Which of the following statements is incorrect?

The correct option is: "isoelectronic ions belong to the same period"

Here is a step-by-step explanation of why this statement is incorrect (making it the correct answer to the question), along with an analysis of the other statements:

1. Analysis of the Incorrect Statement:
Isoelectronic species are atoms or ions that contain the same total number of electrons. However, they do not have to belong to the same period in the periodic table.
For example, consider the following isoelectronic series, each containing exactly 10 electrons:
Nitride ion (N^3-): 7 + 3 = 10 electrons (Period 2)
Oxide ion (O^2-): 8 + 2 = 10 electrons (Period 2)
Fluoride ion (F^-): 9 + 1 = 10 electrons (Period 2)
Sodium ion (Na⁺): 11 - 1 = 10 electrons (Period 3)
Magnesium ion (Mg²⁺): 12 - 2 = 10 electrons (Period 3)
Since N^3-, O^2-, and F^- belong to the second period, while Na⁺ and Mg²⁺ belong to the third period, the statement that isoelectronic ions must belong to the same period is incorrect.

2. Verification of the Other Statements (Which are Correct):
• First Ionization Enthalpy (I.E.₁) of O vs. N: Nitrogen has a stable, half-filled valence electronic configuration:
$N=1s^{2}2s^{2}2p^3$
Oxygen has the configuration:
$O=1s^{2}2s^{2}2p^4$
Because removing an electron from a stable half-filled subshell requires more energy, the first ionization energy of nitrogen is higher than that of oxygen (I.E.₁ of O < I.E.₁ of N).
However, for the second ionization enthalpy (I.E.₂), we remove an electron from the monopositive ions:
$N^{+}=1s^{2}2s^{2}2p^2$
$O^{+}=1s^{2}2s^{2}2p^3$ (stable half-filled configuration)
Here, O⁺ has the stable half-filled configuration, making it much harder to remove the second electron from oxygen than from nitrogen. Thus, I.E.₂ of O is higher than that of N.

• Electron Gain Enthalpy of N vs. P: Nitrogen has a highly stable, half-filled 2p³ subshell and a very small atomic size, resulting in high electron-electron repulsion if an extra electron is added. Consequently, its electron gain enthalpy is close to zero (or slightly positive). Phosphorus, being larger, accommodates the incoming electron more easily.

• Covalent Radius vs. Van der Waals Radius of Iodine: The covalent radius measures the distance between two bonded atoms sharing electrons, involving orbital overlap. The Van der Waals radius represents the closest distance of approach between non-bonded atoms held only by weak dispersion forces. Since there is no orbital overlap in Van der Waals interactions, the Van der Waals radius of any atom, including iodine, is always larger than its covalent radius.