Find the sum of triangular ${\mathrm{BO}}_3$unit and ${\mathrm{BO}}_4$ tetrahedral unit in borax. [If your answer is 6 ${\mathrm{BO}}_3$ unit and $3{\mathrm{BO}}_4$ unit then write 6 + 3 = 9)

The hybridization of ${\mathrm{BO}}_3$ and ${\mathrm{BO}}_4$ is $s{p}^2$ and $s{p}^3$ respectively and the triangular ${\mathrm{BO}}_3$unit and B04 tetrahedral unit in borax is 2 and 2 respectively.

The obtained value is,

$$\begin{gathered} =2+2 \\ =4 \end{gathered}$$