Question

fill in the name and empirical formula of each ionic compound that could be formed from the ions in this table:

some ionic compounds
$na^{+}$	$bro_{4}^{-}$
$na^{+}$	$bro_{3}^{-}$
$na^{+}$	$bro^{-}$
$na^{+}$	$bro_{2}^{-}$

Explanation:

Step1: Determine formula for $Na^+$ and $BrO_4^-$

The charges of $Na^+$ and $BrO_4^-$ are +1 and - 1 respectively. To form a neutral compound, the ratio of cations to anions is 1:1. So the empirical formula is $NaBrO_4$.

Step2: Name the compound for $Na^+$ and $BrO_4^-$

The name of the compound with $Na^+$ and $BrO_4^-$ (perbromate ion) is sodium perbromate.

Step3: Determine formula for $Na^+$ and $BrO_3^-$

The charges of $Na^+$ and $BrO_3^-$ are +1 and -1 respectively. The empirical formula is $NaBrO_3$.

Step4: Name the compound for $Na^+$ and $BrO_3^-$

The name of the compound with $Na^+$ and $BrO_3^-$ (bromate ion) is sodium bromate.

Step5: Determine formula for $Na^+$ and $BrO^-$

The charges of $Na^+$ and $BrO^-$ are +1 and -1 respectively. The empirical formula is $NaBrO$.

Step6: Name the compound for $Na^+$ and $BrO^-$

The name of the compound with $Na^+$ and $BrO^-$ (hypobromite ion) is sodium hypobromite.

Step7: Determine formula for $Na^+$ and $BrO_2^-$

The charges of $Na^+$ and $BrO_2^-$ are +1 and -1 respectively. The empirical formula is $NaBrO_2$.

Step8: Name the compound for $Na^+$ and $BrO_2^-$

The name of the compound with $Na^+$ and $BrO_2^-$ (bromite ion) is sodium bromite.

Answer:

cation	anion	empirical formula	name of compound
$Na^+$	$BrO_3^-$	$NaBrO_3$	sodium bromate
$Na^+$	$BrO^-$	$NaBrO$	sodium hypobromite
$Na^+$	$BrO_2^-$	$NaBrO_2$	sodium bromite