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Among the \(\mathrm{C}-\mathrm{X}\) bond (where \(\mathrm{X}=\mathrm{Cl}, \mathrm{Br}, \mathrm{I})\) the correct bond energy order is (a) \(\mathrm{C}-\mathrm{I}>\mathrm{C}-\mathrm{Cl}>\mathrm{C}-\mathrm{Br}\) (b) \(\mathrm{C}-\mathrm{Cl}>\mathrm{C}-\mathrm{Br}>\mathrm{C}-\mathrm{I}\) (c) \(\mathrm{C}-\mathrm{Br}>\mathrm{C}-\mathrm{Cl}>\mathrm{C}-\mathrm{I}\) (d) \(\mathrm{C}-\mathrm{Cl}>\mathrm{C}-\mathrm{I}>\mathrm{C}-\mathrm{Br}\)

Short Answer

Expert verified
The correct bond energy order is option (b): \(\mathrm{C}-\mathrm{Cl}>\mathrm{C}-\mathrm{Br}>\mathrm{C}-\mathrm{I}\).

Step by step solution

01

Understand Bond Energy Concept

Bond energy is the measure of bond strength in a chemical bond. It is defined as the amount of energy required to break one mole of bonds in gaseous molecules under standard conditions. The larger the bond energy, the stronger the bond.
02

Compare Atomic Sizes

The bond energy generally decreases as the size of the atoms involved increases. Larger atoms, such as iodine (I), have weaker bonds due to the increased bond length compared to smaller atoms like chlorine (Cl).
03

Analyze Periodic Table Trends

Halogens (group 17 elements: Cl, Br, and I) show a trend in bond energies. As we move down the group from chlorine to iodine, the atomic size increases, and thus the bond energy with carbon decreases. Consequently, the bond energy order will be based on this trend: Cl > Br > I.
04

Determine the Correct Bond Energy Order

Using the trend that bond energy decreases from Cl to Br to I due to increasing atomic size, the bond energy order is: \(\mathrm{C}-\mathrm{Cl}>\mathrm{C}-\mathrm{Br}>\mathrm{C}-\mathrm{I}\).
05

Verify the Correct Option

Based on the comparison of bond energies analyzed in the previous steps, the correct order is \(\mathrm{C}-\mathrm{Cl}>\mathrm{C}-\mathrm{Br}>\mathrm{C}-\mathrm{I}\). This matches option (b).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bond Strength
Bond strength is a key concept in chemistry, as it determines how tightly two atoms are held together in a molecule. This strength is quantified by bond energy, the amount of energy needed to break the bond in gaseous molecules under standard conditions. Imagine bond strength as the adhesive force between two bricks in a building. The stronger the adhesive, the more difficult it is to separate the bricks. Similarly, in chemical bonds, stronger bonds require more energy to break.
  • Strong bonds have high bond energies.
  • Weak bonds have low bond energies.
For carbon-halogen bonds, bond strength is affected by the size of the halogen atom. Smaller atoms, like chlorine, form stronger bonds with carbon due to the shorter distance between the nuclei of the two bonded atoms. This short distance means stronger interactions and, therefore, higher bond energy. Given these insights, understanding the nature of bond strength helps in predicting reactivity and stability of molecules.
Halogens
Halogens are fascinating elements belonging to group 17 of the periodic table. These elements include fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At). They are known for their high reactivity, mainly due to their desire to gain one electron to achieve a full outer electron shell.
  • Halogens have seven valence electrons, needing just one more to fulfill the octet rule.
  • This high electronegativity makes them eager participants in forming covalent or ionic bonds.
In the context of bond strength, the size of the halogen greatly influences the bond energy. As we move down the group from fluorine to iodine, each halogen atom gets larger. Larger atoms like iodine will have weaker bonds to carbon due to increased atomic size and subsequently longer bond length. In everyday applications, compounds containing halogens are vital in creating disinfectants, medications, and even in the field of photography. Understanding halogens can provide insight into why they form such a diverse array of useful, reactive compounds across different industries.
Periodic Table Trends
Periodic table trends are essential for making predictions about the properties of elements. One important trend is that as you proceed down a group in the periodic table, the atomic radius increases. This increase is due to the addition of electron shells, which outweighs the effect of increasing nuclear charge. The impact of increased atomic size on bond energy includes:
  • Longer bond lengths, which generally lead to weaker bonds.
  • Lower bond energies due to increased separation between bonded nuclei.
In the case of carbon-halogen bonds, the trend is evident. Chlorine, being higher up in the group, forms a stronger bond with carbon as compared to bromine and iodine. This is because chlorine's smaller atomic radius allows for a shorter bond length, translating to stronger interatomic forces and higher bond energy. Recognizing these periodic trends is crucial for understanding patterns in elements' reactions and bonding behaviors, explaining why certain elements behave similarly while others differ dramatically.

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Most popular questions from this chapter

Following statements regarding the periodic trends of chemical reactivity of the alkali metals and the halogens are given. Which one of these statements gives the correct picture? [2006] (a) the reactivity decreases in the alkali metals but increases in the halogens with increase in atomic number down the group (b) in both the alkali metals and the halogens the chemical reactivity decreases with increase in atomic number down the group (c) chemical reactivity increases with increase in atomic number down the group in both the alkali metals and halogens (d) in alkali metals the reactivity increases but in the halogens it decreases with increase in atomic number down the group

The correct order of solubility of fluorides of alkaline earth metals is (a) \(\mathrm{BeF}_{2}>\mathrm{MgF}_{2}>\mathrm{CaF}_{2}>\mathrm{SrF}_{2}>\mathrm{BaF}_{2}\) (b) \(\mathrm{MgF}_{2}>\mathrm{BaF}_{2}>\mathrm{SrF}_{2}>\mathrm{CaF}_{2}>\mathrm{BeF}_{2}\) (c) \(\mathrm{BaF}_{2}>\mathrm{SrF}_{2}>\mathrm{CaF}_{2}>\mathrm{MgF}_{2}>\mathrm{BeF}_{2}\) (d) \(\mathrm{BeF}_{2}^{2}>\mathrm{Mg}^{2} \mathrm{~F}_{2}>\mathrm{SrF}_{2}^{2}>\mathrm{BaF}_{2}^{2}>\mathrm{CaF}_{2}^{2}\)

The correct order of hydration energy of alkaline earth metal ion is (a) \(\mathrm{Mg}^{2+}>\mathrm{Be}^{2+}>\mathrm{Ba}^{2+}>\mathrm{Ca}^{2+}>\mathrm{Sr}^{2+}\) (b) \(\mathrm{Be}^{2+}>\mathrm{Mg}^{2+}>\mathrm{Ca}^{2+}>\mathrm{Sr}^{2+}>\mathrm{Ba}^{2+}\) (c) \(\mathrm{Ba}^{2+}>\mathrm{Be}^{2+}>\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}<\mathrm{Sr}^{2+}\) (d) \(\mathrm{Be}^{2+}>\mathrm{Ca}^{2+}>\mathrm{Sr}^{2+}>\mathrm{Ba}^{2+}>\mathrm{Mg}^{2+}\)

The van der Waals forces in halogen decrease in the order (a) \(\mathrm{I}_{2}>\mathrm{Br}_{2}>\mathrm{Cl}_{2}>\mathrm{F}_{2}\) (b) \(\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2}\) (c) \(\mathrm{I}_{2}>\mathrm{F}_{2}>\mathrm{Br}_{2}>\mathrm{Cl}_{2}\) (d) \(\mathrm{Cl}_{2}>\mathrm{I}_{2}>\mathrm{Br}_{2}>\mathrm{F}_{2}^{2}\)

The catenation tendency of \(\mathrm{C}, \mathrm{Si}\) and Ge is in the order \(\mathrm{Ge}<\mathrm{Si}<\mathrm{C}\). The bond energies (in \(\mathrm{kJ} \mathrm{mol}^{-1}\) ) of \(\mathrm{C}-\mathrm{C}\), Si-Si and Ge-Ge bonds, respectively are (a) \(348,180,167\) (b) \(180,167,348\) (c) \(348,167,180\) (d) \(167,180,348\)

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