Justify that the reaction

$bold space bold space bold 2 bold Na bold left parenthesis bold s bold right parenthesis bold space bold plus bold space bold H subscript bold 2 bold left parenthesis bold g bold right parenthesis bold space bold space bold space bold rightwards arrow bold space bold space bold space bold 2 bold NaH bold space bold left parenthesis bold s bold right parenthesis bold space bold is bold space bold a bold space bold redox bold space bold change bold.$

Solution

Writing the oxidation number of each atom above its symbol,

stack bold 2 bold Na bold left parenthesis bold s bold right parenthesis with bold 0 on top bold space bold space bold plus stack bold H subscript bold 2 bold left parenthesis bold g bold right parenthesis with bold 0 on top bold space bold rightwards arrow stack bold space bold 2 bold Na with bold plus bold 1 on top bold H with bold minus bold 1 on top

Since, O.N. of hydrogen decreases from 0 in H₂to -1 in NaH, therefore hydrogen (H₂) is reduced. Further, the oxidation number of Na increases from in 0 to Na to +1 in NaH, therefore, sodium is oxidised. Hence the above reaction is a redox reaction.