#6
energy - color by numbers name: bahansidental
directions: answer each question by circling the correct choice. after you have completed all the questions color the picture using the assigned colors. for tiny sections without numbers, just follow the pattern

| question | a | b | c |
| --- | --- | --- | --- |
| 1. which has the lowest ionization energy? | li (light green) | be (dark blue) | b (pink) |
| 2. which has the higher ionization energy? | li (light green) | be (dark blue) | b (pink) |
| 3. which has the lowest electronegativity? | li (dark blue) | na (light blue) | k (purple) |
| 4. which has the highest electronegativity? | li (dark blue) | na (purple) | k (light blue) |
| 5. which has the higher ionization energy? | n (dark blue) | o (light blue) | p (purple) |
| 6. which has the lowest ionization energy? | f (red) | s (orange) | as (yellow) |
| 7. which has the highest ionization energy? | f (red) | s (orange) | as (yellow) |
| 8. which has the lowest ionization energy? | cl (red) | se (orange) | s (yellow) |
| 9. which has the highest ionization energy? | os (red) | rh (orange) | ni (dark green) |
| 10. which has the lowest electronegativity? | os (gray) | rh (orange) | ni (dark green) |

Question

#6
energy - color by numbers	name: bahansidental
directions: answer each question by circling the correct choice. after you have completed all the questions color the picture using the assigned colors. for tiny sections without numbers, just follow the pattern

| question | a | b | c |
| --- | --- | --- | --- |
| 1. which has the lowest ionization energy? | li (light green) | be (dark blue) | b (pink) |
| 2. which has the higher ionization energy? | li (light green) | be (dark blue) | b (pink) |
| 3. which has the lowest electronegativity? | li (dark blue) | na (light blue) | k (purple) |
| 4. which has the highest electronegativity? | li (dark blue) | na (purple) | k (light blue) |
| 5. which has the higher ionization energy? | n (dark blue) | o (light blue) | p (purple) |
| 6. which has the lowest ionization energy? | f (red) | s (orange) | as (yellow) |
| 7. which has the highest ionization energy? | f (red) | s (orange) | as (yellow) |
| 8. which has the lowest ionization energy? | cl (red) | se (orange) | s (yellow) |
| 9. which has the highest ionization energy? | os (red) | rh (orange) | ni (dark green) |
| 10. which has the lowest electronegativity? | os (gray) | rh (orange) | ni (dark green) |

Accepted Answer

### Question 1
# Explanation:
## Step1: Recall ionization energy trend
Ionization energy increases across a period (left to right) and decreases down a group (top to bottom). Li, Be, B are in the same period (period 2). Li is the leftmost, so it has the lowest ionization energy.
# Answer: A. Li (light green)

### Question 2
# Explanation:
## Step1: Apply ionization energy trend
In period 2, from Li to Be to B, ionization energy increases (across period). Be is to the right of Li, so Be has higher ionization energy than Li; B is to the right of Be, but wait, wait—wait, Li, Be, B: Li (group 1), Be (group 2), B (group 13). Ionization energy: group 2 has higher than group 1, and group 13 is lower than group 2 (due to electron configuration: Be has full 2s², B has 2s²2p¹, so Be has higher IE than B? Wait, no—wait, across period, IE generally increases, but there's a dip at group 13 (B) compared to group 2 (Be). So between Li, Be, B: Be has higher IE than Li, and B has lower than Be? Wait, the question is which has higher ionization energy. So among Li, Be, B: Be (group 2) has higher than Li (group 1), and B (group 13) has lower than Be? Wait, no, let's check: Li (IE ~520 kJ/mol), Be (900 kJ/mol), B (800 kJ/mol). So Be has higher than both Li and B. So between the options, Be (B option) has higher than Li (A) and B (C)? Wait, the options are A: Li, B: Be, C: B. So Be (B) has higher ionization energy than Li (A) and B (C)? Wait, no, B (the element) has lower IE than Be. So the higher IE among Li, Be, B is Be.
# Answer: B. Be (dark blue)

### Question 3
# Explanation:
## Step1: Recall electronegativity trend
Electronegativity decreases down a group (top to bottom). Li, Na, K are in group 1 (alkali metals). K is at the bottom (period 4), Na (period 3), Li (period 2). So K has the lowest electronegativity.
# Answer: C. K (purple)

### Question 4
# Explanation:
## Step1: Apply electronegativity trend
In group 1, electronegativity decreases down the group. So Li (top) has the highest electronegativity among Li, Na, K.
# Answer: A. Li (dark blue)

### Question 5
# Explanation:
## Step1: Analyze elements (N, O, P)
N and O are in period 2, P is in period 3. Ionization energy increases across a period and decreases down a group. N (period 2, group 15), O (period 2, group 16), P (period 3, group 15). Comparing N and O: N has higher IE than O (due to half - filled p - orbital in N: 2p³, O has 2p⁴, so N is more stable, higher IE). Comparing N and P: N is in period 2, P in period 3, so N has higher IE than P. So N has higher ionization energy than O and P.
# Answer: A. N (dark blue)

### Question 6
# Explanation:
## Step1: Analyze elements (F, S, As)
F is in period 2, group 17; S is in period 3, group 16; As is in period 4, group 15. Ionization energy decreases down a group and across a period (with some exceptions). As is the lowest in period (4) and lower in group (15) compared to S (group 16) and F (group 17). So As has the lowest ionization energy? Wait, no—wait, F is in period 2, S in 3, As in 4. Down the group, IE decreases. Also, across period, IE increases. So F (period 2) has higher IE than S (period 3), S has higher than As (period 4)? Wait, no, S is in period 3, group 16; As is in period 4, group 15. So As is lower in period and group. So As has the lowest ionization energy? Wait, the options are F (A), S (B), As (C). So As (C) has the lowest? Wait, no, let's check IE values: F (1681 kJ/mol), S (999.6 kJ/mol), As (947 kJ/mol). So As has lower than S, which is lower than F. So As (C) has the lowest ionization energy.
# Answer: C. As (yellow)

### Question 7
# Explanation:
## Step1: Analyze elements (F, S, As)
F is in period 2, group 17. S is in period 3, group 16. As is in period 4, group 15. Ionization energy is highest for the element with the highest effective nuclear charge and smallest atomic radius. F has the smallest atomic radius (period 2) and highest electronegativity, so it has the highest ionization energy among F, S, As.
# Answer: A. F (red)

### Question 8
# Explanation:
## Step1: Analyze elements (Cl, Se, S)
Cl is in period 3, group 17; S is in period 3, group 16; Se is in period 4, group 16. Ionization energy: across period 3, Cl (group 17) has higher than S (group 16). Down group 16, Se (period 4) has lower than S (period 3). So Se has lower ionization energy than S and Cl. So among Cl (A), Se (B), S (C), Se (B) has the lowest? Wait, no—wait, Cl (IE ~1251 kJ/mol), S (999.6 kJ/mol), Se (941 kJ/mol). So Se has lower than S, which is lower than Cl. So Se (B) has the lowest ionization energy.
# Answer: B. Se (orange)

### Question 9
# Explanation:
## Step1: Analyze elements (Os, Rh, Ni)
Ni is in period 4, group 10; Rh is in period 5, group 9; Os is in period 6, group 8. Ionization energy generally increases across a period (left to right) and decreases down a group (top to bottom). Ni is the leftmost and topmost among the three. Wait, no—Ni (period 4), Rh (period 5), Os (period 6). Across the d - block, ionization energy increases with increasing atomic number (across period) and decreases down the group. Ni has atomic number 28, Rh 45, Os 76. Wait, no, ionization energy for Ni: ~737 kJ/mol, Rh: ~720 kJ/mol, Os: ~840 kJ/mol? Wait, no, maybe I got it wrong. Wait, the elements are in the d - block. Let's think about their positions: Ni is in period 4, group 10; Rh in period 5, group 9; Os in period 6, group 8. The trend for ionization energy in transition metals: generally, as we move from left to right across a period, IE increases, and as we move down a group, IE increases slightly (due to lanthanide contraction for Os). Wait, but among these three, Ni is in the highest period (4) and rightmost group (10) compared to Rh (period 5, group 9) and Os (period 6, group 8)? No, period 4 is lower than 5 and 6. Wait, no, period number increases as we go down. So Ni is in period 4 (top), Rh in 5, Os in 6 (bottom). Across the period, group 10 (Ni) has higher IE than group 9 (Rh). Down the group, Os (period 6) has higher IE than Rh (period 5) due to lanthanide contraction (Os has smaller atomic radius than expected, so higher IE). But between Ni and Os: Ni is in period 4, Os in 6. Wait, Ni (IE ~737), Os (IE ~840). So Os has higher than Ni? Wait, but the options are Os (A), Rh (B), Ni (C). Wait, maybe I made a mistake. Alternatively, maybe the trend is that among these three, Ni has the highest ionization energy? Wait, no, let's check the atomic numbers: Ni (28), Rh (45), Os (76). The ionization energy for transition metals: as atomic number increases, IE generally increases (due to increasing nuclear charge, even though electrons are added to d - orbitals). So Os (76) has higher IE than Rh (45) and Ni (28). Wait, but Os is in period 6, group 8; Rh in period 5, group 9; Ni in period 4, group 10. So Os has higher IE than Rh and Ni? But the options: A is Os (red), B is Rh (orange), C is Ni (dark green). So if Os has the highest, then A. But maybe I'm wrong. Alternatively, maybe the question is about the trend in the same period or group. Wait, the problem is a bit unclear, but assuming the trend, let's re - evaluate. Ni is in period 4, group 10; Rh in period 5, group 9; Os in period 6, group 8. The ionization energy order: Ni < Rh < Os? No, that doesn't make sense. Wait, no, the correct order for IE (first ionization energy) for these elements: Ni: ~737 kJ/mol, Rh: ~720 kJ/mol, Os: ~840 kJ/mol. So Os has higher than Ni and Rh. So the highest is Os (A).
# Answer: A. Os (red)

### Question 10
# Explanation:
## Step1: Analyze elements (Os, Rh, Ni)
Electronegativity in transition metals: generally increases across a period (left to right) and decreases down a group (top to bottom). Ni is in period 4, group 10; Rh in period 5, group 9; Os in period 6, group 8. Os is the lowest in period (6) and leftmost in group (8) among the three. So Os has the lowest electronegativity.
# Answer: A. Os (gray)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Question 1

Step1: Recall ionization energy trend

Step1: Apply ionization energy trend

Step1: Recall electronegativity trend

Answer:

Question 2

question	a	b	c
2. which has the higher ionization energy?	li (light green)	be (dark blue)	b (pink)
3. which has the lowest electronegativity?	li (dark blue)	na (light blue)	k (purple)
4. which has the highest electronegativity?	li (dark blue)	na (purple)	k (light blue)
5. which has the higher ionization energy?	n (dark blue)	o (light blue)	p (purple)
6. which has the lowest ionization energy?	f (red)	s (orange)	as (yellow)
7. which has the highest ionization energy?	f (red)	s (orange)	as (yellow)
8. which has the lowest ionization energy?	cl (red)	se (orange)	s (yellow)
9. which has the highest ionization energy?	os (red)	rh (orange)	ni (dark green)
10. which has the lowest electronegativity?	os (gray)	rh (orange)	ni (dark green)