Question

question: how are ionic compounds formed?

1. observe: look at the purple lithium atom and the red oxygen atom. recall that most atoms are stable when their outermost ring has eight electrons. some atoms, such as lithium and beryllium, are stable when their outermost ring has two electrons.

a. how many electrons will the lithium atom give up to become stable?
b. how many electrons does the oxygen atom need to become stable?

Explanation:

Step1: Analyze lithium atom

Lithium has an atomic number of 3. Its electron configuration is \(1s^{2}2s^{1}\). The outermost ring (second shell) has 1 electron. Since lithium is stable when its outermost ring has 2 electrons (but it has only 1 in the outer - most ring in its neutral state). To get a stable configuration (either losing or gaining, but it's easier to lose 1 electron to have a full inner shell), it will give up 1 electron.

Step2: Analyze oxygen atom

Oxygen has an atomic number of 8. Its electron configuration is \(1s^{2}2s^{2}2p^{4}\). The outermost ring (second shell) has 6 electrons. Since atoms are stable with 8 electrons in the outermost ring (octet rule for most atoms like oxygen), it needs \(8 - 6=2\) electrons.

Answer:

A. 1
B. 2