consider the lewis electron formulas for the following four\nhelium has __1__ dots and a charge of…

consider the lewis electron formulas for the following four\nhelium has __1__ dots and a charge of __2__\nradon has __3__ dots and a charge of __4__\nthe lead (ii) ion has __5__ dots and a charge of __6__\nthe lead (iv) ion has __7__ dots and a charge of __8__\na. -2 b. -4 c. +2 or 2 d. +4 or 4 e. +3 or 3 f. 8 g. 0\nh. 6 i. 10

consider the lewis electron formulas for the following four\nhelium has __1__ dots and a charge of __2__\nradon has __3__ dots and a charge of __4__\nthe lead (ii) ion has __5__ dots and a charge of __6__\nthe lead (iv) ion has __7__ dots and a charge of __8__\na. -2 b. -4 c. +2 or 2 d. +4 or 4 e. +3 or 3 f. 8 g. 0\nh. 6 i. 10

Answer

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