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

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This question is about sodium and some of its compounds. 08.1 Use your knowledge of structure and bonding to explain why sodium bromide has a melting point that is ... show full transcript

Worked Solution & Example Answer:This question is about sodium and some of its compounds - AQA - A-Level Chemistry - Question 8 - 2018 - Paper 1

Step 1

Use your knowledge of structure and bonding to explain why sodium bromide has a melting point that is higher than that of sodium, and higher than that of sodium iodide.

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Answer

Sodium bromide has a higher melting point than sodium due to ionic bonding. In sodium bromide (NaBr), there are strong electrostatic forces of attraction between the positively charged sodium ions (Na⁺) and the negatively charged bromide ions (Br⁻). This results in a stable crystal lattice structure that requires a significant amount of energy to break apart.

In comparison, sodium (a metal) has metallic bonding, which is weaker than ionic bonds and allows for more mobility of electrons, making it easier to melt. Sodium iodide (NaI) also has ionic bonding but the larger iodide ion results in a weaker attraction than bromide, hence its melting point is lower than that of NaBr.

Step 2

Give an equation for the reaction that occurs.

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Answer

The balanced equation for the reaction of sodium with water is:

ightarrow ext{2 NaOH (aq) + H}_2 ext{ (g)} $$

Step 3

Calculate the volume, in cm³, of the gas formed at 101 kPa.

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Answer

Using the ideal gas law, we can find the volume of hydrogen gas produced:

PV=nRTPV = nRT

Where, P = 101 kPa (converted to 100,000 Pa), R = 8.31 J K⁻¹ mol⁻¹, t = 298 K, n = 0.0109 mol (from sodium amount).

Plugging in the values:

V=nRTP=(0.0109)(8.31)(298)1000002.68 dm3=2680 cm3V = \frac{nRT}{P} \\ = \frac{(0.0109)(8.31)(298)}{100000} \\ \approx 2.68 \text{ dm}^3 = 2680 \text{ cm}^3

Step 4

Calculate the concentration, in mol dm⁻³, of sodium ions in the solution produced in the reaction in Question 08.2.

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Answer

To find the concentration:

= \frac{0.0109}{0.5} = 0.0218 \text{ mol dm}^{-3}$$

Step 5

Draw the shape of the NH₂⁻ ion.

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Answer

The NH₂⁻ ion has a shape that is bent (or V-shaped). The Lewis structure shows nitrogen at the center, with two hydrogen atoms bonded to it and two lone pairs of electrons on nitrogen affecting the geometry.

  H
  |
H-N
  |  ..

Where .. denotes the lone pairs.

Step 6

Predict the bond angle.

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Answer

The predicted bond angle for NH₂⁻ is approximately 116 degrees, which is slightly less than the ideal tetrahedral angle of 109.5 degrees due to the repulsion exerted by the two lone pairs which pushes the bonding pairs closer together.

Step 7

Justify your prediction.

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Answer

The bond angle in NH₂⁻ is influenced by the lone pairs of electrons on the nitrogen atom. Lone pairs occupy more space than bonding pairs, causing greater repulsion. As a result, the bond angle decreases from the ideal tetrahedral angle. The repulsion between the lone pairs pushes the hydrogen atoms closer together, necessitating a bond angle of approximately 116 degrees.

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