Question

In which of the compounds does hydrogen have oxidation have an oxidation state of $-1$
A. $C{{H}_{4}}$
B. $N{{H}_{3}}$
C. HCl
D. $Ca{{H}_{2}}$

Answer 1

Hint: Oxidation number of an atom is the charge that an atom would have if the respective compound was composed of ions. Hydrogen in most compounds has an oxidation number $(+1)$ except for metallic hydrides as the oxidation number of hydrogen in this compound is $(-1)$.

Complete Step by Step Answer:
Generally, oxidation state or oxidation number defines the gain or loss of electrons of any atom in a chemical compound. Before we move to the calculation part first we have to follow the guidelines written below:
(a) The oxidation state of a neutral compound is always zero.
(b) The oxidation number is also zero for free elements.
(c) Elements of the group $1$ have $(+1)$and the group $2$ has $(+2)$oxidation state.
(d) Group $14$ elements have an oxidation number$(-4)$
And group $15$ elements have an oxidation number $(-3)$
And group $16$ elements have an oxidation state $(-2)$
And group $17$ elements have an oxidation state $(-1)$

Now we can calculate the oxidation state of hydrogen from each compound by using the above guidelines.
Let, the oxidation state of hydrogen be $\mathbf{x}$
From (A), the compound $(C{{H}_{4}})$methane is given. Carbon has an oxidation number $(-4)$ as it is the element from the group $14$.
$(-4)+4\times (x)=0$ (since the oxidation number of any neutral compound is zero)
Or,$x=(+1)$
From (B), N in the compound $N{{H}_{3}}$has an oxidation number $(-3)$as N is the group's $15$element.
$(-3)+3\times (x)=0$
Or,$x=(+1)$
From (C), Cl has an oxidation number $(-1)$ as it belongs to the group $17$.
$(x)+(-1)=0$
Or,$x=(+1)$
Finally, from (D) we have a metallic hydride, $Ca{{H}_{2}}$ . Here group $2$ element Ca has an oxidation number $(+2)$ .
$(+2)\times 2(x)=0$
Or, $x=(-1)$
From (A), (B), and (C) it is seen that hydrogen is in the (+1) oxidation state except for $Ca{{H}_{2}}$ , where H is in $(-1)$ oxidation state.
Thus, option (D) is correct.

Note: To calculate the oxidation number, we must strictly follow the given guidelines. Because these simple guidelines will help us to determine the oxidation number of any atom in a compound. It is merely a construction that can be used to explain the behaviour of the redox system.