The first, second and third successive molar ionization enthalpies of a gaseous element M are IP1, IP2 and X, respectively. Here X0 has units kJ/mol. Given IP1/IP2=1/2 and the total enthalpy for M(g)

Question

The first, second and third successive molar ionization enthalpies of a gaseous element M are IP1, IP2 and X, respectively. Here X>0 has units kJ/mol. Given IP1/IP2=1/2 and the total enthalpy for M(g)→M³⁺(g)+3e⁻ is 2X, find IP1 and IP2, respectively. All options are in kJ/mol.

TestHub · Accepted Answer

IP1+IP2+X=2X, so IP1+IP2=X. Since IP1/IP2=1/2, IP2=2IP1. Thus 3IP1=X, giving IP1=X/3 and IP2=2X/3 (both in kJ/mol). For X>0, X/3<2X/3<X, so the successive ionization enthalpies increase. Option C.

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JEE Advanced 2028 (Class 11)