Question

Explain how we know that charge is conserved in this reaction: Li+ CI → Lici

Answer 1

Answer:

Charge is conserved due to the groups in which Lithium and Chlorine are located in the periodic table of the elements.

Explanation:

In the reaction $Li + Cl - > LiCl$, we can examine the groups in which Li and Cl are found in the periodic table of the elements. Lithium appears in Group 1A, or the alkali metals group, indicating that it carries a charge of +1. Chlorine appears in Group 7A, or the halogen group, indicating that it carries a charge of -1. Because LiCl's constituent elements carry the same charges as previously mentioned, LiCl will have an overall charge of 0.

The chemical equation can then be rewritten as $Li^{+} + Cl^{-} - > LiCl$, which, if looking at the individual charges of Li and Cl in lithium chloride, becomes $Li^{+} + Cl^{-} - > Li^{+}Cl^{-}$. Adding the charges on the reactant and product sides of this chemical equation gives us zero in both locations, meaning that we have a charge of 0 on the reactant side and a charge of 0 on the product side. This indicates that charge is conserved in this reaction.

Another way to look at this is expressed in the valence electrons of Li and Cl. Li has an electron configuration of $1s^{2}2s^{1}$, where the n = 2 electron shell has one of eight total electrons needed to fill the valence shell. This means that Li will easily lose one electron in order to have an electron configuration where the n = 1 electron shell is full, $1s^{2}$, and become the $Li^{+}$ ion. Similarly, Cl has an electron configuration of $1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}$ (or $[Ne]3s^{2}3p^{5}$), meaning that the n = 3 electron shell is one electron away from becoming complete. Cl will easily gain one electron to have the electron configuration $[Ne]3s^{2}3p^{6}$ (or $[Ar]$) in order to have an electron configuration where the n = 3 electron shell is full, $3s^{2}3p^{6}$, and become the $Cl^{-}$ ion. Thus, when Li and Cl bond, Li will lose the electron $[1, 0, 0, +\frac{1}{2}]$ and transfer it to Cl, where it will become the electron $[3, 1, 1, -\frac{1}{2}]$, thus conserving charge, as there is an equal total number of electrons before and after the reaction.