MCQ
When $H_2SO_4$ react with $PCl_5$ the obtained product is/are
- ✓$SO_2Cl_2$
- B$POCl_3$
- C$HCl$
- DAll of these
$H _{2} O \,\,\,\,\,\,\,\,\,\,\,\,\,\, \underbrace{ H _{2} S \,\,\, H _{2} Se \,\,\, H _{2} Te}$
$\,\,$H-Bond $\,\,\,\,\,\,\,\,\,\,$ VMA $\propto$ mol.wt.
$\text { B.P. } \rightarrow H _{2} S < H _{2} Se < H _{2} Te < H _{2} O$
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Reason : Iodine is a polar compound.
$CH_3CH_2CH_2CONH_2$ $\mathop {\xrightarrow{{Ca{{(OH)}_2},C{l_2}}}}\limits_\Delta X\xrightarrow{{HN{O_2}}}Z$


