for each of the following pairs of elements, give the correct molecular formula, explain how you arrived at this answer( you may use the Bohr model notation, or the orbital model notation, or explain linguistically) and note whether the bonding is covalent or ionic.

( I tried all of them and got the wrong answers so i really need help with these)

Lithium and bromine

radium and fluorine

strontium and oxygen

gallium and chlorine

germanium and oxygen

nitrogen and hydrogen

calcium and hydrogen

sulfur and carbon

IF you missed them all, you have some serious stinkin' thinking.

watch this:

http://www.youtube.com/watch?v=afnA223Pyz4

Here is how you do the simpler ones.

Li and Br. Look at the periodic table. Li is in group I which means it has 1 electron in its outside shell. Br is in group 17 (or VIIA depending upon the system you are using) which means it has 7 electrons in its outside shell. Li wants to lose 1, Br wants to gain 1 so the compound is formed, the electron actually transfers from Li (making it a Li^+ ion) to Br(making it a Br^- ion). The compound is ionic since the electron is transferred and not shared.
Ra is in group II meaning it has two electrons to give away. F is in group 17 meaning it has 7 electrons in the outside shell and needs to gain 1; therefore the Ra atom loses its two electrons to TWO F atoms to make the compound RaF2. It, too, is ionic. When atoms on opposite sides of the periodic table (group I, II, or III metals combine with group 15, 16, 17 non-metals) generally we call those ionic compounds. That's the case with RaF2 an LiBr. When atoms on the same side combine OR combination of elements in the middle with either side, those generally are covalent. Examples are Cl2, H2, CH4, NH3, etc. See if this will get you started.

To determine the correct molecular formula and identify the type of bonding (covalent or ionic) for each pair of elements, we can utilize some basic principles of chemical bonding.

1. Lithium and Bromine:
Lithium is an alkali metal with one valence electron, while bromine is a halogen with seven valence electrons. By transferring the electron from lithium to bromine, both elements achieve a stable electron configuration. Thus, the molecular formula is LiBr, and the bonding is ionic.

2. Radium and Fluorine:
Radium is an alkaline earth metal, and fluorine is a halogen. Similar to the previous pair, the electron from radium can transfer to fluorine to achieve a stable electron configuration. Therefore, the molecular formula is RaF, and the bonding is ionic.

3. Strontium and Oxygen:
Strontium is also an alkaline earth metal, while oxygen is a non-metal. The electronegativity difference between them is not significant enough to cause an electron transfer. Therefore, strontium and oxygen will form a covalent bond. The molecular formula is SrO.

4. Gallium and Chlorine:
Gallium is a metal, and chlorine is a non-metal. Since gallium has a lower electronegativity than chlorine, an electron transfer is likely to occur. Thus, the molecular formula is GaCl3, and the bonding is ionic.

5. Germanium and Oxygen:
Germanium is a metalloid, and oxygen is a non-metal. Germanium has a moderate electronegativity, and the electronegativity difference with oxygen is not significant enough for an electron transfer. Therefore, germanium and oxygen will form a covalent bond. The molecular formula is GeO2.

6. Nitrogen and Hydrogen:
Nitrogen is a non-metal, and hydrogen is also a non-metal. These two elements will typically form a covalent bond rather than an ionic bond. The molecular formula is NH3.

7. Calcium and Hydrogen:
Calcium is an alkaline earth metal, and hydrogen is a non-metal. Again, the electronegativity difference between them is not large enough for an electron transfer. Thus, calcium and hydrogen will form a covalent bond. The molecular formula is CaH2.

8. Sulfur and Carbon:
Both sulfur and carbon are non-metals, and they will form a covalent bond between them. The molecular formula is CS2.

Remember that these answers are based on typical situations and electronegativity trends. In some cases, elements may form multiple compounds or exhibit exceptions to the general bonding patterns.