1. We need to find the arrangement that is NOT in accordance with the stated periodic property.

2. Option A: $\text{Li} < \text{Na} < \text{K} < \text{Rb}$ (Increasing metallic radius)

• Down Group 1, atomic/metallic radius increases due to addition of shells.
• Therefore, $\text{Li} < \text{Na} < \text{K} < \text{Rb}$ is correct.

So, Option A follows the property correctly.

3. Option B: $\text{I} < \text{Br} < \text{F} < \text{Cl}$ (Increasing electron gain enthalpy with negative sign)

• Electron gain enthalpy becomes more negative across a period.
• Among halogens, the order of electron gain enthalpy (more negative value) is: $\text{Cl} > \text{F} > \text{Br} > \text{I}$ if we compare magnitude with negative sign.
• So in increasing order of negativity: $\text{I} < \text{Br} < \text{F} < \text{Cl}$ which is correct.

So, Option B is correct.

4. Option C: $\text{B} < \text{C} < \text{N} < \text{O}$ (Increasing first ionization enthalpy)

Let us check the actual trend in Period 2:

• Generally, ionization enthalpy increases across a period.
• But there is an exception: $\text{O} < \text{N}$ because nitrogen has extra stability due to half-filled $2p^3$ configuration.

Thus actual order is: $\text{B} < \text{C} < \text{O} < \text{N}$ not $\text{B} < \text{C} < \text{N} < \text{O}$

So, Option C is NOT correct.

5. Option D: $Al^{3+} < Mg^{2+} < Na^+ < F^-$ (Increasing ionic size)

These ions are isoelectronic; each has 10 electrons.

• In an isoelectronic series, ionic radius decreases with increasing nuclear charge.
• Nuclear charges are:
  • $Al^{3+}: Z=13$
  • $Mg^{2+}: Z=12$
  • $Na^+: Z=11$
  • $F^-: Z=9$

Higher nuclear charge means smaller ion. Therefore increasing ionic size is: $Al^{3+} < Mg^{2+} < Na^+ < F^-$ which is correct.

So, Option D is correct.

6. Therefore, the only arrangement that is not according to the indicated property is: $\boxed{C}$

7. Comparison with stored correct answer

Stored correct answer: $C$

My derived answer is also $C$, so I agree with the stored answer.