Question

write the ions first then write the chemical formula 1) mg+1, o-2 → mg1o-2 2) lithium bromide 3) calcium nitride 4) aluminum sulfide 5) potassium iodide 6) strontium chloride 7) sodium sulfide 8) radium bromide 9) magnesium sulfide 10) aluminum nitride 11) cesium sulfide 12) potassium chloride 13) strontium phosphide 14) barium iodide 15) sodium fluoride 16) calcium bromide 17) beryllium sulfide 18) gallium fluoride 19) aluminum phosphide 20) aluminum phosphide 21) rubidium oxide 22) calcium iodide 23) cesium oxide 24) magnesium iodide 25) lithium chloride 26) beryllium chloride 27) sodium oxide 28) calcium fluoride 29) indium phosphide 30) aluminum oxide

Explanation:

Step1: Recall ion - charge rules

For main - group metals, the charge is determined by their group number in the periodic table. Non - metals usually have negative charges based on the number of electrons needed to achieve a noble - gas configuration.

Step2: Determine the ions for lithium bromide

Lithium (Li) is in Group 1, so it forms a \(Li^{+}\) ion. Bromine (Br) is in Group 17, so it forms a \(Br^{-}\) ion. The chemical formula is \(LiBr\).

Step3: Determine the ions for calcium nitride

Calcium (Ca) is in Group 2, so it forms a \(Ca^{2 + }\) ion. Nitrogen (N) is in Group 15 and forms an \(N^{3-}\) ion. To balance the charges, the formula is \(Ca_{3}N_{2}\).

Step4: Determine the ions for aluminum sulfide

Aluminum (Al) is in Group 13 and forms an \(Al^{3+}\) ion. Sulfur (S) is in Group 16 and forms an \(S^{2 - }\) ion. The formula is \(Al_{2}S_{3}\).

Step5: Determine the ions for potassium iodide

Potassium (K) is in Group 1, so it forms a \(K^{+}\) ion. Iodine (I) is in Group 17, so it forms an \(I^{-}\) ion. The formula is \(KI\).

Step6: Determine the ions for strontium chloride

Strontium (Sr) is in Group 2, so it forms a \(Sr^{2+}\) ion. Chlorine (Cl) is in Group 17, so it forms a \(Cl^{-}\) ion. The formula is \(SrCl_{2}\).

Step7: Determine the ions for sodium sulfide

Sodium (Na) is in Group 1, so it forms a \(Na^{+}\) ion. Sulfur (S) is in Group 16, so it forms an \(S^{2 - }\) ion. The formula is \(Na_{2}S\).

Step8: Determine the ions for rubidium bromide

Rubidium (Rb) is in Group 1, so it forms a \(Rb^{+}\) ion. Bromine (Br) is in Group 17, so it forms a \(Br^{-}\) ion. The formula is \(RbBr\).

Step9: Determine the ions for magnesium sulfide

Magnesium (Mg) is in Group 2, so it forms a \(Mg^{2+}\) ion. Sulfur (S) is in Group 16, so it forms an \(S^{2 - }\) ion. The formula is \(MgS\).

Step10: Determine the ions for aluminum nitride

Aluminum (Al) is in Group 13 and forms an \(Al^{3+}\) ion. Nitrogen (N) is in Group 15 and forms an \(N^{3-}\) ion. The formula is \(AlN\).

Step11: Determine the ions for cesium sulfide

Cesium (Cs) is in Group 1, so it forms a \(Cs^{+}\) ion. Sulfur (S) is in Group 16, so it forms an \(S^{2 - }\) ion. The formula is \(Cs_{2}S\).

Step12: Determine the ions for potassium chloride

Potassium (K) is in Group 1, so it forms a \(K^{+}\) ion. Chlorine (Cl) is in Group 17, so it forms a \(Cl^{-}\) ion. The formula is \(KCl\).

Step13: Determine the ions for strontium phosphide

Strontium (Sr) is in Group 2, so it forms a \(Sr^{2+}\) ion. Phosphorus (P) is in Group 15 and forms a \(P^{3-}\) ion. The formula is \(Sr_{3}P_{2}\).

Step14: Determine the ions for barium iodide

Barium (Ba) is in Group 2, so it forms a \(Ba^{2+}\) ion. Iodine (I) is in Group 17, so it forms an \(I^{-}\) ion. The formula is \(BaI_{2}\).

Step15: Determine the ions for sodium fluoride

Sodium (Na) is in Group 1, so it forms a \(Na^{+}\) ion. Fluorine (F) is in Group 17, so it forms a \(F^{-}\) ion. The formula is \(NaF\).

Step16: Determine the ions for calcium bromide

Calcium (Ca) is in Group 2, so it forms a \(Ca^{2+}\) ion. Bromine (Br) is in Group 17, so it forms a \(Br^{-}\) ion. The formula is \(CaBr_{2}\).

Step17: Determine the ions for beryllium sulfide

Beryllium (Be) is in Group 2, so it forms a \(Be^{2+}\) ion. Sulfur (S) is in Group 16, so it forms an \(S^{2 - }\) ion. The formula is \(BeS\).

Step18: Determine the ions for gallium fluoride

Gallium (Ga) is in Group 13 and forms a \(Ga^{3+}\) ion. Fluorine (F) is in Group 17, so it forms a \(F^{-}\) ion. The formula is \(GaF_{3}\).

Step19: Determine the ions for aluminum phosphide

Aluminum (Al) is in Group 13 and forms an \(Al^{3+}\) ion. Phosphorus (P) is in Group 15 and forms a \(P^{3-}\) ion. The formula is \(AlP\).

Step20: Determine the ions for rubidium oxide

Rubidium (Rb) is in Group 1, so it forms a \(Rb^{+}\) ion. Oxygen (O) is in Group 16, so it forms an \(O^{2 - }\) ion. The formula is \(Rb_{2}O\).

Step21: Determine the ions for calcium iodide

Calcium (Ca) is in Group 2, so it forms a \(Ca^{2+}\) ion. Iodine (I) is in Group 17, so it forms an \(I^{-}\) ion. The formula is \(CaI_{2}\).

Step22: Determine the ions for cesium oxide

Cesium (Cs) is in Group 1, so it forms a \(Cs^{+}\) ion. Oxygen (O) is in Group 16, so it forms an \(O^{2 - }\) ion. The formula is \(Cs_{2}O\).

Step23: Determine the ions for magnesium iodide

Magnesium (Mg) is in Group 2, so it forms a \(Mg^{2+}\) ion. Iodine (I) is in Group 17, so it forms an \(I^{-}\) ion. The formula is \(MgI_{2}\).

Step24: Determine the ions for lithium chloride

Lithium (Li) is in Group 1, so it forms a \(Li^{+}\) ion. Chlorine (Cl) is in Group 17, so it forms a \(Cl^{-}\) ion. The formula is \(LiCl\).

Step25: Determine the ions for beryllium chloride

Beryllium (Be) is in Group 2, so it forms a \(Be^{2+}\) ion. Chlorine (Cl) is in Group 17, so it forms a \(Cl^{-}\) ion. The formula is \(BeCl_{2}\).

Step26: Determine the ions for sodium oxide

Sodium (Na) is in Group 1, so it forms a \(Na^{+}\) ion. Oxygen (O) is in Group 16, so it forms an \(O^{2 - }\) ion. The formula is \(Na_{2}O\).

Step27: Determine the ions for calcium fluoride

Calcium (Ca) is in Group 2, so it forms a \(Ca^{2+}\) ion. Fluorine (F) is in Group 17, so it forms a \(F^{-}\) ion. The formula is \(CaF_{2}\).

Step28: Determine the ions for indium phosphide

Indium (In) is in Group 13 and forms an \(In^{3+}\) ion. Phosphorus (P) is in Group 15 and forms a \(P^{3-}\) ion. The formula is \(InP\).

Step29: Determine the ions for aluminum oxide

Aluminum (Al) is in Group 13 and forms an \(Al^{3+}\) ion. Oxygen (O) is in Group 16, so it forms an \(O^{2 - }\) ion. The formula is \(Al_{2}O_{3}\).

Answer:

1. \(MgO\) (Note: Magnesium forms \(Mg^{2 + }\) not \(Mg^{+1}\), so the correct formula is \(MgO\))
2. \(LiBr\)
3. \(Ca_{3}N_{2}\)
4. \(Al_{2}S_{3}\)
5. \(KI\)
6. \(SrCl_{2}\)
7. \(Na_{2}S\)
8. \(RbBr\)
9. \(MgS\)
10. \(AlN\)
11. \(Cs_{2}S\)
12. \(KCl\)
13. \(Sr_{3}P_{2}\)
14. \(BaI_{2}\)
15. \(NaF\)
16. \(CaBr_{2}\)
17. \(BeS\)
18. \(GaF_{3}\)
19. \(AlP\)
20. \(AlP\)
21. \(Rb_{2}O\)
22. \(CaI_{2}\)
23. \(Cs_{2}O\)
24. \(MgI_{2}\)
25. \(LiCl\)
26. \(BeCl_{2}\)
27. \(Na_{2}O\)
28. \(CaF_{2}\)
29. \(InP\)
30. \(Al_{2}O_{3}\)