Answer to Question #138733 in General Chemistry for Monae

Question

12. Calculate Zeff of a 4s electron and a 3d electron for Vanadium. Using your results, explain why the electronic configuration is more likely to be 3d34s1 as oppose to 3d24s2.

Answer 1

The rules

1.

An amount of 0.35 from each other electron within the same group (n) except for the [1s] group, where the other electron contributes only 0.30.

2.

If the group is of the [(n–1)s, (n–1)p] type, an amount of 0.85 from each electron with principal quantum number (n–1), and an amount of 1.00 for each electron with principal quantum number (n–2) or less.

3.

If the selected electron is in [d] or [f], type,

Then an amount of 1.00 for each electron from (n-1) group have to consider

Vanadium(23), electron configuration

= 1s2, 2s2, 2p6, 3s2, 3p6, 3d3, 4s2

Slater’s Rule: group the configuration for 4S electron

(1s2), (2s2, 2p6), (3s2, 3p6), (3d3),( 4s2)

Following the rules,

Therefore, since selected group is 4s, and there is 1 electron besides the one for which we are calculating the screening constant,

1×0.35=.35

11 electron in ( n-1 ) level (right next to the level of select electron) contributes 0.85 each

Therefore, 11×0.85= 9.35

Rest of the groups, 1 each

Therefore, 10×1 = 10

Total screening constant for 4S electron

= 10+9.35+0.35

=19.7

Hence Z effective for 4s electron

= 23–19.7

=3.3

Slater’s Rule: group the configuration for 3d electron

(1s2), (2s2, 2p6), (3s2, 3p6), (3d3)

Following the rules,

Therefore, since selected group is 3d, and there are 2 electron besides the one for which we are calculating the screening constant,

2×0.35=0.7

Rest of the groups, 1 each

And there are 18 electrons

Therefore, 18×1 = 18

Total screening constant for 3d electron

= 18+0.7

=18.7

Hence Z effective for 3d electron

= 23–18.7

=4.3

As in the 3d shell electrons are attracted by the nuclious much more due to the high

Z effective value than the 4S shell where there is lower Z effective.

Higher the attraction Higher will be the stability of the electronic configuration.

So,

the electronic configuration is more likely to be 3d34s1 where 3 electron is present in 3d as oppose to 3d24s2 where only 2 electron are present in 3d.

That's why 3d34s1 is most likely to be oppose to the 3d24s2.