11. refer to the bohr models above: element c is in period ___. element b is in period ___. element b has…
11. 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
Answer
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