Qstnid -1- Assign Oxidation Numbers To The Underlined Elements In Each Of The Following Species: (A) Nah<sub>2</sub><u>p</u>o<sub>4</sub> (B) Nah<u>s</u>o<sub>4</sub> (C) H<sub>4</sub><u>p</u><sub>2</sub>o<sub>7</sub> (D) K<sub>2</sub><u>mn</u>o<sub>4</sub> (E) Ca<u>o</u><sub>2</sub> (F) Na<u>b</u>h<sub>4</sub> (G) H<sub>2</sub><u>s</u><sub>2</sub>o<sub>7</sub> (H) Kal(<u>s</u>o<sub>4</sub>)<sub>2</sub>.12 H<sub>2</sub>o - 01: Redox Reactions - Neet-xi-chemistry

Home Learn Fast select Course

Show Unit/Chapter Change Unit Change Subject Last Menu

NEET-XI-Chemistry

01: Redox Reactions

Qstn# 1 Prvs-Qstn Next-Qstn

#1
Assign oxidation numbers 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
Ans : null (a)

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

Then, we have

Hence, the oxidation number of P is +5.
(b)

Then, we have

Hence, the oxidation number of S is + 6.
(c)

Then, we have

Hence, the oxidation number of P is + 5.
(d)

Then, we have

Hence, the oxidation number of Mn is + 6.
(e)

Then, we have

Hence, the oxidation number of O is - 1.
(f)

Then, we have

Hence, the oxidation number of B is + 3.
(g)

Then, we have

Hence, the oxidation number of S is + 6.
(h)

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.

#1-a
NaH₂PO₄
Ans :

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

Then, we have

Hence, the oxidation number of P is +5.

#1-b
NaHSO₄
Ans :

Then, we have

Hence, the oxidation number of S is + 6.

#1-c
H₄P₂O₇
Ans :

Then, we have

Hence, the oxidation number of P is + 5.

#1-d
K₂MnO₄
Ans :

Then, we have

Hence, the oxidation number of Mn is + 6.

#1-e
CaO₂
Ans :

Then, we have

Hence, the oxidation number of O is - 1.

#1-f
NaBH₄
Ans :

Then, we have

Hence, the oxidation number of B is + 3.

#1-g
H₂S₂O₇
Ans :

Then, we have

Hence, the oxidation number of S is + 6.

#1-h
KAl(SO₄)₂.12 H₂O
Ans :

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.