- ABoth are rendered passive by conc. HNO3.
- BCarbides of both give methane on hydrolysis.
- CBoth give hydroxides which are basic.
- DBoth give covalent chlorides.
Explanation:
Both form amphoteric hydroxide/oxide not either basic or acidic.
Because as Aluminium is in 3rd group it can either accept electron or donate hence they form oxides which are amphoteric.
The reason why beryllium forms is aluminium and beryllium has diagonal relationship.
Hence both will have common properties.
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$\Delta_{f} \mathrm{H}^{\ominus}$ for $\mathrm{KCl}=-436.7 \,\mathrm{~kJ}\, \mathrm{~mol}^{-1}$
$\Delta_{\text {sub }} \mathrm{H}^{\ominus}$ for $\mathrm{K}=89.2 \,\mathrm{~kJ}\, \mathrm{~mol}^{-1}$
$\Delta_{\text {ionization }} \,\mathrm{H}^{-}$ for $\mathrm{K}=419.0\, \mathrm{~kJ}\, \mathrm{~mol}^{-1}$
$\Delta_{\text {electron gain }} \mathrm{H}^{\ominus}$ for $\mathrm{Cl}_{(\text {e) }}=-348.6 \,\mathrm{~kJ} \,\mathrm{~mol}^{-1}$
$\Delta_{\mathrm{bond}} \mathrm{H}^{-}$ for $\mathrm{Cl}_{2}=243.0 \,\mathrm{~kJ} \,\mathrm{~mol}^{-1}$
The magnitude of lattice enthalpy of $\mathrm{KCl}$ in $\mathrm{kJ} \mathrm{mol}^{-1}$ is ..... . (Nearest integer)
$N _2 O _3, N _2 O _5, P _4 O _6, P _4 { O }_7, H _4 P _2 O _5, H _5 P _3 O _{10}, H _2 S _2 O _3, H _2 S _2 O _5$
$ NH_3(g) \rightleftharpoons \frac{1}{2}\,{N_2}\left( g \right) + \frac{3}{2}{H_2}\left( g \right);\,{K_2}$
$\frac{1}{2}\,{N_2}\left( g \right) + \frac{3}{2}{H_2}\left( g \right) \rightleftharpoons NH_3(g); K_3$
$2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g); K_4$
If $K_1 = K_2^x = K_3^y = K_4^z$ then correct values of $x, y$ and $z$ are respectively