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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 (NaBr) has a higher melting point compared to sodium and sodium iodide due to the ionic bonding present in its structure. The strong electrostatic forces of attraction between the oppositely charged ions (Na⁺ and Br⁻) in a crystal lattice structure require more energy to overcome compared to the metallic bonds in sodium or the weaker ion-dipole interactions in sodium iodide (NaI).

In sodium iodide, the larger iodide ion results in a weaker ionic bond because the larger size leads to a greater distance between the ions, resulting in lower lattice energy. Therefore, NaBr, with smaller ions, possesses higher lattice energy and subsequently a higher melting point.

Step 2

Give an equation for the reaction that occurs.

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Answer

The equation for the reaction of sodium with water is:

ightarrow 2 ext{NaOH} + ext{H}_2$$

Step 3

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

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Answer

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

extMoles=25.0g22.99g mol1=0.25 mol 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=nRTPV = 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=nRTP=(0.25)(8.31)(298)101000=0.00613extm3=6.13extL=6130extcm3V = \frac{nRT}{P} = \frac{(0.25) (8.31)(298)}{101000} = 0.00613 ext{ m}^3 = 6.13 ext{ L} = 6130 ext{ 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

The molarity (C) can be calculated as:

C=moles of Na+volume in dm3C = \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=0.250.5=0.5 mol dm3C = \frac{0.25}{0.5} = 0.5 \text{ mol dm}^{-3}

Step 5

Draw the shape of the NH₂⁻ ion.

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Answer

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.

Step 6

Predict the bond angle.

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Answer

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

Step 7

Justify your prediction.

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Answer

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.

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