Problem 128 - A Complete Resource Book in Chemistry for JEE Main

Question

Which one of the following arrangements do not truely represent the property indicated against it? (a) \(\mathrm{Br}_{2}<\mathrm{Cl}_{2}<\mathrm{F}_{2}-\) Electronegativity (b) \(\mathrm{Br}_{2}<\mathrm{Cl}_{2}<\mathrm{F}_{2}-\) Bond energy (c) \(\mathrm{Br}_{2}<\mathrm{Cl}_{2}<\mathrm{F}_{2}-\) Electron affinity (d) \(\mathrm{Br}_{2}<\mathrm{Cl}_{2}<\mathrm{F}_{2}-\) oxidizing power

Step-by-Step Solution

Answer

Option (b) is incorrect because bond energy order should be \( \mathrm{Cl}_{2} > \mathrm{F}_{2} > \mathrm{Br}_{2} \).

Step 1: Understand the Definitions

First, understand what each property is: Electronegativity is the tendency of an atom to attract electrons. Bond energy is the energy required to break a bond. Electron affinity is the ability of an atom to accept an electron. Oxidizing power is the ability of an element to oxidize others by gaining electrons.

Step 2: Evaluate Electronegativity Order

Fluorine has the highest electronegativity, followed by chlorine and bromine. Therefore, option (a) \( \mathrm{Br}_{2}<\mathrm{Cl}_{2}<\mathrm{F}_{2}- \text{Electronegativity} \) is correct.

Step 3: Evaluate Bond Energy Order

Generally, bond energy increases as atomic size decreases due to shorter bond lengths. Fluorine, however, forms a weaker bond due to electron repulsion, making \( \mathrm{Cl}_{2} > \mathrm{F}_{2} > \mathrm{Br}_{2} \) in bond energy. Thus, option (b) is incorrect.

Step 4: Evaluate Electron Affinity Order

Electron affinity decreases from fluorine to iodine, but chlorine has a higher electron affinity due to less electron repulsion. Thus, \( \mathrm{Cl}_{2} > \mathrm{F}_{2} > \mathrm{Br}_{2} \) in electron affinity. So (b) does not match \( \mathrm{Br}_{2}<\mathrm{Cl}_{2}<\mathrm{F}_{2} \) correctly.

Step 5: Evaluate Oxidizing Power Order

The oxidizing power decreases down the group. Fluorine has the highest oxidizing power, followed by chlorine and bromine. Hence, \( \mathrm{Br}_{2}<\mathrm{Cl}_{2}<\mathrm{F}_{2} \) is correct for oxidizing power, verifying (d).

Step 6: Identify the Incorrect Arrangement

Option (b), Bond Energy, does not correctly follow \( \mathrm{Br}_{2}<\mathrm{Cl}_{2}<\mathrm{F}_{2} \). It should be \( \mathrm{Cl}_{2} > \mathrm{F}_{2} > \mathrm{Br}_{2} \). Hence, (b) is the incorrect arrangement.

Key Concepts

Electronegativity

Electronegativity is a fundamental property of atoms that describes their ability to attract and hold on to electrons within a chemical bond. The higher the electronegativity, the stronger the pull on electrons. In the periodic table, electronegativity generally increases across a period from left to right and decreases down a group.
Fluorine is the most electronegative element in the periodic table. This makes it very effective at attracting electrons in a bond. Following fluorine are elements like chlorine and bromine, which have lower, but still significant electronegativity. This trend explains why in the arrangement of \( \mathrm{Br}_{2}<\mathrm{Cl}_{2}<\mathrm{F}_{2} \) for electronegativity, it holds true. This means out of the halogens, fluorine has the highest tendency to attract electrons, followed by chlorine, and lastly bromine.

Bond Energy

Bond energy is a measure of the strength of a chemical bond and the energy required to break one mole of bonds in a gaseous state. A strong bond means more energy is required to break it. Typically, as atoms in a molecule get larger, the bond lengths increase, leading to generally weaker bonds.
In the case of halogens, a unique exception arises with fluorine. Despite its high electronegativity, fluorine forms weaker bonds compared to chlorine. This is due to the small size of fluorine atoms, which creates an effect known as electron-electron repulsion. This repulsion weakens the bond, making \( \mathrm{Cl}_{2} > \mathrm{F}_{2} > \mathrm{Br}_{2} \) the correct order for bond energy. Therefore, the initial assumption of \( \mathrm{Br}_{2}<\mathrm{Cl}_{2}<\mathrm{F}_{2} \) for bond energy is incorrect.

Electron Affinity

Electron affinity refers to the change in energy when an atom gains an electron and becomes an anion. It generally reflects how strongly an atom can acquire an additional electron. Like electronegativity, electron affinity increases across a period and decreases down a group, though there are exceptions.
One notable exception is between fluorine and chlorine. Even though fluorine comes before chlorine in the period, chlorine actually has a higher electron affinity. This is due to reduced repulsion in chlorine's larger atomic size, allowing it to accommodate an extra electron more comfortably. Therefore, the order of electron affinity should actually be \( \mathrm{Cl}_{2} > \mathrm{F}_{2} > \mathrm{Br}_{2} \).

Oxidizing Power

Oxidizing power is the ability of an element to oxidize another, essentially its ability to gain electrons itself during a redox reaction. In halogens, oxidizing power decreases down the group, making fluorine the most potent oxidizer due to its high electronegativity and ability to strongly attract electrons.
This trend of decreasing oxidizing power as we go down a group is observed in the arrangement \( \mathrm{Br}_{2}<\mathrm{Cl}_{2}<\mathrm{F}_{2} \), which holds true for oxidizing power. Fluorine's strong oxidizing power is such that it can often react with nearly every element, showcasing its ability to accept electrons and thus oxidize other substances.