$\mathrm{NH}_3, \mathrm{SO}_2, \mathrm{SiO}_2, \mathrm{BeCl}_2, \mathrm{CO}_2, \mathrm{H}_2 \mathrm{O}, \mathrm{CH}_4, \mathrm{BF}_3$
- ✓$4$
- B$10$
- C$15$
- D$5$
$\mathrm{NH}_3, \mathrm{SO}_2, \mathrm{SiO}_2, \mathrm{BeCl}_2, \mathrm{CO}_2, \mathrm{H}_2 \mathrm{O}, \mathrm{CH}_4, \mathrm{BF}_3$
$\mathrm{SO}_2 \rightarrow \mathrm{sp}^2$
$\mathrm{SiO}_2 \rightarrow \mathrm{sp}^3$
$\mathrm{BeCl}_2 \rightarrow \mathrm{sp}$
$\mathrm{CO}_2 \rightarrow \mathrm{sp}$
$\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{sp}^3$
$\mathrm{CH}_4 \rightarrow \mathrm{sp}^3$
$\mathrm{BF}_3 \rightarrow \mathrm{sp}^2$
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$XeF _2, I _3^{+}, C _3 O _2, I _3^{-}, CO _2, SO _2, BeCl_2 \text { and } BCl _2^{\Theta}$
$C{H_3} - \mathop {\mathop {C = }\limits_{|\,\,\,\,} }\limits_{C{l_{}}\,} \mathop {\mathop {C\, - }\limits_{|\,\,\,\,\,\,} }\limits_{C{H_3}\,} \mathop {\mathop {CH - \,}\limits_{|\,\,\,\,\,\,\,\,\,} }\limits_{{C_2}{H_5}\,} C{H_2} - C \equiv CH$