Question

c. what group does the element o belong? group 16
12.) which of these elements has the highest ionization energy? a. li b. k c. fr d. na
13.) which of these elements has the lowest ionization energy? a. o b. b c. c d. f
14.) which of the following lists of the atoms in order of increasing ionization energy?
a. calcium, iron, copper b. copper, iron, calcium c. calcium, copper, iron d. iron, copper, calcium
15.) across a period, ionization energy a

Explanation:

Question 12: Which of these elements has the highest ionization energy? (Options: a. Li, b. K, c. Fr, d. Na)

Step 1: Recall ionization energy trend

Ionization energy generally increases up a group (decreases down a group) in the periodic table. Lithium (Li), Potassium (K), Francium (Fr), and Sodium (Na) are all in Group 1 (alkali metals).

Step 2: Determine position in group

Li is at the top of Group 1, followed by Na, then K, then Fr as we go down the group. Since ionization energy decreases down a group, the element at the top (Li) has the highest ionization energy among these. Wait, no—wait, correction: Wait, actually, in Group 1, as we go down, ionization energy decreases. So Li (period 2), Na (period 3), K (period 4), Fr (period 7). So Li is highest? But wait, no—wait, maybe I mixed up. Wait, no, the question's options: Li, K, Fr, Na. So Li is the topmost in Group 1, so it has the highest ionization energy among these. Wait, but let's confirm: Ionization energy (IE) is the energy required to remove an electron. For alkali metals, as we go down the group, the atomic radius increases, so the outermost electron is farther from the nucleus, so IE decreases. So Li (smallest radius in Group 1 here) has the highest IE. So the answer is a. Li? Wait, no—wait, maybe the original question was different? Wait, no, the user's image: question 12 is "Which of these elements has the highest ionization energy? a. Li b. K c. Fr d. Na". So based on group trend, Li is highest.

Step 1: Recall ionization energy trend across a period

Ionization energy generally increases from left to right across a period (due to increasing effective nuclear charge, smaller atomic radius). Boron (B), Carbon (C), Oxygen (O), Fluorine (F) are in Period 2, with B (Group 13), C (Group 14), O (Group 16), F (Group 17).

Step 2: Determine position in period

From left to right: B, C, O, F. Since IE increases left to right, the leftmost element (B) has the lowest IE? Wait, no—wait, B is in Group 13, C in 14, O in 16, F in 17. Wait, but there's a slight dip at Group 13 (B) compared to Group 2, but in this case, among B, C, O, F: B is the leftmost, so it has the lowest IE. Wait, but let's check: B has electron configuration [He] 2s² 2p¹, C is [He] 2s² 2p², O is [He] 2s² 2p⁴, F is [He] 2s² 2p⁵. The effective nuclear charge increases from B to F, so IE increases. So B has the lowest IE. So the answer is b. B? Wait, no—wait, maybe I made a mistake. Wait, the options are a. O, b. B, c. C, d. F. So from left to right: B (Group 13), C (14), O (16), F (17). So B is leftmost, so lowest IE. So answer is b. B.

Step 1: Recall ionization energy trends (period and group)

Calcium (Ca) is in Group 2, Period 4; Iron (Fe) is in Group 8, Period 4; Copper (Cu) is in Group 11, Period 4. All in Period 4. Across a period, ionization energy generally increases (with some exceptions, but for these metals: Ca is an alkali earth metal, Fe and Cu are transition metals. Transition metals have more complex IE trends, but generally, as we move from left to right in a period, IE increases (due to increasing effective nuclear charge, even with d-electrons).

Step 2: Compare positions

Ca is leftmost in Period 4 (Group 2), then Fe (Group 8), then Cu (Group 11). So increasing IE would be Ca (lowest), then Fe, then Cu? Wait, no—wait, the options: a. calcium, iron, copper; b. copper, iron, calcium; c. calcium, copper, iron; d. iron, copper, calcium. Wait, let's check IE values: Ca has lower IE than Fe, which has lower IE than Cu? Wait, no—wait, actually, in Period 4, from Ca (Group 2) to Zn (Group 12), the IE generally increases. So Ca (Group 2) < Fe (Group 8) < Cu (Group 11)? Wait, but let's confirm: Ca is a metal, loses electrons easily (low IE). Fe is a transition metal, has higher IE than Ca. Cu is also a transition metal, with higher IE than Fe? Wait, maybe. So the order of increasing IE would be Ca, Fe, Cu. So option a: calcium, iron, copper.

Answer:

a. Li

Question 13: Which of these elements has the lowest ionization energy? (Options: a. O, b. B, c. C, d. F)