Question

name the following type iii compounds:

1. n₂o₅ dinitrogen pentoxide
2. pcl₅ phosphorus pentachloride
3. of₂ oxygen difluoride
4. cs₂

write formulas for the following type iii compounds:

1. dinitrogen tetroxide
2. sulfur trioxide
3. carbon tetrafluoride
4. xenon hexafluoride

write formulas for the following compounds that contain polyatomic ions:

1. ammonium nitrate
2. potassium dichromate
3. calcium carbonate
4. lead (iv) sulfate
5. cobalt (ii) phosphate

Explanation:

Step1: Naming type - III compounds

Type - III compounds are binary molecular compounds. We use prefixes to indicate the number of each atom. For $N_2O_5$, 'di -' for 2 nitrogen atoms and 'penta -' for 5 oxygen atoms, so it is dinitrogen pentoxide. For $PCl_5$, 'phosphorus' is the first element and 'penta -' and 'chloride' for 5 chlorine atoms, so it is phosphorus pentachloride. For $OF_2$, 'oxygen' first and 'di -' and 'fluoride' for 2 fluorine atoms, so oxygen difluoride. For $CS_2$, 'carbon' first and 'di -' and'sulfide' for 2 sulfur atoms, so carbon disulfide.

Step2: Writing formulas for type - III compounds

For dinitrogen tetroxide, 'di -' means 2 nitrogen atoms and 'tetra -' means 4 oxygen atoms, so the formula is $N_2O_4$. For sulfur trioxide,'sulfur' is S and 'tri -' means 3 oxygen atoms, so $SO_3$. For carbon tetrafluoride, 'carbon' is C and 'tetra -' means 4 fluorine atoms, so $CF_4$. For xenon hexafluoride, 'xenon' is Xe and 'hexa -' means 6 fluorine atoms, so $XeF_6$.

Step3: Writing formulas for compounds with polyatomic ions

For ammonium nitrate, ammonium ion is $NH_4^+$ and nitrate ion is $NO_3^-$, so the formula is $NH_4NO_3$. For potassium dichromate, potassium ion is $K^+$ and dichromate ion is $Cr_2O_7^{2 -}$, so the formula is $K_2Cr_2O_7$. For calcium carbonate, calcium ion is $Ca^{2+}$ and carbonate ion is $CO_3^{2 -}$, so the formula is $CaCO_3$. For lead(IV) sulfate, lead(IV) ion is $Pb^{4+}$ and sulfate ion is $SO_4^{2 -}$, so the formula is $Pb(SO_4)_2$. For cobalt(II) phosphate, cobalt(II) ion is $Co^{2+}$ and phosphate ion is $PO_4^{3 -}$, using the criss - cross method, the formula is $Co_3(PO_4)_2$.

Answer:

1. Dinitrogen pentoxide
2. Phosphorus pentachloride
3. Oxygen difluoride
4. Carbon disulfide
5. $N_2O_4$
6. $SO_3$
7. $CF_4$
8. $XeF_6$
9. $NH_4NO_3$
10. $K_2Cr_2O_7$
11. $CaCO_3$
12. $Pb(SO_4)_2$
13. $Co_3(PO_4)_2$