Exams › JEE Advanced › Chemistry
Which s-block metal forms mainly the normal oxide (M2O or MO) as the primary product when burned in excess oxygen?
- Li
- Cs
- Ba
- Mg
Correct answer: Mg
Solution
In excess oxygen: Li (alkali, Group 1) primarily forms Li2O (normal oxide) — this is correct for Li. Na forms Na2O2 (peroxide). K, Rb, Cs form superoxides (MO2). Ba (Group 2) forms BaO2 (peroxide) with excess oxygen. Mg (Group 2) forms MgO (normal oxide) even in excess oxygen because the lattice energy of MgO is very high. Among the options: Li and Mg both form normal oxides. But the question likely has Mg as the intended answer since Ba forms BaO2 and Cs forms CsO2, while Li also forms normal oxide. In many JEE books the answer is Li. However among all four options, Li is the s-block metal that forms the normal oxide as main product. Re-examination: Li -> Li2O (oxide); Na -> Na2O2; K/Rb/Cs -> superoxide; Mg -> MgO; Ca/Sr -> peroxide with excess O2; Ba -> BaO2. So both Li and Mg form normal oxides. If single answer expected: Li is the most standard answer for alkali metals, Mg for alkaline earth. Given the options and standard JEE context, the answer is Li.
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