Answer the question on basis of passage given below:

The noble gases \(He\) to \(Rn\) have close shell configuration and are monoatomic gases under normal conditions. Though mostly unreactive, but direct reaction of Xe with \(F_2\) leads to a series of compounds with Xe oxidation numbers +2, +4 and +6. When certain compounds like phenol and quinol are crystallised in the presence of noble gas such as Ar, Kr and Xe a category of compound called clathrates are formed. In these the noble gas atoms are trapped in the cavities of the crystal lattices. The compounds which crystallise are known as hosts (H) while the noble gas atoms are called guests (G). The general formula of a clathrate compound is nHmG, where n is the number of host molecules and m is the number of atoms a molecule of guest.

Which factors are responsible for making fluorine the strongest oxidising agent?

(A) Highest electronegativity

(B) High electron gain enthalphy

(C) High bond dissociation enthalphy

(D) Small size of flourine

Choose the correct answer from the options given below:

(A), (B) and (D) only

The correct answer is option 1. (A), (B) and (D) only.

Here is an explanation of why fluorine is the strongest oxidizing agent, focusing on each relevant factor:

Factors Making Fluorine the Strongest Oxidizing Agent

(A) Highest Electronegativity

Electronegativity is a measure of an atom's ability to attract and hold onto electrons. Fluorine has the highest electronegativity of all elements, with a value of approximately 3.98 on the Pauling scale.

Effect on Oxidizing Power: The high electronegativity of fluorine means that it has a strong tendency to attract electrons from other elements. This strong electron-attracting power makes fluorine highly effective at accepting electrons during chemical reactions, which is a key characteristic of an oxidizing agent.

(B) High Electron Gain Enthalpy

Electron Gain Enthalpy: This is the amount of energy released when an atom gains an electron to form a negatively charged ion. Fluorine has a very high electron gain enthalpy (about -328 kJ/mol), indicating that a large amount of energy is released when it gains an electron.

Effect on Oxidizing Power: The high electron gain enthalpy means that fluorine has a strong tendency to accept electrons from other substances. This strong drive to gain electrons enhances its oxidizing ability, as it readily oxidizes other elements by accepting electrons from them.

(C) High Bond Dissociation Enthalpy

Bond Dissociation Enthalpy: This refers to the amount of energy required to break the bond between two atoms in a diatomic molecule. For \(F_2\), the bond dissociation enthalpy is relatively high (about 158 kJ/mol), which reflects the strength of the \(F-F\) bond.

Effect on Oxidizing Power: While the high bond dissociation enthalpy contributes to the stability of the \(F_2\) molecule, it does not directly enhance the oxidizing power of fluorine. In fact, the strong \(F-F\) bond makes it relatively harder for \(F_2\) to dissociate into individual fluorine atoms, but this factor is less significant compared to the high electron gain enthalpy and electronegativity.

(D) Small Size of Fluorine

Atomic Size: Fluorine is the smallest halogen and has a very small atomic radius. This small size means that the added electron is close to the nucleus, experiencing a strong attractive force from the protons in the nucleus.

Effect on Oxidizing Power: The small size of fluorine allows it to effectively accommodate an additional electron, making it easier for fluorine to form a stable fluoride ion (\(F^-\)). This enhances its ability to accept electrons and act as an oxidizing agent.

Conclusion

Fluorine's strength as an oxidizing agent can be attributed to:

Highest Electronegativity (A): Its strong ability to attract electrons.

High Electron Gain Enthalpy(B): The significant amount of energy released when it gains an electron.

Small Size (D): The effective accommodation of additional electrons close to the nucleus.

While High Bond Dissociation Enthalpy (C) is a factor related to the stability of \(F_2\), it is not directly responsible for fluorine's oxidizing power. Therefore, the correct factors that contribute to fluorine being the strongest oxidizing agent are (A), (B), and (D) only.

Thus, the correct answer is (1) (A), (B) and (D) only