Question

Be and Al show diagonal relationship and thus :

• A. their oxides are soluble in alkali solution forming $\left[Be\right(OH{)}_4{]}^{3-}$
• B. they form complex anion $Be{F}_4^{2-}$ and $Al{F}_6^{3-}$
• C. $BeC{l}_2$ and $AlC{l}_3$ are Lewis acid
• D. oxides are basic

Answer 1

Answer: (A), (B), (C)

their oxides are soluble in alkali solution forming $\left[Be\right(OH{)}_4{]}^{3-}$
they form complex anion $Be{F}_4^{2-}$ and $Al{F}_6^{3-}$
$BeC{l}_2$ and $AlC{l}_3$ are Lewis acid

1) $B{e}^{2+}$ and $A{l}^{3+}$ have almost same and smaller size and thus favour for covalent bonding.

2) Both these form covalent compounds having low melting point and soluble in organic solvent.

3) Both have same value of electronegativity.

4) Their oxides are soluble in alkali solution forming $\left[Be\right(OH{)}_4{]}^{3-}$

5) Both form many stable complexes for example $Be{H}_3^{-},Al{H}_4^{-}$.

6) They form complex anion $Be{F}_4^{2-}$ and $Al{F}_6^{3-}$.

7) $BeC{l}_2$ and $AlC{l}_3$ are Lewis acid.

Hence options A,B & C are correct.