Chlorine-35 and chlorine-37 are isotopes of each other. Isotopes are atoms of the same element that have different masses due to different numbers of neutrons.

We know that the element is iron and that it is missing two electrons. This means that the atom has two less electrons than it has protons resulting in a charge of 2+. This can be written as

\(\displaystyle Fe^{2+}\)

If an ion is positively charged it is a cation.  If it is negatively charged (more electrons than protons) it is an anion.

The ion is a cation since it has a charge of 2+.

We should first remember the difference between sulfate, sulfite, and sulfide. Sulfate is \(\displaystyle SO_4^{-2}\), sulfite is \(\displaystyle SO_3^{-2}\)` and sulfide is \(\displaystyle S^{-2}\).

The only answer choices that could be right must have \(\displaystyle SO_4^{-2}\) in them. We then need to see that barium usually has a charge of \(\displaystyle +2\), as the periodic table shows us, and so we need a charge of \(\displaystyle -2\) to cancel that out. The answer is \(\displaystyle BaSO_4\).

Ionic chemical compounds are always named with the metal before the non-metal. The correct way to name sodium chloride would be \(\displaystyle NaCl\), and not \(\displaystyle ClNa\), because sodium is the metal and chloride is the non-metal.

All the other examples have the metal before the non-metal. The other answer choices correspond to potassium chloride, magnesium sulfate, hydrogen chloride (hydrochloric acid), and calcium fluoride.

Because iron has multiple oxidation states, we need to follow the cation with the roman numeral that signifies how many electrons each iron ion has donated to oxygen. Since oxygen gains two electrons in order to satisfy its octet, the three oxygens will accept a total of six electrons from two iron atoms. This means that each iron must donate three electrons, so that each atom donates electrons equally. As a result, the name of the ionic compound is iron(III) oxide.

Another acceptable name for the compound would be ferric oxide. The term "ferric" refers to iron(III) and the term "ferrous" refers to iron (II).

When naming covalent compounds, it is important to use a prefix before each element in order to designate how many atoms are in the compound. One key exception is when you only have one atom at the beginning of the compound (such as in this question). You will never start a covalent compound with "mono-". The prefix for five is "penta-", so the name of this covalent compound is phosphorus pentafluoride. The name indicates that there is one phosphorus atom bound to five fluorine atoms.

When naming molecular compounds, the rule is to put the element farthest to the left of the period table and closest to the bottom of the periodic table first. 

For example, in sulfur dioxide, since sulfur and oxygen are both in group 6A, we put sulfur first since it is below oxygen in the group: \(\displaystyle SO_2\). Similarly, in carbon dioxide we put carbon first because it is farther to the left in the period: \(\displaystyle CO_2\).

\(\displaystyle Cl_2N_2\) is written incorrectly because nitrogen is farther left on the periodic table than chlorine. Written correctly, this compound would be dinitrogen dichloride: \(\displaystyle N_2Cl_2\).

 The molecule \(\displaystyle S_2Cl_2\) is a molecular compound because it contains two non-metals; therefore, it contains no ionic bonds. We use prefixes to describe the subscript for each element in molecular compounds. The molecule in this question has a subscript of two for each of the elements, so the prefix di- should be added to both sulfur and chlorine.

For molecular compounds, we change the ending of the final element to -ide, giving us the name "disulfur dichloride."

It is important to note that "chlorate" has a separate meaning, and refers to the complex ion \(\displaystyle ClO_3^-\).

We know that sodium oxide is a binary ionic compound because it contains one metal cation and one non-metal anion. To write the ionic formula we must write the cation first, followed by the anion, and make sure the net charge of the molecule is zero.

From the periodic table, we know that sodium (Na) has a charge of \(\displaystyle \small +1\) and oxygen (O) has a charge of \(\displaystyle \small -2\). We know that the final charge on the molecule must be zero. To balance the charge, there must be two sodium atoms for every oxygen atom.

\(\displaystyle Na^{_{^{+}}} + Na^{_{^{+}}} + O^{2-}= 0\)

\(\displaystyle Na_{2}O\)

The other answer choices do not have a net charge of zero.

\(\displaystyle Na_{2}O_{2}\) 

\(\displaystyle Na^{_{^{+}}}+Na^{_{^{+}}}+O^{2-} +O^{2-}= -2\)

\(\displaystyle NaO_{2}\)

\(\displaystyle Na^{_{^{+}}}+O^{2-} +O^{2-}= -3\)

\(\displaystyle NaO\)

\(\displaystyle Na^{_{^{+}}} + O^{2-}= -1\)

\(\displaystyle Na_{3}O_{}2\)

\(\displaystyle Na^{_{^{+}}}+Na^{_{^{+}}}+Na^{_{^{+}}}+O^{2-} +O^{2-}= -1\)

Ionic compounds are written with the cation (metal) first, followed by the anion (non-metal). All of the answer choices satisfy this requirement.

Next, we must look at the net charge on each molecule. All of the ionic compounds have a net charge of zero except for \(\displaystyle LiSe\), which has a charge of \(\displaystyle \small -1\) as written.

\(\displaystyle LiSe\)

\(\displaystyle Li^{+}+Se^{2-}=-1\)

The correct ionic formula for a compound containing these two elements is:

\(\displaystyle Li_{2}Se\)

\(\displaystyle Li^{+}+Li^{+}+Se^{2-}=0\)