Question

Predict the formula of the ionic compounds formed when each of the following pairs of elements react:
a) LiandBr b) MgandCl c) FrandO

Answer 1

The ionic compounds here will be formed by the electrostatic attraction between oppositely charged ions. That is, for any two elements reacting:

-One will look to lose one or more of its outer shell (valence) electrons in order to obtain the highly stable noble gas configuration (i.e. a full outer shell of electrons). In doing so, it will become a positively charged species or a positive ion, as it now has more protons than electrons.
-The other reacting element is looking to take a certain number of electrons, as this is the easiest way for it to obtain a noble gas configuration and a full valence shell of electrons. So, the electrons the positive ion had lost are transferred in such a way that the other element achieves this stable configuration, in doing so becoming a negatively charged ion (with more electrons than protons).
-The two oppositely charged ions will be attracted to one another to form an ionic bond. The compound formed with the ionic bond is generally neutral (i.e. it has no overall charge), so the positive charged ions present must cancel out the negatively charged ions.

http://images.tutorcircle.com/cms/images/44/p-table.png

The easiest and quickest way of finding how many electrons an atom of an element loses or gains when reacting is by looking at what group it is in (vertical rows down) in the periodic table. This is also called the 'oxidation number' of the element, and form this we can predict the chemical formula of the product that is formed.

Answer 2

Let's take a look at the first example:

a. ) Li and Br

- On the periodic table, lithium is located in Group 1. If you look at the link I left in my previous post, you will see that all elements in Group 1 tend to have an oxidation number of +1. This means that these elements like to exist as ions with a +1 charge, having lost one electron. For example, if lithium loses one electron, it will have two remaining, meaning it will have the same electronic configuration as helium, a very stable noble gas in Group 18 (or Group 8, as it's sometimes called). So, lithium will look to lose one electron to become Li+.

- Bromine, on the other hand, is in Group 17 (Group 7) of the periodic table. This means, as shown in the link above, that it will look to gain one electron to acquire the same electronic configuration as that of Krypton in Group 18. Thus, it has an oxidation state of -1, as it will look to gain a single electron to form a negatively charged Br- ion.

So, one atom of lithium can give its electron to one atom of bromine so that they both get the charges the want and are attracted to each other to form the LiBr (Lithium Bromide) ionic compound. As the +1 charge on lithium cancels the -1 charge on bromine, the compound is overall neutral.
\[Li ^{+} + Br ^{-} \rightarrow LiBr\]

Answer 3

The second example is slightly less straightforward:

b. ) Mg and Cl

Let's follow the same process as before:

- Mg, magnesium, is in Group 2 of the periodic table. So, it has an oxidation number of +2 and likes to lose 2 electrons from its outer shell to obtain a noble gas configuration and form the Mg2+ ion.
- Cl, chlorine, is in Group 17 (Group 7) of the periodic table. That is, it has an oxidation number of -1 and likes to gain one electron to obtain a noble gas electronic configuration, in doing so forming a Cl- ion.

This is where we now have a problem. We have a magnesium atom, which wants to give away two of its electrons, reacting with chlorine, which will only want to accept one of these electrons. So, what do we do? Well, we react one atom of magnesium with two atoms of chlorine, so that each chlorine can accept one of Mg's electrons. This means that the ionic compound formed will now include 1 magnesium and 2 chlorines, thus forming MgCl2 (magnesium dichloride). The compound is neutral, as the +2 charge on Mg will cancel with the two -1 chagnes from the two chlorines present.

\[Mg ^{2+} + 2Cl ^{-} \rightarrow MgCl _{2}\]

Answer 4

The method for finding the answer to c.) is pretty similar to that of b.). Remember that Fr (francium) is in Group 1 of the periodic table and O (oxygen) is in Group 16 (sometimes called Group 6), and we must have enough atoms of both Fr and O in the ionic compound formed such that it is neutral (i.e. we have the right number of electrons for oxygen to accept that it wants to).
Best of luck and hope this helped you out! :)