In which of the following options the order of arrangement does not agree with the variation of property indicated against it?

Ionisation enthalpy – B < C < O < N
In 2^nd period, first ionisation enthalpy increases with atomic number from left to right when the elements with half or fullyfilled electronic configuration have more  ionisation enthalpy.
“N“ ([He]2s² 2p³) has half-filled 2p orbital, so they acquire extra stability and thus have higher ionisation enthalphy than their next one.
Hence, correct order: B < C < O < N
(1) Electron gain enthalpy order: – I < Br < F < Cl
Generally, electron gain enthalpy increases with decrease in size, i.e., I<Br<Cl<F. As a result of the very small
size of fluorine, the inter-electronic repulsions are more, and fl uorine is reluctant to gain an electron. Correct
electron gain enthalpy order is I < Br < F < Cl.
Metallic radius – Li < Na < K < Rb As we move down the group, atomic number increases, and number of
shells increases. Thus, metallic radius increases - Li < Na < K < Rb.
Ionic size – Al⁺³ < Mg+2 < Na+ < F–
In case of isoelectronic species, Ionic size  magnitude of negative charge
$\text{ Ionic size }\propto \frac{1}{\text{ magnitude of }+\text{ ve charge }}$
Thus, correct ionic size order is:
${\mathrm{Al}}^{+3}<{\mathrm{Mg}}^{+2}<{\mathrm{Na}}^{+1}<{\mathrm{F}}^{-}.$
(2) Metallic radius - Li < Na < K < Rb Down the group, atomic number increases, number of shells increases.
Thus, metallic radius increases Li < Na < K < Rb.
(3) Ionic size - Al⁺³ < Mg⁺³ < Na+ < F In case of isoelectric species, ionic size α magnitude of negative charge.
$\text{ lonic size }\propto \frac{1}{\text{ magnitude of }+\text{ ve charge }}$
Thus, correct ionic size order is:
${\mathrm{Al}}^{+3}<{\mathrm{Mg}}^{+3}<{\mathrm{Na}}^{+}<\mathrm{F}$