The oxidation number of sulphur in Na₂S₄O₆ is

Solution

Na₂S₄O₆
Let O.N. of S be x then 2 ×(+1) + 4× (x) +6 ×(–2) = 0
∴x= 2.5.

The oxidation number of nitrogen in (N₂H₅)⁺ is:

+ 2
– 2
+ 3
– 3

Solution

Let O.N. of Nitrogen be x.

2x – 5 = + 1 ⇒x=⁶⁄₂=3

Oxidation numbers of chlorine atoms in CaOCl₂ are

0, 0
–1, –1
–1, +1
None of these

Solution

CaOCl₂ or Ca(OCl)Cl is the mixed salt of Ca(OH)₂ with HCl and HOCl.

Oxidation number of nitrogen in (NH₄)₂SO₄ is

–1/3
–1
+1
–3

Solution

(NH₄)₂SO₄ is split into ions. \(NH_{4}^{+}\). Let O.N. of N be x
then, 1× (x) + 4 ×(+1) = 1∴ x = –3

What is the oxidation state of P in Ba(H₂PO₂)₂?

+ 1
+ 2
+ 3
– 1

Solution

Let oxidation state of Pin Ba(H₂PO₂)2 be x
∴(+ 2)+ 2 [2 (+ 1) + x + 2 (– 2)] =0
⇒2 + 2 [2 +x –4]= 0⇒2 + 2 [x –2]= 0
⇒x – 2 =– 1 ⇒x = + 1

The oxidation number of sulphur in S₈, S₂F₂, H₂S respectively, are

0, +1 and –2
+2, +1 and –2
0, +1 and +2
–2, +1 and –2

Solution

(i) Oxidation state of element in its free state is zero.
(ii) Sum of oxidation states of all atoms in compound is zero.O.N. of S in S₂₈= 0; O.N. of S in S₂F₂= + 1;O.N. of S in H₂S = –2;

The reaction

\(Ag_{(aq)}^{2+}+Ag_{(s)}\rightleftharpoons 2Ag_{(aq)}^{+}\)

is an example of :

reduction
oxidation
disproportionation
none of these

Solution

The given reaction can be written as

As shown above, one of the reacting species under goes reduction and other undergoes oxidation to give a same speices. These type of reactions are called comproportion reactions.

The brown ring complex compound is formulated as[Fe(H₂O)₅(NO)]SO₄. The oxidation state of iron is:

1
2
3
Zero

Solution

Sum of oxidation state of all atoms in neutral compound is zero.Let the oxidation state of iron in the complex ion
\([Fe(H_{2}O)_{5}(NO)]^{2+}SO_{4}^{2-}\); then x+ 5 × 0 + 0= + 2.
∴x= +2

A compound contains atoms of three elements A, B and C.If the oxidation number of A is +2, B is +5, and that of C is –2, the possible formula of the compound is :

A₂(BC₃)₂
A₃(BC₄)₂
A₃(B₄C)₂
ABC₂

Solution

Oxidation number of a compound must be 0. Using the values for A,B and C in the four options, we find that A₃(BC₄)₂ is the answer.Check :
(+2)3 + [(+5)+4(–2)]2 = 6 + (5–8)2 = 0

The metal that does not displace hydrogen from an acid is :

Solution

Hg which lies below the hydrogen in electrochemical series cannot displace hydrogen from acid.

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