assign oxidation number to the following underlined elements

What Are the Rules for Assigning Oxidation Numbers?

Redox Reactions and Electrochemistry

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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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Assign oxidation numbers to the underlined elements in each of the following species: N a H 2 P O 4 N a H S – – O 4 H 4 P – – 2 O 7 K 2 M n – –– – O 4 C a O – – 2 N a B – – H 4 H 2 S – – 2 O 7 K A l S – – O 4 ) 2 .12 H 2 O

N a h 2 p o 4 let the oxidation number of p be x. we know that, oxidation number of na = +1 oxidation number of h = +1 oxidation number of o = –2 + 1 n a + 1 h 2 x − 2 p o 4 then, we have 1 ( + 1 ) + 2 ( + 1 ) + 1 ( x ) + 4 ( − 2 ) = 0 ⇒ 1 + 2 + x − 8 = 0 ⇒ x = + 5 hence, the oxidation number of p is +5. n a h s – – o 4 then, we have 1 ( + 1 ) + 1 ( + 1 ) + 1 ( x ) + 4 ( − 2 ) = 0 ⇒ 1 + 1 + x − 8 = 0 ⇒ x = + 6 hence, the oxidation number of s is + 6. h 4 p – – 2 o 7 + 1 h 4 x − 2 p 2 o 7 then, we have 4 ( + 1 ) + 2 ( x ) + 7 ( − 2 ) = 0 ⇒ 4 + 2 x − 14 = 0 ⇒ 2 x = + 10 ⇒ x = + 5 hence, the oxidation number of p is + 5. k 2 m n – –– – o 4 + 1 k 2 x − 2 m n o 4 then, we have 2 ( + 1 ) + x + 4 ( − 2 ) = 0 ⇒ 2 + x − 8 = 0 ⇒ x = + 6 hence, the oxidation number of mn is + 6. c a o – – 2 then, we have + 2 c a x o 2 ( + 2 ) + 2 ( x ) = 0 ⇒ 2 + 2 x = 0 ⇒ x = − 1 hence, the oxidation number of o is -1. n a b – – h 4 + 1 n a x − 1 b h 4 then, we have 1 ( + 1 ) + 1 ( x ) + 4 ( − 1 ) = 0 ⇒ 1 + x − 4 = 0 ⇒ x = + 3 hence, the oxidation number of b is + 3. h 2 s – – 2 o 7 + 1 h 2 x − 2 s 2 o 7 then, we have 2 ( + 1 ) + 2 ( x ) + 7 ( − 2 ) = 0 ⇒ 2 + 2 x − 14 = 0 ⇒ 2 x = 12 ⇒ x = + 6 hence, the oxidation number of s is + 6. k a 1 ( s – – o 4 ) 2 .12 h 2 o + 1 3 + k a 1 ( x 2 − s o 4 ) 2 .12 + 1 − 2 h 2 o then, we have 1 ( + 1 ) + 1 ( + 3 ) + 2 ( x ) + 8 ( − 2 ) + 24 ( + 1 ) + 12 ( − 2 ) = 0 ⇒ 1 + 3 + 2 x − 16 + 24 − 24 = 0 ⇒ 2 x = 12 ⇒ x = + 6 or, we can ignore the water molecule as it is a neutral molecule. then, the sum of the oxidation numbers of all atoms of the water molecule may be taken as zero. therefore, after ignoring the water molecule, we have 1 ( + 1 ) + 1 ( + 3 ) + 2 ( x ) + 8 ( − 2 ) = 0 ⇒ 1 + 3 + 2 x − 16 = 0 ⇒ 2 x = 12 ⇒ x = + 6 hence, the oxidation number of s is + 6..

Assign oxidation numbers to the underlined elements in each of the following species:

(a) NaH 2 P O 4 (b) NaH S O 4 (c) H 4 P 2 O 7 (d) K 2 Mn O 4

(e) Ca O 2 (f) Na B H 4 (g) H 2 S 2 O 7 (h) KAl( S O 4 ) 2 .12 H 2 O