Question

i) Write the electronic configuration of an element with atomic number 24.
ii) Find the number of unpaired electrons in the Co²⁺ ion.
iii) Why is the range of oxidation states in the actinoid series greater than in the lanthanoid series?

Hint:

Cr uses half-filled stability (3d5 4s1); Co2+ is 3d7 with 3 unpaired electrons; actinoids have close 5f, 6d, 7s energies allowing more oxidation states than the tightly held 4f lanthanoids.

Answer 1

Part i) Element with Z = 24 (Chromium).
Step 1: The expected filling would be \([Ar]\,3d^4\,4s^2\).
Step 2: But a half-filled \(3d^5\) set and a half-filled \(4s^1\) set give extra stability (exchange energy). So one 4s electron shifts into 3d.
Step 3: Correct configuration:
\[ Cr\ (Z=24): [Ar]\,3d^5\,4s^1 \]

Part ii) Unpaired electrons in Co²⁺.
Step 1: Cobalt has \(Z = 27\): \([Ar]\,3d^7\,4s^2\).
Step 2: To make \(Co^{2+}\), remove 2 electrons; the 4s electrons are lost first, giving \([Ar]\,3d^7\).
Step 3: Fill 5 d-orbitals with 7 electrons by Hund's rule: five orbitals get one electron each (5 unpaired), then 2 orbitals get a second electron (pairing up). That leaves \(5 - 2 = 3\) unpaired electrons.
\[\boxed{Co^{2+}:\ 3d^7,\ \text{number of unpaired electrons} = 3}\]

Part iii) Wider oxidation-state range in actinoids.
Step 1: In actinoids the \(5f, 6d\) and \(7s\) sub-shells lie very close in energy, so all these electrons can take part in bonding.
Step 2: Because more electrons are available for bonding, actinoids show many oxidation states (for example +3 up to +7 for elements like uranium and neptunium).
Step 3: In lanthanoids the \(4f\) electrons are buried deep inside and are tightly held (poor participation in bonding), so mostly only the +3 state (with a little +2 and +4) is seen. Hence actinoids have a much greater range of oxidation states than lanthanoids.