Find the Oxidation Numbers NaBH_4

$N a B H_{4}$

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

$H = 4 (- 7) = - 28$

Step 3

To find the oxidation state of $B$ , set up an equation of each oxidation state found earlier and set it equal to $0$ .

$1 - 28 + 1 x = 0$

Step 4

Solve the equation for $x$ .

Simplify $1 + - 28 + 1 x$ .

Multiply $x$ by $1$ .

$1 + - 28 + x = 0$

Add $1$ and $- 28$ .

$- 27 + x = 0$

Add $27$ to both sides of the equation.

$x = 27$

Step 5

$B$ is using an oxidation state of $27$ .

$27$

Step 6

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

$N a = 1, H = - 28, B = 27$