Question

What is the formula of titanium(IV) bromide?

Answer 1

`\boxed{\text{SnBr}_(4)}`

Step-by-step 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)}`