Give the electron filling for the following ions: Using slater\'s rules, calcula

Question

Give the electron filling for the following ions: Using slater's rules, calculate Z* for the following electrons What trend is evident here? Now calculate Z* for a Zp electron in each of the following: Compare this trend to the one above and also compare the different values for each element and its most stable ion. Calculate Z* for a a 3d electron in Cu, with is usual electron filling of |Ar|3d^10 4s^t a 4s electron in Cu, with its usual electron filling of |Ar|3d^10 4s^1 a 3d electron in Cu, with the "expected" electron filling |Ar|3d^0 4s^2 a 4 s electron in Cu, with "expected" electron filling |Ar|3d^0 4s^2 Do these value help to explain why Cu has the electron filling of |Ar|3d^10 4s^1? Why or why not?

Explanation / Answer

1. Electronic configuration for,

Zn2+ = 1s2 2s2 2p6 3s2 3p6 3d10

V3+ = 1s2 2s2 2p6 3s2 3p6 3d2

Rh+ = 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d8

Ce4+ = 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6

2. Z* by Slater's rule

a. 1 3s electron of Na = 11 - [1(2) + 0.85(8)] = 2.2

b. a 3s electron in Mg = 12 - [0.35(1) + 0.85(8) + 1(2)] = 2.85

c. a 3p electron in Al = 13 - [0.35(2) + 0.85(8) + 1(2)] = 3.5

3. Z* calculation for,

a. Na+ = (1s2)(2s2 2p6)

Z* = 11 - 0.35(7) + 0.85(2) = 6.85

b. Mg2+ = (1s2)(2s2 2p6)

Z* = 12 - [0.35(7) + 0.85(2)] = 7.85

c. Al3+

Z* = 13 - [0.35(7) + 0.85(2)] = 8.85

4. electronic configuration of Cu = (1s2)(2s2 2p6)(3s2 3p6)(3d10)(4s1)

a. 3d electron

Z* = 29 - [0.35(9) + 0.35(8) + 0.85(8) + 1(2)] = 14.25

b. a 4s electron

Z* = 29 - [0.85(18) + 1(10)] = 3.7

electronic configuration of Cu = (1s2)(2s2 2p6)(3s2 3p6)(3d9)(4s2)

c. 3d electron

Z* = 29 - [0.35(16) + 0.85(8) + 1(2)] = 14.6

d. a 4s electron

Z* = 29 - [0.35(1) + 1(17) + 1(10)] = 1.65

As expected the (n-1)d configuation is more effective in shielding than nd configuration.

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