name:______ pd:______ chemistry u3l13_q2 ct module day 2 hw\n* 6.\nbased on periodic trends and the data in…

Question 6

Explanation:

Step1: Analyze periodic trends for atomic radius

Potassium (K) is in period 4, group 1; Calcium (Ca) is in period 4, group 2. Atomic radius increases down a group and decreases across a period. K is to the left of Ca in the same period? No, K is in period 4, Ca is also period 4, but K is group 1, Ca group 2. Wait, actually, K is in period 4, group 1; Ca is period 4, group 2. But also, K is below Na, and Ca is below Mg. Wait, the trend for atomic radius: as we move from left to right across a period, atomic radius decreases. But K is in period 4, group 1; Ca is period 4, group 2. But also, K has one more shell than, say, Na, but Ca is same period as K. Wait, no: K is [Ar]4s¹, Ca is [Ar]4s². The atomic radius of K should be larger than Ca because K has less effective nuclear charge (same period, left of Ca). So Ca has atomic radius 194 pm, so K should be larger than 194 pm? Wait, no, wait: Potassium is in period 4, group 1; Calcium is period 4, group 2. But also, Potassium is below Sodium, and Calcium is below Magnesium. Wait, the atomic radius of K: let's recall, atomic radius increases down a group. K is in group 1, period 4; Na is group 1, period 3. So K has larger radius than Na. Ca is group 2, period 4. So across period 4, from K (group 1) to Ca (group 2), atomic radius decreases. So K's atomic radius should be larger than Ca's 194 pm. So options with 242 pm (larger than 194) are A or B.

Step2: Analyze first ionization energy trend

First ionization energy (IE) is the energy to remove the outermost electron. IE decreases down a group and increases across a period. K is in group 1, period 4; Ca is group 2, period 4. So K is to the left of Ca in the same period, so IE of K should be less than IE of Ca (590 kJ/mol). So we need IE less than 590. So 419 kJ/mol is less than 590, 633 is more. So IE of K should be less than 590, so 419 kJ/mol. And atomic radius larger than 194 pm, so 242 pm. So the values are 242 pm, 419 kJ/mol, which is option B.

Answer: B. 242 pm, 419 kJ/mol

Question 7

Explanation:

Step1: Analyze photoelectron spectrum (PES)

In PES, the peak with higher binding energy (left on the graph) is the inner shell (lower n, higher binding energy). The peak with lower binding energy (right) is outer shell (higher n, lower binding energy). The relative number of electrons: peak X and peak Y. Let's look at the binding energy: Peak X is at higher binding energy (around 12-13 MJ/mol), Peak Y at lower (around 1-2 MJ/mol). The number of electrons: Peak X has a certain height, Peak Y has a height. Let's check the elements: Li and Be.

Li: electron configuration 1s² 2s¹. So 1s has 2 electrons, 2s has 1 electron.

Be: electron configuration 1s² 2s². 1s has 2, 2s has 2.

Now, binding energy: 1s orbitals have higher binding energy than 2s (since they are closer to the nucleus). So Peak X (higher binding energy) should be the inner shell (1s), Peak Y (lower binding energy) outer shell (2s). Wait, no: binding energy is higher for inner shells. So the peak with higher binding energy (left) is 1s, lower (right) is 2s? Wait, no: the x-axis is binding energy (MJ/mol), with 0 on the right. So higher binding energy is to the left (higher numbers). So Peak X is at higher binding energy (left), Peak Y at lower (right).

Now, for Li: 1s² (2 electrons), 2s¹ (1 electron). So the peak for 1s should have twice as many electrons as 2s? Wait, no, the relative number of electrons: the height of the peak is proportional to the number of electrons in the orbital. So for Li: 1s has 2 electrons, 2s has 1. So the 1s peak (higher binding energy) should be taller than 2s? Wait, no, in the graph, Peak X and Peak Y: Peak X is at higher binding energy (left), Peak Y at lower (right). Let's check the options:

Option A: Li, Peak X 1s, Peak Y 2s. But Li has 1s² (2 e⁻) and 2s¹ (1 e⁻). So 1s peak should have more electrons (taller) than 2s. But in the graph, Peak X and Peak Y: are their heights? Wait, the graph shows Peak X and Peak Y with similar heights? Wait, no, the graph: Peak X is at binding energy ~12, Peak Y ~1. Wait, maybe I got the binding energy direction wrong. Wait, binding energy is the energy required to remove an electron, so higher binding energy means the electron is more tightly bound (closer to nucleus, lower energy level, like 1s). Lower binding energy means less tightly bound (higher energy level, like 2s or 3s).

Wait, for Be: electron configuration 1s² 2s². So 1s has 2 e⁻, 2s has 2 e⁻. So both peaks would have the same number of electrons (same height). For Li: 1s² 2s¹, so 1s peak (2 e⁻) taller than 2s (1 e⁻). In the graph, Peak X and Peak Y have similar heights? Wait, the graph: Peak X is at binding energy ~12, Peak Y ~1. Let's check the number of electrons: the y-axis is relative number of electrons. So if Peak X and Peak Y have the same height, that means same number of electrons. So Be has 1s² (2 e⁻) and 2s² (2 e⁻), so both peaks same height. Li has 1s² (2) and 2s¹ (1), so different heights. So the element should be Be? Wait, no, wait the options:

Option C: Be, Peak X 1s, Peak Y 2s. But binding energy: 1s has higher binding energy (left), 2s lower (right). So Peak X (left, higher binding energy) is 1s, Peak Y (right, lower) is 2s? But for Be, 1s has 2 e⁻, 2s has 2 e⁻, so same height. But in the graph, Peak X and Peak Y: are their heights same? The graph shows two peaks, Peak X and Peak Y, with similar heights? Wait, maybe I mixed up the binding energy axis. Wait, the x-axis is binding energy (MJ/mol), with 0 on the right, so higher binding energy is to the left (higher numbers). So Peak X is at ~12 MJ/mol (high binding energy, 1s), Peak Y at ~1 MJ/mol (low binding energy, 2s). Now, for Be: 1s² (2 e⁻) and 2s² (2 e⁻), so both peaks have 2 e⁻, same height. For Li: 1s² (2) and 2s¹ (1), so 1s peak (2) taller than 2s (1). In the graph, Peak X and Peak Y look like same height? Wait, the graph: Peak X is at binding energy ~12, Peak Y ~1. Let's check the options again:

Option D: Be, Peak X 2s, Peak Y 1s. No, because 1s has higher binding energy (should be left, Peak X is left, so Peak X should be 1s, Peak Y 2s. Wait, no: binding energy is higher for 1s, so 1s is at higher binding energy (left), 2s at lower (right). So Peak X (left) is 1s, Peak Y (right) is 2s. But for Be, 1s and 2s both have 2 e⁻, so same height. For Li, 1s has 2, 2s has 1, different heights. So the element is Be? Wait, but option C is Be, Peak X 1s, Peak Y 2s. But wait, the binding energy of 1s is higher than 2s, so 1s should be at higher binding energy (left), so Peak X (left) is 1s, Peak Y (right) is 2s. But for Be, 1s² and 2s², so both peaks have 2 e⁻, same height. In the graph, Peak X and Peak Y have similar heights? Let's check the options:

Option C: Be, Peak X 1s, Peak Y 2s. But wait, the binding energy of 1s is higher, so 1s is at higher binding energy (left), so Peak X (left) is 1s, Peak Y (right) is 2s. But for Be, the electron configuration is 1s² 2s², so 1s has 2 electrons, 2s has 2 electrons. So both peaks should have the same number of electrons (same height), which matches the graph (Peak X and Peak Y same height). For Li, 1s² 2s¹, so 1s peak (2 e⁻) taller than 2s (1 e⁻), which doesn't match. So the correct option is C? Wait, no, wait the options:

Wait the table:

A: Li, 1s, 2s

B: Li, 2s, 1s

C: Be, 1s, 2s

D: Be, 2s, 1s

Wait, binding energy: 1s has higher binding energy than 2s, so 1s is at higher binding energy (left), so Peak X (left) is 1s, Peak Y (right) is 2s. So Peak X is 1s, Peak Y is 2s. Now, the number of electrons: for Be, 1s² (2 e⁻) and 2s² (2 e⁻), so both peaks have same number of electrons (same height), which matches the graph. For Li, 1s² (2 e⁻) and 2s¹ (1 e⁻), so 1s peak taller than 2s, which doesn't match. So the element is Be, Peak X 1s, Peak Y 2s: option C.

Wait, but let's check the binding energy values. The x-axis is binding energy (MJ/mol), with 0 on the right. So higher binding energy is to the left (higher numbers). So 1s orbitals have higher binding energy than 2s, so 1s is at left (Peak X), 2s at right (Peak Y). For Be, 1s² and 2s², so both peaks have 2 electrons, same height. So option C is correct.

Answer: C. C

Question 7 (recheck)

Wait, maybe I made a mistake. Let's think again. The photoelectron spectrum: the peak with higher binding energy (more left) is the inner shell. For Be, electron configuration is 1s² 2s². So the 1s electrons are more tightly bound (higher binding energy), so Peak X (left, higher binding energy) is 1s (2 electrons), Peak Y (right, lower binding energy) is 2s (2 electrons). So both peaks have the same number of electrons (same height), which matches the graph. For Li, 1s² 2s¹: 1s has 2, 2s has 1, so 1s peak taller than 2s, which doesn't match. So the correct option is C (Be, 1s, 2s).

Answer: C (Option C: C. Be, 1s, 2s)

Final Answers

6. B. 242 pm, 419 kJ/mol
7. C. C (Option C: C. Be, 1s, 2s)