What is the formula of titanium(IV) bromide?

Chemistry

1 answer:

Anna [14]3 years ago

3 0

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}}$

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