Which of the following compounds is both paramagnetic and coloured?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → VOSO4
A compound is paramagnetic only if its metal ion has unpaired electrons, and a d–d colour needs a partly filled d subshell. A d0 or d10 ion has neither.
\( \ce{(NH4)2[TiCl6]} \): 2(+1) + x + 6(−1) = 0, so Ti is +4; Ti (Z = 22) is [Ar]3d24s2, so \( \ce{Ti^4+} \) is 3d0: no unpaired electron, so it is diamagnetic.
\( \ce{VOSO4} \): the ions are \( \ce{VO^2+} \) and \( \ce{SO4^2-} \); x + (−2) = +2, so V is +4; V (Z = 23) is [Ar]3d34s2, so \( \ce{V^4+} \) is 3d1: one unpaired electron, so it is paramagnetic, and the d electron gives a blue colour.
\( \ce{K2Cr2O7} \): 2(+1) + 2x + 7(−2) = 0, so Cr is +6; Cr (Z = 24) is [Ar]3d54s1, so \( \ce{Cr^6+} \) is 3d0: the orange crystals are coloured but have no unpaired electron, so they are diamagnetic.
\( \ce{K3[Cu(CN)4]} \): 3(+1) + x + 4(−1) = 0, so Cu is +1; Cu (Z = 29) is [Ar]3d104s1, so \( \ce{Cu^+} \) is 3d10: all electrons paired, diamagnetic and colourless.
Only VOSO4 is both paramagnetic and coloured.
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.”
4.3.10 Formation of Coloured Ions – “When an electron from a lower energy d orbital is excited to a higher energy d orbital, the energy of excitation corresponds to the frequency of light absorbed (Unit 5). This frequency generally lies in the visible region. … In aqueous solutions where water molecules are the ligands, the colours of the ions observed are listed in Table 4.8.”
Table 4.8: Colours of Some of the First Row (aquated) Transition Metal Ions
