Question

What is the formula of titanium(IV) bromide?

Answer 1

Answer:

$\boxed{\text{SnBr}_{4}}$

Explanation:

The name tells you that this is a binary compound (contains two elements).

It contains a metal and a nonmetal, so it is a binary ionic compound. The general rule is:

Name of compound = name of metal  name of ion (two words)

Name of metal = tin(IV), so the tin ion has a charge of 4+

Name of ion = bromide. Br is in Group 17, so bromide ion has charge of 1-.

$\rm Sn ^{4+} + 4Br ^{-} \longrightarrow \, SnBr _{4} \\\\\text{The formula is} \boxed{\textbf{SnBr}_{4}}$