Question

How many electrons should x2 h4 liberate so that in the new compound x shows oxidation number of minus half in which electronegativity of x is greater than hydrogen?

Answer 1

"Here, H= +1 

X= ? 

Arrange it in an equation equal to zero since it's a neutral compound: 

[2 x ?] + [4 x +1] = 0 

2? + (+4) = 0 

2? = -4 

? = -2 

So in X2H4 the oxidation number for X = -2 and H = +1. 

If we liberate three electrons then the compound X2H4 is no longer neutral. 

Instead of equalling zero, it will equal +3 

So we set it up like we did above: 

H= +1 

X= -1/2 or ? 

[2 x ?] + [4 x +1] = +3 

Solve for ? 

2? + (+4) = +3 

2? = -1 

? = -1/2 

After liberating 3 electrons the what was once a neutral compound X2H4 is now a compound with a total charge of +3.

Hence, the answer is 3.

"

Answer 2

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