Question #8f19d

3 Answers
Oct 19, 2017

Well, you need another reactant to write a reaction...

$\text{PbO"_2(s) + 4"HCl"(aq) -> "PbCl"_4(aq) + 2"H"_2"O} \left(l\right)$

${\text{PbCl}}_{4}$ rapidly decomposes in water to give off a gas.

${\text{PbCl"_4(aq) -> "PbCl"_2(s) + "Cl}}_{2} \left(g\right)$

You cannot write a reaction with just lead(IV) oxide in water. It is very insoluble unless you add strong acid.

On the other hand, if you wish to write a chemical FORMULA, then you must know that oxygen atom typically assumes a $- 2$ oxidation state. Aside from a few exceptions, this is generally true, and in this case it is.

Oxygen is more electronegative than lead by far, and thus it assumes a negative oxidation state because it wants the electrons more to a first approximation.

It CANNOT be assumed that lead it is a $+ 4$ a priori, because its common oxidation state is NOT a $+ 4$ but a $+ 2$!! But from the name, we know that $I V = 4$ in English terms, and thus, lead has a $+ 4$ oxidation state from the given name.

That automatically means a $+ 4$ balances out with a $- 4$ from one or more identical oxygen anions. We already stated the $- 2$ oxidation state for each oxygen, so...

$\textcolor{b l u e}{{\text{PbO}}_{2}}$ is lead(IV) oxide.

Oct 19, 2017

The reaction to form Lead Oxide

$P b + 1 {O}_{2} = P b {O}_{2}$

Explanation:

Lead Pb has a common oxidation number of +4
Oxygen O has a common oxidation number of -2
It takes 2 Oxygen atoms to balance one Lead atom

so the formula for Lead Oxide is $P b {O}_{2}$

The equation for forming Lead Oxide is

$P b + 1 {O}_{2} = 1 P b {O}_{2}$

I hope this is what your meant by your question.

I think you might have meant to ask how to write lead (IV) oxide as a chemical formula .

Explanation:

The chemical formula for lead (IV) oxide is: $P b {O}_{2}$

Formulas describe one compound.

Reactions describe changes that involve multiple chemical substances which can include both compounds and elements.

The Roman numeral IV indicates the oxidation state of the lead ion in this compound.

Lead (IV) means the $P {b}^{4 +}$ ion
Oxide ion is always ${O}^{2 -}$

To balance the charges you need two ${O}^{2 -}$ ions to pair with the $P {b}^{4 +}$ ion.

Here are some other examples of compounds that use Roman numerals to tell you about the oxidation state of metals.

Hope this helps!