MCQ
The compound in which ${C^*}$ uses $s{p^3}$ hybrids for bond formation is
- A$H\mathop C\limits^ + OOH$
- B${(N{H_2})_2}\mathop C\limits^ + O$
- ✓${(N{H_3})_3}\mathop C\limits^ + OH$$HgC{l_2}$
- D$C{H_3}\mathop C\limits^ + HO$
✓
Answer
Correct option: C.
${(N{H_3})_3}\mathop C\limits^ + OH$$HgC{l_2}$
c
$\left( NH _3\right)_3 C^*OH HgCl _2$ in this molecule the $C ^*$ has to have all single bonds, therefore, it is $sp ^3$ hybridized.
$\left( NH _3\right)_3 C^*OH HgCl _2$ in this molecule the $C ^*$ has to have all single bonds, therefore, it is $sp ^3$ hybridized.
Where as in $HC^*OOH,\left( NH _2\right)_2{ }C^*O$ and $CH _3 C^*HO$ in that all ${ }_{ C }^*$ having double bonds so the are showing $sp^2$ hybridization
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