MCQ
The state of hybridization of the central atom is not the same as in the others
- ✓$B$ in $BF_3$
- B$O$ in $H_3O^+$
- C$N$ in $NH_3$
- D$P$ in $PCl_3$
One electron from one 's' orbital and two electrons from 'p' orbital overlaps with 'p' orbital electrons of Flourine to form three $\mathrm{sp}^{2}$ hybridised bonds. So, $\mathrm{BF}_{3}$ has sp $^{2}$ hybridisation.
Flourine (a) But in $\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{NH}_{3}$ and $\mathrm{P} \mathrm{Cl}_{3}$, the hybridisation involves one electron from 's $^{\prime}$ orbital, 3 electrons for ${ }^{\prime} \mathrm{p}^{\prime}$ orbitals in central orbital forming $\mathrm{sp}^{3}$ hybridisation.
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Assertion $A :$- Carbon forms two important oxides $- CO$ and $CO _2 . CO$ is neutral whereas $CO _2$ is acidic in nature.
Reason $R :$- $CO _2$ can combine with water in a limited way to form carbonic acid, while $CO$ is sparingly soluble in water.
In the light of the above statements, choose the most appropriate answer from the options given below :-