What Are the Rules for Assigning Oxidation Numbers?

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Electrochemical reactions involve the transfer of electrons . Mass and charge are conserved when balancing these reactions, but you need to know which atoms are oxidized and which atoms are reduced during the reaction. Oxidation numbers are used to keep track of how many electrons are lost or gained by each atom. These oxidation numbers are assigned using the following rules.

Rules for Assigning Oxidation Numbers

  • The convention is that the cation is written first in a formula, followed by the anion . For example, in NaH, the H is H-; in HCl, the H is H+.
  • The oxidation number of a free element is always 0. The atoms in He and N 2 , for example, have oxidation numbers of 0.
  • The oxidation number of a monatomic ion equals the charge of the ion. For example, the oxidation number of Na + is +1; the oxidation number of N 3- is -3.
  • The usual oxidation number of hydrogen is +1. The oxidation number of hydrogen is -1 in compounds containing elements that are less ​ electronegative than hydrogen, as in CaH 2 .
  • The oxidation number of oxygen in compounds is usually -2. Exceptions include OF 2 because F is more electronegative than O, and BaO 2 , due to the structure of the peroxide ion, which is [O-O] 2- .
  • The oxidation number of a Group IA element in a compound is +1.
  • The oxidation number of a Group IIA element in a compound is +2.
  • The oxidation number of a Group VIIA element in a compound is -1, except when that element is combined with one having a higher electronegativity. The oxidation number of Cl is -1 in HCl, but the oxidation number of Cl is +1 in HOCl.
  • The sum of the oxidation numbers of all of the atoms in a neutral compound is 0.
  • The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion. For example, the sum of the oxidation numbers for SO 4 2- is -2.
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What is the oxidation number of H3PO4?

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In H3PO4 the oxidation number for

The oxidation numbers for H, O, and P in H3PO4 are +1, -2, and +5, respectively. This can be determined by assigning variable oxidation states to the atoms in the molecule based on known rules and maintaining overall charge neutrality.

The oxidation number of the entire ion is -1. The oxidation number for phosphorus atom is +3. Hydrogen atoms and all oxygen atoms have +1 and -2 oxidation numbers respectively.

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What is the oxidation number of H in H3PO4?

The oxidation number of H in H3PO4 is +1. In a compound, the oxidation numbers of all the elements add up to the overall charge of the compound, which is 0 in this case. Since the oxidation number of oxygen is typically -2 and there are four oxygen atoms in H3PO4, the oxidation number of hydrogen must be +1 to balance the charge.

What is the oxidation number of acetate?

The oxidation number of acetate (CH3COO-) is -1. The carbon atom has an oxidation number of +3, each hydrogen atom has an oxidation number of +1, and the oxygen atoms have an oxidation number of -2.

What is the oxidation number of H2CO2?

The oxidation number of each hydrogen in H2CO2 is +1, while the oxidation number of each carbon in CO2 is +4. This is because hydrogen usually has an oxidation number of +1, and oxygen usually has an oxidation number of -2.

What is the oxidation number of nitrosyl?

The oxidation number of nitrosyl (NO) is +1. Nitrogen typically has an oxidation number of -3, and oxygen typically has an oxidation number of -2. In NO, nitrogen has a -3 oxidation number and oxygen has a -2 oxidation number, leading to an overall oxidation number of +1 for the nitrosyl ion.

What is the oxidation number for Nb in NbO2?

The oxidation number for Nb in NbO2 is +4. Oxygen has an oxidation number of -2, so the overall charge of the compound must be balanced by the oxidation number of niobium.

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How to draw the structure and find the oxidation number of Phosphorous in H2P2O7 [closed]

Find the Oxidation number of Phosphorus in $\ce{H2P2O7^2-}$ .

So I thought of drawing the lewis structure to find the oxidation numbers. Because I wasn't sure if Oxygen was bonded to a Hydrogen or not. Then I realised that the two Hydrogens could be ionised, so I'd be left with $\ce{P2O7^4-}$ .

But then when I tried drawing this Lewis structure I got really confused, As I tried to draw it with the two Phosphorus as central atoms and I couldn't make it. (I know I can just use general rules to find oxidation numbers but I feel like lewis structures often give more accurate answers). So could you please tell me how to draw the structure and more importantly guide me through the process . Any help is appreciated, thanks!

  • inorganic-chemistry
  • molecular-structure
  • oxidation-state
  • lewis-structure

Aniruddha Deb's user avatar

  • 1 $\begingroup$ Please use $\LaTeX$ and also see How to ask a good question $\endgroup$ –  Aniruddha Deb Commented Jun 18, 2020 at 7:38
  • $\begingroup$ You have O and P, both in their most common valence states. You start connecting them until there is nothing more to connect. What could possibly go wrong? $\endgroup$ –  Ivan Neretin Commented Jun 18, 2020 at 9:18
  • 2 $\begingroup$ Take 2 dihydrogenphosphate anions and condense 2 OH groups ( 1 from each ion ) into -O- bridge, eliminating water. $\endgroup$ –  Poutnik Commented Jun 18, 2020 at 13:36

If you didn't understand what Poutnik has sain in his comment, this is how it should be done:

You have realized your parent compound is $\ce{H4P2O7}$ (adding two $\ce{H+}$ ). Correct? If so, concentrate on following hypothetical reaction:

$$\ce{2H3PO4 -> H4P2O7 + H2O} \tag1$$

I believe you know what $\ce{H3PO4}$ is looks like. Now, draw two $\ce{H3PO4}$ side by side. Choose two closest $\ce{OH}$ groups (that would be $\ce{-P-\color{red}{OH} \space \space \color{tequish}{HO}-P -}$ ). Then, remove one $\ce{\color{red}{-OH}}$ and the $\ce{\color{tequish}{-H}}$ from the second $\ce{\color{tequish}{-OH}}$ you chose to eliminate a $\ce{H2O}$ molecule. Finally combine two remaining molecules together such that final structure forms $\ce{-P\color{red}{-}\color{tequish}{O}-P -}$ bond. This is your $\ce{H4P2O7}$ . Eliminate any two hydrogens to get $\ce{H2P2O7^2-}$ structure.

Note: Since above hypothetical equation $(1)$ does not require oxidation or reduction to form $\ce{H4P2O7}$ , it is safe to assume that oxidation state of $\ce{P}$ in both $\ce{H3PO4}$ and $\ce{H4P2O7}$ is same.

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assign oxidation numbers to h3po4

Find the oxidation number of elements in each case : P in H 3 P O 4 , H 3 P O 3 , H P O 3 , P 2 O 5 and P H 3 . + 4 , + 2 , + 3 , + 5 , − 2 + 5 , + 3 , + 5 , + 5 , − 3 + 5 , + 3 , + 4 , + 5 , − 3 None of the above

The oxidation numbers of p in h 3 p o 4 , h 3 p o 3 , h p o 3 , p 2 o 5 are + 5 , + 3 , + 5 , + 5 , 3 respectively. let x be the oxidation number of p in h 3 p o 4 . 3 + x + 4 ( − 2 ) = 0 x = + 5 let x be the oxidation number of p in h 3 p o 3 3 + x + 3 ( − 2 ) x = + 3 let x be the oxidation number of p in h p o 3 1 + x + 3 ( − 2 ) x = + 5 let x be the oxidation number of p in p 2 o 5 2 x + 5 ( − 2 ) = 0 x = + 5 let x be the oxidation number of p in p h 3 x + 3 = 0 x = − 3.

assign oxidation numbers to h3po4

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  2. Oxidation number of P in H3PO2 H3PO3 & H3PO4 class 11th ncert

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  3. 4.Oxidation number of phosphorus in H3po4

    assign oxidation numbers to h3po4

  4. How to calculate the oxidation state of Phosphorus in H3PO4

    assign oxidation numbers to h3po4

  5. what is oxidation number of P in H3PO4

    assign oxidation numbers to h3po4

  6. Assigning Oxidation Numbers Practice

    assign oxidation numbers to h3po4

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  1. H3PO4 Oxidation Number

    To calculate the oxidation numbers for H3PO4, count the number of atoms, draw the lewis structure by adding bonds, assign electrons from each bond, and count the number of electrons assigned to each atom. Count Atoms Use the chemical formula, H 3 PO 4, to count the number of atoms of each element.

  2. How to find the Oxidation Number for P in H3PO4 (Phosphoric acid)

    To find the correct oxidation state of P in H3PO4 (Phosphoric acid), and each element in the molecule, we use a few rules and some simple math.First, since t...

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  6. Solved Assign oxidation numbers to each of the elements in

    Step 1. Three hydrogen ( H) atoms, one phosphorus ( P) atom, and four oxygen ( O) atoms make up the compound H A 3 PO A 4. Hydr... View the full answer Step 2. Unlock.

  7. Rules for Assigning Oxidation Numbers

    The convention is that the cation is written first in a formula, followed by the anion. For example, in NaH, the H is H-; in HCl, the H is H+. The oxidation number of a free element is always 0. The atoms in He and N 2, for example, have oxidation numbers of 0. The oxidation number of a monatomic ion equals the charge of the ion.

  8. PDF 9.2 Oxidation Numbers

    The sum of the oxidation numbers for the atoms in an uncharged formula is equal to zero. The sum of the oxidation numbers for the atoms in a polyatomic ion is equal to the overall charge on the ion. example See Example 9.1. ObjeCtive 3 Sample Study Sheet 9.1 Assignment of Oxidation Numbers Table 9.2 Guidelines for Assigning Oxidation Numbers ...

  9. Balancing redox reactions by oxidation number change method

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  10. H3PO4{-} Oxidation Number

    H3PO4{-} Oxidation Number. Oxidation State and Numbers of H 3 PO 4-This result is an approximation, as it is not always possible to determine the exact oxidation numbers based on a molecular formula alone. +1 +4-2: H 3: P: O 4-I: IV-II: Net Ionic Charge: -1. Element Oxidation Number (Avg) Count

  11. Solved Use the rules (in order) to assign oxidation numbers

    Here's the best way to solve it. Ans : In H3PO4 , the overall charge of the compound = 0 t …. Use the rules (in order) to assign oxidation numbers to each of the elements in the compounds below. phosphoric acid H3PO4 H - P - 0 - sulfur dioxide SO2 nitrous acid HNO2 H - N - 0 -.

  12. What is the oxidation number of H3PO4?

    The oxidation numbers for H, O, and P in H3PO4 are +1, -2, and +5, respectively. This can be determined by assigning variable oxidation states to the atoms in the molecule based on known rules and ...

  13. How to draw the structure and find the oxidation number of Phosphorous

    Find the Oxidation number of Phosphorus in $\ce{H2P2O7^2-}$.. So I thought of drawing the lewis structure to find the oxidation numbers. Because I wasn't sure if Oxygen was bonded to a Hydrogen or not. Then I realised that the two Hydrogens could be ionised, so I'd be left with $\ce{P2O7^4-}$.. But then when I tried drawing this Lewis structure I got really confused, As I tried to draw it with ...

  14. Find the oxidation number of elements in each case

    Find the oxidation number of elements in each case : P in H 3 P O 4, H 3 P O 3, H P O 3, P 2 O 5 and P H 3. View Solution. Q2. The ratio of the oxidation state of P in H 3 P O 3 and H 3 P O 4 is. View Solution. Q3. The maximum oxidation state shown by the elements V, C r, M n and S c is respectively: View Solution. Q4.

  15. Assign oxidation number to H3PO4

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    ASSIGN OXIDATION NUMBERS TO EACH ATOM IN THE COMPOUND: UF6. ASSIGN OXIDATION NUMBERS TO EACH ATOM IN THE COMPOUND: H2SO4. ASSIGN OXIDATION NUMBERS TO EACH ATOM IN THE COMPOUND: ClO3^-1. ASSIGN OXIDATION NUMBERS TO EACH ATOM IN THE COMPOUND: H2O2. WHAT IS THE OXIDATION NUMBER OF PHOSPHORUS IN H3PO4?

  17. Solved Use the rules (in order) to assign oxidation numbers

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    The oxidation number of silver (Ag) is typically +1, as it is a Group 1 element, and bromine (Br) is typically assigned a -1 oxidation number. For phosphoric acid (H3PO4), we can assign the oxidation numbers as follows: H: +1, P: +5, O: -2. Again, hydrogen (H) is typically assigned a +1 oxidation number, oxygen (O) is typically assigned a -2 ...

  19. Solved Assign oxidation numbers to every element in each of

    See Answer. Question: Assign oxidation numbers to every element in each of the following compounds. a. H3PO4 b. CH2Cl2 c. SrCl2 d.SO2 e. NH3. Assign oxidation numbers to every element in each of the following compounds. a. H 3 PO 4 b.