Whenever a reaction between an oxidisina adent and a reducina aqent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. J ustify this statement giving three illustrations. Justify the above statement with three examples.

Home » Whenever a reaction between an oxidisina adent and a reducina aqent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. J ustify this statement giving three illustrations. Justify the above statement with three examples.

Whenever a reaction between an oxidisina adent and a reducina aqent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. J ustify this statement giving three illustrations. Justify the above statement with three examples.

Solution:

When there is a response between lessening specialist and oxidizing specialist, a compound is framed which has lower oxidation number if the diminishing specialist is in abundance and a compound is shaped which has higher oxidation number if the oxidizing specialist is in overabundance.

(I) $P_{4}$ and $F_{2}$ are decreasing and oxidizing specialist individually.

In an overabundance measure of $P_{4}$ is responded with $F_{2}$ , then, at that point, $P F_{3}$ would be created, where the oxidation no. of $P$ is $+3 .$

$P_{4 \text { (overabundance })}+F_{2} \rightarrow P F_{3}$

Assuming $P_{4}$ is responded with overabundance of $F_{2}$ , $P F_{5}$ would be delivered, where the oxidation no. of $P$ is $+5$ .

$P_{4}+F_{2(\text { abundance })} \rightarrow P F_{5}$

(ii) $\mathrm{K}$ and $\mathrm{O}_{2}$ goes about as a lessening specialist and oxidizing specialist individually.

On the off chance that an overabundance of $\mathrm{K}$ responds with $O_{2}$ , it produces $\mathrm{K}_{2} \mathrm{O}$ . Here, the oxidation number of 0 is $-2$ .

$4 K_{(\text {excess })}+\mathrm{O}_{2} \rightarrow 2 \mathrm{~K}_{2} \mathrm{O}^{-2}$

On the off chance that $\mathrm{K}$ responds with an overabundance of $O_{2}$ , it produces $\mathrm{K}_{2} \mathrm{O}_{2}$ , where the oxidation number of $\mathrm{O}$ is $-1$ .

$2 \mathrm{~K}+\mathrm{O}_{2}($ overabundance $) \rightarrow \mathrm{K}_{2} \mathrm{O}_{2}^{-1}$

(iii) $\mathrm{C}$ and $O_{2}$ goes about as a decreasing specialist and oxidizing specialist individually.

On the off chance that an overabundance measure of $C$ is responded with inadequate measure of $O_{2}$ , then, at that point, it produces $C 0$ , where the oxidation

number of $\mathrm{C}$ is $+2$ .

$C_{(\text {excess })}+\mathrm{O}_{2} \rightarrow \mathrm{CO}$

In the event that $\mathrm{C}$ is singed in abundance measure of $O_{2}$ , $\mathrm{CO}_{2}$ is created, where the oxidation number of $\mathrm{C}$ is $+4$ .

$C+O_{2}$ (overabundance $) \rightarrow \mathrm{CO}_{2}$

i

More Material

Write chromyl chloride test with equation

What do you understand by lanthanide contraction

What are lanthanide elements?

Explain oxidization properties of potassium permanganate in acidic medium.

Show that the lines

Compute the shortest distance between the lines

Find the shortest distance between the given lines.

Find the shortest distance between the given lines.

Find the shortest distance between the given lines.

Find the shortest distance between the given lines.

Find the shortest distance between the given lines.

Find the shortest distance between the given lines.

Sign up now and study all subjects for free

Class

Syllabus

Question Papers

Social Media

Apps