Problem 118
Question
Match the following $$ \begin{array}{ll} \hline \text { List-I } & \text { List-II } \\ \hline \text { (a) Coloured ion } & \text { (p) } \mathrm{Cu}^{+} \\ \text {(b) } \mu=1.73 \mathrm{~B} \cdot \mathrm{M} & \text { (q) } \mathrm{Cu}^{2+} \\ \text { (c) } \mathrm{d}^{10} \text { configuration } & \text { (r) } \mathrm{Fe}^{2+} \\ \text { (d) More than } 3 \text { un-paired } & \text { (s) } \mathrm{Mn}^{2+} \\\ \text { electrons } & \end{array} $$
Step-by-Step Solution
Verified Answer
(a) \( \mathrm{Cu}^{2+} \), (b) \( \mathrm{Fe}^{2+} \), (c) \( \mathrm{Cu}^{+} \), (d) \( \mathrm{Mn}^{2+} \).
1Step 1: Understanding Coloured Ions
Many transition metal ions are coloured due to d-d electron transitions. Of the options given, \( \mathrm{Cu}^{2+} \) typically displays a blue or green colour in solutions. Therefore, the correct match for \( \text{(a) Coloured ion} \) is \( \text{(q) } \mathrm{Cu}^{2+} \).
2Step 2: Calculating Magnetic Moment
The magnetic moment \( \mu \) in Bohr Magnetons (B.M.) is often calculated with the formula \( \mu = \sqrt{n(n+2)} \), where \( n \) is the number of unpaired electrons. For \( \mu = 1.73 \mathrm{~B} \cdot \mathrm{M} \), solving \( 1.73 = \sqrt{n(n + 2)} \) gives \( n = 1 \). Thus, the ion with one unpaired electron is \( \text{(r) } \mathrm{Fe}^{2+} \).
3Step 3: Identifying d^10 Configuration
A \( \mathrm{d}^{10} \) configuration means a full d-subshell, typically resulting in no unpaired electrons, making the ion diamagnetic. \( \mathrm{Cu}^{+} \) has a \( \mathrm{d}^{10} \) configuration because it originates from an atom of copper \( [\mathrm{Ar}] 3\mathrm{d}^{10} 4\mathrm{s}^{1} \) losing one electron, thus \( \text{(c) } \mathrm{d}^{10} \text{ configuration} \) matches \( \text{(p) } \mathrm{Cu}^{+} \).
4Step 4: Determining Unpaired Electrons
For more than 3 unpaired electrons, an element like Manganese in the 2+ oxidation state \( \mathrm{Mn}^{2+} \) with a configuration of \( [\mathrm{Ar}] 3\mathrm{d}^{5} \) having 5 unpaired electrons fits. Thus, \( \text{(d) More than 3 un-paired electrons} \) matches \( \text{(s) } \mathrm{Mn}^{2+} \).
Key Concepts
Coloured IonsMagnetic Momentd10 ConfigurationUnpaired Electrons
Coloured Ions
Transition metals often form ions that are vividly colored. This is due to the presence of partially filled d-orbitals. When light hits these ions, electrons can jump between different d-orbitals, absorbing certain wavelengths and giving off colors.
The specific color of these ions occurs due to what we call "d-d transitions." In simple terms, it's like electrons dancing between levels. Each transition can absorb specific light portions, and the colors we see are the wavelengths that aren't absorbed.
For instance, Cupric ions (\( \mathrm{Cu}^{2+} \)) are typically blue or green in solution. Why? Because the specific arrangement of electrons in the d-orbitals of \( \mathrm{Cu}^{2+} \) reflects these colors, while other parts of the spectrum get absorbed.
The specific color of these ions occurs due to what we call "d-d transitions." In simple terms, it's like electrons dancing between levels. Each transition can absorb specific light portions, and the colors we see are the wavelengths that aren't absorbed.
For instance, Cupric ions (\( \mathrm{Cu}^{2+} \)) are typically blue or green in solution. Why? Because the specific arrangement of electrons in the d-orbitals of \( \mathrm{Cu}^{2+} \) reflects these colors, while other parts of the spectrum get absorbed.
Magnetic Moment
The magnetic moment of an ion gives information about its magnetic properties, and it's expressed in "Bohr Magnetons" (B.M.).
The formula for calculating a magnetic moment is \( \mu = \sqrt{n(n+2)} \), where \( n \) is the number of unpaired electrons. This formula assumes a simple spin-only contribution from unpaired electrons in the d-orbitals. A detected magnetic moment of \( 1.73 \mathrm{~B} \cdot \mathrm{M} \) corresponds to having one unpaired electron.
In practical terms, if you have a magnetic moment of \( 1.73 \mathrm{~B} \cdot \mathrm{M} \), your ion is likely to be like \( \mathrm{Fe}^{2+} \). This suggests there is one lazy electron, spinning all on its own without a pair, making its magnetic presence known.
The formula for calculating a magnetic moment is \( \mu = \sqrt{n(n+2)} \), where \( n \) is the number of unpaired electrons. This formula assumes a simple spin-only contribution from unpaired electrons in the d-orbitals. A detected magnetic moment of \( 1.73 \mathrm{~B} \cdot \mathrm{M} \) corresponds to having one unpaired electron.
In practical terms, if you have a magnetic moment of \( 1.73 \mathrm{~B} \cdot \mathrm{M} \), your ion is likely to be like \( \mathrm{Fe}^{2+} \). This suggests there is one lazy electron, spinning all on its own without a pair, making its magnetic presence known.
d10 Configuration
A \( \mathrm{d}^{10} \) configuration means that all d-orbitals are completely filled. Such a configuration typically leads to a condition known as "diamagnetism," where no unpaired electrons are present. Consequently, the ion does not exhibit magnetic properties.
Taking the example of \( \mathrm{Cu}^{+} \), this ion arises when metallic copper loses one electron from its \( 4s \) sublevel. This results in the electronic configuration of \( [\mathrm{Ar}] 3\mathrm{d}^{10} \), filling all d-orbitals. Thus, the \( \mathrm{Cu}^{+} \) ion has its d-subshell full, and no colorful electronic transitions can occur.
Taking the example of \( \mathrm{Cu}^{+} \), this ion arises when metallic copper loses one electron from its \( 4s \) sublevel. This results in the electronic configuration of \( [\mathrm{Ar}] 3\mathrm{d}^{10} \), filling all d-orbitals. Thus, the \( \mathrm{Cu}^{+} \) ion has its d-subshell full, and no colorful electronic transitions can occur.
Unpaired Electrons
The presence of unpaired electrons in an atom or ion can reveal interesting properties, especially in the field of magnetism. More than three unpaired electrons imply a significant degree of paramagnetism, contributing to more robust magnetic characteristics.
Taking \( \mathrm{Mn}^{2+} \) as an example, it has an electronic configuration that includes \( [\mathrm{Ar}] 3\mathrm{d}^{5} \), housing five unpaired electrons. Unpaired electrons in the d-subshell do not just sit idly. They interact with each other and with external magnetic fields, making the ion very magnetic.
These interactions can help in several applications, from developing new materials to understanding biological systems where such ions are involved.
Taking \( \mathrm{Mn}^{2+} \) as an example, it has an electronic configuration that includes \( [\mathrm{Ar}] 3\mathrm{d}^{5} \), housing five unpaired electrons. Unpaired electrons in the d-subshell do not just sit idly. They interact with each other and with external magnetic fields, making the ion very magnetic.
These interactions can help in several applications, from developing new materials to understanding biological systems where such ions are involved.
Other exercises in this chapter
Problem 114
The d-orbitals participating in hybridization of central metal atom may be from the outermost shell or the penultimate shell. This depends on the nature of meta
View solution Problem 115
The d-orbitals participating in hybridization of central metal atom may be from the outermost shell or the penultimate shell. This depends on the nature of meta
View solution Problem 120
Match the following $$ \begin{array}{ll} \hline \text { List-I } & \text { List-II } \\ \hline \text { (a) } \mathrm{Ti}^{3+} & \text { (p) Paramagnetic } \\ \t
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Match the following $$ \begin{array}{ll} \hline \text { Column-I } & \text { Column-II } \\ \hline \text { (a) } \mathrm{Ce}^{4+} & \text { (p) Oxidizing agent
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