Question

The metal tin is readily oxidized to 2+ and 4+ unsing the electron configurations and effective nuclear charges. Using electron configuration and effective nuclear charge, explain which specific electrons are being removed to form these ions. Include the calculations of effective nucear charge for the valence electrons in your answer.

Answer 1

the electronic configuration of Sn is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 5s2 5p2

Sn get oxidised to +2 and +4 by loosing 5s and 5p electrons. to go to +2, Sn has to lose 2 electorns and to become +4, it has to 4 electrons. +4 state is gained by removing two 5s and two 5p electrons, which leads to stable configuration.

To check which electrons will go away to get +2 state can be determined by checking the Z* of the electrons

we know Z*=Z-S, S is shielding constant

we have to calculate S

the ordering and grouping of orbitals we can write as (1s2)(2s2p6)(3s2 3p6) (3d10) (4s2 4p6) (4d10) (5s2 5p2)

S for a electron of 5p is = (1*2)+(8*1)+ (8*1)+ (10*1) + (8*0.85)+(10*0.85) +(3*0.35)= 44.35

S for 2nd electron of 5p is = (1*2)+(8*1)+ (8*1)+ (10*1) + (8*0.85)+(10*0.85) +(2*0.35)= 44.0

S for a electron of 5s is = (1*2)+(8*1)+ (8*1)+ (10*1) + (8*0.85)+(10*0.85) +(1*0.35) =43.65

S for a electron of 5p is = (1*2)+(8*1)+ (8*1)+ (10*1) + (8*0.85)+(10*0.85)=43.30

Z* for 5p2 is = 50-44.35=5.65

Z* for 5p1 is= 50-44=6.0

Z* for 5s2 is = 50-43.65=6.35

Z* for 5s1 = 50-43.30=6.70

so both 5p electrons will be removed as the Z* is less for them as compared to that of 5s electrons.