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Chapter 21: Problem 23

Write electron configurations for the following transition metals. a. Sc b. Ru c. Ir d. Mn

Short Answer

Expert verified
The electron configurations for the transition metals Sc, Ru, Ir, and Mn are as follows: a. Sc: \(1s^2 \> 2s^2 \> 2p^6 \> 3s^2 \> 3p^6 \> 4s^2 \> 3d^1\) b. Ru: \(1s^2 \> 2s^2 \> 2p^6 \> 3s^2 \> 3p^6 \> 4s^2 \> 3d^{10} \> 4p^6 \> 5s^2 \> 4d^7\) c. Ir: \(1s^2 \> 2s^2 \> 2p^6 \> 3s^2 \> 3p^6 \> 4s^2 \> 3d^{10} \> 4p^6 \> 5s^2 \> 4d^{10} \> 5p^6 \> 6s^2 \> 4f^{14} \> 5d^7\) d. Mn: \(1s^2 \> 2s^2 \> 2p^6 \> 3s^2 \> 3p^6 \> 4s^2 \> 3d^5\)

Step by step solution

01

Determine the atomic number

Scandium (Sc) is a transition metal with an atomic number of 21.
02

Fill the orbitals following the Aufbau principle

To find the electron configuration, we will fill up the orbitals in the order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p. We have 21 electrons to place, so the electron configuration for Scandium (Sc) is: \[1s^2 \> 2s^2 \> 2p^6 \> 3s^2 \> 3p^6 \> 4s^2 \> 3d^1\] #b. Ruthenium (Ru) Electron Configuration#
03

Determine the atomic number

Ruthenium (Ru) is a transition metal with an atomic number of 44.
04

Fill the orbitals following the Aufbau principle

We will follow the same order as before. We have 44 electrons to place, so the electron configuration for Ruthenium (Ru) is: \[1s^2 \> 2s^2 \> 2p^6 \> 3s^2 \> 3p^6 \> 4s^2 \> 3d^{10} \> 4p^6 \> 5s^2 \> 4d^7\] #c. Iridium (Ir) Electron Configuration#
05

Determine the atomic number

Iridium (Ir) is a transition metal with an atomic number of 77.
06

Fill the orbitals following the Aufbau principle

We will follow the same order as before. We have 77 electrons to place, so the electron configuration for Iridium (Ir) is: \[1s^2 \> 2s^2 \> 2p^6 \> 3s^2 \> 3p^6 \> 4s^2 \> 3d^{10} \> 4p^6 \> 5s^2 \> 4d^{10} \> 5p^6 \> 6s^2 \> 4f^{14} \> 5d^7\] #d. Manganese (Mn) Electron Configuration#
07

Determine the atomic number

Manganese (Mn) is a transition metal with an atomic number of 25.
08

Fill the orbitals following the Aufbau principle

We will follow the same order as before. We have 25 electrons to place, so the electron configuration for Manganese (Mn) is: \[1s^2 \> 2s^2 \> 2p^6 \> 3s^2 \> 3p^6 \> 4s^2 \> 3d^5\]

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Most popular questions from this chapter

Write electron configurations for the following transition metals and their ions. a. \(\mathrm{Co}, \mathrm{Co}^{2+}, \mathrm{Co}^{3+}\) b. \(\mathrm{Pt}, \mathrm{Pt}^{2+}, \mathrm{Pt}^{4+}\) c. \(\mathrm{Fe}, \mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}\)

Name the following coordination compounds. a. $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Br}\right] \mathrm{Br}_{2}$ b. \(\mathrm{Na}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\) c. $\left[\mathrm{Fe}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{2}\left(\mathrm{NO}_{2}\right)_{2}\right] \mathrm{Cl}$ d. $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{2}\right]\left[\mathrm{PtI}_{4}\right]$

A metal ion in a high-spin octahedral complex has two more unpaired electrons than the same ion does in a low-spin octahedral complex. Name some possible metal ions for which this would be true.

Which of the following ions is(are) expected to form colored octahedral aqueous complex ions? a. \(Z n^{2+}\) b. \(C u^{2+}\) c. \(M n^{3+}\) d. \(T i^{4+}\)

Name the following coordination compounds. a. \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{2}\) b. $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{I}_{3}$ c. \(\mathrm{K}_{2}\left[\mathrm{PtC}_{4}\right]\) d. \(\mathrm{K}_{4}\left[\mathrm{Pt} \mathrm{C}_{6}\right]\) e. $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$ f. \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{NO}_{2}\right)_{3}\right]\)
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