Nomenclature Dry Lab Read all three introductions on pp 85-94 prior to lab. No p

Question

Nomenclature Dry Lab Read all three introductions on pp 85-94 prior to lab. No pre-lab is due. The assigned problems are due one week after the lab session. Submit answers to the problems individually. Part A Oxidation Numbers For elements in free state, oxidation number is zero. For monatomic ions, oxidation number is the same as the charge. Oxygen almost always has an oxidation number of-2. The exceptions are peroxides, such as H O, where oxygen is -1, and compounds with oxygen attached to fluorine, where oxygen can be zero or positive. Hydrogen almost always has an oxidation number of +1 Exceptions are metal hydrides, where hydrogen has a -1 oxidation number Main group metals (outer columns of the periodic table, left side), normally have oxidation 1 for column 1A (Na and K), +2 for column 2A (Ca and Mg), and +3 for column 3A (Al and Ga). Main group nonmetals (outer columns of the periodic table, right side) often have oxidation numbers that are found by column number minus eight. That is-1 for column VIIA (CI and Br), and-2 for column VIA (0 and S). Note that the oxidation number in polyatomic ions, such as CIO and SO4, may be different (see next item). Polyatomic ions have a charge equal to sum of oxidation numbers of its atoms. For instance, SO. has +6+(4X-2)--2. Transition metals and other heavy metals often have more than one possible oxidation number each, and the numbers are difficult to predict. (Fe can be +2 or +3, and Cu can be +1 or +2.) The combined sum of oxidation numbers for all of the atoms in a neutral compound is zero. Problems for Part A 1. Show how to find the oxidation number of S in SO Note that SO, is a neutral compound, not the sulfite anion. 2. Show how to find the oxidation number of P in PO. 3. Using the periodic table, state the oxidation number of K in K CrO Then, show how to find the oxidation number of Cr. 4. Determine the formulas for the four compounds formed from Fe with P1 S2, OH', and CIOs

Explanation / Answer

1) Oxygen being highly electronegative and belonging to colum V1 A will exists in -2 oxidiation

Let the oxidation state of Sulphur be x

Since the compound is neutral, hence the overall charge must be zero

x + 3(-2) = 0

x = 6

Hence oxidation number of Sulphur is +6

2) Oxygen being highly electronegative and belonging to colum V1 A will exists in -2 oxidiation

Let the oxidation state of Phosphours be x

Since the compound is having a charge of -3, hence the overall charge must be -3

x + 4(-2) = -3

x = 5

Hence oxidation number of Phosphorus is +5

3)

Oxygen being highly electronegative and belonging to colum V1 A will exists in -2 oxidiation

Potassium belongs to alkaline metals, hence its oxidation state will be +1

Let the oxidation number of Chromium be x

Since the compound is neutral, hence the overall charge must be 0

2 + x + 4(-2) = 0

x = 6

Hence oxidation number of Chromium is +6

4)

With P(3-) compound will be Fe3P2

With S(2-) compound will be FeS

With OH- compound will be Fe(OH)2

With ClO4- compound will be Fe(ClO4)2

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.