The standard heat of combustion of solid boron is equal to:
(A) $\Delta H_{f}^{o}\left( {{B}_{2}}{{O}_{3}} \right)$
(B) ${1}/{2}\;\Delta H_{f}^{o}\left( {{B}_{2}}{{O}_{3}} \right)$
(C) $2\Delta H_{f}^{o}\left( {{B}_{2}}{{O}_{3}} \right)$
(D) $-{1}/{2}\;\Delta H_{f}^{o}\left( {{B}_{2}}{{O}_{3}} \right)$

Hint: When the solid boron undergoes combustion boron trioxide is being formed as the product. The standard heat of combustion of solid boron will be equal to the one half of the value of standard heat of formation of boron trioxide.

Complete step by step solution:
- Let’s start with the concept of combustion. It’s a high temperature exothermic redox chemical reaction between an oxidant, (commonly atmospheric oxygen) and a fuel which produces oxidized, gaseous products.
- The Standard heat or enthalpy of combustion can be defined as the enthalpy change when one mole of a compound is fully burnt in presence of oxygen with all the products and reactants in their standard state under standard conditions such as 1 bar pressure and 298K.
- When the solid boron undergoes combustion boron trioxide is being formed as the product and the balanced chemical equation for the formation of ${{B}_{2}}{{O}_{3}}$ can be written as follows
\[2B+\dfrac{3}{2}{{O}_{2}}\to {{B}_{2}}{{O}_{3}};{{\Delta }_{f}}{{H}^{o}}of\text{ }{{B}_{2}}{{O}_{3}}\]
The balanced chemical equation for the combustion of one mole of boron can be shown as follows
\[B+\dfrac{3}{4}{{O}_{2}}\to \dfrac{1}{2}{{B}_{2}}{{O}_{3}};{{\Delta }_{C}}{{H}^{o}}~of\text{ }~B\]
Hence we can write as ${{\Delta }_{comb}}{{H}^{o}}of\text{ boron }=\dfrac{1}{2}{{\Delta }_{f}}{{H}^{o}}of\text{ }{{\text{B}}_{2}}{{O}_{3}}$.
Therefore the standard heat of combustion of solid boron will be equal to the one half of the value of standard heat of formation of boron trioxide ( ${{B}_{2}}{{O}_{3}}$)

Thus the correct option is (B) ${1}/{2}\;\Delta H_{f}^{o}\left( {{B}_{2}}{{O}_{3}} \right)$ .

Note: Keep in mind that the combustion won’t always result in fire, but when it occurs, a flame will be the characteristic indicator of the reaction. Also, while the activation energy must be overcome to initiate combustion, the heat from a flame can offer enough energy to make the reaction self-sustaining.