Question

Write the condensed ground-state electron configurations of these transition metal ions, and state which are paramagnetic: (a) \(\mathrm{V}^{3+}\) (b) \(\mathrm{Cd}^{2+}\) (c) \(\mathrm{Co}^{3+}\) (d) \(\mathrm{Ag}^{+}\)

Short answer

(a) [Ar] 3d^2 (paramagnetic), (b) [Kr] 4d^{10} (not paramagnetic), (c) [Ar] 3d^6 (paramagnetic), (d) [Kr] 4d^{10} (not paramagnetic)

Step-by-step solution

Determine the electron configuration of the neutral atoms

First, write the electron configurations for the neutral atoms: V: \[ \text{1s}^2 \text{2s}^2 \text{2p}^6 \text{3s}^2 \text{3p}^6 \text{4s}^2 \text{3d}^3 \] Cd: \[ \text{1s}^2 \text{2s}^2 \text{2p}^6 \text{3s}^2 \text{3p}^6 \text{4s}^2 \text{3d}^3 \text{4p}^6 \text{4d}^{10} \text{5s}^2 \] Co: \[ \text{1s}^2 \text{2s}^2 \text{2p}^6 \text{3s}^2 \text{3p}^6 \text{4s}^2 \text{3d}^7 \] Ag: \[ \text{1s}^2 \text{2s}^2 \text{2p}^6 \text{3s}^2 \text{3p}^6 \text{4s}^2 \text{3d}^3 \text{4p}^6 \text{4d}^{10} \text{5s}^1 \]

Remove electrons for the given ionization states

Next, remove electrons based on the charge of each ion. For \( \text{V}^{3+} \) : Remove 3 electrons from \( \text{V} \) -- first from \( 4s \) and then from the \( 3d \) subshell. i.e., \[ \text{1s}^2 \text{2s}^2 \text{2p}^6 \text{3s}^2 \text{3p}^6 \text{3d}^2 \] Condensed: [Ar] \(3d^2 \) For \( \text{Cd}^{2+} \) : Remove 2 electrons from \( \text{Cd} \) -- both from \( 5s \) subshell. i.e., \[ \text{1s}^2 \text{2s}^2 \text{2p}^6 \text{3s}^2 \text{3p}^6 \text{3d}^{10} \text{4s}^2 \text{4p}^6 \text{4d}^{10} \] Condensed: [Kr] \(4d^{10} \) For \( \text{Co}^{3+} \) : Remove 3 electrons from \( \text{Co} \) -- first from \( 4s \) and then from the \( 3d \) subshell. i.e., \[ \text{1s}^2 \text{2s}^2 \text{2p}^6 \text{3s}^2 \text{3p}^6 \text{3d}^6 \] Condensed: [Ar] \(3d^6 \) For \( \text{Ag}^{+} \) : Remove 1 electron from \( \text{Ag} \). i.e., \[ \text{1s}^2 \text{2s}^2 \text{2p}^6 \text{3s}^2 \text{3p}^6 \text{3d}^{10} \text{4s}^2 \text{4p}^6 \text{4d}^{10} \] Condensed: [Kr] \(4d^{10} \)

Determine if the ions are paramagnetic

Now, determine if the ions are paramagnetic (have unpaired electrons): \( \text{V}^{3+} \) : \[ \text{ [Ar] 3d}^2 \] has unpaired electrons. \( \text{Cd}^{2+} \): \[ \text{ [Kr] 4d}^{10} \] has no unpaired electrons, it is not paramagnetic. \( \text{Co}^{3+} \) : \[ \text{ [Ar] 3d}^6 \] has unpaired electrons. \( \text{Ag}^{+} \) : \[ \text{ [Kr] 4d}^{10} \] has no unpaired electrons, it is not paramagnetic.

Key concepts

paramagnetism

Paramagnetism occurs when an atom or ion contains one or more unpaired electrons. These unpaired electrons cause the material to be attracted to a magnetic field.

For example, in the case of transition metal ions such as \(\mathrm{V}^{3+}\) and \(\mathrm{Co}^{3+}\), they exhibit paramagnetism because they have unpaired electrons in their d orbitals. Conversely, ions like \(\mathrm{Cd}^{2+}\) and \(\mathrm{Ag}^{+}\) do not exhibit paramagnetism because all their electrons are paired.

Unpaired electrons align with the magnetic field, contributing to the magnetic properties. This alignment enhances the magnetic susceptibility of the material. This trait is essential in understanding magnetic materials and their applications.

ground-state electron configuration

The ground-state electron configuration refers to the most stable arrangement of electrons in an atom or ion.

For transition metal ions, understanding their ground-state configuration helps in predicting their chemical behavior and properties. The electron configuration is typically represented in a condensed form using noble gas notation. For example, the electron configuration for a neutral V (Vanadium) atom is \[\mathrm{[Ar] 4s^2 3d^3}\].

When forming ions, electrons are removed first from the outermost shell, usually the s-orbitals, and then from the d-orbitals if necessary. This is crucial for understanding how transition metal ions adopt stable configurations.

ionization states

Ionization states refer to the number of electrons removed from or added to an atom to form an ion.

The charge on the ion will determine its ionization state. For instance, \(\mathrm{V}^{3+}\) means three electrons have been removed from Vanadium.

Transition metals often have multiple ionization states due to their ability to lose different numbers of electrons from their s and d orbitals. This property is significant when predicting the reactivity and formation of coordination compounds. For example, Cobalt (\text{Co}) can exist in \(\mathrm{Co}^{2+}\) or \(\mathrm{Co}^{3+}\) ionization states, affecting its chemical applications.

unpaired electrons

Unpaired electrons are electrons located in an orbital alone, without a corresponding paired electron.

These electrons play a crucial role in the magnetic properties of atoms and ions. Unpaired electrons create paramagnetic materials, which are attracted to magnets.

Using transition metal ions as examples: \(\mathrm{V}^{3+}\) has two unpaired electrons in \[3d^2\]; while \(\mathrm{Cd}^{2+}\) has no unpaired electrons as its \[4d^{10}\] configuration is completely filled.

This distinction is vital for understanding why certain materials are magnetic while others are not.