Question

The Energy needed For $Li\left(g\right) \rightarrow \left(Li\right)^{3 +}\left(g\right)+3e^{-}$ , is $19600kJmole^{- 1}$ . The first ionisation energy of $Li\left(g\right)= \, 520 \, \left(kJmole\right)^{- 1}$ . Calculate $IE_{2}$ For $Li\left(g\right)$ .
(ionisation energy of $H=13.6eV$ )

• A. $\text{75.3 eV/species}$
• B. $\text{25.30 eV/species}$
• C. $\text{30.45 eV/species}$
• D. $\text{62.40 eV/species}$

Answer 1

Answer: (A)

$Li\left(\right.g\left.\right) \rightarrow Li^{3 +}\left(\right.g\left.\right)+3e;\Delta H=1.96\times \left(10\right)^{4}kJmol^{- 1}\ldots .\left(\right.i\left.\right)$
$Li\left(\right.g\left.\right) \rightarrow Li^{+}\left(\right.g\left.\right)+e;IE_{1}=520kJmol^{- 1}\ldots .\left(\right.ii\left.\right)$
$Li^{+}\left(g\right) \rightarrow Li^{2 +}\left(g\right)+e;IE_{2}=akJmol^{- 1}\ldots ..\left(i i i\right)$
$Li^{2 +}\left(\right.g\left.\right) \rightarrow Li^{3 +}\left(\right.g\left.\right)+e;IE_{3}=bkJmol^{- 1}\ldots ..\left(\right.iv\left.\right)$
Also,
$b=E_{1} \, for \, Li^{2 +}\times N_{A}$
$=H\times Z^{2}\times N_{A}=2.18\times 10^{- 18}\times 3^{2}\times 6.023\times 10^{23}Jmol^{- 1}=11.81\times 10^{3}kJmol^{- 1}$
$\therefore eqs \, \left(i\right),\left(i i\right),\left(i i i\right),\left(i v\right)$
$1.96\times 10^{4}=520+a+11.81\times 10^{3}$
$a=7270\text{kJmol}^{- 1}$
$a=75.3eV/species$
$\left[1 \text{ ev/atom } = 96.5 \, \text{kJ/mol}\right]$