Question

Enthalpy of formation $A{B}_{(g)}$ from $A_{2(g)}$ and $B_{2(g)}$ is represented by ${\Delta }_{f}H(AB)$ in $kJ/mol$. Thus, electronegativity difference $(\Delta X)$ between $A$ and $B$ is related with ${\Delta }_{f}H(AB)$ as

• A. $-{\Delta }_{f}H=129.13\times (\Delta X)$
• B. ${\Delta }_{f}H=129.13\times (\Delta X)$
• C. $-{\Delta }_{f}H=129.13\times (\Delta X{)}^2$
• D. ${\Delta }_{f}H=129.13\times (\Delta X{)}^2$

Answer 1

Answer: (C)

$A_{2(g)}+B_{2(g)}\to A{B}_{(g)}$

As per Pauling's electronegativity difference, $\Delta X$ is proportional to the ${\Delta }_{f}H$ as

$\therefore X_A-X_B=0.088\sqrt{{\Delta }_{f}H},\therefore {\Delta }_{f}H=\frac{(\Delta X{)}^2}{0.088\times 0.088}$

$-{\Delta }_{f}H=129.13\times (\Delta X{)}^2$

Negative sign shows that bond formation is exothermic process.