Question

What is the oxidation number of the following elements when they form ions?

1. Se

2. Li

3. Al

4. P

Answer 1

Answer 1) Se has -2 oxidation number in ion form (Se^2-)

Se (selenium ) is an element of group 16  

For selenium n = 4

it has outer electronic configration  ns² np⁴

To achieve a stable octet it needs to gain two electrons in its p orbital

then the configuration will be ns²np⁶, thus the last shell has 8 electrons (octet) , so stable configuration.

Se + 2e- ---> Se^2-

Due to gain of 2 electron having negative charge , the Se gets two negative charge .

The oxidation number of the ion is the charge on the ion

Therefore , Se^2- has  -2 oxidation state.

Answer 2) Li has +1 oxidation number in ion form (Li⁺)

Lithium has atomic number = 3

Its electronic configuration : 1s²2s¹

It can lose 1 electron to get a stable duplet configuration.

Li ----> Li⁺ + 1e-

Answer 3) Al has + 3 Oxidation number in its Ion form (Al³⁺)

Atomic number of Aluminium is 13

Its electronic configuration 1s² 2s² 2p⁶ 3s² 3p¹

Stable electronic configuration of octet that is 8 electrons in its outer orbital . aluminium loses 3 electrons thus attaining + 3 Oxidation State

Al ---> Al³⁺ +3e-

Answer4) P has -3 Oxidation number ( P^3-)

Phosphorus belongs to nitrogen family that is element of 15 group so it has outer electronic configuration of ns2 np3

It gains three electron to get filled outer electronic configuration that is octet .

P + 3e- ----> P^3-