Oxidation Numbers????
Look at the composition of the substance. AgNO3 is made up of Ag+ and NO3- ions, so since Ag+ is a monatomic ion, its charge is its oxidation number, +1. NO3-, two elements N and O, the addition of the oxidation numbers of N and O must equal the charge, -1. Oxygen is usually -2, (very few exceptions and this ion is not one of them), so using this knowledge: -1 = 3 (oxid. # O) + oxid. # N -1 = 3 (-2) + x ==> -1 = -6 + x, thus x = +5 for N. Fe(HCO3)2, HCO3- is the ion, you have two, so Fe is +2. H is always +1 when combined with non metals, O is -2 as in the previous example. So: -1 = +1 + 3(-2) + x -1 = +1 -6 + x, x = +4, which is the oxidation number of C in HCO3-.
Finding oxidation number of Fe in K4Fe(CN)6.?
K4Fe(CN)6 Potassium is +1, and cyanide is -1 The sum of the oxidation numbers must be zero for a compound. Write it as an algebra expression where x is the oxidation number of Fe: 4(+1) +x +6(-1) = 0 Solve for x, which is equal to +2 Therefore, iron has an oxidation number of +2. ======= Follow up ======= The oxidation number of cyanide ion is -1. Why? Because in nature we find the compound KCN which we call potassium cyanide. It has one atom each of K, C and N. Potassium has an oxidation number of +1. That is well established. Therefore, CN must have an oxidation number of -1 since the convention is that the oxidation numbers must add up to zero. Remember that oxidation numbers are ARBITRARY numbers that are ASSIGNED to elements and radicals (aka polyatomic ions).
Oxidation numbers??
I understand your trouble. In most cases it is easy to find the oxidation numbers. For example; Al2S3 ; Al = +3 , S = -2 OR MgO ; Mg = +2 , O = -1 The problem arises for the elements having more than one oxidation numbers. For example; CrCO3 , Cr2(CO3)3, NH4NO3, (NH4)3PO4, Ca(ClO4)2 In order to determine the oxidation numbers of the elements, sometimes the oxidation numbers of the polyatomic ions are used. The common polyatomic ions and their charges are given below. (OH)^- = hydroxide (NO3)^- = nitrate (NO2)^- = nitrite (CO3)^2- = carbonate (SO4)^2- = sulfate (SO3)^2- = sulfite (PO4)^3- = phosphate (PO^3)^3- = phosphite Now consider HNO3. This is simple, H is +1 and O is -2. Since we have 3 O = 3 x (-2) = -6 1 H = +1 and 3 O = -6 ıf we add them we get -5. To make the compound neutral N should be +5. Now consider Cu(NO3)2. Cu can be either +1 and +2 (transition element) To determine this first make use of the charge of nitrate ion. As it is stated above nitrate ion is (NO3)^-. Since there is 2 nitrate ion total charge is -2. To make the compound neutral Cu should be +2. Now for N in the nitrate ion. O is -2. there are 3 O atoms = -6. But this time we will not make it neutral, because it is a poly atomic ion. To leave -1 charge to the ion, N should ne +5. One more example; NH4NO3 We know that NO3 is a polyatomic ion having -1 charge. Therefore NH4 (ammonium polyatomic ion) should have +1 charge. We have already calculated the charge of N in NO3^- ion. What will be the charge of N in NH4^+ ion? Since H = 41, 4 H = +4, to leave 1 + charge to the ion , N should be -3. The last example; K2Cr2O7: Potassium is in Group IA (alkali metal) always +1 2 K = +2 Oxygen is -2. 7 O = -14. To make the compound neutral 2Cr should be +12. Single Cr atom should be 12 / 2 = +6 I think this explanation is satisfactory for you.
How to find oxidation number of for example SO4 2-?
Hey dude its easy. As you know the valency of oxygen is 2 : outermost orbit contains 6 electrons so valency = 6-8= -2 in SO4 -2 it has 2 negative charge so: lets suppose O.N of S be x, x + 4*(-2) = -2 x = -2 + 8 x= +6 SO got it now man .its easy
Help with oxidation numbers?
Start with the periodic table to make things easy. Group 1 = +1, group 2 = +2, group 3 = +3, group 14 = + or - 4, group 15 = -3, group 16 = -2, group 17 = -1, group 18 has full valence shell and bonds with no one so it has no oxidation number. Groups 4-13 have various oxidation numbers, some with more than one, and you can often get a clue by what it's bonded with. Let's look at your list. Mg is in group 2 so it's +2. O is in group 16 and is -2. So they balance at MgO (=2 = -2) Al2S3. This one you can tell by who is bonded with who. You have 2 Al's and 3 S's. S is in group 16 so it's -2. Since it is behind the Al and the 3 is behing the S, you can assume the oxidation number of Al is +3. Check: Al is +3 and is given a subscript of 2 (Al2). Multiplied together, they total 6. We know S is -2 and it is given a subscript of 3 (S3) so multiplied together they total 6 also - balanced. H is in group 1 and is +1. NO3 is onle of the ones you just have to memorize. It's nitrate and its oxidation number is -1 so it balances with H's +1 so they fit together without subscripts. HNO3. H2SO4. H has a +1 and there are 2 of them so the oxidation number totals +2. SO4 is another one to memorize. It's sulfate with an oxidation number of -2 so they fit together (+2 & - 2) without subscripts. Ca(OH)2. Ca is in group 2 and is +2. OH is another memorization called hydroxide with an oxidation number of -1 When you have a difference take Ca's 2 and give it to (OH) and give (OH)'s 1 to Ca (don't write the 1 when writing the compound) = Ca(OH)2. Fe(NO3)2. We already know that nitrate is -1. Since there are two of them to balance Fe we know Fe must be +2. This is iron (ll) or ferrous (ll). It's an important distinction because iron is one of the metals that has two states with different oxidation numbers: the one above and Fe (lll) or ferric (lll) with a +3 oxidation number. There are several metals with the same thing. If you see the "ous" ending you know it is the +2 and the "ic" is +3. If you want an easy way to name your compounds of write the formula, go to this web page and in the empty box in the middle of the left side of the page, type in your formula or name and press enter and it will give you the name or formula. If you type in sulfate, it will tell you that it is -2 etc. so it's a very useful site. http://en.wikipedia.org/wiki/Portal:Contents
What is the oxidation number of S in FeSO4?
The oxidation number of Fe in FeSO4 is +2, O is-2, so S is +6. See anion SO4 (2-) as a whole.
How do you find the oxidation of PCl3?
Alright, so today, I've been learning about oxidation numbers. I find that they are very much like naming binary compounds etc. However, in this molecular compound, I don't see how you can make it balance out with oxidation numbers. I mean, Chlorine needs to gain 1 electron to have a full outer energy level, and Phosphorus needs 3. Also, there aren't any rules that specify like fluorine what one of the oxidation numbers must be. I am quite stumped. If any background and explanation about this topic, and this question could be provided, thanks!
How do you find oxidation numbers for compounds? ?
You don't. A compound does not have an oxidation number. It's the individual elements within a compound to which oxidation numbers may be assigned. It is possible to assign an oxidation number to a polyatomic ion. The oxidation number of a polyatomic ion will be equal to the charge on that ion. For instance, consider the nitrate ion, NO3^-. The ion has a charge of -1. The nitrate ion has an oxidation number of -1. Since oxygen usually has an oxidation number of -2, and since the sum of the oxidation numbers of a polyatomic ion must add up to the charge of the ion, then we can easily compute the oxidation number of nitrogen in the nitrate ion. It is +5. +5 -6 = -1 ? -2 NO3^- Since H is usually +1, then AsO3 in H3AsO3 has an oxidation number of -3. Knowing that oxygen is -2, we can compute the oxidation number of As in H3AsO3. It is +3. In Na2SO3, sodium is +1, sulfur is +4 and oxygen is -2, and the sulfite ion has an oxidation number of -2.