
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
161.4k+ views
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.
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.
Recently Updated Pages
Two pi and half sigma bonds are present in A N2 + B class 11 chemistry JEE_Main

Which of the following is most stable A Sn2+ B Ge2+ class 11 chemistry JEE_Main

The enolic form of acetone contains a 10sigma bonds class 11 chemistry JEE_Main

The specific heat of metal is 067 Jg Its equivalent class 11 chemistry JEE_Main

The increasing order of a specific charge to mass ratio class 11 chemistry JEE_Main

Which one of the following is used for making shoe class 11 chemistry JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2026 Syllabus PDF - Download Paper 1 and 2 Syllabus by NTA

JEE Main Eligibility Criteria 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

Degree of Dissociation and Its Formula With Solved Example for JEE
