Question

1. for each of the positive ions listed in column 1, use the periodic table to find in column 2 the to number of electrons that ion contains. the same answer may be used more than once.

____ 1. ( mathrm{al}^{+3} ) a. 2
____ 2. ( mathrm{fe}^{+3} ) b. 10
____ 3. ( mathrm{mg}^{+2} ) c. 21
____ 4. ( mathrm{sn}^{+2} ) d. 23
____ 5. ( mathrm{co}^{+2} ) e. 24
____ 6. ( mathrm{co}^{+3} ) f. 25
____ 7. ( mathrm{li}^{+1} ) g. 36
____ 8. ( mathrm{cr}^{+3} ) h. 48
____ 9. ( mathrm{rb}^{+1} ) i. 76
____ 10. ( mathrm{pt}^{+2} ) j. 81

Explanation:

Step1: Recall the relationship between atomic number and electrons in ions

The atomic number of an element is equal to the number of protons in its neutral atom. For a positive ion (\(M^{n+}\)), the number of electrons is \(N_{e}=N_{p}-n\), where \(N_{p}\) is the number of protons (equal to the atomic number) and \(n\) is the charge of the ion.

Step2: Find the atomic number of each element

• For \(Al\): Atomic number \(Z = 13\). For \(Al^{+3}\), \(N_{e}=13 - 3=10\)
• For \(Fe\): Atomic number \(Z = 26\). For \(Fe^{+3}\), \(N_{e}=26- 3=23\)
• For \(Mg\): Atomic number \(Z = 12\). For \(Mg^{+2}\), \(N_{e}=12 - 2=10\)
• For \(Sn\): Atomic number \(Z = 50\). For \(Sn^{+2}\), \(N_{e}=50- 2=48\)
• For \(Co\): Atomic number \(Z = 27\). For \(Co^{+2}\), \(N_{e}=27- 2=25\)
• For \(Co\): Atomic number \(Z = 27\). For \(Co^{+3}\), \(N_{e}=27- 3=24\)
• For \(Li\): Atomic number \(Z = 3\). For \(Li^{+1}\), \(N_{e}=3 - 1=2\)
• For \(Cr\): Atomic number \(Z = 24\). For \(Cr^{+3}\), \(N_{e}=24- 3=21\)
• For \(Rb\): Atomic number \(Z = 37\). For \(Rb^{+1}\), \(N_{e}=37- 1=36\)
• For \(Pt\): Atomic number \(Z = 78\). For \(Pt^{+2}\), \(N_{e}=78- 2=76\)

Answer:

1. B. 10
2. D. 23
3. B. 10
4. H. 48
5. F. 25
6. E. 24
7. A. 2
8. C. 21
9. G. 36
10. I. 76