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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. 1. Use your knowledge of structure and bonding to explain why sodium bromide has a melting point that is hi... 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 than sodium (Na) due to its ionic bond formation which arises from the strong electrostatic forces between the positively charged sodium ions (Na⁺) and negatively charged bromide ions (Br⁻). In contrast, sodium is a metal that forms metallic bonds, which are generally weaker than ionic bonds.

Comparing sodium bromide with sodium iodide (NaI), bromide ions are smaller than iodide ions, leading to a stronger ionic bond in NaBr. The smaller size of the Br⁻ ion allows for closer packing and a stronger attraction between it and the Na⁺ ion compared to the Na⁺ and I⁻ ions.

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:

2Na+2H2O2NaOH+H22 Na + 2 H_2O \rightarrow 2 NaOH + H_2

Step 3

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

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Answer

To find the volume of hydrogen gas produced, we can use the ideal gas law equation:

PV=nRTPV = nRT

Where:

  • P = pressure (101 kPa)
  • V = volume (in cm³)
  • n = number of moles of gas produced
  • R = gas constant (8.31 J K⁻¹ mol⁻¹)
  • T = temperature in Kelvin (T = 25 °C = 298 K)

First, calculate the number of moles of sodium reacted:

extMolesofNa=0.250g23g/mol=0.0109 moles ext{Moles of } Na = \frac{0.250 g}{23 g/mol} = 0.0109 \text{ moles}

Thus, 0.0109 moles of Na produces 0.0109 moles of H₂ gas, so substituting into the ideal gas law:

101kPaV=0.0109mol8.31JK1mol1298K101 kPa \cdot V = 0.0109 mol \cdot 8.31 J K^{-1} mol^{-1} \cdot 298 K

Converting kPa to Pa (1 kPa = 1000 Pa):

101000V=27.066m3101000 \cdot V = 27.066\, m^3

Rearranging for V,

V=27.0661010001000000=267.8cm3V = \frac{27.066}{101000} \cdot 1000000 = 267.8\, cm³

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 concentration of sodium ions (Na⁺) can be calculated using the formula:

C=nVC = \frac{n}{V}

Where:

  • n = moles of Na⁺ produced (0.0109 moles)
  • V = volume of solution in dm³ (500 cm³ = 0.5 dm³)

Substituting:

C=0.01090.5=0.0218moldm3C = \frac{0.0109}{0.5} = 0.0218 \, mol \, dm^{-3}

Step 5

Draw the shape of the NH₂⁻ ion.

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Answer

The NH₂⁻ ion has a bent shape due to the presence of lone pairs of electrons on the nitrogen atom. The drawing should show the nitrogen atom in the center with two hydrogen atoms bonded to it, and one lone pair of electrons represented on the nitrogen as dots.

H | H--N: (lone pairs represented as dots)

Step 6

Predict the bond angle.

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Answer

The bond angle in the NH₂⁻ ion is approximately 105 degrees due to the repulsion between the lone pair and the bonding pairs, which is less than the ideal bond angle of 109.5 degrees for a tetrahedral arrangement. This reduction in angle occurs because lone pairs occupy more space and result in greater repulsion.

Step 7

Justify your prediction.

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Answer

The presence of the lone pair of electrons on the nitrogen atom causes greater repulsion compared to bonding pairs. This, in turn, compresses the bond angle between the hydrogen atoms to approximately 105 degrees instead of the ideal 109.5 degrees associated with tetrahedral shapes without lone pairs.

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