QJEEM21Ch40ae: JEE Main 2021 Chemistry - Class 11 JEEM21Ch40ae Solution: The first ionization energy of magnesium is smaller as compared …
Multiple Choice Options
- A neon, sodium and chlorine
- B argon, chlorine and sodium
- C chlorine, lithium and sodium
- D argon, lithium and sodium
Step-by-Step Verified Solution
We are told:
is smaller than and .
is greater than .
Hence we want:
\mathrm{IE}(\mathrm{Mg}) < \mathrm{IE}(X), \quad \mathrm{IE}(\mathrm{Mg}) < \mathrm{IE}(Y), \quad \mathrm{IE}(\mathrm{Mg}) > \mathrm{IE}(Z).
Approximate first ionization energies (in ) of the relevant elements:
(not listed but for reference)
(not listed but for reference)
(not listed but for reference)
(not listed but for reference)
We test each option:
(A) neon, sodium, chlorine
: \mathrm{IE}(\mathrm{Ne})=2080 > 738 ✓
: \mathrm{IE}(\mathrm{Na})=496 < 738 ✗ (contradicts “ < ”)
: \mathrm{IE}(\mathrm{Cl})=1251 > 738 ✗ (contradicts “ > ”)
So (A) does not fit the requirement.
(B) argon, chlorine, sodium
: \mathrm{IE}(\mathrm{Ar})=1521 > 738 ✓
: \mathrm{IE}(\mathrm{Cl})=1251 > 738 ✓
: \mathrm{IE}(\mathrm{Na})=496 < 738 ✓
Thus \mathrm{IE}(\mathrm{Mg}) < \mathrm{IE}(\mathrm{Ar}), \,\mathrm{IE}(\mathrm{Cl}) and \mathrm{IE}(\mathrm{Mg}) > \mathrm{IE}(\mathrm{Na}) .
This matches perfectly.
(C) chlorine, lithium, sodium
: \mathrm{IE}(\mathrm{Cl})=1251 > 738 ✓
: \mathrm{IE}(\mathrm{Li})=520 < 738 ✗ (contradicts “ < ”)
: \mathrm{IE}(\mathrm{Na})=496 < 738 , but the mismatch in is enough to discard it.
(D) argon, lithium, sodium
: \mathrm{IE}(\mathrm{Ar})=1521 > 738 ✓
: \mathrm{IE}(\mathrm{Li})=520 < 738 ✗ (contradicts “ < ”)
: \mathrm{IE}(\mathrm{Na})=496 < 738 is fine for , but fails.
Hence only (B) satisfies all conditions: