MCQ
The third line in Balmer series corresponds to an electronic transition between which Bohr’s orbits in hydrogen
- A$5 → 3$
- ✓$5 → 2$
- C$4 →3$
- D$4 → 2$
$O \to L$, ${n_2} = 5 \to {n_1} = 2$
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$(I)$ $\begin{array}{*{20}{c}}
{{{\left( {C{H_3}} \right)}_3}Si - NB{H_2}}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Si{{(C{H_3})}_3}}
\end{array}$ $(II)$ $\begin{array}{*{20}{c}}
{{{\left( {C{H_3}} \right)}_3}C - NB{H_2}}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{{(C{H_3})}_3}}
\end{array}$
$\begin{array}{*{20}{c}}{ - OAc}\\{\rm{I}}\end{array}$$\begin{array}{*{20}{c}}{ - OMe}\\{{\rm{II}}}\end{array}$$\begin{array}{*{20}{c}}{ - OS{O_2}Me}\\{{\rm{III}}}\end{array}$$\begin{array}{*{20}{c}}{ - OS{O_2}C{F_3}}\\{{\rm{IV}}}\end{array}$
The order of leaving group ability is
|
LIST $I$ (Hybridization) |
LIST $II$ (Orientation in Space) |
| $A$ $\mathrm{sp}^3$ | $I$ Trigonal bipyramidal |
| $B$ $\mathrm{dsp}^2$ | $II$ Octahedral |
| $C$ $\mathrm{sp}^3 \mathrm{~d}$ | $III$ Tetrahedral |
| $D$ $s p^3 d^2$ | $IV$ Square planar |
Choose the correct answer from the options given below :