Answer:
The electronic configuration of ions formed are:
Li+ (2) => 1s2
Cl- (18) => 1s2 2s2 2p6 3s2 3p6
Explanation:
Lithium metal, Li will lose 1 electron to lithium ion Li+. Chlorine atom, Cl will receive the 1 electron to form the chloride ion Cl- as shown by the following equation below:
Li —> Li+ + e
Cl+ e —> Cl-
Combine both equation
Li + Cl + e —> Li+ + Cl- + e
Cancel out 'e'
Li + Cl —> Li+ Cl-
Thus, we can write the electronic configuration for the reaction as follow:
Before reaction:
Li (3) => 1s2 2s1
Cl (17) => 1s2 2s2 2p6 3s2 3p5
After reaction
Li+ (2) => 1s2
Cl- (18) => 1s2 2s2 2p6 3s2 3p6
Therefore, the electronic configuration of the ions formed are:
Li+ (2) => 1s2
Cl- (18) => 1s2 2s2 2p6 3s2 3p6
