Write the ground-state electron configuration of Sb³⁺. A)[Kr]4d⁸5s¹5p³ B)[Kr]4d⁵5s¹5p⁶ C)[Kr]4d¹⁰5p² D)[Kr]4d¹⁰5s¹5p¹ E)[Kr]4d¹⁰5s²

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First, let's determine the ground-state electron configuration of Sb (antimony) and S (sulfur): Sb: [Kr]4d10 5s2 5p3 S: [Ne]3s2 3p4 To form Sb³⁺, we must add the electron configuration of Sb and S together, while accounting for the positive charge on the ion. Starting with Sb, we remove three electrons from the 5p subshell to account for the charge of 3+. This gives us: Sb: [Kr]4d10 5s2 5p0 Next, we add the electron configuration of S, starting with the 3p subshell. However, since we are dealing with a cation, we need to remove an electron before adding the configuration. This gives us: S: [Ne]3s2 3p3 Removing one electron from S gives us: S: [Ne]3s2 3p2 Adding this to the electron configuration of Sb, we get: [Kr]4d10 5s2 5p0 3s2 3p2 Finally, we must account for the charge of 1+ on the ion. This means we must remove one electron from the highest energy level, which is the 5s subshell. This gives us the ground-state electron configuration of: [Kr]4d10 5s1 5p0 3s2 3p2 Option E is the only choice that matches this configuration.

Write the Ground-State Electron Configuration of Sb3+

Write the Ground-State Electron Configuration of Sb³⁺