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Zinc has the lowest enthalpy of atomisation in the first transition series because:
AIt has the smallest atomic radius
BIts $3d^{10}\,4s^2$ configuration has no unpaired d-electrons available for metallic bonding
CIt is the most metallic of the series
DIt forms strong covalent bonds with itself
Answer & Solution
Correct answer: B. Its $3d^{10}\,4s^2$ configuration has no unpaired d-electrons available for metallic bonding
Strength of metallic bonding in transition metals is governed largely by the number of unpaired d-electrons that can participate in delocalised bonding. Zn has the configuration $[\mathrm{Ar}]3d^{10}\,4s^2$ — all d-electrons paired and unavailable. Only the 4s electrons contribute to metallic bonding, giving Zn the weakest interatomic bonding of the first series and hence the lowest enthalpy of atomisation ($\approx 126\,\text{kJ mol}^{-1}$).
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