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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Transition Metals
Transition metals are elements found in the central block of the periodic table, specifically in the d-block. These metals have partially filled d sub-shells, which contributes to many of their unique properties, such as variable oxidation states and the formation of colorful compounds. Transition metals are essential for various industries due to their durability and ability to act as catalysts in chemical reactions.
  • Scandium (Sc), Ruthenium (Ru), Iridium (Ir), and Manganese (Mn) are all transition metals.
  • These metals are involved in important technological applications, like electronics and catalysis.
  • Their electron configuration often involves filling d orbitals, which leads to complex electron activity.
Atomic Orbitals
Atomic orbitals are regions in an atom where there is a high probability of finding electrons. These orbitals come in different shapes and sizes and are designated as s, p, d, and f. Understanding these orbitals is crucial for determining how electrons are arranged in atoms.
  • s orbitals are spherical and can hold up to 2 electrons.
  • p orbitals are dumbbell-shaped and can hold up to 6 electrons.
  • d orbitals are more complex, capable of holding up to 10 electrons, which is important for transition metals.
  • f orbitals, even more intricate, can contain up to 14 electrons, though less involved in transition metals.
Aufbau Principle
The Aufbau principle is a guideline for determining the electron configuration of atoms. It states that electrons will fill the lowest energy levels first before moving to higher ones. This principle helps in predicting the order in which atomic orbitals are filled, lending insight into the arrangement of electrons in transition metals.
  • Electrons fill orbitals starting from the 1s orbital, moving through 2s, 2p, 3s, and so on.
  • This order of filling is represented by the Aufbau diagram, which visually maps out the sequence.
  • For transition metals, the 3d orbitals are crucial and usually filled after the 4s orbital.
Atomic Number
The atomic number of an element is the number of protons in its nucleus and also determines the number of electrons in a neutral atom. It's pivotal for defining an element's identity and its behavior in chemical reactions. For example, determining the electron configuration requires knowing the atomic number so one knows how many electrons to place in orbitals.
  • Scandium has an atomic number of 21, which affects how its electrons are configured.
  • Ruthenium has an atomic number of 44, allowing for more complex configurations involving higher energy orbitals.
  • Iridium, with an atomic number of 77, requires accommodating even more electrons, filling multiple shells and sub-shells.
  • Manganese has an atomic number of 25, leading to its unique electron arrangement.

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Most popular questions from this chapter

A compound related to acetylacetone is 1,1,1-trifluoroacetylacetone (abbreviated Htfa): Htfa forms complexes in a manner similar to acetylacetone. (See Exercise 49.) Both Be \(^{2+}\) and \(\mathrm{Cu}^{2+}\) form complexes with tfa - having the formula \(\mathrm{M}(\mathrm{tfa})_{2}\) . Two isomers are formed for each metal complex. a. The Be \(^{2+}\) complexes are tetrahedral. Draw the two isomers of \(\mathrm{Be}(\mathrm{tfa})_{2} .\) What type of isomerism is exhibited by \(\mathrm{Be}(\mathrm{tfa})_{2} ?\) b. The \(\mathrm{Cu}^{2+}\) complexes are square planar. Draw the two isomers of \(\mathrm{Cu}(\mathrm{tfa})_{2} .\) What type of isomerism is exhibited by \(\mathrm{Cu}(\mathrm{tfa})_{2} ?\)

The complex ion \(\operatorname{Ru}(\text { phen })_{3}^{2+}\) has been used as a probe for the structure of DNA. (Phen is a bidentate ligand.) a. What type of isomerism is found in \(\operatorname{Ru}(\text { phen })_{3}^{2+} ?\) b. \(\operatorname{Ru}(\text { phen })_{3}^{2+}\) is diamagnetic (as are all complex ions of \(\mathrm{Ru}^{2+} \)). Draw the crystal field diagram for the \(d\) orbitals in this complex ion.

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. \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{NO}_{2}\right)_{3}\right]\)

The equilibrium constant \(K_{\mathrm{a}}\) for the reaction $$\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad \mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})^{2+}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)$$ is \(1.0 \times 10^{-5}\) a. Calculate the pH of a 0.10\(M\) solution of \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}\) b. Will a 1.0\(M\) solution of cobalt(Il) nitrate have a higher or lower pH than a 1.0\(M\) solution of cobalt (III) nitrate? Explain. c. \(\mathrm{Co}^{3+}\) complex ions are generally low-spin cases, whereas \(\mathrm{Co}^{2+}\) complex ions are generally high-spin cases. Explain. If this is the situation, how many unpaired electrons are present in \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) and \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+} ?\)

Ammonia and potassium iodide solutions are added to an aqueous solution of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} .\) A solid is isolated (compound A), and the following data are collected: i. When 0.105 g of compound A was strongly heated in excess \(\mathrm{O}_{2}, 0.0203 \mathrm{g} \mathrm{CrO}_{3}\) was formed. ii. In a second experiment it took 32.93 \(\mathrm{mL}\) of 0.100\(M \mathrm{HCl}\) to titrate completely the \(\mathrm{NH}_{3}\) present in \(0.341 \mathrm{g} \mathrm{com}-\) pound A. iii. Compound A was found to contain 73.53\(\%\) iodine by mass. iv. The freezing point of water was lowered by \(0.64^{\circ} \mathrm{C}\) when 0.601 \(\mathrm{g}\) compound A was dissolved in 10.00 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O}\left(K_{\mathrm{f}}=\right.\) \(1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol} )\) What is the formula of the compound? What is the structure of the complex ion present? (Hints: \(\mathrm{Cr}^{3+}\) is expected to be six-coordinate, with \(\mathrm{NH}_{3}\) and possibly I- - as ligands. The I- ions will be the counterions if needed.)
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