This question is about Group 7 elements - AQA - GCSE Chemistry - Question 7 - 2018 - Paper 1

Chlorine is more reactive than iodine due to several factors:

1. Electron Configuration: Chlorine has fewer electron shells than iodine, with its outer electrons closer to the nucleus. This configuration leads to a stronger attraction between the nucleus and the outer electrons.

2. Nuclear Charge: The nucleus of chlorine has a greater effective nuclear charge which allows it to gain an electron more easily compared to iodine.

3. Shielding Effect: Chlorine experiences less shielding compared to iodine, making it easier for chlorine to attract additional electrons.

Hydrogen chloride is a gas at room temperature because it consists of simple molecular structures. The intermolecular forces present in hydrogen chloride are weak van der Waals forces, which require little energy to overcome. This allows hydrogen chloride to remain in a gaseous state under standard conditions.

To calculate the bond energy X for the C–Br bond, we refer to the given information:

1. Bond energies for bonds broken: C–H bonds = 4 × 412 kJ/mol and Br–Br bond = 193 kJ/mol.

   Total energy for bonds broken =

   $4(412) + 193 = 1841 \, \text{kJ/mol}$

2. Bond energies for bonds formed: C–H bonds = 3 × 412 kJ/mol, H–Br bond = 366 kJ/mol, and the unknown C–Br bond = X.

   Total energy for bonds formed =

   $3(412) + 366 + X = 1602 + X \, \text{kJ/mol}$

3. Overall energy change: Given as -51 kJ/mol.

   Therefore, we can use the equation:

   $-51 = 1841 - (1602 + X)$

   Rearranging gives:

   $-51 + 1602 = 1841 - X$

   Simplifying:

   $X = 290 \, \text{kJ/mol}$

   Consequently, the bond energy X for the C–Br bond is 290 kJ/mol.