Atomic Structure and Properties

2.a) The mass spectrum of a pure sample of Si is shown below.

![](https://zustagstorage.blob.core.windows.net/community/files_questions/frq/ap24-frq-chemistry_661403751e0f4bde804cb8ba1ebde946)

    2.a.i) How many protons and how many neutrons are in the nucleus of an atom of the most abundant isotope of Si? (1 point)

    14 protons and 14 neutrons

    2.a.ii) Write the ground-state electron configuration of Si. (1 point)

    <math-inline>1s^22s^22p^63s^23p^2</math-inline>

2.b) Two compounds that contain Si are $SiO_2$ and $SiH_4$.

At 161 K, <math-inline>SiH\_4</math-inline> boils but <math-inline>SiO\_2</math-inline> remains as a solid. Using principles of interparticle forces, explain the difference in boiling points. (2 points)

<math-inline>SiH\_4</math-inline> is a nonpolar molecule and has only London dispersion forces. <math-inline>SiO\_2</math-inline> is a network covalent solid with strong covalent bonds between Si and O atoms. Therefore, it requires much more energy to overcome the covalent bonds in <math-inline>SiO\_2</math-inline> than the weak London dispersion forces in <math-inline>SiH\_4</math-inline>.

2.c) At high temperatures, $SiH_4$ decomposes to form solid silicon and hydrogen gas.

    2.c.i) Write a balanced equation for the reaction. (1 point)

    <math-inline>SiH\_4(g) \rightarrow Si(s) + 2H\_2(g)</math-inline>

2.d) A table of absolute entropies of some substances is given below.

|Substance  | <math-inline>S°</math-inline> (J/(mol × K)) |
|---|---|
|<math-inline>H\_2(g)</math-inline>  | 131 |
|<math-inline>Si(s)</math-inline>  | 18 |
|<math-inline>SiH\_4(g)</math-inline>  | 205 |

    2.d.i) Explain why the absolute molar entropy of <math-inline>Si(s)</math-inline> is less than that of <math-inline>H\_2(g)</math-inline>. (1 point)

    The absolute molar entropy of <math-inline>Si(s)</math-inline> is less than that of <math-inline>H\_2(g)</math-inline> because solids have less freedom of motion than gases.

    2.d.ii) Calculate the value, in J/(mol× K), of <math-inline>ΔS°</math-inline> for the reaction. (2 points)

    <math-inline>ΔS° = ΣS°\_{products} - ΣS°\_{reactants}</math-inline>
    <math-inline>ΔS° = (18 + 2(131)) - 205 = 75 J/(mol \times K)</math-inline>

    2.d.iii) The reaction is thermodynamically favorable at all temperatures. Explain why the reaction occurs only at high temperatures. (1 point)

    The reaction is thermodynamically favorable at all temperatures but occurs only at high temperatures because the reaction has a high activation energy.

2.e) A partial photoelectron spectrum of pure Si is shown below. On the spectrum, draw the missing peak that corresponds to the electrons in the 3p sublevel. (1 point)

![](https://zustagstorage.blob.core.windows.net/community/files_questions/frq/ap24-frq-chemistry_373e9bc9e5be46d88b5ccb19ec6306e3)

2.f) Using principles of atomic structure, explain why the first ionization energy of Ge is lower than that of Si. (1 point)

The first ionization energy of Ge is lower than that of Si because Ge has electrons in the 4p subshell, which are farther from the nucleus and are more shielded by the inner electrons than the 3p electrons in Si. Thus, the outer electrons in Ge are easier to remove.

2.g) A single photon with a wavelength of $4.00 \times 10^{-7}$ m is absorbed by the Si sample. Calculate the energy of the photon in joules. (2 points)

<math-inline>E = \frac{hc}{λ} = \frac{(6.626 \times 10^{-34} Js)(2.998 \times 10^8 m/s)}{4.00 \times 10^{-7} m} = 4.97 \times 10^{-19} J</math-inline>

3.a. Write the complete ground-state electron configuration of an Al atom.

$1s^22s^22p^63s^23p^1$

3.b. Based on principles of atomic structure, explain why the radius of the Al atom is larger than the radius of the $Al^{3+}$ ion.

The $Al^{3+}$ ion has three fewer electrons than the Al atom. The remaining electrons in the ion experience a stronger effective nuclear charge, which pulls them closer to the nucleus, resulting in a smaller radius.

3.c. A student plans to combine solid aluminum with an aqueous solution of silver ions. The student determines the mass of solid $AgNO_3$ needed to prepare the solution with a specific concentration.

3.c.i. In the following table, briefly list the steps necessary to prepare 200.0 mL of an aqueous solution of $AgNO_3$ using only equipment selected from the choices given. Assume that all appropriate safety measures are already in place. Not all equipment or lines in the table may be needed.

• Solid $AgNO_3$ · Weighing paper and scoop · 250 mL beakers
• Distilled water · 200.00 mL volumetric flask · Pipet
• Balance · 50.0 mL graduated cylinder

3.d. After preparing the solution, the student places some of the solution into a beaker and adds a sample of aluminum. The reaction represented by the following equation occurs.

$Al(s) + 3Ag^+(aq) \rightarrow Al^{3+}(aq) + 3Ag(s)$

3.d.i. The following diagram gives an incomplete particulate representation of the reaction. The beaker on the left represents the system before the mixture reacts. Complete the drawing on the right to represent the system after the reaction has occurred. Be sure to include 1) the correct type and number of particles based on the number shown on the left and 2) the relative spacing to depict the appropriate phases.

3.e. The student finds the standard reduction potentials given in the table, which are related to the reaction that occurs.

3.e.i. Using the standard reduction potentials, calculate the value of $E^\circ$ for the reaction.

$E^\circ = E^\circ_{cathode} - E^\circ_{anode} = 0.80 - (-1.66) = 2.46 V$

3.e.ii. Based on the value of $E^\circ$, would the standard free energy change of the reaction under standard conditions, $\Delta G^\circ$, be positive, negative, or zero? Justify your answer.

$\Delta G^\circ$ would be negative. Since $E^\circ$ is positive, $\Delta G^\circ = -nFE^\circ$ would be negative.

3.e.iii. Once the reaction appears to stop progressing, would the change in free energy, $\Delta G$, be positive, negative, or zero? Justify your answer.

$\Delta G$ would be zero. At equilibrium, $\Delta G$ = 0