Question

1. refer to the bohr models above: element c is in period _. element b is in period _. element b has (lower, higher) shielding effect compared to element c. element a is in group _. element b is in group _. element c is in group ___. 12. what is the oxidation number (charge) for each of the following? rb -1 br -3 n +3 al +2 mg -2 s sulfur, tellurium, oxygen

Explanation:

Step1: Recall oxidation number rules

Oxidation number of elements in their elemental form is 0. For simple ions, it is equal to the charge of the ion.

Step2: Determine oxidation number of Rb

Rubidium (Rb) is an alkali - metal. In compounds, it typically has an oxidation number of + 1. Here, if we assume a simple ionic situation, for Rb, the oxidation number is + 1.

Step3: Determine oxidation number of Br

Bromine (Br) in a simple binary ionic compound with a metal like Rb has an oxidation number of - 1 as it gains one electron to achieve a stable electron configuration.

Step4: Determine oxidation number of N

Nitrogen in compounds can have various oxidation numbers. In many cases, when it forms an anion, like in some nitrogen - containing compounds, it can have an oxidation number of - 3 as it can gain 3 electrons to achieve a stable octet.

Step5: Determine oxidation number of Al

Aluminum (Al) is a metal in group 13. It typically loses 3 electrons in compounds, so its oxidation number is + 3.

Step6: Determine oxidation number of Mg

Magnesium (Mg) is a group 2 metal. It loses 2 electrons in compounds, so its oxidation number is + 2.

Step7: Determine oxidation number of S

Sulfur (S) can have different oxidation numbers. In some compounds, it can gain 2 electrons to form an anion, so it can have an oxidation number of - 2.

Answer:

Rb: + 1
Br: - 1
N: - 3
Al: + 3
Mg: + 2
S: - 2