Question

“Compounds of transition elements are paramagnetic and coloured.” Explain.

Hint:

Unpaired d-electrons = colour + magnetism. Full d$^{10}$ configuration → colourless and diamagnetic.

Answer 1

Step 1: Paramagnetic nature.
The compounds of transition elements show paramagnetism due to the presence of one or more unpaired electrons in their (n−1)d orbitals.
The magnetic moment ($\mu$) of such compounds is given by: \[ \mu = \sqrt{n(n+2)} \text{ Bohr Magneton (B.M.)} \] where \( n \) = number of unpaired electrons.
Step 2: Explanation.
- When an element or its ion has unpaired electrons, it is attracted by a magnetic field (paramagnetic).
- If all electrons are paired, the substance becomes diamagnetic.
Step 3: Examples.
- Fe$^{2+}$: [Ar] 3d$^6$ → 4 unpaired electrons → paramagnetic.
- Zn$^{2+}$: [Ar] 3d$^{10}$ → all paired → diamagnetic.
Step 4: Coloured nature.
Transition metal compounds are coloured due to the presence of incompletely filled d-orbitals.
When light falls on these compounds, electrons in the d-orbitals absorb specific wavelengths of light and jump from lower energy levels to higher ones (known as d–d transitions). The remaining transmitted light imparts colour.
Step 5: Examples.
- [Ti(H$_2$O)$_6$]$^{3+}$ is purple due to d–d transition.
- [Cu(H$_2$O)$_6$]$^{2+}$ is blue due to the absorption of red light.
Step 6: Conclusion.
Hence, the presence of unpaired d-electrons makes transition metal compounds both paramagnetic and coloured.