JEE Main 2026 Apr 6 Shift 1, Chemistry Q21: Bohr's Atomic Model

JEE Main2026Apr 6, Shift 1Chemistry

First and second ionization enthalpies of lithium are $520$ kJ mol $^{- 1}$ and $7297$ kJ mol $^{- 1}$ respectively. Energy required to convert $3.5$ mg lithium $(g)$ into $Li^{2 +} (g)$ [ $Li (g) \to Li^{2 +} (g)$ ] is _____ kJ mol $^{- 1}$ . (nearest integer)

[Molar mass of Li $= 7$ g mol $^{- 1}$ ]

Your answer

Solution

Converting Li $(g) \to$ Li $^{2 +} (g)$ requires removal of two electrons, so per mole of Li the energy needed is $IE_{1} + IE_{2} = 520 + 7297 = 7817$ kJ.

Moles of Li in $3.5$ mg $= \frac{3.5 \times 1 0 ^{- 3}}{7} = 5 \times 1 0^{- 4}$ mol.

Total energy $= 5 \times 1 0^{- 4} \times 7817 = 3.9085$ kJ $\approx 4$ kJ.

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JEE Main 2026 Apr 6 Shift 1, Chemistry Q21: Bohr's Atomic Model

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