Question #8e83b

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Question #8e83b

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Answer 1 · 2017-02-21T02:12:53

$Mercuric bromide.........$ $"Mercuric bromide........."$

Explanation:

Or $mercury(II) bromide$ $"mercury(II) bromide"$ . And $H g_{2} B r_{2}$ $Hg_2Br_2$ is $mercurous bromide$ $"mercurous bromide"$ . What are the oxidation states of mercury in each salt?

Answer 2 · 2017-02-21T02:13:27

${HgBr}_{2}$ $"HgBr"_2"$ is mercury (II) bromide.

Explanation:

Since bromide anions always have a $1^{-}$ $1^(-)$ charge, and there are two bromide ions in the formula, the overall charge for the bromide part of the compound is $2^{-}$ $2^(-)$ . This compound is neutral, which means that the mercury cation must have a $2^{+}$ $2^(+)$ charge, and is therefore the mercury (II) cation.

Therefore, ${HgBr}_{2}$ $"HgBr"_2"$ is mercury (II) bromide.

If you go to the following site, you will see that ${HgBr}_{2}$ $"HgBr"_2"$ has many names, but usually it is called mercury (II) bromide. https://pubchem.ncbi.nlm.nih.gov/compound/24612#section=Top

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