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Chapter 21: Problem 24

Write electron configurations for the following transition metal ions. a. \(\mathrm{Sc}^{3+}\) b. \(\mathrm{Ru}^{2+}, \mathrm{Ru}^{3+}\) c. \(\operatorname{Ir}^{+}, \operatorname{Ir}^{3+}\) d. \(M n^{2+}\)

Short Answer

Expert verified
The electron configurations for the given transition metal ions are: a. Sc³⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^0\) b. Ru²⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^4\) b. Ru³⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^3\) c. Ir¹⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^8\) c. Ir³⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 4f^{14} 5d^5\) d. Mn²⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\)

Step by step solution

01

Reminder: Aufbau Principle and Orbital Notations

Keep in mind the Aufbau Principle, which states that electrons fill orbitals in order of increasing energy, starting from the lowest energy orbitals. The typical order is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, etc.
02

a. Sc³⁺ Electron Configuration

Step 1: Write the electron configuration for neutral Scandium (Sc) Sc (atomic number 21) has the electron configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1\) Step 2: Remove 3 electrons from the highest energy orbitals for Sc³⁺ Since we have a 3+ ion, we need to remove 3 electrons. We take the electrons from the highest energy orbitals, which are 4s and 3d in this case. Electron configuration for Sc³⁺ is: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^0\)
03

b. Ru²⁺ and Ru³⁺ Electron Configurations

Step 1: Write the electron configuration for neutral Ruthenium (Ru) Ru (atomic number 44) has the electron configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^6\) Step 2: Remove 2 electrons for Ru²⁺ Electron configuration for Ru²⁺ is: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^4\) Step 3: Remove an additional electron for Ru³⁺ Electron configuration for Ru³⁺ is: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^3\)
04

c. Ir¹⁺ and Ir³⁺ Electron Configurations

Step 1: Write the electron configuration for neutral Iridium (Ir) Ir (atomic number 77) has the electron configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^7\) Step 2: Add 1 electron for Ir¹⁺ Electron configuration for Ir¹⁺ is: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^8\) Step 3: Remove 3 electrons for Ir³⁺ Electron configuration for Ir³⁺ is: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 4f^{14} 5d^5\)
05

d. Mn²⁺ Electron Configuration

Step 1: Write the electron configuration for neutral Manganese (Mn) Mn (atomic number 25) has the electron configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5\) Step 2: Remove 2 electrons for Mn²⁺ Electron configuration for Mn²⁺ is: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\)

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Most popular questions from this chapter

Qualitatively draw the crystal field splitting for a trigonal bi- pyramidal complex ion. (Let the \(z\) axis be perpendicular to the trigonal plane.)

Give formulas for the following complex ions. a. tetrachloroferrate(III) ion b. pentaammineaquaruthenium(III) ion c. tetracarbonyldihydroxochromium(III) ion d. amminetrichloroplatinate(II) ion

Draw all geometrical and linkage isomers of square planar \(\operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{SCN})_{2}\) .

Draw the \(d\) -orbital splitting diagrams for the octahedral complex ions of each of the following. a. \(\mathrm{Fe}^{2+}\) (high and low spin) b. \(\mathrm{Fe}^{3+}(\text { high spin })\) c. \(\mathrm{Ni}^{2+}\)

Both \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) and \(\mathrm{Ni}(\mathrm{SCN})_{4}^{2-}\) have four ligands. The first is paramagnetic, and the second is diamagnetic. Are the complex ions tetrahedral or square planar? Explain.
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