This question is about sodium and some of its compounds - AQA - A-Level Chemistry - Question 8 - 2018 - Paper 1

The equation for the reaction of sodium with water is:

First, we need to find the number of moles of sodium used:

$ext{Moles} = \frac{25.0\, \text{g}}{22.99\, \text{g mol}^{-1}} = 0.25 \text{ mol}$

Using the ideal gas equation, we can find the volume (V) of the gas produced:

$PV = nRT$

Where:

• P = 101 kPa = 101000 Pa
• n = 0.25 mol
• R = 8.31 J K⁻¹ mol⁻¹
• T = 298 K

Solving for V:

$V = \frac{nRT}{P} = \frac{(0.25) (8.31)(298)}{101000} = 0.00613 ext{ m}^3 = 6.13 ext{ L} = 6130 ext{ cm}^3$

The molarity (C) can be calculated as:

$C = \frac{\text{moles of Na}^+}{\text{volume in dm}^3}$

From the previous calculation, the moles of Na⁺ produced = 0.25 mol. The volume of solution = 500 cm³ = 0.5 dm³.

Thus,

$C = \frac{0.25}{0.5} = 0.5 \text{ mol dm}^{-3}$

The shape of the NH₂⁻ ion can be represented as follows:

   H
    \
     N
    / \
   H   : (lone pair)

The NH₂⁻ ion has a bent shape due to the presence of one lone pair of electrons on the nitrogen atom.

The predicted bond angle in the NH₂⁻ ion is approximately 104.5°.

The bond angle of approximately 104.5° is influenced by the lone pair of electrons on the nitrogen atom, which repels the bonding pairs of electrons. This results in a bond angle that is slightly less than the ideal tetrahedral angle of 109.5°, characteristic of such a geometry.