How to find the combining valence of an anion when it exhibits variable valency?

I'm doing an online course and can't for the life of me figure out how to write chemical formulas from the chemical name.

Take the example of calcium chloride.

I know the the valence for calcium is +2 because my periodic table shows me this, but I cannot figure out the valence of chlorine, (or any other element) with multiple valences. My periodic table lists chlorine as having a possible valence of -1, +1, +3, +5, or +7. How am I to know which one is to be used?

• See, you have to combine plus (Ca) with minus (Cl). – Ivan Neretin May 17 '18 at 21:08
• Per Ivan the compound has to have a neutral charge. So if calcium is +2, then there must be -2 somehow. The rub here is that there isn't a -2 charge on a single chloride anion, but two chloride anions each with a -1 charge. The name "calcium chloride" doesn't give you a clue as to how many chloride atoms there are. I can't imagine any chemist saying calcium dichloride. Truthfully if you study chemistry or a while you just learn how inorganic salts are named and the appropriate valences. The naming convention for inorganic "molecules" just isn't as formulaic as the naming of organic molecules. – MaxW May 18 '18 at 0:07
• chemistry.stackexchange.com/questions/4547/… – Mithoron May 18 '18 at 18:34
• Generally, $\ce{Cl}$ shows a valency of -1. As $\ce{CaCl2}$ is an ionic bond, also neutral therefore, valency of $\ce{Ca}$ * 1 + valency of $\ce{Cl}$ * 2 = 0 which gives 2 + valency of $\ce{Cl}$ * 2 = 0. Therefore, valency of $\ce{Cl}$ = -2 / 2 = (-1). – Rahul Verma Jun 17 '18 at 2:55

First of all, sorry for the (very!) long post!

Determining the Number of Valence Electrons

The number of valence electrons of atom can be determined by the group in which the atom is in. For example, Na is in group 1A, and thus has one valence electron (1+ charge). Chlorine is in group 7A and thus has 7 valence electrons (1- charge). A quick (but very simplified) way of determining if an element has a positive or negative charge is to look again at its group. Groups 1A, 2A, and 3A are positive; while groups 5A, 6A, and 7A are negative. An element in group 4A can be both.

Ionic Compounds:

If the charges are the same, no "balancing" is necessary:

• if Na (1+) were to bond with Cl (1-) it would form NaCl.

If the Charges are different, a quick way to determine the formula unit is to assign each element the charge of the other:

• if B (3+) were to bind with Cl (1-) it would form BCl3: Boron get the 1 from Chlorine; Chlorine gets the 3 from Boron.

Covalent Bonding

Covalent Bonds are a little trickier and has more contributing factors. It seems like you're having trouble with Lewis-Dot diagrams so that's what I'll address. In order to correctly draw the Lewis-Dot diagram of a compound, you have to determine the formal charge of each atom. For demonstration purposes I'm going to use CH4.

The formal charge of an atom can be determined by:

• Number of Valence Electrons - non bonding electrons - number of bonds

So in our example, the formal charge of Carbon is :

• 4 Valence electrons - 0 non bonding electrons - 4 bonds = formal charge of 0.

Then Hydrogen:

• 1 Valence electron - 0 non bonding electrons - 1 bond = formal charge of 0.

So if we add the formal charged together we get 0, which makes sense because the charge of the compound is 0.

In general, the more compounds that can have a formal charge of 0, the more likely it is that that particular structure will occur, as it will be more stable. If the compound is a poly atomic ion, the more electronegative atom should get the negative formal charge (electronegativity increases as you move up the group and to the right in a period).

One more significant detail is the ability of some atoms to expand the octet, which can occur if the element is in the third period or beyond. In this case, the charge of the atom can be made 0 (or whatever appropriate) if more bonds are made than typically seen.

• i.e. in ICl4-, Iodine, though typically only making one bond, bonds to four Chlorine atoms. In this case formal charge = 7 valence electrons - 4 bonds - 4 non bonding electrons = 1-, the charge of the polyatomic ion. While Cl is more electronegative, only I can expand its octet, thus I has the negative formal charge. The 4 non bonding electrons on Iodine were determined by counting the total number of valence electrons (7 from I + 4*7 for Cl + 1 from the charge = 36) and adjusting accordingly.

There are other exceptions such as free radicals and atoms that can have less than a full octet, but the "rules" above should be sufficent guides for basic Lewis Dot structures.

Hope that covered everything :)