No It is not correct.
Correct oxidation number of hydrogen in NaH,H2 and H2S are:
- Let x be the Oxidation state of ’H’ by oxidation state rule.
x+1 =0 (Sodium has +1 Oxidation state )
x=−1
Therefore, -1 is oxidation number of Hydrogen in NaH.
- The oxidation number of an element's atom in its standard state is zero. As a result, the oxidation number for H2 is zero.
- The hydrogen oxidation number in H2 is +1. As a result, that hydrogen atom can be reduced to lower oxidation states of hydrogen, such as 0 and -1. However, because +1 is the highest oxidation state of hydrogen, the hydrogen atom cannot be reduced any further.
<h3>What is oxidation number?</h3>
In simple terms, the oxidation number is the number assigned to elements in a chemical combination. The oxidation number is the number of electrons that atoms in a molecule can share, lose, or gain when forming chemical bonds with atoms of a different element.
Learn more about oxidation number in:
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[H₃O⁺] = 8.86 x 10³ M
<h3>Further explanation</h3>
The dissociation of water :
H₂O(l)⇒H⁺+OH⁻
Hydrogen ions H⁺ bond with H₂O to form H₃O⁺ ions(hydronium ion)
When acid(Hbr acid) is in the presence of water(dissociates), the H⁺ ions bond with water molecules to form hydronium, the reaction :
HBr(aq)+H₂O→H₃O+(aq)+Br−(aq
So Ionization of HBr :
[HBr]= 8.86 x 10³,
From equation ratio [H₃O⁺] : HBr = 1 : 1, so
there is a <em><u>covalent </u></em> bonding in uranium
hope this helps you xx