Question

The ionisation enthalpy of $H{e}^{+}$ ion is $19.60\times {10}^{-18}Jato{m}^{-1}$. The ionisation enthalpy of $L{i}^{2+}$ ion will be:

• A. $84.2\times {10}^{-18}Jato{m}^{-1}$
• B. $44.10\times {10}^{-18}Jato{m}^{-1}$
• C. $63.20\times {10}^{-18}Jato{m}^{-1}$
• D. $21.20\times {10}^{-18}Jato{m}^{-1}$
• E. $2.17\times {10}^{-19}Jato{m}^{-1}$

Answer 1

Answer: (B)

$44.10\times {10}^{-18}Jato{m}^{-1}$

Ionisation enthalpy of $L{i}^{2+}=\frac{9}{4}\times ionisationenthalpyofH{e}^{+}$
$=\frac{9}{4}\times 19.60\times {10}^{-18}$
$=44.1\times {10}^{-18}Jato{m}^{-1}$.