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A common reason given on why 3rd ionization energy > 2nd > 1st is because of increasing effective nuclear charge.

As per my book $Z_\mathrm{eff}$ = Atomic number $-$ Number of inner electrons.

Now let us consider a Carbon atom. C has an Electronic configuration of $\ce{[He] 2s^2 2p^2}$ or $2,4$

Right now the effective nuclear charge is $6 - 2 = 4$

Now if we remove the first electron the first ionization energy will be $=\pu{1086.45 kJ/mol}$

After doing this effective nuclear chrarge is still $6 - 2 = 4$

Now if we remove the second electron the second ionization energy will be $=\pu{2352.62 kJ/mol}$

If the effective nuclear charge in removal of 1st and 2nd electron is same then why is 2nd ionization energy > 1st.

note : these values are taken from here

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  • $\begingroup$ The website you link to gives a 2nd ionization energy that is more than twice that of the first. Also, you should regard effective nuclear charges as aids in interpretation. As soon as you remove one electron you can expect repulsion between the remaining electrons to be reduced. $\endgroup$ – Buck Thorn Nov 25 '19 at 16:26
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Effective nuclear charge $Z_{\text{eff}}$ is given by Slater's Rules.

According to these rules, for $\ce{C}$, effective nuclear charge experienced by an outer shell electron is: $$Z_{\text{eff}} = 6 - 0.85\times 2 - 0.35\times 3 = 3.25$$

For $\ce{C+}$, effective nuclear charge is $$Z_{\text{eff}} = 6 - 0.85\times 2 - 0.35\times 2=3.6$$

Since the effective charge for $\ce{C+}$ is more than that on $\ce{C}$, the second ionisation energy is more than that of the first.

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