Boron is a non-metal, i.e. it accepts electrons and acids donate electrons, so boron should react with acid but it doesn't. Why?


To illustrate Oscar's answer further, it should be noted that boron, despite existing in more than ten allotropic modifications, is in general quite chemically inert (especially in crystalline form).

For example, boron doesn't react with hydrogen directly and all boranes are synthesized by other means. Although boron may be oxidized by fluorine at room temperature, other halogens (even chlorine) require elevated temperatures above $\pu{400 °C}$:

$$\ce{2 B(s) + 3 E2(g) -> 2 BE3}\qquad(\pu{30 °C}: \ce{E} = \ce{F}; \quad\text{above}~\pu{400 °C}: \ce{E} = \ce{Cl}, \ce{Br}, \ce{I})$$

There is no preference in reactivity towards acids or bases either. Boron indeed doesn't appreciably react either with non-oxidizing acids or with alkali solutions under ambient conditions.

However, concentrated oxidizing acids such as nitric and sulfuric acids as well as aqua regia react all right — especially at elevated temperatures — yielding boric acid:

$$\ce{B(s) + 3 HNO3(aq) ->[Δ] B(OH)3(aq) + 3 NO2(g)}$$

Amorphous boron reacts with concentrated alkali solutions:

$$\ce{2 B (am) + 2 NaOH (conc) + 6 H2O(l) -> 2 Na[B(OH)4](aq) + 3 H2(g)}$$

When fused in air with alkali, $\ce{Na2O2}$ or $\ce{KNO3}$ + $\ce{Na2CO3}$ mix, borates are formed:

$$\ce{4 B(s) + 4 NaOH(s) + 3 O2(g) ->[\pu{350-400 °C}] 4 NaBO2(s) + 2 H2O(g)}$$

All chemical reactions have been adapted from [1, p. 74].


  1. R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)

Who said boron is a base? The fact that it does not weaken or neutralize most acids ought to tip you off that it isn't.

Boron is slowly oxidized when powdered and exposed to nitric acid (see here), and we may call it "basic" by virtue of the products being much weaker acids than the nitric acid. But usually and with better precision the boron (or any other element that reacts similarly) is instead interpreted as a reducing agent.

  • 1
    $\begingroup$ Wellllll....... If you are using the Usanovich definition of acids and bases en.wikipedia.org/wiki/… the distinction disappears. But that is far from mainstream $\endgroup$
    – Ian Bush
    Jul 27 '19 at 7:43

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