JEE Main levelSingle correct optionDifficulty 2 of 3pt070
Use the following rule set. Slater's rules: write the configuration in groups (1s)(2s,2p)(3s,3p)(3d)(4s,4p)(4d)(4f)(5s,5p)… For an electron in an ns or np group, every other electron in the same group contributes 0.35 (0.30 for the 1s group), every electron in the (n−1) shell contributes 0.85, every electron in the (n−2) or lower shells contributes 1.00, and electrons in groups to the right contribute 0. For an electron in an nd or nf group, every other electron in the same group contributes 0.35, every electron in all groups to its left contributes 1.00, and groups to the right contribute 0. Z∗=Z−S. The ions KX+ and ClX− are both (1s)2(2s,2p)8(3s,3p)8. The Slater Z∗ values for a 3p electron of KX+ and ClX− respectively are:
A5.75 and 7.75
B7.75 and 5.75
C7.40 and 5.40
D11.00 and 9.00
Answer: B
Both have the same screening: S=7(0.35)+8(0.85)+2(1.00)=11.25. Hence Z∗(KX+)=19−11.25=7.75 and Z∗(ClX−)=17−11.25=5.75.