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Academic session 2026-27, Chapter 7 in the current curriculum.
Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter Redox Reactions. This page offers a step-by-step solution to the specific question from Exercise 1, Question 1:
Assign oxidation number to the underlined elements in each of the following species:
(a) NaH2PO4
(b) NaHSO4
(c) H4P2O7
(d) K2MnO4
(e) CaO2
(f) NaBH4
(g) H2S2O7
(h) KAl(SO4)2.12 H2O
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.Assign oxidation number to the underlined elements in each of the following species:
(a) NaH2PO4
(b) NaHSO4
(c) H4P2O7
(d) K2MnO4
(e) CaO2
(f) NaBH4
(g) H2S2O7
(h) KAl(SO4)2.12 H2O
(a) NaH2PO4
Let's assume oxidation number of P is x.
We know that,
Oxidation number of Na = +1
Oxidation number of H = +1
Oxidation number of O = -2
Then we have
1(+1) + 2(+1) + 1 (x) + 4(-2) = 0
⇒ 1 + 2 + x - 8 = 0
⇒ x - 5 = 0
⇒ x = + 5
Hence, oxidation number of P is +5
(b) NaHSO4
Let's assume oxidation number of S is x.
Oxidation number of Na = +1
Oxidation number of H = +1
Oxidation number of O = -2
Then we have:
1(+1) + 1(+1) + 1 (x) + 4(-2) = 0
⇒ 1 + 1 + x - 8 = 0
⇒ x-6 = 0
⇒ x = +6
Hence, oxidation number of S is +6
(c) H4P2O7
Let's assume oxidation number of P is x.
Oxidation number of H = +1
Oxidation number of O = -2
Then we have:
4(+1) + 2(x) + 7 (-2) = 0
⇒ 4 + 2x - 14 = 0
⇒ 2x - 10 = 0
⇒ 2x = +10
⇒ x = +5
Hence, Oxidation number of P is +5
(d) K2MnO4
Let's assume oxidation number of Mn is x.
Oxidation number of K = +1
Oxidation number of O = -2
Then we have:
2(+1) + 1(x) + 4 (-2) = 0
⇒ 2 + x - 8 = 0
⇒ x - 6 = 0
⇒ x = +6
Hence, Oxidation number of Mn is +6
(e) CaO2
Let's assume oxidation number of O is x.
Oxidation number of Ca = +2
Then we have:
1(+2) + 2(x) = 0
⇒ 2 + 2x = 0
⇒ 2x = -2
⇒ x = -1
Hence, Oxidation number of O is -1
(f) NaBH4
Let's assume oxidation number of B is x.
Oxidation number of Na = +1
Oxidation number of H = -1
Then we have:
1(+1) + 1(x) + 4(-1) = 0
⇒ 1 + x -4 = 0
⇒ x - 3 = 0
⇒ x = +3
Hence, Oxidation number of B is +3.
(g) H2S2O7
Let's assume oxidation number of S is x.
Oxidation number of O = -2
Oxidation number of H = +1
Then we have:
2(+1) + 2(x) + 7(-2) = 0
⇒ 2 + 2x - 14 = 0
⇒ 2x - 12 = 0
⇒ x = +6
Hence, Oxidation number of S is +6.
(h) KAl(SO4)2.12 H2O
Let's assume oxidation number of S is x.
Oxidation number of K = +1
Oxidation number of Al = +3
Oxidation number of O = -2
Oxidation number of H = +1
Then we have:
1(+1) + 1 (+3) + 2(x) + 8(-2) + 24(+1) + 12 (-2) = 0
⇒ 1 + 3 + 2x -16 +24 -24 = 0
⇒ 2x - 12 = 0
⇒ 2x = +12
⇒ x = +6
Hence, Oxidation number of S is +6.
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 1: Assign oxidation number to the underlined elements in each of the following species: (a) NaH2PO4 ....
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Gud to disclose all the answers .anyway thanks