Step 1: Frame the question. Both properties trace back to how tightly the metal atoms are bonded in the lattice; stronger bonding means greater hardness and higher boiling point (lower volatility).
Step 2: Electron count argument. A transition atom offers its ns electrons plus a variable number of unpaired (n-1)d electrons to the sea of delocalised electrons. More bonding electrons per atom means a denser, stronger metallic bond.
Step 3: Effect on properties. Such strong bonds resist scratching and deformation (high hardness) and require large energy to pull atoms into the gas phase, so the metals boil at high temperature and vaporise with difficulty (low volatility).
Step 4: Where it fails. In Zn, Cd and Hg the d-shell is full (\(d^{10}\)); these paired d-electrons stay out of the bonding, leaving only weak ns-based bonds, which is why this group is soft and relatively volatile. \[ \boxed{\text{Unpaired d-electrons} \Rightarrow \text{strong bonds} \Rightarrow \text{hard and non-volatile}} \]