JEE Advanced levelSingle correct optionDifficulty 3 of 3pt096
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. Neutral iron is (1s)2(2s,2p)8(3s,3p)8(3d)6(4s)2 and FeX3+ is (1s)2(2s,2p)8(3s,3p)8(3d)5 with the 4s electrons removed. The value of Z∗(3d in FeX3+)−Z∗(3d in Fe) is:
A1.05
B0.00
C3.00
D0.35
Answer: D
In FeX3+: S=4(0.35)+18=6.60, Z∗=6.60. In Fe: Z∗=6.25. The two 4s electrons are in a group to the right of 3d and never screen it, so only the loss of one 3d electron (0.35) matters: difference =0.35.