AP Chemistry 1.7 Periodic Trends- Exam Style questions - FRQs- New Syllabus

(a)
For a correct electron configuration:
Accept one of the following:

• \( 1s^2\,2s^2\,2p^6\,3s^2\,3p^1 \)
• \( [\mathrm{Ne}]\,3s^2\,3p^1 \)

Aluminum has atomic number \( 13 \), so its \( 13 \) electrons fill orbitals in order up to \( 3p^1 \).

(b)
For a correct explanation:
The highest occupied electron shell \( (n = 3) \) of \( \mathrm{Al} \) is at a greater average distance from the nucleus than the highest occupied electron shell \( (n = 2) \) of \( \mathrm{Al^{3+}} \).

When aluminum forms \( \mathrm{Al^{3+}} \), it loses its three valence electrons, so the entire \( n = 3 \) shell is removed. That makes the ion smaller.

(c)
For the correct steps to dissolve the solute in water \( (\text{steps may be consolidated}) \):
\( 2 \). Partially fill the volumetric flask with some distilled water.
\( 3 \). Add the weighed \( \mathrm{AgNO_3}(s) \) to the volumetric flask.
\( 4 \). Swirl to dissolve the solid.

For the correct step to ensure quantitative dilution:
\( 5 \). After the solid is dissolved, fill the flask to the calibration \( (200.0\ \mathrm{mL}) \) mark and mix.

The volumetric flask is used because it gives the most accurate final solution volume.

(d)
For a drawing that shows product formation and indicates the conservation of matter:
\( 4 \) \( \mathrm{Al} \) atoms and \( 8 \) \( \mathrm{Ag} \) particles in the beaker on right \( (\text{see sample drawing below}) \)

For a drawing that shows product formation and conservation of charge:
\( 2 \) \( \mathrm{Ag^{+}} \) ions and \( 2 \) \( \mathrm{Al^{3+}} \) ions in the beaker on the right \( (\text{see sample drawing below}) \)

For a drawing that shows product formation and correct phases of matter for all species:
\( 6 \) \( \mathrm{Ag} \) atoms that are solid and \( 2 \) \( \mathrm{Al^{3+}} \) ions that are aqueous in the beaker on the right

A quick count check: starting with \( 4 \) aluminum atoms and \( 8 \) silver ions, the reaction uses \( 2 \) aluminum atoms and \( 6 \) silver ions, leaving \( 2 \) aluminum atoms and \( 2 \) silver ions unreacted.

(e)
For the correct calculated value:
Accept one of the following:

• \( E^\circ = 0.80\ \mathrm{V} + 1.66\ \mathrm{V} = 2.46\ \mathrm{V} \)
• \( E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{red}} – E^\circ_{\mathrm{ox}} = 0.80\ \mathrm{V} – (-1.66\ \mathrm{V}) = 2.46\ \mathrm{V} \)

Silver ion is reduced, and aluminum is oxidized. The large positive cell potential shows the reaction is strongly favorable.

(f)
For the correct answer and a valid justification:
Negative. The reaction has a positive value of \( E^\circ \), indicating that it is thermodynamically favorable and would therefore have a negative value of \( \Delta G^\circ \). \( \left( \Delta G^\circ = -nFE^\circ \right) \)

Since \( E^\circ_{\mathrm{cell}} > 0 \), the sign of \( \Delta G^\circ \) must be negative.

(g)
For the correct answer and a valid justification:
Accept one of the following:

• Zero. The observation that the reaction stops progressing implies that \( E_{\mathrm{cell}} = 0 \), indicating that there is no longer a driving force for the reaction.

• Zero. The observation that reaction stops progressing implies that equilibrium is established, and \( \Delta G = 0 \) at equilibrium.

Once the system reaches equilibrium, forward and reverse processes balance, so there is no net free-energy change.