Problem 47 - Chemistry: The Central Science

Question

$$ \begin{aligned} &\text { Write the electron configurations for the following ions: }\\\ &\text { (a) } \mathrm{In}^{3+} \text { , (b) } \mathrm{Sb}^{3+} \text { , (c) } \mathrm{Te}^{2-} \text { , (d) } \mathrm{Te}^{6+} \text { , (e) } \mathrm{Hg}^{2+} \text { , (f) } \mathrm{Rh}^{3+} \text { . } \end{aligned} $$

Step-by-Step Solution

Answer

The electron configurations for the given ions are: (a) In³⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}\) (b) Sb³⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}\) (c) Te²⁻: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\) (d) Te⁶⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2\) (e) Hg²⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 4f^{14} 5d^8\) (f) Rh³⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^7\)

Step 1: Identify the atomic numbers of the neutral elements

Consult the periodic table and identify the atomic numbers (number of protons and electrons in a neutral atom) of the elements in question: (a) In: 49 (b) Sb: 51 (c) Te: 52 (d) Te: 52 (e) Hg: 80 (f) Rh: 45

Step 2: Write the electron configuration of the neutral elements

Using the periodic table, find how many electrons in each subshell for the neutral elements: (a) In: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^1\) (b) Sb: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^3\) (c) Te: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^4\) (d) Te: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^4\) (e) Hg: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10}\) (f) Rh: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^1 4d^{8}\)

Step 3: Determine the electron configuration for the ions

Consider the addition or removal of electrons based on the ion's charge: (a) In³⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}\) (Remove 3 electrons from 5s² and 5p¹) (b) Sb³⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}\) (Remove 3 electrons from 5s² and 5p³) (c) Te²⁻: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\) (Add 2 electrons to 5p^4) (d) Te⁶⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2\) (Remove 6 electrons from 5p^4 and 4d¹⁰) (e) Hg²⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 4f^{14} 5d^8\) (Remove 2 electrons from 6s²) (f) Rh³⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^7\) (Remove 3 electrons from 5s¹ and 4d⁸) The electron configurations for the given ions are: (a) In³⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}\) (b) Sb³⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}\) (c) Te²⁻: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\) (d) Te⁶⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2\) (e) Hg²⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 4f^{14} 5d^8\) (f) Rh³⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^7\)

Key Concepts

Atomic Numbers

Understanding the atomic number of an element is crucial when considering electron configurations, particularly of ions. The atomic number corresponds to the number of protons in the nucleus of an atom and, by extension, the number of electrons in a neutrally charged atom. This simple number determines an element's position in the periodic table and is foundational for predicting how it will bond and interact with other elements.

For example, indium (In) has an atomic number of 49, which means a neutral indium atom has 49 protons and, to balance the charge, 49 electrons. When indium forms an In³⁺ ion, it loses three electrons, leading to a modified electron configuration. Atomic numbers serve as a starting point to determine the electron configuration, which changes accordingly when atoms gain or lose electrons to form ions.

Periodic Table

The periodic table is an organized chart of elements arranged by increasing atomic number, and it holds vital information for determining electron configurations. Elements in the same column (group) have similar electron configurations in their outer shells, influencing their chemical properties. As we move across a period (row), each element has one more proton and one more electron than the previous one, with electrons filling into subshells in a predictable pattern.

The periodic table helps us visualize these patterns, helping students understand that the electron configuration of an element is built by adding electrons to the previously filled subshells. This organization is particularly important when it comes to ions; after writing down the electron configuration of a neutral atom as a reference, students can use the charge of the ion to determine how many electrons to add or remove to get the final electron configuration.

Subshell Electron Distribution

Each atom has electrons organized in shells and subshells, following rules of quantum mechanics. The most common notation for electron configurations involves a sequence of numbers and letters (e.g., 1s², 2s², 2p⁶) where the number indicates the shell (or energy level), the letter indicates the subshell (s, p, d, f), and the superscript indicates the number of electrons within that subshell.

Subshells fill according to the Aufbau principle, with electrons occupying the lowest available energy level first. In a neutral atom, subshells fill up in a 'ground state' configuration. When creating ions, it is typically the outermost (highest-energy-level) electrons that are removed or added. Thus, understanding how subshells are organized and filled is pivotal to predicting how an ion's electron configuration might look. For instance, removing three electrons from a neutral Rh atom, which initially has a single electron in the 5s subshell and eight in the 4d subshell, results in a Rh³⁺ ion with the highest energy electrons removed, leaving seven electrons in the 4d subshell.