JEE Main 2014 — s Block Elements Question with Solution

(i) % Ionic character = 16(X_A – X_B) + 3.5 (X_A – X_B)²

${\text{X}}_{\text{Rb}}=0·8$ ,    ${\text{X}}_{\text{Cl}}=3·0$,  ${\mathrm{X}}_{\mathrm{Be}}=2$

For $\mathrm{Rb}$ and $\mathrm{Cl}$ the electronegativity difference is highest so having more ionic character. For $\mathrm{Be} \text{and} \mathrm{Cl}$, electronegativity difference is the lowest so having less ionic character

Covalent character in compound depends upon the charge and size of ions. As charge in cation increases, covalent character increases and as size of cation increases covalent character decreases. Charge factor dominate over size factor. Since ${\mathrm{Be}}^{2+}$ has higher charge and smallest size therefore, ${\mathrm{BeCl}}_2$ has maximum covalent character and least ionic character