Verified Solution • 2025–26 Session moderate

QJEEM21Ch40ae: JEE Main 2021 Chemistry - Class 11 JEEM21Ch40ae Solution: The first ionization energy of magnesium is smaller as compared …

The first ionization energy of magnesium is smaller as compared to that of elements X and Y, but higher than that of Z. The elements X, Y and Z, respectively, are :

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:

IE(Mg)\mathrm{IE}(\mathrm{Mg}) is smaller than IE(X)\mathrm{IE}(X) and IE(Y)\mathrm{IE}(Y) .

IE(Mg)\mathrm{IE}(\mathrm{Mg}) is greater than IE(Z)\mathrm{IE}(Z) .

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 kJ mol−1\mathrm{kJ\,mol^{-1}} ) of the relevant elements:

IE(Ne)≈2080\mathrm{IE}(\mathrm{Ne}) \approx 2080

IE(Na)≈496\mathrm{IE}(\mathrm{Na}) \approx 496

IE(Mg)≈738\mathrm{IE}(\mathrm{Mg}) \approx 738

IE(Al)≈578\mathrm{IE}(\mathrm{Al}) \approx 578 (not listed but for reference)

IE(Si)≈787\mathrm{IE}(\mathrm{Si}) \approx 787 (not listed but for reference)

IE(P)≈1012\mathrm{IE}(\mathrm{P}) \approx 1012 (not listed but for reference)

IE(S)≈1000\mathrm{IE}(\mathrm{S}) \approx 1000 (not listed but for reference)

IE(Cl)≈1251\mathrm{IE}(\mathrm{Cl}) \approx 1251

IE(Ar)≈1521\mathrm{IE}(\mathrm{Ar}) \approx 1521

IE(Li)≈520\mathrm{IE}(\mathrm{Li}) \approx 520

We test each option:

(A) neon, sodium, chlorine

X=NeX=\mathrm{Ne} : \mathrm{IE}(\mathrm{Ne})=2080 > 738 ✓

Y=NaY=\mathrm{Na} : \mathrm{IE}(\mathrm{Na})=496 &lt; 738 ✗ (contradicts “ Mg\mathrm{Mg} < YY ”)

Z=ClZ=\mathrm{Cl} : \mathrm{IE}(\mathrm{Cl})=1251 &gt; 738 ✗ (contradicts “ Mg\mathrm{Mg} > ZZ ”)

So (A) does not fit the requirement.

(B) argon, chlorine, sodium

X=ArX=\mathrm{Ar} : \mathrm{IE}(\mathrm{Ar})=1521 &gt; 738 ✓

Y=ClY=\mathrm{Cl} : \mathrm{IE}(\mathrm{Cl})=1251 &gt; 738 ✓

Z=NaZ=\mathrm{Na} : \mathrm{IE}(\mathrm{Na})=496 &lt; 738 ✓

Thus \mathrm{IE}(\mathrm{Mg}) &lt; \mathrm{IE}(\mathrm{Ar}), \,\mathrm{IE}(\mathrm{Cl}) and \mathrm{IE}(\mathrm{Mg}) &gt; \mathrm{IE}(\mathrm{Na}) .

This matches perfectly.

(C) chlorine, lithium, sodium

X=ClX=\mathrm{Cl} : \mathrm{IE}(\mathrm{Cl})=1251 &gt; 738 ✓

Y=LiY=\mathrm{Li} : \mathrm{IE}(\mathrm{Li})=520 &lt; 738 ✗ (contradicts “ Mg\mathrm{Mg} < YY ”)

Z=NaZ=\mathrm{Na} : \mathrm{IE}(\mathrm{Na})=496 &lt; 738 , but the mismatch in YY is enough to discard it.

(D) argon, lithium, sodium

X=ArX=\mathrm{Ar} : \mathrm{IE}(\mathrm{Ar})=1521 &gt; 738 ✓

Y=LiY=\mathrm{Li} : \mathrm{IE}(\mathrm{Li})=520 &lt; 738 ✗ (contradicts “ Mg\mathrm{Mg} < YY ”)

Z=NaZ=\mathrm{Na} : \mathrm{IE}(\mathrm{Na})=496 &lt; 738 is fine for ZZ , but YY fails.

Hence only (B) satisfies all conditions:

(B) argon, chlorine, and sodium.\boxed{\text{(B) argon, chlorine, and sodium.}}

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