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) NaH2
P
O4

(b) NaH
S
O4

(c) H4
P2
O7

(d) K2
Mn
O4

(e) Ca
O2

(f) Na
B
H4

(g) H2
S2
O7

(h) KAl(
S
O4
)
2
·12H2O
Solution:  
Let the oxidation no. of underlined element in all the given compounds = x
(a)
+1
Na
+1
H2
x
P
2
O4

Since the sum of oxidation number of various atoms in NaH2PO4 (neutral)is zero.
1(+1) + 2(+1) + (x) + 4(–2) = 0 or x = +5
Thus, the oxidation number of P in NaH2PO4 = +5.
(b) In
+1
Na
+1
H
S
2
O4
: 1(+1) + 1(+1) +x + 4 (–2)= 0 or x = + 6
Thus, the oxidation number of S in NaHSO4 = +6.
(c) In
+1
H4
x
P2
2
O7
: 4(+1) + 2(x) + 7(–2) = 0 or x = +5
Thus, the oxidation number of P in H4P2O7 = +5.
(d) In
+1
K2
x
Mn
2
O4
: 2(+1) +1(x) + 4(–2) = 0 or x = +6
Thus, the oxidation number of Mn in K2MnO4 = +6.
(e) In
+2
Ca
x
O2
: 2 + 2x = 0 or x = –1
Thus, the oxidation number of oxygen in CaO2 = –1.
(f) In NaBH4, hydrogen is present as hydride ion. Therefore, its oxidation number is –1. Thus,
In
+1
Na
x
B
1
H4
: 1(+1) + x + 4(–1) = 0 or x = +3
Thus, the oxidation number of B in NaBH4 = +3
(g) In
+1
H2
x
S2
2
O7
: 2(+1)+2(x) + 7(–2) = 0 or x = +6
Thus, the oxidation number of S in H2S2O7 = +6.
(h) In
+1
K
+3
Al
(
x
S
2
O4
)
2
.12H2O
: +1 + 3 + 2x + 8(–2) + 12 × 0 = 0 or x = +6
Thus, the oxidation number of S in KAl(SO4)2.12H2O = +6
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