Question

putting it all together: complete the following chart. you may need to reference a periodic table to complete some portions of the chart.

	atomic #	mass #	p⁺	n⁰	e⁻	charge	nuclide notation
1	16	32	16	16	16	0	$_{16}^{32}\text{s}$
2	22	48	22	26	22	0	$_{22}^{48}\text{ti}^{0}$
3	35	80	35	45	36	-1	$_{35}^{80}\text{br}^{-1}$
4	86	222	86	136	86	0	$_{86}^{222}\text{rn}^{0}$
5	53	127	53	74	54	-1	$_{53}^{127}\text{i}^{-1}$
6	31					+3	$\text{ga}$
7			20		18
8	34						$\text{se}^{2-}$
9		55	25			+2
10			15		18

reference the problems above. which numbers describe cations?

reference the problems above. which numbers describe neutral atoms?

reference the problems above. which numbers describe anions?

Explanation:

Step1: Recall definitions of ions

Cations are positively charged ions (lose electrons), neutral atoms have \( p^+ = e^- \), anions are negatively charged ions (gain electrons).

Step2: Analyze each item

• Cations: Check charge > 0. Item 6 (charge +3), 7 (charge: \( p^+=20, e^-=18 \), charge \( +2 \)), 9 (charge +2), 10 (charge: \( p^+=15, e^-=18 \), charge \( -3 \)? Wait, no: 7: \( p^+=20, e^-=18 \), so charge \( +2 \); 6: +3; 9: +2. Wait, original items:
• Item 6: charge +3 (cation)
• Item 7: \( p^+=20, e^-=18 \), charge \( +2 \) (cation)
• Item 9: charge +2 (cation)

Wait, maybe I misread. Let's recheck:

• Neutral atoms: \( p^+ = e^- \). Items 1 (\( p^+=16, e^-=16 \)), 2 (\( p^+=22, e^-=22 \)), 4 (\( p^+=86, e^-=86 \)) → neutral.
• Anions: charge < 0. Items I Do (\( -3 \)), 3 (\( -1 \)), 5 (\( -1 \)), 8 (\( Se^{2-} \), charge -2), 10 (\( p^+=15, e^-=18 \), charge -3? Wait, 10: \( p^+=15, e^-=18 \), so charge \( -3 \)? But let's list:
• Cations: 6 (charge +3), 7 (charge \( +2 \) as \( e^- = 18, p^+=20 \)), 9 (charge +2)
• Neutral: 1, 2, 4 ( \( p^+ = e^- \))
• Anions: I Do (\( -3 \)), 3 (\( -1 \)), 5 (\( -1 \)), 8 (\( Se^{2-} \)), 10 (\( e^-=18, p^+=15 \), charge \( -3 \))

Wait, the problem has three sub-questions:

1. Cations: Ions with positive charge ( \( e^- < p^+ \) or charge > 0). Items 6 (charge +3), 7 ( \( p^+=20, e^-=18 \) → charge +2), 9 (charge +2). Wait, item 7: \( p^+=20, e^-=18 \), so charge \( +2 \) (cation). Item 6: +3, item 9: +2. Also, item 10: \( e^-=18, p^+=15 \) → charge -3 (anion). Item 8: \( Se^{2-} \) → anion.

1. Neutral atoms: \( p^+ = e^- \) (charge 0). Items 1 (16p,16e), 2 (22p,22e), 4 (86p,86e) → so 1, 2, 4.

1. Anions: Ions with negative charge ( \( e^- > p^+ \) or charge < 0). Items I Do (\( -3 \)), 3 (\( -1 \)), 5 (\( -1 \)), 8 (\( Se^{2-} \)), 10 (\( e^-=18, p^+=15 \) → charge -3) → I Do, 3, 5, 8, 10.

Wait, maybe the original table's items:

• Cations: 6 (charge +3), 7 (charge +2), 9 (charge +2)
• Neutral: 1, 2, 4
• Anions: I Do, 3, 5, 8, 10

But let's confirm with the table:

• Cations:
• Item 6: Charge +3 (cation)
• Item 7: \( p^+ = 20 \), \( e^- = 18 \) → charge \( +2 \) (cation)
• Item 9: Charge +2 (cation)
• Neutral Atoms:
• Item 1: \( p^+ = 16 \), \( e^- = 16 \) (neutral)
• Item 2: \( p^+ = 22 \), \( e^- = 22 \) (neutral)
• Item 4: \( p^+ = 86 \), \( e^- = 86 \) (neutral)
• Anions:
• "I Do": Charge -3 (anion)
• Item 3: Charge -1 (anion)
• Item 5: Charge -1 (anion)
• Item 8: \( \text{Se}^{2-} \) (anion, charge -2)
• Item 10: \( p^+ = 15 \), \( e^- = 18 \) → charge -3 (anion)

Answer:

• Cations: 6, 7, 9
• Neutral Atoms: 1, 2, 4
• Anions: I Do, 3, 5, 8, 10

(Note: Depending on exact charge calculations, but this is the general approach based on proton-electron balance and charge sign.)