IB DP Chemistry Topic 2.2 Electron configuration SL Paper 2

a. \(63x + 65(1 – x) = 63.55\);

(or some other mathematical expression).

\(^{63}{\text{Cu}} = 72.5\% \) and \(^{65}{\text{Cu}} = 27.5\% \);

Allow ⁶³Cu = 0.725 and ⁶⁵Cu = 0.275.

Award [2] for correct final answer.

⁶⁰Co /¹³¹I /¹²⁵I;

Must contain correct mass numbers.

Allow other formats such as cobalt-60, Co-60 etc.

Award no marks if a correct radioisotope is given with an incorrect radioisotope.

Allow any other radioisotope if you can verify its use.

\({\text{2Na(s)}} + {\text{2}}{{\text{H}}_2}{\text{O(l)}} \to {\text{2NaOH(aq)}} + {{\text{H}}_2}{\text{(g) / Na(s)}} + {{\text{H}}_2}{\text{O(l)}} \to {\text{NaOH(aq)}} + \frac{1}{2}{{\text{H}}_2}{\text{(g)}}\)

Award [1] for correct balanced equation.

Award [1] for correct state symbols for sodium, water, sodium hydroxide and hydrogen.

Second mark is not dependent on equation being correctly balanced.

(Rb more reactive because) electron lost further from nucleus so less tightly held;

Rb electron is in 5th energy level and (Na less reactive) as electron lost in 3rd energy level / OWTTE;

Allow [1 max] for electron arrangements of Na (e.g. 2,8,1) and Rb if second mark is not scored.

(i)     solution becomes yellow/orange/brown/darker;

chlorine is more reactive than iodine (and displaces it from solution) / OWTTE;

Allow correct equation (KI + Cl₂ \( \to \) KCl + I₂) for second mark or stating that iodine/I₂ is formed.

(ii)     no colour change/nothing happens as fluorine is more reactive than chlorine / OWTTE;

Detailed Explanation:

a. Relative atomic mass = Sum of products of isotope mass and fraction abundance

Let fraction abundance of Cu-63 be x. 

63.55 = 63x + 65(1-x)

x=0.725

Cu-63 = 72.5% and Cu-65 = 27.5%.

b. Radioisotopes are widely used in medicine for various diagnostic and therapeutic purposes. Some commonly used radioisotopes in medicine include:Iodine-131, Cobalt-60,Gallium-67 etc.

c.

The balanced equation for the reaction of sodium (Na) with water (H2O) is as follows:

2Na + 2H2O -> 2NaOH + H2

In this reaction, sodium reacts with water to produce sodium hydroxide (NaOH) and hydrogen gas (H2). The equation is balanced, as there are equal numbers of atoms of each element on both sides of the equation.

d. The reaction between rubidium (Rb) and water is more vigorous than that between sodium (Na) and water due to the larger size and lower ionization energy of rubidium compared to sodium. Rubidium is located below sodium in Group 1 of the periodic table. As you move down the group, the size of the atoms increases due to the addition of more electron shells. Rubidium atoms are larger than sodium atoms, which means they have more electron-electron repulsion and a more loosely held outer electron.

e. 

i.

When chlorine gas (Cl2) is bubbled through a solution of potassium iodide (KI), a redox reaction occurs. The chlorine gas oxidizes iodide ions (I-) to iodine (I2) while being reduced itself. The balanced chemical equation for this reaction is as follows:

Cl2 + 2KI -> 2KCl + I2

In this reaction, the chlorine gas reacts with potassium iodide to form potassium chloride (KCl) and elemental iodine (I2). The chlorine gas is reduced from a zero oxidation state to a -1 oxidation state in potassium chloride, while iodide ions are oxidized from a -1 oxidation state to a zero oxidation state in iodine. The liberated iodine can be observed as a characteristic brown color, indicating the formation of I2. The potassium ions (K+) remain in solution as spectator ions and do not participate in the redox reaction.

ii. if chlorine gas is bubbled through a solution of potassium fluoride, no colour change/nothing happens as fluorine is more reactive than chlorine.