/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Write out the ground-state elect... [FREE SOLUTION] | 91Ó°ÊÓ

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Write out the ground-state electron configurations of (a) \(\mathrm{Ti}^{3+}\), (b) \(\mathrm{Ru}^{24}\), (c) \(\mathrm{Au}^{3+}\), (d) \(\mathrm{Mn}^{4+}\).

Short Answer

Expert verified
The ground-state electron configurations of the given ions are: (a) \(\mathrm{Ti}^{3+}\): \([Ar]\) (b) \(\mathrm{Ru}^{2+}\): \([Kr]\ 4d^2\) (c) \(\mathrm{Au}^{3+}\): \([Xe]\ 4f^{14}\ 5d^{7}\) (d) \(\mathrm{Mn}^{4+}\): \([Ar]\ 3d^1\)

Step by step solution

01

Find atomic number and the neutral atom's electron configuration

First, identify the atomic numbers of each of the elements since they determine the total number of electrons in a neutral atom. Then, use the periodic table to find their electron configurations. (a) \(\mathrm{Ti}\ (Titanium)\): Atomic number 22 Electron configuration: \([Ar]\ 4s^2 \ 3d^2\) (b) \(\mathrm{Ru}\ (Ruthenium)\): Atomic number 44 Electron configuration: \([Kr]\ 5s^2 \ 4d^4\) (c) \(\mathrm{Au}\ (Gold)\): Atomic number 79 Electron configuration: \([Xe]\ 6s^2 \ 4f^{14}\ 5d^{10}\) (d) \(\mathrm{Mn}\ (Manganese)\): Atomic number 25 Electron configuration: \([Ar]\ 4s^2 \ 3d^5\)
02

Remove electrons for each ion

Consider the charges of each ion, remove the appropriate number of electrons. Remember that electrons are removed starting from the highest energy level (n value): (a) \(\mathrm{Ti}^{3+}\): Remove 3 electrons \([Ar]\ \cancel{4s^2} \ \cancel{3d^2}\) = \([Ar]\) (b) \(\mathrm{Ru}^{2+}\): Remove 2 electrons \([Kr]\ 5s^2 \ \cancel{4d^4}\) = \([Kr]\ 4d^2\) (c) \(\mathrm{Au}^{3+}\): Remove 3 electrons \([Xe]\ 6s^2 \ 4f^{14}\ 5d^{10}\) = \([Xe]\ 4f^{14}\ 5d^{7}\) (d) \(\mathrm{Mn}^{4+}\): Remove 4 electrons \([Ar]\ \cancel{4s^2}\ \cancel{3d^5}\) = \([Ar]\ 3d^1\)
03

Write final ground-state electron configurations

Now, write out the electron configurations for each ion: (a) \(\mathrm{Ti}^{3+}\) ground-state electron configuration: \([Ar]\) (b) \(\mathrm{Ru}^{2+}\) ground-state electron configuration: \([Kr]\ 4d^2\) (c) \(\mathrm{Au}^{3+}\) ground-state electron configuration: \([Xe]\ 4f^{14}\ 5d^{7}\) (d) \(\mathrm{Mn}^{4+}\) ground-state electron configuration: \([Ar]\ 3d^1\)

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Key Concepts

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

Atomic Number
The atomic number of an element is a fundamental concept in chemistry. It is the total number of protons present in the nucleus of an atom and is denoted by the symbol \( Z \). Every element in the periodic table is defined by its unique atomic number, which also tells us the number of electrons in a neutral atom. This is because atoms are electrically neutral, meaning the number of protons (positive charge) and electrons (negative charge) are equal.

For example, consider the element Titanium (Ti) mentioned in the problem. Its atomic number is 22. This means a neutral Titanium atom has 22 protons and 22 electrons. Knowing the atomic number allows chemists to predict how an atom will interact chemically, as well as determine the electron configuration of the element. Through the atomic number, you can quickly retrieve the standard electron configuration from the periodic table and then make adjustments based on ionization. This forms the foundation for understanding electron configurations.
Ground-state
The ground-state electron configuration of an atom is the arrangement of electrons around the nucleus of the atom at the lowest possible energy level. When electrons fill the atomic orbitals, they do so in a specific order guided by the Aufbau principle, the Pauli Exclusion Principle, and Hund's Rule.

  • The Aufbau Principle dictates that electrons fill orbitals starting from the lowest energy level to higher energy levels.
  • The Pauli Exclusion Principle states that each orbital can hold a maximum of two electrons with opposite spins.
  • Hund's Rule explains that electrons will fill degenerate orbitals (orbitals of the same energy) singly first, to maximize spin multiplicity, and then pair up.

For instance, the electron configuration for Titanium in its neutral ground state is \[ [Ar]\, 4s^2 \, 3d^2 \, \]. This indicates that the first 18 electrons hold the same configuration as Argon (Ar), a noble gas referenced as a core. Following Argon, the next electrons occupy the 4s and then 3d orbitals, which are not fully filled. Understanding ground-state configurations is critical as it reflects the most stable, naturally occurring form of an element.
Ion Electron Removal
Ions are atoms or molecules that have gained or lost electrons, leading to a net charge. When atoms lose electrons, they form positively charged ions known as cations. Ion electron removal refers to the process of removing electrons, often described in connection with oxidation and reduction reactions.

To determine the electron configuration of an ion, electrons are typically removed starting from the outermost orbitals, as they are the highest in energy. It's important to remember that the order of electron removal might not strictly follow the Aufbau principle, especially for transition metals. Instead, electrons are often taken from the \( s \) orbital before the \( d \) orbital in these cases.

For example, in the exercise, Titanium (Ti) becomes \( \mathrm{Ti}^{3+} \) by losing 3 electrons. These electrons are removed from the highest energy orbitals, which follows this order: from the 4s, then the 3d orbital. The final configuration for \( \mathrm{Ti}^{3+} \) is \[ [Ar] \, \] as the 4s and 3d electrons are removed.

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Most popular questions from this chapter

The lanthanide contraction explains which of the following periodic trends? (a) The atomic radii of the transition metals first decrease and then increase when moving horizontally across each period. (b) When forming ions the transition metals lose their valence s orbitals before their valence \(d\) orbitals. (c) The radii of the period 5 transition metals (Y-Cd) are very similar to the radii of the period 6 transition metals (Lu-Hg).

Carbon monoxide is toxic because it binds more strongly to the iron in hemoglobin (Hb) than does \(\mathrm{O}_{2}\), as indicated by these approximate standard free-energy changes in blood: $$ \begin{array}{ll} \mathrm{Hb}+\mathrm{O}_{2} \longrightarrow \mathrm{HbO}_{2} & \Delta G^{\mathrm{e}}=-70 \mathrm{~kJ} \\ \mathrm{Hb}+\mathrm{CO} \longrightarrow \mathrm{HbCO} & \Delta G^{\mathrm{a}}=-80 \mathrm{~kJ} \end{array} $$ Using these data, estimate the equilibrium constant at \(298 \mathrm{~K}\) for the equilibrium $$ \mathrm{HbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HbCO}+\mathrm{O}_{2} $$

Carbon monoxide, \(\mathrm{CO}\), is an important ligand in coordination chemistry. When \(\mathrm{CO}\) is reacted with nickel metal the product is \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\), which is a toxic, pale yellow liquid. (a) What is the oxidation number for nickel in this compound? (b) Given that \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) is diamagnetic molecule with a tetrahedral geometry, what is the electron configuration of nickel in this compound? (c) Write the name for \(\left[\mathrm{Nu}(\mathrm{CO})_{4}\right]\) using the nomenclature rules for coordination compounds.

Write the formula for each of the following compounds, being sure to use brackets to indicate the coordination sphere: (a) tetraaquadibromemanganese(III) perchlorate (b) bis(bipyridyl)cadmium(II) chloride (c) potassium tetrabromo(ortho-phenanthroline)-cobaltate (III) (d) cesium diamminetetracyanochromate(III) (e) tris(ethylenediamine)rhodium(III) tris(oxalato)cobaltate(III)

When Alfred Werner was developing the field of coordination chemistry, it was argued by some that the optical activity he observed in the chiral complexes he had prepared was because of the presence of carbon atoms in the molecule. To disprove this argument, Werner synthesized a chiral complex of cobalt that had no carbon atoms in it, and he was able to resolve it into its enantiomers. Design a cobalt(III) complex that would be chiral if it could be synthesized and that contains no carbon atoms. (It may not be possible to synthesize the complex you design, but we will not worry about that for now.)

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