See the answer below.
Work Step by Step
a) Period 4, group 8: Iron, Fe b) The ion lost the 4s electrons, the atom has 7 with the ones in the 3d orbitals, group 7, period 4: Mn, it's paramagnetic since it has 5 unpaired electrons. c) Ni is in group 10, losing 2 electrons would leave the configuration as $[Ar]\ 3d^8$ so two unpaired electrons. d) It has 15 electrons past the noble gas core, group 15, outermost shell is 4 (period) so it's As. It's paramagnetic since there are unpaired electrons. No unpaired electrons in the 3- ion, the 4p orbitals would be filled. e) Kr is in period 4, so this atom is in the fifth. 6 electrons past the noble gas core, group 6 It's molybdenum, Mo. 1: 4, 2,-2,-1/2 2: 4,2,+1,-1/2 3: 5,0,0,-1/2