MCQ
The distance between two adjuscent carbonn atoms is maximum in
- ✓Diamond
- BGraphite
- CBenzene
- DEthene
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$(I)$ It is easier to remove $2 \mathrm{p}$ electron than $2 \mathrm{s}$ electron
$(II)$ $2 \mathrm{p}$ electron of $\mathrm{B}$ is more shielded from the nucleus by the inner core of electrons than the $2 s$ electrons of $Be.$
$(III)\; 2 s$ electron has more penetration power than $2 \mathrm{p}$ electron.
$(IV)$ atomic radius of $\mathrm{B}$ is more than $\mathrm{Be}$
(Atomic number $\mathrm{B}=5, \mathrm{Be}=4$ )
The correct statements are
$2{C_4}{H_{10(g)}}\, + \,13{O_{2(g)}}\, \to \,8C{O_{2(g)}}\, + \,10{H_2}{O_{(l)}}\,;$ $\Delta {H^o}\, = \, - \,5756\,KJ$