Home
Class 11
CHEMISTRY
How many electrons should A(2)H(3)libera...

How many electrons should `A_(2)H_(3)`liberate so that in the new compound .A shows oxidation number of `-(1)/(2)`?

Text Solution

Verified by Experts

Let `A_(2)H_(2) ` will liberate x electrons
Therfore ,`2xx(1)/(2)+3xx(+1)=+x `
or `-1+3 =+x or,x=2 `
Promotional Banner

Topper's Solved these Questions

  • REDOX REACTIONS

    CHHAYA PUBLICATION|Exercise ENTRANCE QUESTION|20 Videos
  • REDOX REACTIONS

    CHHAYA PUBLICATION|Exercise MCQ HOTSPOT (SINGLE CORRECT TYPE)|40 Videos
  • REDOX REACTIONS

    CHHAYA PUBLICATION|Exercise SOLVED NCERT EXERCISE|64 Videos
  • CLASSIFICATION OF ELEMENTS & PERIODICITY IN PROPERTIES

    CHHAYA PUBLICATION|Exercise PRACTICE SET|12 Videos
  • s-BLOCK ELEMENTS

    CHHAYA PUBLICATION|Exercise PRACTICE SET|16 Videos

Similar Questions

Explore conceptually related problems

1 mol N_(2)H_(4) loses 10 mol of electrons with th electrons with the formation of I mol of a new compound y. of the new compound contains same number of N- atoms then what will be the oxidation number of nitrogen in the new compound ? (Assume that the oxidation number of H -atom does not change)

Determine the oxidation number of S in H_2SO_4 .

How many electrons with l =2 are there in on atom having atomic number 23?

2 mole of N_2H_4 loses 16 mole of election is being converted to a new compound X . Assuming that all of the N-appears in the new compound. What is the oxidation state of 'N' in X?

The oxidation number of P in Ba(H_(2)PO_(2))_(2) is-

The oxidation number of S in (CH_(3))_(2)SO 0's ____.

Balance the reaction by oxidation number method CuO+NH_(3)toCu+N_(2)+H_(2)O

Find the oxidation no. of S in H_2SO_5

In an experiment, 50 ml of 0.1 M solution of a salt reacted with 25 ml of 0.1 M solution of sodium sulphite. The half equation for the oxidation of sulphite ion : SO_(3(aq))H_2OrarrSO_(4(aq))^(2-)+2H_((aq))^+ If the oxidation number of the metal in the salt was 3, what would be the new oxidation number of the metal ?

Determine the equivalent masses of the following underline compounds by both oxidation number and electrons methods: underline (HNO_(3))toNO_(2)+H_(2)O