When oxygen combines with any alkali metal, M, what is the formula of the compound produced usually?

1 Answer
Jun 15, 2017

Answer:

The basic formula is #color(red)("M"_2"O"#.

Explanation:

To answer this, we can take a look at the ionic charges of the elements involved.

Since oxygen is in group 16, it will have #color(red)(6# valence electrons:

upload.wikimedia.org

Recall that an atom will want to gain or lose electrons to fulfill the octet rule; it wants to do whatever is easiest to obtain #8# electrons in its outermost shell.

For an oxygen atom, the easiest way to get #8# electrons in the outer shell is to obtain #2# more electrons. This will in turn cause the #sfcolor(blue)("oxygen"# to have an ionic charge of #color(blue)(2-#, because there are #color(blue)(2# more electrons than protons now.

#-----#

For the alkali metals, which are located in group 1 of the periodic table, they each have #1# valence electron. Here's the electrons-by-shell for potassium, one of the alkali metals:

upload.wikimedia.org

In order to obtain an octet for each of the alkali metals, the easiest way to do it is by removing its one valence electron. This will cause each of the atoms (symbol #color(red)("M"# for the alkali metal) to have an ionic charge of #color(red)(1+#, because there is #color(red)(1# less electron than the number of protons.

The formula:

The compound the alkali metal #color(red)("M"# and oxygen #color(blue)("O"# form must be electrically neutral. That is to say, the charges must balance out to #0#.

The easiest way to write a formula of any ionic compound if you know the ionic charges of each species is to put the charge of the anion (the negative ion, in this case is #"O"^(2-)#), #2-# as the subscript of the cation, and the charge of the cation (the positive ion, in this case #"M"^+#) as the subscript of the *anion*. Here's what I mean, using sodium (#"Na"#) as an example:

http://web.fscj.edu

#sfcolor(green)("Therefore, the formula for any alkali metal M and oxygen (O)"#
#sfcolor(green)(" is M"_2"O"#.