In ionic hydrides like NaH, the oxidation state of hydrogen is:
A$-1$, since H gains an electron from the active metal here
B$+1$, since hydrogen always behaves like a proton donor
CZero, since the metal and hydrogen share equally always
D$+2$, since hydrogen behaves like an alkaline-earth here
Answer & Solution
Correct answer: A. $-1$, since H gains an electron from the active metal here
In NaH the active metal donates an electron to H, giving H$^-$ with oxidation state $-1$.
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