Figure shows the radial probability distribution functions for the 2s orbitals and 2p orbitals, (a) Which orbital. 2s or 2p, has more electron density close to the nucleus? (b) How would you modify Slater's rules to adjust for the difference in electronic penetration of the nucleus for the 2s and 2p orbitals? A Figure Comparison of 1s. 2s. and 2p radial probability functions. Based on this figure, is it possible for an electron in a 2s orbital to be closer to the nucleus than an electron in a 1 s orbital? (a) If the core electrons were totally effective at screening the valence electrons and the valence electrons provided no screening for each other, what would be the effective nuclear charge acting on the 3s and 3p valence electrons in P? (b) Repeat these calculations using Slater's rules, (c) Detailed calculations indicate that the effective nuclear charge is 5.6+ for the 3s electrons and 4.9+ for the 3p electrons. Why are the values for the 3s and 3p electrons different? (d) If you remove a single electron from a P atom, which orbital will it come from?
Solution Step 1 The effective nuclear charge Zeff=Z-S Where Z is atomic number S is screening constant Assuming that core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant S for 3s electron in P can be found as given below For P,electronic configuration is 1s 2s 2p 3s 3p2 3 Thus the number of core electron are 10 Thus Zeff=15-10=5 The effective nuclear charge on 3p electrons will also be 5 c)