Find the Oxidation Numbers NaHCO_3

$N a H C O_{3}$

Step 1

Since $N a$ is in column $1$ of the periodic table, it will share $1$ electrons and use an oxidation state of $1$ .

$N a = 1 (1) = 1$

Step 2

Since $O$ is in column $6$ of the periodic table, it will share $6$ electrons and use an oxidation state of $6$ .

$O = 3 (- 2) = - 6$

Step 3

Since $H$ is in column $1$ of the periodic table, it will share $1$ electrons and use an oxidation state of $1$ .

$H = 1 (1) = 1$

Step 4

Since $C$ is in column $4$ of the periodic table, it will share $4$ electrons and use an oxidation state of $4$ .

$C = 1 (- 4) = - 4$

Step 5

This is the full list of oxidation states for this molecule.

$N a = 1, O = - 6, H = 1, C = - 4$