Give the full electron configuration for the calcium ion, Ca²⁺ - AQA - A-Level Chemistry - Question 4 - 2018 - Paper 1

The second ionisation energy of calcium is lower than that of potassium because calcium's second electron is removed from a higher principal energy level (n=3) compared to potassium (n=4). Additionally, calcium has more electron shielding than potassium, which reduces the effective nuclear charge experienced by the electron in calcium, making it easier to remove.

The s-block metal with the highest first ionisation energy is beryllium (Be). This is due to its smaller atomic size and higher effective nuclear charge compared to other s-block metals.

The formula of the hydroxide that is least soluble in water is:

$\text{Ba(OH)}_2$

Barium hydroxide is more soluble than magnesium hydroxide, which is considered the least soluble of the group.

Firstly, calculate the moles of barium chloride (BaCl₂) added:

For BaCl₂: $6 \text{ cm}^3 \times 2.50 \text{ mol dm}^{-3} = 0.015 \text{ mol}$

Now calculate the moles of sodium sulfate (Na₂SO₄):

For Na₂SO₄: $8 \text{ cm}^3 \times 0.15 \text{ mol dm}^{-3} = 0.0012 \text{ mol}$

Using the reaction:

$ext{Ba}^{2+} + ext{SO}_4^{2-} \rightarrow ext{BaSO}_4 \text{ (s)}$

1 mole of Ba²⁺ reacts with 1 mole of SO₄²⁻. The limiting reagent is sodium sulfate, showing that barium chloride is in excess.

The reagent in excess is barium chloride (BaCl₂).

The total volume of sodium sulfate (Na₂SO₄) is 8 cm³.

Isotopes of strontium have identical chemical properties because they have the same number of electrons and, therefore, the same electronic configuration. The chemical behavior is determined by the electron arrangement, not by the number of neutrons.

Let the abundance of $^{86}$Sr be x.

Given that the sample consists of $^{84}$Sr, $^{86}$Sr, and $^{88}$Sr:

1. The relative masses are: $84x + 87.0y + 88(1-y) = 87.7$ where $y = \text{abundance of } ^{88}Sr$

2. Solving gives: $x + y + (1-y) = 1$ $x = 0.2$ or 20%.

Thus, percentage abundance of $^{86}$Sr is 20%.

The ion with the longest time of flight is $^{138}$Ba. Heavier ions travel slower through the flight tube, meaning that the lighter isotope with less mass $^{138}$Ba will have the longest time before detection.

Given:

• Kinetic energy (KE) = $3.65 \times 10^{-19}$ J
• Time = $2.71 \times 10^{-6}$ s

1. Calculate mass using:

$KE = \frac{1}{2}mv^2$

with velocity ($v$) calculated as:

$v = \frac{d}{t}$

2. Reorganizing and substituting values:

• Length of tube L can be calculated giving:

$L = v \times t$

Final calculation gives:

$L = 1.54 \text{ m}$