Question

1. fill in the table below for ionic compounds with multivalent ions.

formula	name

|v(iii)br₃|
|ni(ii)₃p₂|
|fe(ii)o|
|fe(iii)₂o₃|

fe(iii)₂f	iron(ii) fluoride
	mercury(i) sulfide
	manganese(iv) nitrate
	vanadium(iii) phosphide

Explanation:

Step1: Recall naming rules for ionic compounds

For ionic compounds with multivalent (variable - charge) metals, we use Roman numerals in parentheses to indicate the metal's oxidation state. The non - metal part is named as an anion.

Step2: Name $V(III)Br_3$

The metal is vanadium with an oxidation state of + 3 and the non - metal is bromine. The bromide ion is $Br^-$. So the name is Vanadium(III) Bromide.

Step3: Name $Ni(II)_3P_2$

The metal is nickel with an oxidation state of + 2 and the non - metal is phosphorus. The phosphide ion is $P^{3 - }$. So the name is Nickel(II) Phosphide.

Step4: Name $Fe(II)O$

The metal is iron with an oxidation state of + 2 and the non - metal is oxygen. The oxide ion is $O^{2 - }$. So the name is Iron(II) Oxide.

Step5: Name $Fe(III)_2O_3$

The metal is iron with an oxidation state of + 3 and the non - metal is oxygen. The oxide ion is $O^{2 - }$. So the name is Iron(III) Oxide.

Step6: Write formula for Iron(II) Fluoride

Iron(II) has a charge of $Fe^{2+}$ and fluoride ion is $F^-$. To balance the charges, the formula is $FeF_2$.

Step7: Write formula for Mercury(I) Sulfide

Mercury(I) exists as $Hg_2^{2+}$ and sulfide ion is $S^{2 - }$. The formula is $Hg_2S$.

Step8: Write formula for Manganese(IV) Nitrate

Manganese(IV) has a charge of $Mn^{4+}$ and nitrate ion is $NO_3^-$. To balance the charges, the formula is $Mn(NO_3)_4$.

Step9: Write formula for Vanadium(III) Phosphide

Vanadium(III) has a charge of $V^{3+}$ and phosphide ion is $P^{3 - }$. The formula is $VP$.

Answer:

Formula	Name
$Ni(II)_3P_2$	Nickel(II) Phosphide
$Fe(II)O$	Iron(II) Oxide
$Fe(III)_2O_3$	Iron(III) Oxide
$FeF_2$	Iron(II) Fluoride
$Hg_2S$	Mercury(I) Sulfide
$Mn(NO_3)_4$	Manganese(IV) Nitrate
$VP$	Vanadium(III) Phosphide