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
$pb^{4 + }$	$cn^{-}$
$na^{+}$	$no_{2}^{-}$
$fe^{2+}$	$clo_{3}^{-}$

Explanation:

Step1: Determine formula for $Pb^{4 +}$ and $CN^-$

For the cation $Pb^{4+}$ and anion $CN^-$, to balance the charges, we need 4 $CN^-$ ions for 1 $Pb^{4+}$ ion. So the empirical - formula is $Pb(CN)_4$. The name of the compound is lead(IV) cyanide.

Step2: Determine formula for $Na^+$ and $NO_2^-$

For the cation $Na^+$ and anion $NO_2^-$, the charges are already balanced in a 1:1 ratio. So the empirical - formula is $NaNO_2$. The name of the compound is sodium nitrite.

Step3: Determine formula for $Fe^{2+}$ and $ClO_3^-$

For the cation $Fe^{2+}$ and anion $ClO_3^-$, to balance the charges, we need 2 $ClO_3^-$ ions for 1 $Fe^{2+}$ ion. So the empirical - formula is $Fe(ClO_3)_2$. The name of the compound is iron(II) chlorate.

Answer:

cation	anion	empirical formula	name of compound
$Na^+$	$NO_2^-$	$NaNO_2$	sodium nitrite
$Fe^{2+}$	$ClO_3^-$	$Fe(ClO_3)_2$	iron(II) chlorate