Question

9.23 Using Slater's rules, calculate and compare the Zef for both the 4s and 3d electrons of a copper atom. Discuss the results relative to the known fact that the 4s electrons are the first to be ionized from the first-row transition metals.

Answer 1

Dear Student,

While applying Slater's rule we need to first know the electronic configuration and then write the different multiplicative values for different electrons in different orbitals. It would be of two types

1.s or p orbital (n electrons should be multiplied by 0.35,n-1 by 0.85 ,n-2 by 1, (n-3,n-4 etc by 1) in calculation of s

2. d or f orbital(n electrons should be multiplied by 0.35,n-1 by 1 ,n-2 by 1, (n-3,n-4 etc by 1) in calculation of s

I have solved the question in the following pages.If you can not understand something or something is not clear feel free to comment here.

> SLATER'S RULE:- numerical emistry, slaters rule gives the for effective nuclear charge values Electronic configuration of copper: 2= Atomic number [Ar]3d1oys' Copperceu)2=29 reff for both 3d as electrons = gd electron] Inner shell electrons. reff = 2-5 = -15.as Z eft = 13.05 S. for 3d electron. [Ar]= 15²/5² 2861 35² 3881 for d-orbital 1 & everal [cu]= 15² / 25² 2861 38² 386 1 300 ys' for ny multiple by 0.35 non mutiply 1 =3 ne n-2 multiply exit 8x1 & 8x0.35+ axo.35 10+ 5.95 BE S9

Tus electron) reff= 2-s Reff =29-25-3 = 3.7 for us electron [ew] = 152/28?286 135238113a Cos & 1o - (n) calculating for this fors electrons) Chal) a 0.85 (n=2) 100 (n-3) - too

S- 18 x 0.854 8x noot 2x1.oo - 15.3t to = 253 AS us electrons nove less effective nuclear charge = 3-7 a ps electrons will be ionized first rather than 3d electrons here having nuclear charge=13.05 - followed The in same trend will be transition metals.

Answer 2

I do mot know