- A$Li > Be > B$
- ✓$Li > B > Be$
- C$Be > Li > B$
- D$B > Be > Li$
Electronic configuration for these ions is:
$Li ^{+}: 1 s ^2$
$Be ^{+}: 1 s ^2\, 2 s ^1$
$B ^{+}: 1 s ^2\, 2 s ^2$
For, lithium and boron, electron is to be ejected from fully filled stable shells.
Further, it is difficult to remove electron from is shell of $Li$ because it is closer to the nucleus as compared to $2 s$ shell of $B$. It is much easier to remove electron from $Be$ because it will attain stable electronic configuration after losing this electron.
Thus, order is $Li \,> \,B\, >\, Be$
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[Atomic mass: $\mathrm{Ag}=108, \mathrm{Br}=80$ ]
$n-$propyl bromide $\xrightarrow{{Na}}(A)\xrightarrow[{A{l_2}{O_3}/773\,K}]{{Cr{O_3}}}(B)\xrightarrow[{500\,K}]{{C{l_2}/UV}}(C)$
$n-$propyl bromide $\xrightarrow{{Na}}(A)\xrightarrow[{A{l_2}{O_3}/773\,K}]{{Cr{O_3}}}(B)$ $\xrightarrow[{}]{{C{H_3} - Cl/AlC{L_3}}}(D)\xrightarrow{{C{l_2}/hv}}(E)$