Problem 55 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Arrange the following : (i) \(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{~F}_{2}, \mathrm{Cl}_{2}\) in increasing order of bond dissociation energy. [1988-1 Mark] (ii) Increasing strength of hydrogen bonding \((\mathrm{X}-\mathrm{H}-\mathrm{X})\) : [1991-1 Mark] \(\mathrm{O}, \mathrm{S}, \mathrm{F}, \mathrm{Cl}, \mathrm{N}\) (iii) In the decreasing order of the \(\mathrm{O}-\mathrm{O}\) bond length present in them $$ \mathrm{O}_{2}, \mathrm{KO}_{2} \text { and } \mathrm{O}_{2}\left[\mathrm{AsF}_{4}\right] $$

Step-by-Step Solution

Answer

(i) \(\mathrm{F}_{2} < \mathrm{Cl}_{2} < \mathrm{O}_{2} < \mathrm{N}_{2}\); (ii) \(\mathrm{Cl} < \mathrm{S} < \mathrm{N} < \mathrm{O} < \mathrm{F}\); (iii) \(\mathrm{KO}_{2} > \mathrm{O}_{2}[\mathrm{AsF}_{4}] > \mathrm{O}_{2}\)."

Step 1: Understanding the concept of bond dissociation energy (BDE)

Bond dissociation energy is the energy required to break one mole of the bond in gaseous molecules. Typically, BDE increases with increasing bond order and decreases with the increasing size of atoms involved.

Step 2: Arranging N2, O2, F2, Cl2 by increasing BDE

- Nitrogen (\(\text{N}_{2}\)) has a triple bond, very high BDE.- Oxygen (\(\text{O}_{2}\)) has a double bond, high BDE.- Fluorine (\(\text{F}_{2}\)) has a single bond; due to its small size and repulsion, it has a lower BDE than chlorine.- Chlorine (\(\text{Cl}_{2}\)) has a single bond, larger atoms than fluorine, moderately high BDE. Thus, the increasing order is \(\mathrm{F}_{2} <\mathrm{Cl}_{2} <\mathrm{O}_{2}<\mathrm{N}_{2}\).

Step 3: Understanding factors affecting hydrogen bonding strength

The strength of hydrogen bonds depends on the difference in electronegativity and size of the atoms involved. Generally, smaller and more electronegative atoms form stronger hydrogen bonds.

Step 4: Arranging O, S, F, Cl, N by increasing hydrogen bonding strength

- Fluorine (\(\mathrm{F}\)) forms the strongest hydrogen bonds due to high electronegativity.- Oxygen (\(\mathrm{O}\)) follows, being very electronegative too.- Nitrogen (\(\mathrm{N}\)) can form substantial hydrogen bonds.- Sulfur (\(\mathrm{S}\)) has less strength than above, being less electronegative.- Chlorine (\(\mathrm{Cl}\)) has the weakest due to large atomic size and lesser electronegativity. Thus, the increasing order is \(\mathrm{Cl}<\mathrm{S}<\mathrm{N}<\mathrm{O}<\mathrm{F}\).

Step 5: Understanding O-O bond lengths in compounds

The bond length is affected by the bond order and the presence of additional interactions or charges. Peroxides generally have longer bond lengths than single bonds like in \(\mathrm{O}_{2}\) due to lower bond order in superoxides.

Step 6: Arranging \( \mathrm{O}_{2}, KO_{2}, O_{2}[AsF_{4}] \) by decreasing bond length

- \(\mathrm{O}_{2}\) has a double bond with a shorter bond length.- \(\mathrm{O}_{2}\) in \(\mathrm{O}_{2}[\mathrm{AsF}_{4}]\) is more of a complex ion with interactions affecting the bond length, making it longer than simple \(\mathrm{O}_{2}\).- In \(\mathrm{KO}_{2}\), the \(\mathrm{O}_{2}^{-}\) is a superoxide ion with the longest bond length due to the lowest bond order.Thus, the decreasing order is \(\mathrm{KO}_{2} >\mathrm{O}_{2}[\mathrm{AsF}_{4}] >\mathrm{O}_{2}\).

Key Concepts

Bond Dissociation Energy

Bond Dissociation Energy (BDE) refers to the amount of energy needed to break a specific chemical bond in a molecule. It is often measured for one mole of a substance in the gas phase. BDE is an essential concept in understanding molecular stability:

• Higher BDE means a stronger bond.
• BDE typically increases with the bond order, meaning triple bonds have higher BDE than double or single bonds.
• It decreases with larger sizes of atoms, as larger atoms tend to share electrons less effectively, leading to weaker bonds.

For example, in a sequence of molecules like ewline\[\text{N}_2, \text{O}_2, \text{F}_2, \text{Cl}_2\],ewline- Nitrogen with a triple bond indicates a high BDE,- Oxygen has a double bond and high BDE,- Chlorine and fluorine, having single bonds, exhibit lower BDE, with fluorine being particularly weak due to its small size causing repulsion. Thus, increasing BDE: \[\text{F}_2 < \text{Cl}_2 < \text{O}_2 < \text{N}_2\] is the order for these molecules.

Hydrogen Bonding

Hydrogen bonding is a special type of dipole-dipole interaction involving hydrogen and electronegative atoms. It significantly impacts the properties of compounds:

• The strength of hydrogen bonding is dictated by the electronegativity and size of the atoms bonded to hydrogen.
• Smaller atoms with high electronegativity form stronger hydrogen bonds.

In the context of elements such as ewline\(\text{O}, \text{S}, \text{F}, \text{Cl}, \text{N}\),ewline- Fluorine forms the strongest hydrogen bonds due to its high electronegativity.- Oxygen, being very electronegative, follows fluorine.- Nitrogen can also form strong hydrogen bonds, although not as strong as oxygen or fluorine.- Sulfur and chlorine are less effective. Sulfur forms weaker bonds compared to nitrogen given its lower electronegativity.The order of increasing hydrogen bond strength is: \[\text{Cl} < \text{S} < \text{N} < \text{O} < \text{F}\].

Bond Length

Bond length is the distance between the nuclei of two bonded atoms. It is influenced by factors such as bond order, atomic size, and additional molecular interactions:

• Shorter bond lengths correlate with higher bond orders.
• Compounds with additional interactions or charges may experience changes in bond lengths.

For example, considering ewline\(\text{O}_2, \text{KO}_2, \text{O}_2[\text{AsF}_4]\),ewline- \(\text{O}_2\) has the shortest bond length due to its double bond.- In \(\text{O}_2[\text{AsF}_4]\), the bond length increases due to interactions within the complex ion.- \(\text{KO}_2\) has the longest bond length, being a superoxide ion with the lowest bond order. The decreasing order of bond length is: \[\text{KO}_2 > \text{O}_2[\text{AsF}_4] > \text{O}_2\].