Chemistry·MCQ Practice

Lattice Enthalpy — MCQ Practice

NEET UG
Updated 13 Sep 2026

Interactive MCQ Practice

Test your knowledge. Click “Solve” to reveal options, select your answer, then check the result. 6 questions available.

Q1medium

Given the following standard enthalpy changes for the formation of potassium iodide (KI): Enthalpy of sublimation of K(s) = +89 kJ/mol+89\,\text{kJ/mol} Ionization enthalpy of K(g) = +419 kJ/mol+419\,\text{kJ/mol} Dissociation enthalpy of I2(g)I_2(g) = +151 kJ/mol+151\,\text{kJ/mol} Electron gain enthalpy of I(g) = −295 kJ/mol-295\,\text{kJ/mol} Standard enthalpy of formation of KI(s) = −327 kJ/mol-327\,\text{kJ/mol} Calculate the lattice enthalpy of KI(s).

Q2medium

Which of the following ionic compounds is expected to have the highest magnitude of lattice enthalpy?

Q3easy

Which of the following statements about lattice enthalpy is INCORRECT?

Q4medium

Arrange the following ionic compounds in increasing order of their lattice enthalpy magnitude: NaCl, CsCl, LiCl, KCl.

Q5easy

Which of the following processes represents the lattice enthalpy of magnesium oxide (MgO)?

Q6hard

Given the following data, calculate the second electron gain enthalpy of oxygen (O−(g)+e−→O2−(g)O^-(g) + e^- \rightarrow O^{2-}(g)): ΔHf∘(MgO(s))=−602 kJ/mol\Delta H_f^\circ (MgO(s)) = -602\,\text{kJ/mol} ΔHsub(Mg(s))=+148 kJ/mol\Delta H_{sub} (Mg(s)) = +148\,\text{kJ/mol} IE1(Mg(g))=+738 kJ/molIE_1 (Mg(g)) = +738\,\text{kJ/mol} IE2(Mg(g))=+1451 kJ/molIE_2 (Mg(g)) = +1451\,\text{kJ/mol} ΔHdiss(O2(g))=+498 kJ/mol\Delta H_{diss} (O_2(g)) = +498\,\text{kJ/mol} EA1(O(g))=−141 kJ/molEA_1 (O(g)) = -141\,\text{kJ/mol} ΔHlattice(MgO(s))=−3791 kJ/mol\Delta H_{lattice} (MgO(s)) = -3791\,\text{kJ/mol}