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

fill in the name and empirical formula of each ionic compound that could be formed from the ions in this table:\n| some ionic compounds | | | |\n|----|----|----|----|\n| cation | anion | empirical formula | name of compound |\n| $pb^{4 + }$ | $cn^{-}$ | | |\n| $na^{+}$ | $no_{2}^{-}$ | | |\n| $fe^{2+}$ | $clo_{3}^{-}$ | | |

fill in the name and empirical formula of each ionic compound that could be formed from the ions in this table:\n| some ionic compounds | | | |\n|----|----|----|----|\n| cation | anion | empirical formula | name of compound |\n| $pb^{4 + }$ | $cn^{-}$ | | |\n| $na^{+}$ | $no_{2}^{-}$ | | |\n| $fe^{2+}$ | $clo_{3}^{-}$ | | |

Answer

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
$Pb^{4+}$ $CN^-$ $Pb(CN)_4$ lead(IV) cyanide
$Na^+$ $NO_2^-$ $NaNO_2$ sodium nitrite
$Fe^{2+}$ $ClO_3^-$ $Fe(ClO_3)_2$ iron(II) chlorate