is not occupied in \( \ce{Cr^{3+}} \).

NCERT (2026-27 edition):
Crystal Field Theory – “… Let us assign electrons in the d orbitals of metal ion in octahedral coordination entities. Obviously, the single d electron occupies one of the lower energy t_2g orbitals. In d² and d³ coordination entities, the d electrons occupy the t_2g orbitals singly in accordance with the Hund’s rule.”