Question

Oxidation number of p in PF6 And l in NalO3

Answer 1

This is the easy way to find the oxidation number.

Answer 2

Oxidation number of P in $PF_6$ is +6

• To find the oxidation number of P in $PF_6$, the oxidation number of each element must be first identified.
• To find the oxidation number of P, the oxidation number of F must be identified.
• The rule to find oxidation number states that an element belonging to group 17 has -1 as the oxidation number in binary condition.
• As Fluorine belongs to group 17, its oxidation number is -1.
• Thus with help of this, the oxidation number of P is calculated through the following steps:

X + (-1)× 6 = 0

X + -6 = 0

X = +6

• Thus the oxidation number of P in $PF_6$ is found to be +6.

Oxidation number of I in $NaIO_3$ is +5

• As sodium is a group 1 element, its oxidation number is +1.
• The oxidation number of oxygen is -2. As 3 oxygen are there, the oxidation number of oxygen is -6.
• Let the oxidation number of I be "X".

∴1 + X + (-6) = 0

1 + X -6 = 0

X - 5 = 0

X = +5

• Thus the oxidation number of I in $NaIO_3$ is found to be +5.

Learn more about oxidation number

Question 8.2 What are the oxidation numbers of the underlined elements in each of the following and how do you rationalize your results?

(a) KI3 (b) H2S4O6 (c) Fe3O4 (d) CH3CH2OH (e) CH3COOH

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