Question

The oxidation number of an element in a compound is evaluated on the basis of certain rules. Which of the following rules is not correct in this respect?

• A. The algebraic sum of all the oxidation numbers in a compound is zero.
• B. The oxidation number of hydrogen is always $+1$
• C. In all its compounds, the oxidation number of fluorine is $-1.$
• D. An element in the free or the uncombined state bears oxidation number zero.

Answer 1

The correct option is B

The oxidation number of hydrogen is always $+1$

$A)$ Oxidation number of hydrogen depends on the adjacent atom attached to Hydrogen.

For example:

In $HCl,$ oxidation number of $H$ is $+1,$ because it is less electronegative than $Cl$ and thus bears positive charge.

In $NaH,$ the oxidation number of $H$ is $-1,$ because it is more electronegative than $Na$ and thus bears negative charge.

Hence, the oxidation number of $H$ can be $+1$ or $-1$ depending on its adjacent atom in given compound, thus this statement is incorrect.

$B)$ Every compound is electrically neutral and thus the algebraic sum of all the oxidation numbers of each atom in a compound is zero.

For example:

In $FeO,$ oxidation number of $Fe=+2,$ oxidation number of $O=-2$

$Algebraicsum=+2+(-2)=0$

In $H_{2}O,$ oxidation number of $H=+1,$

oxidation number of $O=-2$

$Algebraicsum=(1\times 2)+(-2)$

$=2-2=0$

Hence, the given statement is correct.

$C)$ This is also a correct statement.

For example, oxidation number of $H_2=0$

Oxidation number of $Na=0,$ oxidation number of $B{r}_2=0$

(where $H_2,Na\&B{r}_2$ are free elemental state)

$D)$ Due to highest electronegativity and small size, fluorine only shows -1 oxidation number in all compounds.

Thus statement $\left(A\right)$ is incorrect while statement $\left(B\right),\left(C\right)$ and $\left(D\right)$ are correct.

Hence, the correct answer is option $\left(A\right).$