Question #25b8e

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Question #25b8e

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Answer 1 · 2016-06-10T00:12:56

$X_{2} Y_{3}$ $"X"_2"Y"_3$

Explanation:

The idea here is that you need to use the fact that you know the charges on the bromide anion and on the sodium cation to determine the charges on the $X$ $"X"$ cation and on the $Y$ $"Y"$ anion.

So, bromine is located in group 17 of the periodic table, which means that it forms $1 -$ $1-$ anions.

You know that ionic compounds follow the criss cross rule, which states that the charge on the cation becomes the subscript on the anion and vice versa.

This means that the ${XBr}_{3}$ $"XBr"_3$ compound can be broken up into ions like this

${XBr}_{3} \to X^{3 +} + 3 {Br}^{-}$ $"XBr"_color(red)(3) -> "X"^color(red)(3+) + 3"Br"^(-)$

The $X$ $"X"$ cation will thus carry a $3 +$ $3+$ charge.

Sodium is located in group 1 of the periodic table, which means that it forms $1 +$ $1+$ cations. Use the same rule to break up the ${Na}_{2} Y$ $"Na"_2"Y"$ compound

${Na}_{2} Y \to 2 {Na}^{+} + Y^{2 -}$ $"Na"_color(blue)(2)"Y" -> 2"Na"^(+) + "Y"^color(blue)(2-)$

The $Y$ $"Y"$ anion will thus have a $2 -$ $2-$ charge.

Therefore, the compound that is made up of $X^{3 +}$ $"X"^(color(red)(3+))$ cations and $Y^{2 -}$ $"Y"^(color(blue)(2-))$ anions will be

$2 X^{3 +} + 3 Y^{2 -} \to X_{2} Y_{3}$ $2"X"^color(red)(3+) + 3"Y"^color(blue)(2-) -> "X"_color(blue)(2)"Y"_color(red)(3)$

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Question #25b8e

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