The oxidation number of boron in NaBH₄ is

Correct answer: +3.

Explanation: In sodium borohydride, NaBH₄ is best viewed as Na⁺ and BH₄⁻. In the borohydride ion, each hydrogen acts as hydride with oxidation state −1 (H is more electronegative than boron here). Let the oxidation number of boron be x:

\(x + 4(-1) = -1 \;\Rightarrow\; x - 4 = -1 \;\Rightarrow\; x = +3.\)

• −3, −1, +1 — Do not satisfy the charge balance for BH₄⁻ with H = −1.
• +3 — Correct.