Chemistry- Chapter 3, Review Questions: Section 5

Why does an atom have a net zero charge?

An atom has a net zero charge because the amount or number of positively charged protons is equal to the amount or number of negatively charged electrons, so the positives and the negatives cancel/balance each other out.

a. Show how fluorine forms an ion. Does it form a cation or an anion? b. Show how lithium forms an ion. Does it form a cation or an anion?

Fluorine forms an ion by accepting one electron from another element which gives it a -1 charge because it now has one more negatively charged electron than positively charged protons. This is called an anion because the element gained electron and is a negatively charged ion. b. Lithium forms an ion by giving up or losing one electron and giving it to another element, such as fluorine, which gives it a +1 charge because there is now one more positively charged proton than there are negatively charged electrons. This is called a cation because the element gave up electrons and is a positively charged ion.

What evidence do we have that convinces us that sodium chloride is an ionic compound?

Sodium chloride is an ionic compound because of its high melting and boiling points. Sodium chloride in its solid form will not conduct an electric current. However when salt (sodium chloride) is dissolved in water, the solution conducts electricity very easily.

What are the relationships between the group number on the periodic table and the types of ions formed?

The group number determines what charge the element will have (elements in group 2 have a +2 charge) and whether the element is a metal or a nonmetal determines whether the charge will be positive or negative. For example, metals will normally take on positive charges giving up electrons and nonmetals will normally take on negative charges and accept electrons from other elements.

Give the formulas for the compounds that contain the following pairs of ions. a. Ba ²⁺ and N³⁻ c. Li ⁺ and S²⁻ b. Mg ²⁺ and O²⁻ d. Ca²⁺ and Br⁻

a. Ba₃N₂ c. Li₂S b. MgO d. CaBr₂