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Answer the following · Q13

Q.Ionization enthalpy of Li is 520 kJ mol⁻¹ while that of F is 1681 kJ mol⁻¹. Explain.

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✓ Free question

Step 1. Locate the two elements. Li is period 2, group 1 (1 valence electron, 2s12s^1); F is period 2, group 17 (7 valence electrons, 2s22p52s^22p^5) — both in the same period, near opposite ends.

Step 2. Recall the across-period IE trend. Ionization enthalpy increases across a period because screening from core electrons stays essentially constant while effective nuclear charge ZeffZ_{eff} climbs steadily (section 7.5.2.3).

Step 3. Apply it to Li vs F. Li, at the far left, has the lowest ZeffZ_{eff} felt by its single valence electron of any period-2 element — that electron is easily removed. F, near the far right, has one of the highest ZeffZ_{eff} values in period 2, and its atom is also small (radius 64 pm, Table 7.2), so its outer electrons are held unusually tightly.

Step 4. Match the numbers. This matches Table 7.4 exactly: Li = 520 kJ mol⁻¹ (lowest in period 2 after considering the trend) and F = 1681 kJ mol⁻¹ (one of the highest, just below the inert gas Ne at 2080).

✓Final answer

Li's low IE (520 kJ mol⁻¹) reflects its single, loosely-held, weakly-shielded valence electron; F's high IE (1681 kJ mol⁻¹) reflects its small radius and high effective nuclear charge, holding its near-complete octet tightly — both following the standard across-period IE trend.

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