Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
44.3k views
in Chemistry by (30.0k points)
edited by

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

(a) NaH2PO4 (b) NaHSO4 (c) H4P2O7 (d) K2MnO

(e) CaO2 (f) NaBH4 (g) H2S2O7 (h) KAl(SO4)2.12 H2

1 Answer

+1 vote
by (130k points)
selected by
 
Best answer

(a) Na2PO4

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 

Na+1H+12PxO-24

Then, we have 

Hence, the oxidation number of P is +5. 

(b) Na+1H+1SxO-24

Then, we have 

Hence, the oxidation number of S is + 6. 

(c) H+14Px2O-27

Then, we have 

Hence, the oxidation number of P is + 5. 

(d) K+12MnxO-24

Then, we have 

Hence, the oxidation number of Mn is + 6. 

(e) Ca+2Ox2

Then, we have 

Hence, the oxidation number of O is – 1. 

(f) Na+1BxH-14

Then, we have 

Hence, the oxidation number of B is + 3

(g) H+2Sx2O-72

Then, we have 

Hence, the oxidation number of S is + 6. 

(h) K+1Al3+(SxO2-4).12H+12O-2

Then, we have 

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 

Hence, the oxidation number of S is + 6.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...