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Example · Example 5

Q.Arrange Li, Na, K, Rb and Cs in order of increasing first ionization enthalpy and explain the trend.

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First ionization enthalpy is the energy needed to remove the single outermost electron from a gaseous atom. Descending Group 1, each successive element has one additional filled shell of electrons between the nucleus and the outermost electron. This extra shell (a) increases the average distance of the outer electron from the nucleus and (b) increases the shielding (screening) of the nuclear charge by the intervening electrons. Both effects reduce the effective pull the nucleus exerts on the outer electron, so it becomes progressively easier -- i.e. requires less energy -- to remove that electron on descending the group.

The measured values confirm this trend clearly (in kJ mol−1^{-1}): Li=520\text{Li} = 520, Na=496\text{Na} = 496, K=419\text{K} = 419, Rb=403\text{Rb} = 403, Cs=376\text{Cs} = 376. Arranged in order of increasing first ionization enthalpy, this gives Cs<Rb<K<Na<Li\text{Cs} < \text{Rb} < \text{K} < \text{Na} < \text{Li}. …

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