Arrange the following bonds according to their average bond energies in descending order
C-Cl, C-Br,

Q: Why is bond energy inversely proportional to bond length?

A shorter bond means the nuclei of the two bonded atoms are closer together. This proximity allows for a stronger electrostatic attraction between the positively charged nuclei and the shared negatively charged electrons. Stronger attraction means more energy is required to pull the atoms apart, hence a higher bond energy.

Q: Does the electronegativity of halogens affect the bond energy here?

While electronegativity differences do create bond polarity (which can add some ionic character and strengthen the bond), the dominant factor when comparing bonds down a group is the atomic size and the resulting bond length. The drastic increase in size from F to I overshadows the effects of electronegativity differences.

Q: Why does the atomic size increase down the group?

As you move down a group in the periodic table, each successive element has an additional principal energy level (electron shell). Even though the nuclear charge also increases, the shielding effect of the inner electron shells prevents the outermost electrons from being pulled in tightly, resulting in a larger overall atomic radius.

Q: Is there any exception like the F-F bond being weaker than Cl-Cl?

That's a great observation! The $F-F$ bond is indeed exceptionally weak due to strong lone pair-lone pair repulsions between the two small, highly electronegative fluorine atoms. However, in our case, we are looking at $C-X$ bonds. Carbon does not have lone pairs to cause this severe repulsion, so the standard trend of 'shorter bond = stronger bond' holds perfectly true.

Q: What exactly is 'average' bond energy?

In polyatomic molecules (like $CH_3Cl$), breaking the first $C-Cl$ bond might require a slightly different amount of energy than breaking a $C-Cl$ bond in a different molecule (like $CCl_4$). 'Average bond energy' is the mean value of the bond dissociation energies for a specific type of bond across a wide variety of different chemical compounds.

Question

Arrange the following bonds according to their average bond energies in descending order
$C-Cl, C-Br, C-F, C-I$

Options:

(A) $C-F > C-Cl > C-Br > C-I$

(B) $C-Br > C-I > C-Cl > C-F$

(C) $C-I > C-Br > C-Cl > C-F$

(D) $C-Cl > C-Br > C-I > C-F$

Sigma Percentile · Accepted Answer

### The Core Concept: Bond Energy vs. Bond Length When we talk about the strength of a chemical bond, we are essentially discussing its **bond energy**—the amount of energy required to break one mole of that specific bond in the gaseous state. But what determines this energy? Imagine two atoms connected by a spring. If the spring is short and tightly coiled, it takes a lot of effort to pull the atoms apart. Conversely, a long, stretched-out spring is much easier to break. This analogy perfectly describes the relationship between bond energy and bond length: **they are inversely proportional**. Mathematically, we can express this as: $$ ext{Bond Energy} \propto \frac{1}{ ext{Bond Length}}$$ Therefore, to find the order of bond energies, our primary task is to determine the order of their bond lengths. ### The Halogen Family: A Story of Growing Sizes In our problem, we are comparing bonds formed between a central Carbon ($C$) atom and four different halogens: Fluorine ($F$), Chlorine ($Cl$), Bromine ($Br$), and Iodine ($I$). Since the carbon atom is a constant factor in all four bonds, the variation in bond length will depend entirely on the size of the halogen atom. Let's take a trip down Group 17 of the periodic table. As we move from top to bottom (from Fluorine down to Iodine), a new electron shell is added at each step. This addition of shells causes the atomic radius to increase significantly. Thus, the order of atomic radii for these halogens is: $$r_F < r_{Cl} < r_{Br} < r_I$$ ### Putting It Together: The Final Order Now, let's combine our two core concepts. A larger halogen atom means its nucleus is further away from the carbon nucleus, resulting in a longer bond. Based on the atomic sizes, the order of bond lengths will be: $$C-F < C-Cl < C-Br < C-I$$ Since the $C-F$ bond is the shortest, it is the most tightly held together, making it the strongest bond with the highest bond energy. On the other end of the spectrum, the $C-I$ bond is the longest, making it the weakest and easiest to break. Reversing the bond length order gives us our final descending order for average bond energies: **$$C-F > C-Cl > C-Br > C-I$$** This elegant relationship between atomic size, bond length, and bond energy is a fundamental principle in chemistry that helps us predict the stability and reactivity of various molecules.

Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Arrange the following bonds according to their average bond energies in descending order $C - C l, C - B r, C - F, C - I$

Select Answer:

Visualized Solution

$A na l y z in g t h e B o n d s$

$B o n d E n er g y v s B o n d L e n g t h$

$A t o mi c S i z e T r e n d$

$B o n d L e n g t h O r d er$

$F ina l B o n d E n er g y O r d er$

The Sigma Insight: Bond Parameters and Resonance

The Core Concept

The Halogen Family

Putting It Together

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