Question

Transition metals, with a few exceptions, are extremely hard and less volatile. Explain.

Hint:

Unpaired d-electrons give strong metallic bonding; the soft, volatile exceptions (Zn, Cd, Hg) have a filled d¹⁰ shell.

Answer 1

Step 1: Key idea. Hardness and low volatility of a metal both depend on the strength of the metallic bonding that holds the atoms together in the solid.
Step 2: Why the bonding is strong. Transition metals have both (n-1)d and ns electrons available for bonding. Because the d-orbitals contribute many unpaired electrons to the metallic bond, a large number of electrons take part in delocalised bonding. This gives very strong interatomic (metallic) bonds.
Step 3: Consequences. Strong metallic bonds are hard to deform, so the metals are very hard. Breaking these strong bonds to form vapour needs a lot of energy, so the metals have high melting and boiling points and are therefore less volatile.
Step 4: The exceptions. Metals such as Zn, Cd and Hg have a completely filled d-subshell \( (d^{10}) \), so their d-electrons do not take part in metallic bonding. Their bonding is weak; hence they are soft and comparatively volatile (mercury is even a liquid at room temperature). \[ \boxed{\text{Strong metallic bonding from unpaired d-electrons} \rightarrow \text{hard, less volatile}} \]