# What is the oxidation number of the metal in lead hydroxide?

Nov 27, 2016

Lead takes a $+ I I$ oxidation number.

#### Explanation:

The oxidation number is formally the charge left on the central atom when all the bonding pairs of electrons are removed, with the charge assigned to the most electronegative atom.

When we do this for $P b {\left(O H\right)}_{2}$, we get:

$P b {\left(O H\right)}_{2} \rightarrow P {b}^{2 +} + 2 \times H {O}^{-}$, lead is definitely LESS electronegative than oxygen, so oxygen gets the charge. Do this again for hydroxide anion:

$H {O}^{-} \rightarrow {H}^{+} + {O}^{2 -}$

Hydrogen is definitely LESS electronegative than oxygen, so oxygen gets the charge again.

We are left with $P {b}^{2 +}$, ${H}^{+}$, and ${O}^{2 -}$ as FORMAL species, and thus the oxidation numbers are $P b \left(I I +\right) , O \left(- I I\right) , \mathmr{and} H \left(+ I\right)$.

Remember that this exercise is a formalism; an operation that has little real existence. Assignment of oxidation numbers is important when we balance redox equations, and we use electrons as virtual (as formal) particles.