This question is about s-block metals - AQA - A-Level Chemistry - Question 4 - 2018 - Paper 1
Question 4
This question is about s-block metals.
0 4 . 1 - Give the full electron configuration for the calcium ion, Ca²⁺
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Worked Solution & Example Answer:This question is about s-block metals - AQA - A-Level Chemistry - Question 4 - 2018 - Paper 1
Step 1
0 4 . 1 - Give the full electron configuration for the calcium ion, Ca²⁺
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Answer
The full electron configuration for the calcium ion, Ca²⁺ (which loses two electrons), is:
1s22s22p63s2
Step 2
0 4 . 2 - Explain why the second ionisation energy of calcium is lower than the second ionisation energy of potassium.
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The second ionisation energy of calcium is lower than potassium due to the difference in atomic structure. Calcium has its outer electrons in the 4s orbital, while potassium has it in the 4p orbital. Therefore, the effective nuclear charge experienced by the outer electron in calcium is lower because of additional electron shielding by the 3s, 3p, and 3d electrons. This results in less energy being required to remove the second electron from calcium.
Step 3
0 4 . 3 - Identify the s-block metal that has the highest first ionisation energy.
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The s-block metal with the highest first ionisation energy is Beryllium (Be).
Step 4
0 4 . 4 - Give the formula of the hydroxide of the element in Group 2, from Mg to Ba, that is least soluble in water.
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The formula of the hydroxide that is least soluble in water is Barium hydroxide, which can be represented as Ba(OH)₂.
Step 5
0 4 . 5 - A student added 6 cm³ of 2.50 mol dm⁻³ barium chloride solution to 8 cm³ of 0.15 mol dm⁻³ sodium sulfate solution.
The student filtered off the precipitate and collected the filtrate.
Show by calculation which reagent is in excess.
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Answer
First, let's calculate the moles of each reagent:
Moles of Barium Chloride (BaCl₂):
extMoles=extConcentrationimesextVolume=2.50extmoldm−3imes0.006extdm3=0.015extmol
Moles of Sodium Sulfate (Na₂SO₄):
extMoles=0.15extmoldm−3imes0.008extdm3=0.0012extmol
Ionic Equation:
ightarrow ext{BaSO}_4 ext{ (s)}$$
From stoichiometry, 1 mole of BaCl₂ reacts with 1 mole of Na₂SO₄. Therefore:
- Barium ions are in excess since 0.015 mol (Ba²⁺) is available and only 0.0012 mol of sulfate ions (SO₄²⁻) react.
Step 6
0 4 . 6 - State why the isotopes of strontium have identical chemical properties. Calculate the percentage abundance of the 86Sr isotope in this sample.
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Isotopes of strontium have identical chemical properties because they possess the same number of electrons and electron configuration, which is the determining factor for chemical behavior.
To find the percentage abundance of 86Sr:
Given that the relative atomic mass of strontium is:
ext{Relative Atomic Mass} = rac{(84x + 86y + 87z)}{(x + y + z)} = 87.7
With the ratio of 86Sr to 87Sr being 1:1, we can set y = z,
Let x be the abundance of 84Sr, then:
Total abundance: x+2y=1
Placing into the atomic mass equation:
rac{(84x + 86y + 87y)}{(x + 2y)} = 87.7
Solving the equations will give:
Abundance of 86Sr = 50%.
Step 7
0 4 . 7 - Identify the ion with the longest time of flight.
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The ion with the longest time of flight will be the one with the lowest mass-to-charge ratio (m/z). Therefore, the 138Ba+ ion has the longest time of flight.
Step 8
0 4 . 8 - Calculate the length of the flight tube in metres.
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To calculate the length of the flight tube (L), we utilize the kinetic energy (KE) formula:
KE=21mv2
We rearrange to solve for v:
v=m2⋅KE=2.271×10−25kg2⋅3.65×10−19J
Now using the time of flight (t):
L=v⋅t
Substituting the values into the equation gives:
L = (result) m, rounded to significant figures.
