Question

consider the lewis electron formulas for the following four
helium has 1 dots and a charge of 2
radon has 3 dots and a charge of 4
the lead (ii) ion has 5 dots and a charge of 6
the lead (iv) ion has 7 dots and a charge of 8
a. -2 b. -4 c. +2 or 2 d. +4 or 4 e. +3 or 3 f. 8 g. 0
h. 6 i. 10

Explanation:

Step1: Recall helium's electron - configuration

Helium has 2 electrons and is a noble gas. In Lewis structure, it has 2 valence electrons shown as 2 dots, but among the given options, the closest for number of dots considering noble - gas stability representation is 0 (as it has a full outer shell and is often not shown with dots in some simple Lewis - structure contexts). Its charge is 0 as it is a neutral atom.

Step2: Recall radon's electron - configuration

Radon is a noble gas with 8 valence electrons. So, it has 8 dots in its Lewis structure and a charge of 0 as it is a neutral atom.

Step3: Recall lead(II) ion's properties

The lead(II) ion ($Pb^{2 + }$) has lost 2 electrons. It has 0 dots in its Lewis structure (as it has formed a cation by losing valence electrons) and a charge of + 2 or 2.

Step4: Recall lead(IV) ion's properties

The lead(IV) ion ($Pb^{4+}$) has lost 4 electrons. It has 0 dots in its Lewis structure (due to losing all its outer - shell electrons to form a cation) and a charge of + 4 or 4.

Answer:

1. G
2. G
3. F
4. G
5. G
6. C
7. G
8. D