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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