Answer:
The answer to your question is:
Re (III) has 5 electrons
Sc(III) = has 1 electron
Ru(IV) = has 6 electrons
Hg(II) = has 10 electrons
Explanation:
75 Re(III) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d⁵
21 Sc(III) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹
44 Ru(IV) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d⁶
80 Hg(II) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰
