Question

Explain, giving reasons:
i) Transition metals and most of their compounds are paramagnetic.
ii) Transition metals generally form coloured compounds.

Hint:

Unpaired d electrons cause paramagnetism (moment \(\mu = \sqrt{n(n+2)}\) BM); d-d transitions in partly filled d orbitals cause colour. Empty or full d (Sc³⁺, Zn²⁺) are colourless and diamagnetic.

Answer 1

Step 1: Both properties trace back to one feature: partly filled (n-1)d orbitals.
Step 2 (magnetism): A substance is drawn into a magnetic field, that is paramagnetic, when it owns unpaired electrons. Across the transition series the d orbitals fill one electron at a time, so ions such as Ti³⁺ (d¹), Mn²⁺ (d⁵) and Fe²⁺ (d⁶) all keep unpaired electrons and are paramagnetic; the measured moment matches \(\mu = \sqrt{n(n+2)}\) BM. Only the d⁰ and d¹⁰ ends of the series (Sc³⁺, Zn²⁺) are diamagnetic.
Step 3 (colour): When ligands surround the metal ion, the d orbitals no longer stay equal in energy; they divide into a lower t_2g set and an upper e_g set separated by an energy gap that matches the energy of visible light.
Step 4: A d electron absorbs a photon of visible light and is promoted across this gap in a d-d transition. The light that is not absorbed, its complementary colour, reaches our eye and gives the compound its colour. If the d orbitals are empty or completely full, no such jump is possible, so those compounds appear white or colourless.