Question

pls solve and complete q 14 and also q 15​

Answer 1

Here you go:

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Main concept of oxidation:

1. Removal of electron

2. Addition of oxygen

3. Addition of electronegative element

4. Removal of hydrogen

Main concept of reduction:

1. Addition of electron

2. Removal of oxygen

3. Removal of electronegative element

4. Addition of hydrogen

a) Fe is getting reduced, since one electron is added

b) S isgetting oxidised since electrons are lost

c) O is getting reduced since it is gaining 2 electrons

d) Zn is getting oxidised, since two electrons are lost

e) Cu is getting reduced since it gains electrons

a) electronic configuration = 1s² 2s² 2p⁶ 3s² = metal

b) electronic configuration = 1s² 2s² = metal

c) electronic configuration = 1s² 2s² 2p³ = non metal

d) 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² = metal

Answer 2

Answer:

14.

(a)  Here, Fe+3 is converted into Fe+2. So, the decrease in the oxidation state                                .      indicates that Fe+3 is reduced.

(b) S-2 ---------->  S .

     It is oxidized(S-2).

(c)  O  ------------> O-2.  O is reduced.

(d)  Zn ----------->  Zn+2. Zn is oxidized.

(e)  Cu+2 -----------> Cu . Cu+2 is reduced.

15.

(a) $12^{P^{24}}$ . Atomic number = 12 => element is Magnesium(Mg).

    Mg is a metal.

(b) $2^{Q^{4}}$ . Atomic number = 2 => element is helium (He).

     He is an inert gas.

(c) $7^{R^{14}}$ . Atomic number = 7 => element is nitrogen(N).

     N exists as N2 and is gas.

(d) $20^{S^{40}}$ . Atomic number = 20 => element is calcium(Ca).

     Ca is a metal.