Problem 46
Question
Two stable ions of first transition series which have the largest number of unpaired electrons and have the highest magnetic moment are (a) \(\mathrm{Ni}^{2+}\) and \(\mathrm{Co}^{2+}\) (b) \(\mathrm{Cr}^{3+}\) and \(\mathrm{Fe}^{2+}\) (c) \(\mathrm{Fe}^{3+}\) and \(\mathrm{Mn}^{2+}\) (d) \(\mathrm{Ti}^{4+}\) and \(\mathrm{Cr}^{3+}\)
Step-by-Step Solution
Verified Answer
(c) \\( \mathrm{Fe}^{3+} \\) and \\( \mathrm{Mn}^{2+} \\) have the largest number of unpaired electrons and highest magnetic moments.
1Step 1: Understanding Unpaired Electrons
In transition metal ions, the number of unpaired electrons determines their magnetic properties. Unpaired electrons are those that do not have a paired electron in the same orbital.
2Step 2: Rule for Calculating Unpaired Electrons
To find the unpaired electrons, first write the electron configuration of the ion. Then, count the electrons in the d-orbitals. Remember that a higher oxidation state means fewer d-electrons.
3Step 3: Counting Unpaired Electrons in \\( \mathrm{Ni}^{2+} \\) and \\( \mathrm{Co}^{2+} \\)
For \( \mathrm{Ni}^{2+} \), electron configuration is \( [\mathrm{Ar}] \, 3d^8 \). It has 2 unpaired electrons. For \( \mathrm{Co}^{2+} \), electron configuration is \( [\mathrm{Ar}] \, 3d^7 \). It has 3 unpaired electrons.
4Step 4: Counting Unpaired Electrons in \\( \mathrm{Cr}^{3+} \\) and \\( \mathrm{Fe}^{2+} \\)
For \( \mathrm{Cr}^{3+} \), electron configuration is \( [\mathrm{Ar}] \, 3d^3 \). There are 3 unpaired electrons. For \( \mathrm{Fe}^{2+} \), electron configuration is \( [\mathrm{Ar}] \, 3d^6 \), with 4 unpaired electrons.
5Step 5: Counting Unpaired Electrons in \\( \mathrm{Fe}^{3+} \\) and \\( \mathrm{Mn}^{2+} \\)
For \( \mathrm{Fe}^{3+} \), electron configuration is \( [\mathrm{Ar}] \, 3d^5 \). It has 5 unpaired electrons. For \( \mathrm{Mn}^{2+} \), electron configuration is \( [\mathrm{Ar}] \, 3d^5 \), also with 5 unpaired electrons.
6Step 6: Counting Unpaired Electrons in \\( \mathrm{Ti}^{4+} \\) and \\( \mathrm{Cr}^{3+} \\)
For \( \mathrm{Ti}^{4+} \), electron configuration is \( [\mathrm{Ar}] \), with no d-electrons and hence 0 unpaired electrons. While \( \mathrm{Cr}^{3+} \) as already calculated has 3 unpaired electrons.
7Step 7: Conclusion of Unpaired Electrons and Magnetic Moment
The magnetic moment,\(\text{Magnetic Moment} = \sqrt{n(n+2)} \text{ Bohr Magnetons}\)where \(n\) is the number of unpaired electrons. More unpaired electrons mean higher magnetic moments.
8Step 8: Selecting the Option
Option (c) \( \mathrm{Fe}^{3+} \) and \( \mathrm{Mn}^{2+} \) have 5 unpaired electrons each, which is the highest. Therefore, they have the highest magnetic moments.
Key Concepts
Unpaired ElectronsMagnetic PropertiesElectron ConfigurationFirst Transition Series
Unpaired Electrons
In transition metal ions, electrons in d-orbitals play a crucial role. These electrons may or may not pair in their orbitals. **Unpaired electrons** are simply those that stand alone without a partner in their orbital. This occurs because each orbital can hold a maximum of two electrons and follows Hund's rule of maximum multiplicity.
Knowing how to identify unpaired electrons is important because they directly relate to the ion's magnetic properties. For transition metals, these unpaired electrons contribute to magnetism. They also affect how metals can chemically interact and form compounds. To find unpaired electrons, you first need the electron configuration of the ion. While counting, be aware of electron spin and the order of orbital filling because these elements together determine which electrons are unpaired.
Knowing how to identify unpaired electrons is important because they directly relate to the ion's magnetic properties. For transition metals, these unpaired electrons contribute to magnetism. They also affect how metals can chemically interact and form compounds. To find unpaired electrons, you first need the electron configuration of the ion. While counting, be aware of electron spin and the order of orbital filling because these elements together determine which electrons are unpaired.
Magnetic Properties
Magnetic properties of transition metal ions are fascinating. They depend on the number of unpaired electrons. When you encounter a **paramagnetic** material, it means the substance has unpaired electrons. Paramagnetic materials are attracted to magnetic fields.
The opposite is **diamagnetic** substances, which have all paired electrons and are slightly repelled by magnetic fields. Transition metal ions with more unpaired electrons display stronger magnetic properties, categorized as paramagnetic. The measure of magnetism, called the magnetic moment, increases as the number of unpaired electrons rises. This moment is calculated using the formula: \[\text{Magnetic Moment} = \sqrt{n(n+2)} \text{ Bohr Magnetons}\]where \( n \) is the number of unpaired electrons. Thus, ions like \( \mathrm{Fe}^{3+} \) and \( \mathrm{Mn}^{2+} \), with five unpaired electrons, display very high magnetic moments.
The opposite is **diamagnetic** substances, which have all paired electrons and are slightly repelled by magnetic fields. Transition metal ions with more unpaired electrons display stronger magnetic properties, categorized as paramagnetic. The measure of magnetism, called the magnetic moment, increases as the number of unpaired electrons rises. This moment is calculated using the formula: \[\text{Magnetic Moment} = \sqrt{n(n+2)} \text{ Bohr Magnetons}\]where \( n \) is the number of unpaired electrons. Thus, ions like \( \mathrm{Fe}^{3+} \) and \( \mathrm{Mn}^{2+} \), with five unpaired electrons, display very high magnetic moments.
Electron Configuration
To understand the chemical behavior of transition metals, you need to start with their **electron configurations**. This configuration shows how electrons are distributed across different orbitals of an atom. For ions, it additionally reveals changes after the loss or gain of electrons, typical in transition metals.
Typically, transition metals lose electrons from the 4s orbital before the 3d, even though 4s is filled first. This can affect the number of unpaired electrons and, subsequently, the magnetic properties. For instance, an ion such as \( \mathrm{Fe}^{2+} \), with an electron configuration of \([\mathrm{Ar}] \, 3d^6\), implies that the ion has four unpaired electrons, influencing both its chemical behavior and magnetism.
Typically, transition metals lose electrons from the 4s orbital before the 3d, even though 4s is filled first. This can affect the number of unpaired electrons and, subsequently, the magnetic properties. For instance, an ion such as \( \mathrm{Fe}^{2+} \), with an electron configuration of \([\mathrm{Ar}] \, 3d^6\), implies that the ion has four unpaired electrons, influencing both its chemical behavior and magnetism.
First Transition Series
Transition metals from **the first transition series** consist of elements from Scandium (Sc) to Zinc (Zn) in the periodic table. These elements possess unique electron configurations in which the d-orbitals are being filled.
The d-orbital filling leads to a wide range of oxidation states and complex magnetic and chemical properties. Within this series, the presence or absence of unpaired electrons in d-orbitals gives these elements distinctive characteristics. This is noticeable in elements like \( \mathrm{Fe}^{3+} \) and \( \mathrm{Mn}^{2+} \), both residing in this series, with maximum numbers of unpaired electrons. These elements stand out due to their significant paramagnetic properties. Additionally, their wide range of oxidation states make them versatile in forming various compounds and mixtures.
The d-orbital filling leads to a wide range of oxidation states and complex magnetic and chemical properties. Within this series, the presence or absence of unpaired electrons in d-orbitals gives these elements distinctive characteristics. This is noticeable in elements like \( \mathrm{Fe}^{3+} \) and \( \mathrm{Mn}^{2+} \), both residing in this series, with maximum numbers of unpaired electrons. These elements stand out due to their significant paramagnetic properties. Additionally, their wide range of oxidation states make them versatile in forming various compounds and mixtures.
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Problem 45
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