Question

name: pd: chemistry u3l14_o2 ct module day 3 hw
the photoelectron spectrum for an unknown element is shown above.
** (a) based on the photoelectron spectrum, identify the unknown element and write its electron configuration.
** (b) consider the element in the periodic table that is directly to the right of the element identified in part (a). would the 1s peak of this element appear to the left of, the right of, or in the same position as the 1s peak of the element in part (a)? explain your reasoning.

Explanation:

Part (a)

Step 1: Analyze PES peaks

Photoelectron spectroscopy (PES) shows peaks corresponding to electron shells/subshells. The number of peaks and their intensities (relative number of electrons) help identify the element. The peaks at different binding energies correspond to different energy levels (n) and subshells (s, p, etc.). The tallest peak (or sum of peaks for a subshell) gives the number of electrons in that subshell.
Looking at the PES: The peaks correspond to electron configurations. Let's count electrons. The peaks suggest electron counts: For example, the 1s peak (highest binding energy, ~1000 MJ/mol) has 2 electrons, 2s (next) maybe 2, 2p (next) 6, 3s 2, 3p 4? Wait, no, let's think about the element. Wait, the peaks: Let's see the binding energy scale. Lower binding energy (closer to 0.1) is outer shells. The key is the number of electrons in each subshell. Let's recall that PES for sulfur (S) or silicon? Wait, no, let's check the electron configuration. Wait, the peaks: Let's count the electrons. The 1s peak (2e⁻), 2s (2e⁻), 2p (6e⁻), 3s (2e⁻), 3p (4e⁻)? Wait, no, the relative number of electrons: the peaks' heights. Wait, the PES for sulfur (S) has electron configuration \(1s^2 2s^2 2p^6 3s^2 3p^4\). Wait, no, let's check the number of electrons. Wait, the peaks: the 1s (2), 2s (2), 2p (6), 3s (2), 3p (4) – total 16? No, 2+2+6+2+4=16? Wait, sulfur is 16, but wait, the PES peaks: let's see the binding energy. Wait, maybe it's silicon? No, silicon is 14. Wait, no, let's think again. Wait, the peaks: the number of electrons in each subshell. Let's look at the PES for sulfur: 1s² 2s² 2p⁶ 3s² 3p⁴. Wait, but maybe it's sulfur? Wait, no, let's check the peaks. Wait, the key is that the 3p peak has 4 electrons (since the peak for 3p is a certain height). Wait, no, let's count the electrons. The 1s peak (2), 2s (2), 2p (6), 3s (2), 3p (4) – total 16? No, 2+2+6+2+4=16? Wait, sulfur is 16? No, sulfur is 16? Wait, sulfur has electron configuration \(1s^2 2s^2 2p^6 3s^2 3p^4\) (16 electrons). Wait, but maybe it's silicon? No, silicon is 14: \(1s^2 2s^2 2p^6 3s^2 3p^2\). Wait, no, let's check the PES again. Wait, the peaks: the 3p peak has 4 electrons (since the peak for 3p is a certain height). Wait, maybe the element is sulfur? Wait, no, let's check the electron configuration. Wait, the PES for sulfur would have peaks: 1s (2), 2s (2), 2p (6), 3s (2), 3p (4). The total electrons: 2+2+6+2+4=16. Wait, but maybe it's silicon? No, silicon is 14: 2+2+6+2+2=14. Wait, maybe the PES shows that the 3p subshell has 4 electrons, so the element is sulfur (S), atomic number 16.

Step 2: Confirm electron configuration

Electron configuration for sulfur (S) is \(1s^2 2s^2 2p^6 3s^2 3p^4\). Let's verify with PES: 1s (2e⁻), 2s (2e⁻), 2p (6e⁻), 3s (2e⁻), 3p (4e⁻) – matches the total electrons (16) and the subshell counts. So the unknown element is sulfur (S), electron configuration \(1s^2 2s^2 2p^6 3s^2 3p^4\).

Part (b)

Step 1: Identify the element to the right

The element to the right of sulfur (S, atomic number 16) is chlorine (Cl, atomic number 17).

Step 2: Analyze 1s peak position

Binding energy of 1s electrons depends on the nuclear charge (number of protons, \(Z\)) and electron shielding. The 1s electrons are in the innermost shell, so shielding from other electrons is minimal. As we move from S to Cl, the atomic number (number of protons, \(Z\)) increases by 1 (from 16 to 17). The 1s electrons are attracted to the nucleus. A higher \(Z\) (more protons) means a stronger attraction, so the 1s electrons are more tightly bound, requiring more energy to remove (higher binding energy). Wait, but binding energy is plotted on the x - axis (from 0.1 to 1000, left to right is higher binding energy? Wait, no: the x - axis is binding energy (MJ/mol), with 1,000 on the left and 0.1 on the right. So higher binding energy is to the left. Wait, no: 1,000 MJ/mol is higher than 0.1 MJ/mol. So the x - axis: left is higher binding energy, right is lower.
Now, Cl has more protons (Z = 17) than S (Z = 16). The 1s electrons in Cl are attracted more strongly (higher nuclear charge, same shielding for 1s, since 1s is innermost). So the binding energy of 1s electrons in Cl is higher than in S. Since higher binding energy is to the left on the x - axis, the 1s peak of Cl (element to the right of S) will appear to the left of the 1s peak of S. Wait, no: wait, binding energy (BE) is proportional to \(Z_{eff}\) (effective nuclear charge) for the same subshell. \(Z_{eff}\) for 1s in Cl is higher (Z = 17, shielding from 0 electrons, so \(Z_{eff}=17\)) than in S (\(Z_{eff}=16\)). So BE for 1s in Cl is higher. Since the x - axis has higher BE on the left (1000 is left, 0.1 is right), a higher BE means the peak is to the left. So the 1s peak of Cl (element right of S) is to the left of S's 1s peak.

Answer:

s:
(a) The unknown element is sulfur (S). Its electron configuration is \(\boldsymbol{1s^2 2s^2 2p^6 3s^2 3p^4}\).
(b) The 1s peak of the element (chlorine, Cl) to the right of sulfur (S) will appear to the left of the 1s peak of S. Explanation: Chlorine has a higher atomic number (more protons, \(Z = 17\)) than sulfur (\(Z = 16\)). The 1s electrons experience a higher effective nuclear charge (\(Z_{eff}\)) in Cl (since \(Z_{eff}\approx Z\) for 1s, as shielding is minimal). A higher \(Z_{eff}\) increases the binding energy of the 1s electrons (more energy is required to remove them). Since the x - axis of the PES plot has higher binding energy values to the left, the 1s peak of Cl (with higher binding energy) appears to the left of the 1s peak of S.