Question

1. for each compound below, write in the missing formula or name.

name	formula	name	formula
ruthenium (iii) sulfide		*	ti₃p₂
**	zno	*rhodium (iii) phosphide
**	cdi₂	*niobium (vi) sulfide
zirconium (iv) sulfide		*	tci₄
**	hfcl₄		cu₃p₂
iron (iii) nitride		palladium (vi) selenide
**	mno₂	cadmium (ii) oxide
**	wf₆	mercury (i) selenide
rhenium (vi) oxide			mnp

Explanation:

Step1: Analyze vanadium (II) sulfide

Vanadium (II) has a charge of \( +2 \), sulfide (\( S^{2 -} \)). To balance charges, the formula is \( \text{V}_2\text{S}_2 \) (simplify to \( \text{VS} \))? Wait, no: charge of V is +2, S is -2. So ratio 1:1. Wait, vanadium (II) is \( \text{V}^{2 +} \), sulfide is \( \text{S}^{2 -} \). So formula \( \text{VS} \)? Wait, no, maybe I made a mistake. Wait, vanadium (II) sulfide: \( \text{V}^{2 +} \) and \( \text{S}^{2 -} \), so formula \( \text{VS} \)? Wait, no, let's check again. Wait, maybe I confused. Wait, vanadium (II) sulfide: the cation is \( \text{V}^{2 +} \), anion is \( \text{S}^{2 -} \). So the formula is \( \text{VS} \)? Wait, no, maybe I should do criss - cross. Charge of V: +2, S: -2. So \( \text{V}_2\text{S}_2 \) simplifies to \( \text{VS} \).

Step2: Analyze ruthenium (III) sulfide

Ruthenium (III) is \( \text{Ru}^{3 +} \), sulfide is \( \text{S}^{2 -} \). Criss - cross: 2 Ru and 3 S. So formula \( \text{Ru}_2\text{S}_3 \).

Step3: Analyze ZnO

Zn has a charge of +2 (Zn is in group 12, common charge +2), O is -2. So the name is zinc (II) oxide.

Step4: Analyze \( \text{CdI}_2 \)

Cd is +2 (group 12), I is -1. So name is cadmium (II) iodide.

Step5: Analyze zirconium (IV) sulfide

Zirconium (IV) is \( \text{Zr}^{4 +} \), sulfide is \( \text{S}^{2 -} \). Criss - cross: 2 Zr and 4 S? Wait, no: \( \text{Zr}^{4 +} \) and \( \text{S}^{2 -} \). So ratio: 4 and 2, simplify to 2:1? Wait, no, criss - cross: charge of Zr is +4, S is -2. So formula \( \text{ZrS}_2 \)? Wait, no: \( 4\times1 + (- 2)\times2=0 \). Wait, Zr⁴⁺ and S²⁻: so 1 Zr and 2 S: \( \text{ZrS}_2 \)? Wait, no, wait zirconium (IV) sulfide: \( \text{Zr}^{4 +} \), \( \text{S}^{2 -} \). So \( \text{ZrS}_2 \) (because \( 4+2\times(- 2)=0 \)).

Step6: Analyze \( \text{HfCl}_4 \)

Hf is +4 (group 4, common charge +4), Cl is -1. So name is hafnium (IV) chloride.

Step7: Analyze iron (III) nitride

Iron (III) is \( \text{Fe}^{3 +} \), nitride is \( \text{N}^{3 -} \). Criss - cross: 1 Fe and 1 N. So formula \( \text{FeN} \).

Step8: Analyze \( \text{MnO}_2 \)

Mn in \( \text{MnO}_2 \): O is -2, so Mn has charge +4 (because \( x+2\times(- 2)=0\Rightarrow x = + 4 \)). So name is manganese (IV) oxide.

Step9: Analyze \( \text{WF}_6 \)

W in \( \text{WF}_6 \): F is -1, so W has charge +6 ( \( x+6\times(- 1)=0\Rightarrow x = + 6 \) ). Name is tungsten (VI) fluoride.

Step10: Analyze rhenium (VI) oxide

Rhenium (VI) is \( \text{Re}^{6 +} \), oxide is \( \text{O}^{2 -} \). Criss - cross: 2 Re and 6 O? Wait, no: \( \text{Re}^{6 +} \) and \( \text{O}^{2 -} \). So \( \text{Re}_2\text{O}_6 \) simplifies to \( \text{ReO}_3 \)? Wait, no: charge of Re: +6, O: -2. So \( \text{ReO}_3 \) (because \( 6+3\times(- 2)=0 \)).

Step11: Analyze \( \text{ZrS}_2 \)

Zr is +4, S is -2. Wait, no, \( \text{ZrS}_2 \): Zr charge? Wait, the formula is \( \text{ZrS}_2 \), so Zr must be +4 (since 2 S at -2 each: \( 4+2\times(- 2)=0 \)). So name is zirconium (IV) sulfide? Wait, no, the formula is \( \text{ZrS}_2 \), so the name is zirconium (IV) sulfide? Wait, no, the cation is Zr, anion is S²⁻. So Zr has charge +4 (because 2 S²⁻: \( 4 + 2\times(- 2)=0 \)). So name is zirconium (IV) sulfide? Wait, but the first column has zirconium (IV) sulfide. Wait, the second column has \( \text{ZrS}_2 \), so the name is zirconium (IV) sulfide? Wait, no, let's check: Zr⁴⁺ and S²⁻: formula \( \text{ZrS}_2 \), so name is zirconium (IV) sulfide.

Step12: Analyze \( \text{Ti}_3\text{P}_2 \)

Ti: let's find charge. P is -3 (phosphide). So 3 Ti and 2 P. Let charge of Ti be \( x \). \( 3x+2\times(- 3)=0\Rightarrow 3x = 6\Rightarrow x = + 2 \). So Ti is +2. So name is titanium (II) phosphide.

Step13: Analyze rhodium (III) phosphide

Rhodium (III) is \( \text{Rh}^{3 +} \), phosphide is \( \text{N}^{3 -} \) (wait, phosphide is \( \text{P}^{3 -} \)). So formula: \( \text{RhP} \) (since \( 3+(- 3)=0 \)).

Step14: Analyze niobium (VI) sulfide

Niobium (VI) is \( \text{Nb}^{6 +} \), sulfide is \( \text{S}^{2 -} \). Criss - cross: 2 Nb and 6 S? Wait, no: \( \text{Nb}^{6 +} \) and \( \text{S}^{2 -} \). So \( \text{Nb}_2\text{S}_6 \) simplifies to \( \text{NbS}_3 \)? Wait, no: \( 6+3\times(- 2)=0 \). So formula \( \text{NbS}_3 \)? Wait, no, criss - cross: charge of Nb: +6, S: -2. So \( \text{Nb}_2\text{S}_6 \) (simplify to \( \text{NbS}_3 \)).

Step15: Analyze \( \text{TcCl}_4 \)

Tc: let's find charge. Cl is -1. So \( x+4\times(- 1)=0\Rightarrow x = + 4 \). So Tc is +4. Name is technetium (IV) chloride.

Step16: Analyze \( \text{Cu}_3\text{P}_2 \)

Cu: let's find charge. P is -3. \( 3x+2\times(- 3)=0\Rightarrow 3x = 6\Rightarrow x = + 2 \). So Cu is +2. Name is copper (II) phosphide.

Step17: Analyze palladium (VI) selenide

Palladium (VI) is \( \text{Pd}^{6 +} \), selenide is \( \text{Se}^{2 -} \). Criss - cross: 2 Pd and 6 Se? Wait, no: \( \text{Pd}^{6 +} \) and \( \text{Se}^{2 -} \). So \( \text{Pd}_2\text{Se}_6 \) simplifies to \( \text{PdSe}_3 \) (since \( 6+3\times(- 2)=0 \)).

Step18: Analyze cadmium (II) oxide

Cadmium (II) is \( \text{Cd}^{2 +} \), oxide is \( \text{O}^{2 -} \). Formula \( \text{CdO} \).

Step19: Analyze mercury (I) selenide

Mercury (I) is \( \text{Hg}_2^{2 +} \) (mercury (I) is a diatomic cation), selenide is \( \text{Se}^{2 -} \). So formula \( \text{Hg}_2\text{Se} \) (since \( 2+(- 2)=0 \)).

Step20: Analyze \( \text{MnP} \)

Mn: let's find charge. P is -3. So \( x+(- 3)=0\Rightarrow x = + 3 \). So Mn is +3. Name is manganese (III) phosphide.

Answer:

Name	Formula	Name	Formula
ruthenium (III) sulfide	\( \text{Ru}_2\text{S}_3 \)	titanium (II) phosphide	\( \text{Ti}_3\text{P}_2 \)
zinc (II) oxide	\( \text{ZnO} \)	rhodium (III) phosphide	\( \text{RhP} \)
cadmium (II) iodide	\( \text{CdI}_2 \)	niobium (VI) sulfide	\( \text{NbS}_3 \)
zirconium (IV) sulfide	\( \text{ZrS}_2 \)	technetium (IV) chloride	\( \text{TcCl}_4 \)
hafnium (IV) chloride	\( \text{HfCl}_4 \)	copper (II) phosphide	\( \text{Cu}_3\text{P}_2 \)
iron (III) nitride	\( \text{FeN} \)	palladium (VI) selenide	\( \text{PdSe}_3 \)
manganese (IV) oxide	\( \text{MnO}_2 \)	cadmium (II) oxide	\( \text{CdO} \)
tungsten (VI) fluoride	\( \text{WF}_6 \)	mercury (I) selenide	\( \text{Hg}_2\text{Se} \)
rhenium (VI) oxide	\( \text{ReO}_3 \)	manganese (III) phosphide	\( \text{MnP} \)

(Note: Some of the formulas and names may need further verification as the process of determining ionic compound formulas and names depends on accurate charge determination and criss - cross method application. For example, vanadium (II) sulfide: if vanadium (II) is \( \text{V}^{2 +} \) and sulfide is \( \text{S}^{2 -} \), the formula is \( \text{VS} \), but sometimes transition metals can have more complex behaviors, but based on simple charge - balance, this is the approach.)