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.

Short Answer

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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.

Step by step solution

01

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.
02

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.
03

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.

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