## NCERT Class XI Chemistry Redox Reactions Solutions

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Question : 1

Total: 30

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

(a)N a H 2

O 4

(b)N a H

O 4

(c)H 4

O 7

(d)K 2

O 4

(e)C a

(f)N a

H 4

(g)H 2

O 7

(h)K A l (

O 4 ) 2 · 12 H 2 O

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

**Solution:**

Let the oxidation no. of underlined element in all the given compounds = x

(a)

Since the sum of oxidation number of various atoms inN a H 2 P O 4 (neutral)is zero.

1(+1) + 2(+1) + (x) + 4(–2) = 0 or x = +5

Thus, the oxidation number of P inN a H 2 P O 4 = +5.

(b) In

S

: 1(+1) + 1(+1) +x + 4 (–2)= 0 or x = + 6

Thus, the oxidation number of S inN a H S O 4 = +6.

(c) In

: 4(+1) + 2(x) + 7(–2) = 0 or x = +5

Thus, the oxidation number of P inH 4 P 2 O 7 = +5.

(d) In

: 2(+1) +1(x) + 4(–2) = 0 or x = +6

Thus, the oxidation number of Mn inK 2 M n O 4 = +6.

(e) In

: 2 + 2x = 0 or x = –1

Thus, the oxidation number of oxygen inC a O 2 = –1.

(f) InN a B H 4 , hydrogen is present as hydride ion. Therefore, its oxidation number is –1. Thus,

In

: 1(+1) + x + 4(–1) = 0 or x = +3

Thus, the oxidation number of B inN a B H 4 = +3

(g) In

: 2(+1)+2(x) + 7(–2) = 0 or x = +6

Thus, the oxidation number of S inH 2 S 2 O 7 = +6.

(h) In

(

) 2 . 12 H 2 O : +1 + 3 + 2x + 8(–2) + 12 × 0 = 0 or x = +6

Thus, the oxidation number of S inK A l ( S O 4 ) 2. 12 H 2 O = +6

(a)

Since the sum of oxidation number of various atoms in

1(+1) + 2(+1) + (x) + 4(–2) = 0 or x = +5

Thus, the oxidation number of P in

(b) In

Thus, the oxidation number of S in

(c) In

Thus, the oxidation number of P in

(d) In

Thus, the oxidation number of Mn in

(e) In

Thus, the oxidation number of oxygen in

(f) In

In

Thus, the oxidation number of B in

(g) In

Thus, the oxidation number of S in

(h) In

Thus, the oxidation number of S in

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