Problem 6
Question
Consider the series \(\mathrm{HF}, \mathrm{HCl}, \mathrm{HBr}, \mathrm{HI}\). a) What is the bond order for each H-X bond? b) What trend is observed in bond energy? c) Considering the relative size of \(\mathrm{F}, \mathrm{Cl}, \mathrm{Br}\), and \(\mathrm{I}\), what trend would you predict in H-X bond length? Bond length is defined as the distance between the nuclei of two bonding atoms.
Step-by-Step Solution
Verified Answer
a) The bond order for each H-X bond in HF, HCl, HBr, and HI is 1. b) The bond energy decreases from HF to HI. c) The bond length increases from HF to HI.
1Step 1: Determine the bond order
Bond order refers to the number of chemical bonds between a pair of atoms. In this case, hydrogen and halogen form one bond in each of the molecules. Therefore, the bond order for each H-X bond in HF, HCl, HBr, and HI is 1.
2Step 2: Analyze the trend in bond energy
Bond energy is the amount of energy required to break one mole of a chemical bond to produce separated atoms. Going from fluorine (F) to iodine (I) across the period, the bond energy decreases. This is mainly due to the fact that the size of the atoms increases. As a result, the bonding pair of electrons is more distant from the nucleus and less tightly held, so less energy is required to break the bond. Therefore, the bond energy is highest for HF and lowest for HI.
3Step 3: Predict the trend in H-X bond length
Bond length is defined as the distance between the nuclei of two bonded atoms. As the size of the atoms increases (from F to I), the bond length also increases. This is because larger atoms have a larger atomic radius, leading to a greater distance between the two bonded nuclei. Therefore, the bond length is shortest for HF and longest for HI.
Key Concepts
Bond OrderBond EnergyBond LengthHalogen Series
Bond Order
In chemical bonding, bond order is a simple yet crucial concept. It tells us how many chemical bonds exist between a pair of atoms. For example, in a single bond like in the H-X bonds found in the molecules
To sum up, bond order can be understood using the formula\[ \text{Bond Order} = \frac{\text{Number of Bonds}}{\text{Number of Bonded Atom Pairs}} \]For the exercise series, all H-X bonds have been determined to have a bond order of 1.
This means a single chemical bond exists between the hydrogen and each halogen.
- HF
- HCl
- HBr
- HI
To sum up, bond order can be understood using the formula\[ \text{Bond Order} = \frac{\text{Number of Bonds}}{\text{Number of Bonded Atom Pairs}} \]For the exercise series, all H-X bonds have been determined to have a bond order of 1.
This means a single chemical bond exists between the hydrogen and each halogen.
Bond Energy
Bond energy is an important property that reveals how strong a chemical bond is. It represents the amount of energy required to break one mole of bond into its constituent atoms.
In the series
This trend is primarily due to atom size. As the size of the halogen atom increases from fluorine to iodine, the shared pair of electrons is further from the nucleus.
This larger distance doesn't hold the electrons very strongly, meaning less energy is needed to break the bond as we move down the group.
In the series
- HF
- HCl
- HBr
- HI
This trend is primarily due to atom size. As the size of the halogen atom increases from fluorine to iodine, the shared pair of electrons is further from the nucleus.
This larger distance doesn't hold the electrons very strongly, meaning less energy is needed to break the bond as we move down the group.
Bond Length
Bond length is another vital concept which helps in understanding molecular structures. It is essentially the distance between the nuclei of two bonded atoms. When we talk about bond lengths in the halogen series
Why does this happen?
- HF
- HCl
- HBr
- HI
Why does this happen?
- Atomic Size: As we move from fluorine to iodine, the atomic size increases significantly.
- Larger Atoms: Larger atoms naturally tend to have longer radii, thereby increasing the bond length.
Halogen Series
The halogen series is composed of these elements:
As you go down the group from fluorine to iodine, there are several observable trends:
- Fluorine (F)
- Chlorine (Cl)
- Bromine (Br)
- Iodine (I)
As you go down the group from fluorine to iodine, there are several observable trends:
- Atomic Radius: The atomic size increases, leading to changes in bond length and energy.
- Reactivity: Generally, the reactivity decreases due to the increased atomic size, which affects the ability of the atom to attract electrons towards itself.
Other exercises in this chapter
Problem 3
What is the relationship between the bond order of a bond and the designation of single, double, and triple bonds?
View solution Problem 4
What is the relationship between the bond order and the number of electrons shared by two adjacent atoms?
View solution Problem 7
Consider the series \(\mathrm{Cl}_{2}, \mathrm{Br}_{2}, \mathrm{I}_{2}\). a) What is the bond order for each X-X bond? b) What trend is observed in bond energy?
View solution