Chemistry - CHEMICAL BONDING Question with Solution | TestHub

Hybridisation of $\mathrm{Br}$ is ${\mathrm{BrO}}_3^{-}$ Total valence elctrons $=7+3\times 6=25$ Charge $=-1$ Total $=25+1=26$ $=\frac{26}{8}=3(\mathrm{Q})+2(\mathrm{R})=5$ hybridization $=ds{p}^3$ hybridization of $\mathrm{Xe}$ in ${\mathrm{XeO}}_3$ Total valence electrons $=8+3\times 6=26=\frac{26}{8}=3(Q)+2(R)=5$ Hybridization $=ds{p}^3$ The structure ${\mathrm{BrO}}_3^{-}$and ${\mathrm{XeO}}_3$ is trigonal pyramidal. Caution Molecular structure of ${\mathrm{NO}}_3^{-},{\mathrm{NH}}_3,{\mathrm{NF}}_3=$ trigonal pyramidal Molecular structure of ${\mathrm{SO}}_3^{2-}=$ Trigonal Planar Molecular structure of ${\mathrm{SF}}_4=$ Seesaw Molecular structure of ${\mathrm{XeF}}_4=$ Square Planar