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What is the formula of titanium(IV) bromide?

User Aldorado
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1 Answer

5 votes

Answer:


\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)}

User Kevin Vella
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