- ANeutral.
- BAmphoteric.
- ✓Basic.
- DAcidic.
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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
$H _{2}+\frac{1}{2} O _{2} \rightarrow H _{2} O , \cdots \cdots( ii )$ $\Delta H =-\,287.3 \,kJ\,mol ^{-1}$
$2 CO _{2}+3 H _{2} O \rightarrow C _{2} H _{5} OH +3 O _{2} \cdots \cdots ( iii )$; $ \Delta H =1366.8 \,kJ\,mol ^{-1}$
Find the standard enthalpy of formation of $C _{2} H _{5} OH (1)$