In Section 3.6, we estimated the effective nuclear charge on
bery...

Question

In Section 3.6, we estimated the effective nuclear charge on
beryllium’s valence electrons to be slightly greater than 2+. What
would a similar treatment predict for the effective nuclear charge
on boron’s valence electrons? Would you expect the effective
nuclear charge to be different for boron’s 2s electrons compared
to its 2p electron? In what way? (Hint: Consider the shape of the
2p orbital compared to that of the 2s orbital.)

Accepted Answer

Hi everyone here we have a question asking us to identify the statement that would be a good adjustment to slater's rule. To account for the difference in electron penetration of the nucleus for three S three P and three D. Orbital's use the radial probability distribution curve below for reference. So we have three S. Three P. And three D. So R. N. Is going to equal three because we have three sub shells And in equals three. As electrons have some probability of being closer to the nucleus. This probability of being closer to the nucleus results in a smaller peak. P electrons have a similar condition but they are slightly farther than the smaller S. Peak. So this means that S can slightly shield better than P and P can slightly shield better than D. So let's look at our options here, A assign equally larger shielding values for S. P. N. D. B assign a shielding value of S slightly smaller than that of P. And a value for P smaller than that of D. C assign a shielding value of S slightly larger than that of P. And a value for P slightly larger than that of D. Or D. Assigned equally smaller shielding values for S. P. And D. Our answer here is going to be C. A sign, a shielding value of S slightly larger than that of P. And a value for P slightly larger than that of D. Thank you for watching. Bye.

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