Question

name: pd: chemistry usl14 q2 ct module day 3 hw note: annotate all questions & diagrams! cross out part of the answer choices that makes that answer incorrect. 1. which of the following ions has the same number of electrons as br? (a) k+ (b) sr2+ (c) i- (d) cl- 2. if na reacts with chlorine to form nacl, which of the following elements reacts with na to form an ionic compound in a one - to - one ratio, and why? (a) k, because it is in the same group as na. (b) mg, because its mass is similar to that of na. (c) ar, because its mass is similar to that of cl. (d) br, because it has the same number of valence electrons as cl. 3. the complete photoelectron spectrum for an element is shown above. which of the following observations would provide evidence that the spectrum is consistent with the atomic model of the element? (a) a neutral atom of the element contains exactly two electrons. (b) the element does not react with other elements to form compounds. (c) in its compounds, the element tends to form ions with a charge of + 1. (d) in its compounds, the element tends to form ions with a charge of + 3.

Explanation:

Question 1

Step 1: Determine the number of electrons in \( \text{Br}^- \)

Bromine (Br) has an atomic number of 35, so a neutral Br atom has 35 electrons. The \( \text{Br}^- \) ion has gained 1 electron, so it has \( 35 + 1 = 36 \) electrons.

Step 2: Analyze each ion

• Option A: \( \text{K}^+ \)

Potassium (K) has an atomic number of 19. The \( \text{K}^+ \) ion has lost 1 electron, so it has \( 19 - 1 = 18 \) electrons. Not equal to 36.

• Option B: \( \text{Sr}^{2+} \)

Strontium (Sr) has an atomic number of 38. The \( \text{Sr}^{2+} \) ion has lost 2 electrons, so it has \( 38 - 2 = 36 \) electrons. Equal to \( \text{Br}^- \)’s electrons.

• Option C: \( \text{I}^- \)

Iodine (I) has an atomic number of 53. The \( \text{I}^- \) ion has gained 1 electron, so it has \( 53 + 1 = 54 \) electrons. Not equal to 36.

• Option D: \( \text{Cl}^- \)

Chlorine (Cl) has an atomic number of 17. The \( \text{Cl}^- \) ion has gained 1 electron, so it has \( 17 + 1 = 18 \) electrons. Not equal to 36.

To form an ionic compound in a 1:1 ratio with Na, the element must form an ion with a charge opposite to Na⁺ (which has a +1 charge) and with a magnitude of 1 (so the charges cancel in a 1:1 ratio).

• Option A: K

K is in the same group as Na (Group 1) and forms \( \text{K}^+ \) (same charge as \( \text{Na}^+ \)). Two positive ions cannot form an ionic compound (needs a cation and anion). Eliminate.

• Option B: Mg

Mg is in Group 2 and forms \( \text{Mg}^{2+} \). To form a compound with \( \text{Na}^+ \), the ratio would be 2:1 (e.g., \( \text{Na}_2\text{Mg} \) is not typical; actual ratio for \( \text{Na}^+ \) and \( \text{Mg}^{2+} \) would be 2:1 for charge balance). Eliminate.

• Option C: Ar

Ar is a noble gas and is chemically inert (does not form ions easily). Eliminate.

• Option D: Br

Br is in Group 17 (halogens) and forms \( \text{Br}^- \) (charge of -1). \( \text{Na}^+ \) (charge +1) and \( \text{Br}^- \) (charge -1) combine in a 1:1 ratio to form \( \text{NaBr} \), similar to \( \text{NaCl} \). Br has 7 valence electrons (same as Cl), so it forms a -1 ion.

The photoelectron spectrum (PES) shows two peaks. The peak at lower binding energy (closer to 0.1 MJ/mol) corresponds to valence electrons, and the peak at higher binding energy (closer to 10 MJ/mol) corresponds to core electrons.

• The area of the valence peak (right peak) is smaller, and the core peak (left peak) is larger. From the x-axis (binding energy) and relative number of electrons, we infer:
• The left peak (core) has more electrons (larger area) – likely 2 electrons (K shell).
• The right peak (valence) has fewer electrons – likely 1 electron (valence shell). Wait, no—wait, binding energy increases with shell closeness to the nucleus. Wait, actually, the peak with higher binding energy (left peak, ~10 MJ/mol) is core electrons, and the lower binding energy (right peak, ~1 MJ/mol) is valence.

Wait, let’s re-express:

• The peak at ~10 MJ/mol (higher binding energy) has a larger area (more electrons) – this is the core (inner) electrons.
• The peak at ~1 MJ/mol (lower binding energy) has a smaller area – valence electrons.

From the atomic model, if the core has 2 electrons (K shell) and valence has 1 electron (valence shell), the element is Li? No, wait—wait, the options:

• Option A: Neutral atom has exactly two electrons – If the core peak (left) has 2 electrons, and valence has 1, total electrons = 3 (Li). But Li has 3 electrons, but the peak areas: the left peak (core) is larger (more electrons) – if core is 2, valence is 1, total 3. But option A says “exactly two electrons” – no, Li has 3. Eliminate.

• Option B: Element does not react – Noble gases have full valence shells, but the PES here has two peaks (core and valence), so not a noble gas. Eliminate.

• Option C: Forms +1 ion – If valence has 1 electron, the element would lose 1 electron to form +1 ion. But wait, the core peak (left) is larger (more electrons) – maybe the core is 2, valence is 1 (total 3: Li, forms +1). But wait, the other option (D, likely a typo) – wait, original options may have a mistake. Wait, re-reading:

Wait, the correct analysis: The PES has two peaks. The left peak (higher binding energy) has a larger area (more electrons) – this is the inner shell (e.g., n=1, 2 electrons), and the right peak (lower binding energy) has a smaller area (e.g., n=2, 1 electron). So the element has 3 electrons (Li), which forms +1 ions (loses its 1 valence electron). But wait, the options:

Wait, the options given:
(A) Neutral atom has exactly two electrons – No, Li has 3.
(B) Element does not react – No, Li is reactive.
(C) Forms +1 ion – Li forms +1, B forms +3, etc. Wait, maybe the element is B? No, B has 5 electrons. Wait, maybe the peaks: the left peak (core) has 2 electrons, and the right peak (valence) has 3 electrons (total 5: B). Then, B forms +3 ions (loses 3 valence electrons). So:

• If the valence peak (right) has 3 electrons, the element (B) forms +3 ions. So option C or D (typo) – assuming option D was a repeat, the correct answer is that the element forms +3 ions (loses 3 valence electrons), so in compounds, it tends to form +3 ions.

Answer:

B. \( \text{Sr}^{2+} \)

Question 2