For each of the following ground-state ions, predict the type of orbital \((1 \mathrm{~s}, 2 \mathrm{p}, 3 \mathrm{~d}, 4 \mathrm{f}\), etc.) from which an electron will need to be removed to form the ions of one greater positive charge: (a) \(\mathrm{Ti}^{3+} ;\) (b) \(\mathrm{In}^{+}\); (c) \(\mathrm{Te}^{2-}\); (d) \(\mathrm{Ag}^{+}\).

Short Answer

Expert verified
The electrons will be removed from the following orbitals to form ions of one greater positive charge: (a) Ti^3+ from 3d, (b) In^+ from 5s, (c) Te^2- from 5p, (d) Ag^+ from 4d.

Step by step solution

01

Identify the Electron Configuration

To determine which orbital electron will be removed from an ion to form the ion of one greater positive charge, you first need to know the ground-state electron configuration of the neutral atom. Then, consider the ion's existing charge to deduce its electron configuration.
02

Consider the Ions' Charges

Based on the charge of the ion, remove electrons starting from the highest energy level orbital first. This typically corresponds to the electron added last according to the Aufbau principle.
03

Find Electron Removed for Ti^3+

For Ti which has an atomic number of 22, the electron configuration of the neutral atom is [Ar] 4s^2 3d^2. Ti^3+ has three fewer electrons, so the configuration becomes [Ar] 3d^1. Removing an electron to form Ti^4+ will take it from the 3d orbital, since this is its highest energy electron.
04

Find Electron Removed for In^+

In (Indium) has an atomic number of 49; the electron configuration of the neutral atom is [Kr] 5s^2 4d^10 5p^1. In^+ means one electron is removed, which would come from the 5p orbital, resulting in [Kr] 5s^2 4d^10. To move to In^2+, another electron would need to be removed from the 5s orbital, as it is now the highest energy level with electrons present.
05

Find Electron Removed for Te^2-

Te (Tellurium) has an atomic number of 52, and its electron configuration is [Kr] 5s^2 4d^10 5p^4. Te^2- has gained two electrons in the 5p orbital, making the configuration [Kr] 5s^2 4d^10 5p^6. To become Te^1-, an electron would need to be removed from the 5p orbital since it's the outermost filled orbital.
06

Find Electron Removed for Ag^+

Ag (Silver) has an atomic number of 47, the electron configuration is [Kr] 5s^1 4d^10. Losing one electron to become Ag^+ would remove the electron in the 5s orbital. If Ag^+ loses another electron to become Ag^2+, the electron would be removed from the 4d orbital, as it is now the outermost orbital with electrons.

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Key Concepts

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

Electronic Configuration
Understanding an atom's electronic configuration is integral to grasping atomic and chemical behaviors. With electron configuration, we map out the positions of electrons in an atom's electron shells and subshells according to their energy levels. The electrons are distributed across various orbitals such as s, p, d, and f, with each able to hold a specific number of electrons: s can hold 2, p can hold 6, d can hold 10, and f can hold 14.

When we talk about the electronic configuration of ions, it involves adding or subtracting electrons from the neutral atom's electronic configuration depending on whether the ion is positively or negatively charged. Removing electrons to form a more positively charged ion typically involves taking them from the outermost, highest energy level orbital, which reflects the order in which they were added.
Aufbau Principle
The Aufbau Principle is a fundamental concept in chemistry that guides the filling of electrons into an atom's orbitals. The word 'Aufbau' is German for 'building up', and this principle states that electron orbitals are filled with electrons from the lowest energy level to the highest. It helps us predict the electronic configuration of an atom by following a specific sequence. Electrons are first placed in the lowest energy orbital available, which is typically the 1s orbital, before moving on to higher energy levels like 2s, 2p, 3s, and so on.

The removal of an electron to form a higher-charged ion will usually come from the highest energy orbital that contains electrons, effectively reversing the Aufbau process. This is because these electrons are further from the nucleus, experience less electrostatic attraction, and require less energy to remove.
Ground-state Ions
Ground-state ions are ions in their lowest energy state, and they have a configuration that reflects the loss or gain of electrons when compared to the neutral atom. Positive ions, or cations, are formed by the loss of one or more electrons, leading to a deficiency in the electron count. Conversely, negative ions, or anions, are formed by the addition of electrons, resulting in a surplus.

In exercises like the one described, predicting the type of orbital for electron removal to form an ion of one greater positive charge requires understanding both the electronic configuration of the starting ion and the relative energies of the occupied orbitals. The orbital from which the electron is removed corresponds to the highest occupied energy level as per the ion's ground-state configuration.
Valence Electrons
Valence electrons are the electrons located in an atom's outermost shells. They play a key role in chemical bonding and reactions, as they are the ones involved in forming bonds with other atoms. In the context of ions, valence electrons can be either lost or gained, leading to the formation of positively or negatively charged ions, respectively.

For instance, to predict the removal of an electron for Titanium (Ti) to form a Ti^(4+) ion, we start by considering Ti^(3+) ion's valence electrons. Since Ti^(3+) would have lost electrons from its valence shell, the remaining electron in the 3d subshell is the valence electron, and hence its removal would result in Ti^(4+). Understanding the concept of valence electrons is crucial for predicting the formation and reactivity of ions.

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Most popular questions from this chapter

Place the following elements in order of increasing electronegativity: oxygen, carbon, nitrogen, fluorine, silicon, phosphorus, sulfur.

Using only Lewis structures that obey the octet rule, draw the Lewis structures and determine the formal charge on each atom in (a) \(\mathrm{CH}_{3}^{-}\); (b) \(\mathrm{BrO}_{2}\); (c) \(\mathrm{PO}_{3}^{3}\).

The perchlorate ion, \(\mathrm{ClO}_{4}^{-}\), is described by resonance structures. (a) Draw the Lewis structures that contribute to the resonance hyhrid and identify the most plawsible Lewis structures by using formal charge arguments. (b) The average length of a single \(\mathrm{Cl}-\mathrm{O}\) bond is 172 pm and that of a double \(\mathrm{Cl}=\mathrm{O}\) bond can be estimated at \(140 \mathrm{pm}\). The Cl-O bond length in the perchlorate ion is found experimentally to be \(144 \mathrm{pm}\) for all four bonds. Identify the most plausible Lewis structures of the perchlorate ion from these experimental data. (c) What is the oxidation mumber of chlorine in the perchlorate ion? Identify the most plausible Lewis structure by using the coxidation number, assuming that lone pairs belong to the atom to which they are attached but that all electrons shared in a bond helong to the atom of the more negative element. (d) Are these three approaches consistent? Explain why or why not.

Write the most likely charge for the ions formed by each of the following elements: (a) S; (b) Te; (c) Rb; (d) Ga; (c) Cd.

Arrange the anions \(\mathrm{Cl}^{-}, \mathrm{Br}^{-}, \mathrm{N}^{3-}\), and \(\mathrm{O}^{2-}\) in order of increasing polarizability and give reasons for your decisions.

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