Periodic Table & Periodicity

We are told:

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

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

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

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

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

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

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

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

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

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

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

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

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

We test each option:

(A) neon, sodium, chlorine

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

$Y=\mathrm{Na}$: $\mathrm{IE}(\mathrm{Na})=496 < 738$ ✗ (contradicts “$\mathrm{Mg}$ < $Y$”)

$Z=\mathrm{Cl}$: $\mathrm{IE}(\mathrm{Cl})=1251 > 738$ ✗ (contradicts “$\mathrm{Mg}$ > $Z$”)

So (A) does not fit the requirement.

(B) argon, chlorine, sodium

$X=\mathrm{Ar}$: $\mathrm{IE}(\mathrm{Ar})=1521 > 738$ ✓

$Y=\mathrm{Cl}$: $\mathrm{IE}(\mathrm{Cl})=1251 > 738$ ✓

$Z=\mathrm{Na}$: $\mathrm{IE}(\mathrm{Na})=496 < 738$ ✓

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

$X=\mathrm{Cl}$: $\mathrm{IE}(\mathrm{Cl})=1251 > 738$ ✓

$Y=\mathrm{Li}$: $\mathrm{IE}(\mathrm{Li})=520 < 738$ ✗ (contradicts “$\mathrm{Mg}$ < $Y$”)

$Z=\mathrm{Na}$: $\mathrm{IE}(\mathrm{Na})=496 < 738$, but the mismatch in $Y$ is enough to discard it.

(D) argon, lithium, sodium

$X=\mathrm{Ar}$: $\mathrm{IE}(\mathrm{Ar})=1521 > 738$ ✓

$Y=\mathrm{Li}$: $\mathrm{IE}(\mathrm{Li})=520 < 738$ ✗ (contradicts “$\mathrm{Mg}$ < $Y$”)

$Z=\mathrm{Na}$: $\mathrm{IE}(\mathrm{Na})=496 < 738$ is fine for $Z$, but $Y$ fails.

Hence only (B) satisfies all conditions:

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

The ionic radii (in $\mathop A\limits^o $) of $\text{Mg}^{2+}$ and $\text{Al}^{3+}$, respectively, are:

Option B:

$0.72 \mathop A\limits^o$ for $\text{Mg}^{2+}$

$0.54 \mathop A\limits^o$ for $\text{Al}^{3+}$

Among halogens (F₂, Cl₂, Br₂ and I₂), bond dissociation enthalpy ($\Delta$_dissH$^\circ$) of I₂, is minimum because of larger size of I-atom there is a steric repulsion between bonded I-atoms, which makes I-I bond weakest.

Whereas, smaller size and highest electronegativity of F-atom cause highest electron density on F-atom of F₂ molecule. As a result, F-F bond becomes weaker due to electrostatic repulsion between bonded F-atoms.

Thus, the order of $\Delta$_dissH$^\circ$ (in kJ mol^$-$1) is

$\mathop {Cl - Cl}\limits_{(242.6)} > \mathop {Br - Br}\limits_{(192.3)} > \mathop {F - F}\limits_{(158.8)\,Electrostatic\,repulsion} > \mathop {I - I}\limits_{(151.1)\,Steric\,repulsion} $

On moving left to right in a period of the periodic table, Ionisation energy (I.E.) increases due to increase in effective nuclear charge i.e. Z_eff.

But due to extra stability of fully filled and half-filled electronic configuration of Mg and P required ionisation enthalpy is more from neighbouring elements.

i.e. First ionisation enthalpy order is Al < Mg < Si < S < P.

Across a period (left to right) in the periodic table :

Ionization enthalpy tends to increase

Electron gain enthalpy (in magnitude) tends to increase (more negative values)

Electronegativity tends to increase

Atomic radius tends to decrease

Hence, atomic radius shows the opposite trend compared to the other three properties. Therefore, the correct answer is

(D) Atomic radius.