Question

An electron in a multi-electron atom is partially shielded from the full charge of the nucleus by other electrons, and this leads to the idea of an effective nuclear charge Zeff. For example, for the Li atom where the true charge is Z=3, quantum calculations give an effective nuclear charge of 2.69 for the 1s electrons, and 1.28 for the 2s electron. Use the first and second ionization energies given below to estimate Zeff for the 2s and 1s electrons in Li. (Hint: Write the electron configuration for Li and think about which electron comes off in the first and second ionizations. It may help to write the ioniation equations for each step to help.)

First ionization of Li = 520.2 kJ/mol

Second ionization of Li = 7289.1 kJ/mol

I've tried over and over using the equation En = -2.178 * 10^-18 (Zeff^2 / n^2) but I cant get it to work. Thank you

Answer 1

You are using the wrong formula.

E_n= -2.178*10 ^-18(Z_eff ² / n²) J atom ^-1

Ionisation energy given here is in kJ/mol not in J/atom. The above formula is used when you are finding energy in terms of Joule/atom.

As ionisation energy is the amount of energy required to remove an electron from nth orbit to n= , it is always positive. Ionisation energy in kJ/mol can be written as-

∆E = + 1312 Z_eff² / n²  kJ/mol

Li has the electronic configuration of 1s ² 2s ¹. For 1st ionisation energy (520.2kJ/mol) we can write,(n=2)

520.2kJ/mol = 1312 × Z _eff ² ÷ 2 ²

=> Z _eff = √(520.2×4÷1312) =1.259≈ 1.26

Similarly for 2nd ionisation energy ( 7289.1kJ/mol ) we can write (here, n=1) ,

7289.1kJ/mol = 1312× Z _eff² ÷ 1

=> Z _eff = √(7289.1÷1312) ≈ 2.36