unit 1 hw: atomic structure & properties
part i: short response
use the pes spectrum below to answer questions 1–4.
1. what element does this spectrum represent? _______________
2. which peak represents the 2s subshell? place an ‘x’ above this peak on the pes.
3. an electron from which peak (104, 6.84, 4.98, 2.29, 1.76 mj/mol) would have the greatest velocity after ejection? why?
4. how many valence electrons does this atom have?
5. why does an anion of p³⁻ have a larger radius than a neutral atom of p?
section 2 is mcq on college board

Question

Accepted Answer

### Question 1 # Explanation: ## Step1: Analyze PES peaks (electron counts) The PES spectrum has peaks with relative electron counts (from heights, considering binding energy - lower binding energy is outer shells). The peak at 100 MJ/mol (highest binding energy, 1s) has 104? Wait, no, wait: Wait, PES peak areas (or heights, if normalized) represent number of electrons. Wait, typical PES for atomic structure: binding energy increases with shell (1s < 2s < 2p < 3s < 3p... no, wait, higher binding energy is closer to nucleus. So 1s has highest binding energy, then 2s, 2p, 3s, 3p, etc. The number of electrons: 1s², 2s², 2p⁶, 3s², 3p³ for P? Wait, no, let's calculate total electrons. Let's sum the electrons: the peak at ~100 (1s) has 104? Wait, no, maybe the y - axis is relative number, so the heights: 104 (1s: 2 electrons? Wait, no, maybe the numbers are the binding energy (x - axis) and the height is number of electrons. Wait, let's list the peaks with their binding energy (MJ/mol) and relative electron count (height): - 100 MJ/mol: height 104 (1s subshell, 2 electrons? Wait, no, maybe the values are the number of electrons. Wait, no, PES peak intensity (area) is proportional to number of electrons. Let's assume the peaks correspond to subshells: Highest binding energy (leftmost) is 1s: let's say the peak at ~100 MJ/mol (1s) has 2 electrons (but the height is 104? Maybe it's a typo, or maybe the numbers are the number of electrons. Wait, then next peak: around 6.84 MJ/mol (maybe 2s? No, 2s should have higher binding energy than 2p. Wait, no, binding energy: 1s > 2s > 2p > 3s > 3p. So the leftmost peak (highest binding energy, ~100 MJ/mol) is 1s (2 electrons). Then next peak: let's see the binding energy values: 100, then a small peak, then 6.84, 4.98, 2.29, 1.76. Wait, maybe the peak at 6.84 MJ/mol is 2s (2 electrons), 4.98 is 2p (6 electrons), 2.29 is 3s (2 electrons), 1.76 is 3p (3 electrons). Wait, let's sum the electrons: 2 (1s) + 2 (2s) + 6 (2p) + 2 (3s) + 3 (3p) = 15? No, 2 + 2 + 6 + 2 + 3 = 15? Wait, phosphorus (P) has atomic number 15? Wait, no, P has 15 electrons: 1s² 2s² 2p⁶ 3s² 3p³. Wait, but 2 + 2 + 6 + 2 + 3 = 15. Wait, but the peak at 4.98: if 2p has 6 electrons, the height should be 6 times the 2s (which has 2). Wait, the peak at 4.98 has height 4.98, 6.84 has 6.84? Wait, maybe the numbers are the number of electrons. So 1s: 104? No, that can't be. Wait, maybe the x - axis is binding energy (MJ/mol) and the y - axis is relative number of electrons, so the area (or height, if normalized) is number of electrons. Let's re - evaluate: Typical PES for P: 1s (2 e⁻, high binding energy), 2s (2 e⁻, next), 2p (6 e⁻, next), 3s (2 e⁻, next), 3p (3 e⁻, lowest binding energy). So the peaks from left (high binding) to right (low binding): 1. 1s: 2 e⁻ (highest binding, ~100 MJ/mol) 2. 2s: 2 e⁻ (next binding energy, ~6 - 7 MJ/mol) 3. 2p: 6 e⁻ (next, ~5 MJ/mol) 4. 3s: 2 e⁻ (next, ~2 - 3 MJ/mol) 5. 3p: 3 e⁻ (lowest binding, ~1 - 2 MJ/mol) Looking at the given peaks: binding energy (MJ/mol) and height (relative electrons): - 100 MJ/mol: height 104 (maybe 2 e⁻, but scaled? Or maybe the numbers are the number of electrons. Wait, 104 is too big. Wait, maybe the y - axis is relative intensity, and the numbers are the intensity values. So we need to find the ratio of intensities to get the number of electrons. The peak at 4.98 has the highest intensity (so most electrons, 6 e⁻ for 2p). Then 6.84 (2 e⁻ for 2s), 104 (2 e⁻ for 1s), 2.29 (2 e⁻ for 3s), 1.76 (3 e⁻ for 3p). Let's check the ratios: 4.98 (6 e⁻) : 6.84 (2 e⁻) = 4.98/6.84 ≈ 0.728, and 6/2 = 3? No, that's not. Wait, maybe the peak at 4.98 is 2p (6 e⁻), 6.84 is 2s (2 e⁻), 104 is 1s (2 e⁻), 2.29 is 3s (2 e⁻), 1.76 is 3p (3 e⁻). Let's sum the electrons: 2 (1s) + 2 (2s) + 6 (2p) + 2 (3s) + 3 (3p) = 15. Atomic number 15 is phosphorus (P). So the element is Phosphorus (P). ## Step2: Confirm with electron configuration Electron configuration of P: $1s^2 2s^2 2p^6 3s^2 3p^3$. The number of electrons in each subshell: 1s (2), 2s (2), 2p (6), 3s (2), 3p (3). The PES peaks should correspond to these subshells with their electron counts. The peak with the most electrons (highest intensity) is 2p (6 e⁻), which matches the peak at 4.98 (highest height). Then 2s (2 e⁻) at 6.84, 1s (2 e⁻) at 100, 3s (2 e⁻) at 2.29, 3p (3 e⁻) at 1.76. Summing these (2 + 2 + 6 + 2 + 3 = 15) gives atomic number 15, which is P. # Answer: Phosphorus (or P) ### Question 2 # Explanation: ## Step1: Recall subshell binding energy order Subshells have binding energy in the order: $1s > 2s > 2p > 3s > 3p$. So 2s is between 1s (highest binding) and 2p. ## Step2: Identify 2s peak In the PES spectrum, the peak corresponding to 2s should have a binding energy higher than 2p (so to the left of 2p peak) and lower than 1s (to the right of 1s peak). The peak at 6.84 MJ/mol is between the 1s peak (100 MJ/mol) and the 2p peak (4.98 MJ/mol? Wait, no, 6.84 is higher than 4.98? Wait, binding energy on x - axis: lower x - value is lower binding energy. Wait, I had it reversed! Binding energy (MJ/mol) on x - axis: higher value is higher binding energy (closer to nucleus). So left is higher binding energy (1s), then moving right, binding energy decreases (2s, 2p, 3s, 3p). So the order from left (high binding) to right (low binding) is 1s (100 MJ/mol), then 2s (6.84 MJ/mol), then 2p (4.98 MJ/mol), then 3s (2.29 MJ/mol), then 3p (1.76 MJ/mol). Because 6.84 > 4.98 > 2.29 > 1.76 (higher binding energy is larger number). So 2s is the peak at 6.84 MJ/mol. So we place an 'x' above the peak at 6.84 MJ/mol. # Answer: Place 'x' above the peak with binding energy 6.84 MJ/mol. ### Question 3 # Explanation: ## Step1: Relate binding energy to kinetic energy When an electron is ejected, the kinetic energy (KE) of the electron is given by $KE = h u - BE$, where $h u$ is the energy of the photon (assuming photoelectron spectroscopy) and $BE$ is the binding energy of the electron. If we assume the same photon energy (or that the excess energy after overcoming binding energy goes into kinetic energy), then lower binding energy means higher kinetic energy (since $KE\propto\frac{1}{BE}$ in terms of the excess, assuming photon energy is constant and greater than BE). ## Step2: Relate kinetic energy to velocity Kinetic energy is also given by $KE=\frac{1}{2}mv^2$. So higher kinetic energy means higher velocity (since mass of electron is constant). So the electron with the lowest binding energy will have the highest kinetic energy and thus the highest velocity. ## Step3: Identify the lowest binding energy peak Among the peaks (104, 6.84, 4.98, 2.29, 1.76 MJ/mol), the lowest binding energy is 1.76 MJ/mol. So the electron from the peak at 1.76 MJ/mol will have the greatest velocity because it has the lowest binding energy, so the most excess energy (kinetic energy) after ejection, and higher kinetic energy means higher velocity (from $KE = \frac{1}{2}mv^2$). # Answer: The electron from the peak at 1.76 MJ/mol will have the greatest velocity. This is because binding energy is the energy required to eject an electron; a lower binding energy means more of the photon's energy (or the energy used to eject) is converted into kinetic energy ($KE=\frac{1}{2}mv^2$). Since kinetic energy is proportional to the square of velocity (for constant mass), lower binding energy leads to higher kinetic energy and thus higher velocity. ### Question 4 # Explanation: ## Step1: Recall valence electrons definition Valence electrons are the electrons in the outermost shell (highest principal quantum number, n). For phosphorus (P), the electron configuration is $1s^2 2s^2 2p^6 3s^2 3p^3$. The outermost shell is n = 3, which has $3s^2 3p^3$ electrons. ## Step2: Calculate number of valence electrons Sum the electrons in n = 3: $2 + 3=5$. Alternatively, from the PES spectrum, the outermost subshells are 3s (2 electrons, peak at 2.29 MJ/mol) and 3p (3 electrons, peak at 1.76 MJ/mol). Summing these: $2 + 3 = 5$. # Answer: 5 ### Question 5 # Explanation: ## Step1: Recall ion formation and electron - electron repulsion A $P^{3 - }$ anion is formed when a neutral P atom gains 3 electrons. The neutral P atom has an electron configuration of $1s^2 2s^2 2p^6 3s^2 3p^3$. The $P^{3 - }$ anion has an electron configuration of $1s^2 2s^2 2p^6 3s^2 3p^6$. ## Step2: Analyze electron - electron repulsion and radius When electrons are added to form the anion, the number of electrons increases while the number of protons (nuclear charge, +15) remains the same. The increased number of electrons in the same valence shell (n = 3 for $P^{3 - }$, since it's isoelectronic with Ar) leads to greater electron - electron repulsion. This repulsion causes the electron cloud to expand, increasing the distance from the nucleus to the outermost electrons, thus increasing the atomic radius. # Answer: A $P^{3 - }$ anion has 3 more electrons than a neutral P atom. The added electrons increase electron - electron repulsion in the valence shell (n = 3 for $P^{3 - }$), causing the electron cloud to expand. Since the nuclear charge (number of protons) remains the same, the outermost electrons are held less tightly, resulting in a larger atomic radius compared to the neutral P atom.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Question 1

Step1: Analyze PES peaks (electron counts)

Step1: Recall subshell binding energy order

Step2: Identify 2s peak

Step1: Relate binding energy to kinetic energy

Step2: Relate kinetic energy to velocity

Step3: Identify the lowest binding energy peak

Answer:

Question 2