Problem 147
Question
Borax \(\left[\mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7} .10 \mathrm{H}_{2} \mathrm{O}\right]\) when heated on platinum loop it gives a dark transparent glass like bead. The hot bead is dipped in the salt till it reacts with transition metal oxide. It produces characteristic bead of meta borate. $$ \begin{array}{ll} \text { Colour of the bead } & \text { Ion } \\ \text { Blue green or light blue } & \mathrm{Cu}^{2+} \\ \text { Yellow } & \mathrm{Fe}^{2+} \text { or } \mathrm{Fe}^{3+} \\ \text { Green } & \mathrm{Cr}^{3+} \\ \text { Violet } & \mathrm{Mn}^{2+} \\ \text { Dark blue } & \mathrm{Co}^{2+} \\ \text { Brown } & \mathrm{Ni}^{2+} \end{array} $$ The hybridisation of \(\mathrm{B}\) in borax is (a) sp (b) \(\mathrm{sp}^{2}\) (c) \(\mathrm{sp}^{3}\) (d) Both (B) and (C)
Step-by-Step Solution
VerifiedKey Concepts
Transition metal ions
Hybridization in borax
- In a borate ion, typically, each boron connects with three oxygen atoms, forming a trigonal planar structure. This specific arrangement necessitates \(\text{sp}^{2}\) hybridization.
- However, boron can also develop tetrahedral configurations, especially when coordinating with four other atoms, going for \(\text{sp}^{3}\) hybridization.
Borate ion structure
- Tetrahedral and trigonal planar geometries, based on the hybridization of boron.
- Covalent bonds formed with oxygen, allowing borate to exhibit stable, negatively charged polyatomic ion characteristics.