The Energy needed for `Li(g) to Li^(3+) (g) + 3e^-` , is 19600 kJ `"mole"^(-1)` . The first ionisation energy of Li (g) = 520 kJ `"mole"^(-1)`. Calculate `IE_2` For Li(g) (ionisation energy of H = 13.6 eV)

Options

A. 75.3 eV/species
B. 25.30 eV/species
C. 30.45 eV/species
D. 62.40 eV/species

Answer

A

Explanation

`Li(g) to Li^(3+) (g) +e, Delta H = 1.96 xx 10^4 kJ "mol"^(-1)` ….(i)
`Li(g) to Li^(+) (g) + e, IE_1 = 520 kJ "mol"^(-1)` …….(ii)
`Li^+ (g) to Li^(2+) (g) e, IE_2 = akJ"mol"^(-1)` …..(iii)
`Li^(2+) (g) to Li^(3+) (g) + e, IE_3 = bkJ "mol"^(-1) ` ....(iv)
also, `b = E_1 ` for `Li^(2+) xx N_A`
` = H xx Z^2 xx N_A = 2.18 xx 10^(-18) xx 3^2 xx 6.023 xx 10^23 J"mol"^(-1) = 11.81 xx 10^3 kJ "mol"^(-1)`
` therefore eqs(i), (ii) , (iii),(iv)`
`1.96 xx 10^4 = 520 + a + 11.81 xx 10^3 `
` a = 7270 kJ "mol"^(-1)`
` a= 75.3 eV`/ species [ 1ev/atom = 96.5 kJ/atom]