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🧪AP Chemistry
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🧪AP Chemistry

FRQ 1 – Long Answer
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Unit 1: Atomic Structure and Properties
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Practice FRQ 1 of 111/11
1. Answer the following questions about the element boron.
A mass spectrum for a sample of pure boron is shown in Figure 1.

Figure 1. Mass spectrum of boron (relative abundance of boron isotopes).

A clean, black-and-white vertical bar graph.

Axes and layout:
- Horizontal axis label centered below the axis: "Atomic Mass (amu)".
- Horizontal axis numeric range: from 9.5 to 11.5.
- Horizontal axis tick marks and labels: 9.5, 10.0, 10.5, 11.0, 11.5 (tick interval 0.5). Each tick label is directly below its tick.
- Vertical axis label rotated along the left side: "Relative Abundance (%)".
- Vertical axis numeric range: from 0 to 100.
- Vertical axis tick marks and labels: 0, 20, 40, 60, 80, 100 (tick interval 20). 
- The origin is explicitly labeled "0" at the bottom-left intersection of the axes.
- Arrowhead at the positive end of the x-axis (right end) and arrowhead at the positive end of the y-axis (top end).
- No gridlines.

Bars (mass spectrum peaks):
- Exactly two vertical rectangular bars (no slant, no curve). Both bars start at the y=0 baseline and extend upward.
- Bar 1 (boron-10 peak): a single, narrow bar centered exactly above the x-axis value 10.01. The top of the bar ends exactly at the y-axis value 19.9 (just below the 20% tick). Place a small text label directly above the bar top: "19.9%". Place a small text label directly below the x-axis at the bar’s center: "10.01".
- Bar 2 (boron-11 peak): a single, narrow bar centered exactly above the x-axis value 11.01. The top of the bar ends exactly at the y-axis value 80.1 (just above the 80% tick). Place a small text label directly above the bar top: "80.1%". Place a small text label directly below the x-axis at the bar’s center: "11.01".

Relative sizing and alignment constraints:
- The two bars have identical width and identical outline thickness.
- The 11.01 amu bar is visually about four times as tall as the 10.01 amu bar, consistent with 80.1% vs 19.9%.
- No other bars, no legend, and no additional annotations besides the axis labels, tick labels, and the four numeric labels (10.01, 11.01, 19.9%, 80.1%).
A.
i. Using the data in Figure 1, calculate the average atomic mass of boron. Show your work.
ii. Describe the difference in atomic structure that accounts for the difference in mass between the two isotopes shown in Figure 1.
The photoelectron spectrum for boron is shown in Figure 2.

Figure 2. Photoelectron spectrum (PES) of boron showing three peaks.

A black-and-white photoelectron spectrum (line plot with three peaks) on a logarithmic binding-energy axis that increases from right to left.

Axes and layout:
- Horizontal axis label centered below: "Binding Energy (MJ/mol)".
- The x-axis is logarithmic and explicitly reversed: the largest binding energy values are on the far left and the smallest are on the far right.
- Horizontal axis numeric range: from 0.60 on the far right to 25 on the far left.
- Horizontal axis tick marks and labels (log spacing visually, but labeled exactly): 0.60, 0.80, 1.00, 1.36, 2.00, 5.00, 10.0, 19.3, 25.0. Ensure 0.80 and 1.36 appear to the right of 19.3 because the scale is reversed.
- Vertical axis label rotated on the left: "Relative Intensity (arbitrary units)".
- Vertical axis numeric range: from 0 to 2.5.
- Vertical axis tick marks and labels: 0, 1, 2 (tick interval 1). 
- The origin is labeled "0" at the bottom-left corner of the axes.
- Arrowhead at the positive end of the y-axis (top). Arrowhead at the positive end of the x-axis should still appear at the left end (since binding energy increases leftward); include the arrowhead at the left tip of the x-axis.
- No gridlines.

Spectrum trace and peaks:
- Draw a single thin black baseline along y=0 across the entire x-range.
- Draw three smooth, symmetric peaks (Gaussian-like humps) rising from the baseline. Each peak returns to the baseline on both sides (no overlap between peaks).

Peak A (leftmost, highest binding energy):
- Place Peak A in the far-left region of the plot, centered exactly at the x-axis value 19.3 MJ/mol.
- Peak A maximum height reaches exactly the y=2 tick (relative intensity 2).
- Label the peak with a bold letter "A" placed just above the peak apex.
- Place a small numeric label "19.3" directly below the x-axis tick at the peak center (in addition to the axis tick label), ensuring the peak center aligns with that tick.

Peak B (middle binding energy):
- Place Peak B near the middle-right region, centered exactly at the x-axis value 1.36 MJ/mol.
- Peak B maximum height reaches exactly the y=2 tick (relative intensity 2), matching Peak A height.
- Label the peak with a bold letter "B" just above its apex.
- Place a small numeric label "1.36" directly below the x-axis tick at the peak center (in addition to the axis tick label), ensuring alignment.

Peak C (rightmost, lowest binding energy):
- Place Peak C on the far-right region, centered exactly at the x-axis value 0.80 MJ/mol.
- Peak C maximum height reaches exactly the y=1 tick (relative intensity 1), exactly half the height of Peaks A and B.
- Label the peak with a bold letter "C" just above its apex.
- Place a small numeric label "0.80" directly below the x-axis tick at the peak center (in addition to the axis tick label), ensuring alignment.

Relative spacing constraints (to enforce the log and reversed axis):
- The horizontal distance between 1.36 and 0.80 is visibly smaller than the distance between 19.3 and 1.36, consistent with a logarithmic scale.
- Peak order from left to right must be: A at 19.3 (left), then B at 1.36 (middle-right), then C at 0.80 (right).

Line styling:
- All peaks are drawn with the same thin black line.
- No shading under peaks.
- No legend; the letters A, B, C serve as identifiers.
B.
i. Identify the subshell corresponding to the peak at 0.80 MJ/mol in Figure 2.
ii. Explain why the peak at 19.3 MJ/mol is located at a much higher binding energy than the peak at 1.36 MJ/mol using Coulomb's law.
C. A student obtains a pure 5.00 g sample of the isotope boron-11 (molar mass 11.01 g/mol). Calculate the number of boron atoms in the sample.
D. A sample of boron reacts completely with fluorine to form a compound. Elemental analysis reveals that the compound contains 1.50 g of boron and 7.90 g of fluorine. Determine the empirical formula of the compound.
E.
i. Predict whether the atomic radius of boron is larger than, smaller than, or equal to the atomic radius of fluorine.
ii. Justify your prediction in part E(i) using principles of atomic structure.
The first ionization energies for Period 2 elements are plotted in Figure 3.

Figure 3. First ionization energies of Period 2 elements (Li through Ne).

A black-and-white scatter plot (discrete points only) showing first ionization energy versus atomic number for Period 2 elements.

Axes (all required features):
- Horizontal axis label centered below: "Atomic Number".
- Horizontal axis numeric range: from 2 to 11.
- Horizontal axis tick marks and labels: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 (tick interval 1). Emphasize that the data points will fall on the integer ticks 3 through 10.
- Vertical axis label rotated along the left: "First Ionization Energy (kJ/mol)".
- Vertical axis numeric range: from 0 to 2200.
- Vertical axis tick marks and labels: 0, 200, 400, 600, 800, 1000, 1200, 1400, 1600, 1800, 2000, 2200 (tick interval 200).
- The origin is explicitly labeled "0" at the bottom-left intersection of the axes.
- Arrowhead on the positive end of the x-axis (right end) and arrowhead on the positive end of the y-axis (top end).
- No gridlines.

Data points (exact values must be reflected by placement relative to y ticks):
- Plot eight filled circular markers, all the same size.
- Each marker is placed directly above its correct atomic-number tick (integer) and at the exact vertical level of the stated ionization energy value.
- Next to each marker, place the element symbol as visible text, positioned slightly above and to the right of the marker so it does not overlap the dot. The symbol text must be exactly: "Li", "Be", "B", "C", "N", "O", "F", "Ne".

Exact point values to encode (as visible labels or exact placement):
- Li at atomic number 3 with ionization energy 520 kJ/mol (dot slightly above the 500 midpoint between 400 and 600).
- Be at atomic number 4 with ionization energy 900 kJ/mol (dot exactly halfway between 800 and 1000).
- B at atomic number 5 with ionization energy 800 kJ/mol (dot exactly on the 800 tick).
- C at atomic number 6 with ionization energy 1086 kJ/mol (dot slightly above the 1000 tick and clearly below 1200, closer to 1100 than 1000).
- N at atomic number 7 with ionization energy 1400 kJ/mol (dot exactly on the 1400 tick).
- O at atomic number 8 with ionization energy 1314 kJ/mol (dot above 1200 and below 1400, clearly closer to 1300 than 1400).
- F at atomic number 9 with ionization energy 1680 kJ/mol (dot between 1600 and 1800, clearly closer to 1700 than 1600).
- Ne at atomic number 10 with ionization energy 2080 kJ/mol (dot between 2000 and 2200, clearly closer to 2100 than 2000).

Connecting lines and trend depiction:
- Do NOT connect the points with lines.
- No best-fit curve.

Explicit trend features that must be visually obvious from the plotted values:
- From Li to Be, the points rise.
- From Be to B, the plot drops (B lower than Be).
- From B to N, points rise steadily.
- From N to O, the plot drops (O lower than N).
- From O through Ne, points rise to the highest point at Ne.

No additional text beyond axis labels, tick labels, and element symbols.
F. Based on the data in Figure 3, the first ionization energy of boron (800 kJ/mol) is lower than the first ionization energy of beryllium (900 kJ/mol), even though boron has a higher nuclear charge. Explain this phenomenon using the electron configurations of the two elements.






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