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The correct order of magnetic moments (spin only values in B.M.) among the following is (Atomic number of \(\mathrm{Mn}=25, \mathrm{Fe}=26, \mathrm{Co}=27\) ) (a) \(\left[\mathrm{MnCl}_{4}\right]^{2-}>\left[\mathrm{CoCl}_{4}\right]^{2-}>\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4}\) (b) \(\left[\mathrm{MnCl}_{4}\right]^{2-}>\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4}>\left[\mathrm{CoCl}_{4}\right]^{2-}\) (c) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4}>\left[\mathrm{MnCl}_{4}\right]^{2->}\left[\mathrm{CoCl}_{4}\right]^{2-}\) (d) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4}>\left[\mathrm{CoCl}_{4}\right]^{2-}>\left[\mathrm{MnCl}_{4}\right]^{2-}\)

Short Answer

Expert verified
Option (a) is the correct order of magnetic moments.

Step by step solution

01

Determine Electronic Configurations

List the electronic configurations of \( \text{Mn} \), \( \text{Fe} \), and \( \text{Co} \) when they form the respective complexes. For \( \left[\text{MnCl}_4\right]^{2-} \), manganese is in the +2 oxidation state (d^5), \( \left[\text{CoCl}_4\right]^{2-} \), cobalt is in the +2 oxidation state (d^7) and \( \left[\text{Fe(CN)}_6\right]^{4-} \), iron is in the +2 oxidation state (d^6). Since \( \text{CN}^- \) is a strong field ligand, it causes pairing.
02

Calculate Spin-Only Magnetic Moment

Apply the formula for spin-only magnetic moment: \( \mu = \sqrt{n(n+2)} \) where \( n \) is the number of unpaired electrons. For \( \left[\text{MnCl}_4\right]^{2-} \) (with 5 unpaired electrons): \( \mu_{\text{Mn}} = \sqrt{5(5+2)} = \sqrt{35} = 5.92 \text{ B.M.} \). For \( \left[\text{CoCl}_4\right]^{2-} \) (with 3 unpaired electrons): \( \mu_{\text{Co}} = \sqrt{3(3+2)} = \sqrt{15} = 3.87 \text{ B.M.} \). For \( \left[\text{Fe(CN)}_6\right]^{4-} \) (with 0 unpaired electrons, due to pairing): \( \mu_{\text{Fe}} = 0 \text{ B.M.} \).
03

Order the Magnetic Moments

Based on the calculated magnetic moments, order them from highest to lowest: \( \mu_{\text{Mn}} = 5.92 \text{ B.M.} > \mu_{\text{Co}} = 3.87 \text{ B.M.} > \mu_{\text{Fe}} = 0 \text{ B.M.} \).
04

Match the Correct Option

Option (a) shows the correct order according to the calculated magnetic moments: \( \left[\text{MnCl}_4\right]^{2-} > \left[\text{CoCl}_4\right]^{2-} > \left[\text{Fe(CN)}_6\right]^{4-} \).

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Key Concepts

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

Electronic Configurations
Electronic configurations are fundamental to understanding how elements behave in different chemical environments. They describe the arrangement of electrons around an atom's nucleus and are crucial for predicting chemical properties. When forming complexes, transition metals like manganese (Mn), iron (Fe), and cobalt (Co) can change their electronic configurations based on their oxidation states. In a +2 oxidation state, these elements lose two electrons from their 3d and 4s orbitals.
  • For \( ext{Mn}^{2+}\), the configuration is \(3d^5\).
  • For \( ext{Fe}^{2+}\), it is \(3d^6\).
  • For \( ext{Co}^{2+}\), you have \(3d^7\).
Knowing these configurations allows us to determine the number of unpaired electrons, which directly impacts the magnetic properties of the complexes.
Oxidation States
The oxidation state of an element in a compound provides insight into its electron count relative to a neutral atom. For transition metals, oxidation states significantly affect their magnetic and electronic properties. An element's oxidation state is determined by the charge it carries in a given compound.
In the given complexes:
  • In \([ ext{MnCl}_4]^{2-}\), manganese is in a +2 state.
  • For \([ ext{Fe(CN)}_6]^{4-}\), iron adopts a +2 state.
  • Similarly, cobalt in \([ ext{CoCl}_4]^{2-}\) is also in a +2 state.
Each metal achieves this through the removal of electrons, which directly affects the magnetic moments by altering the electronic configurations. Understanding oxidation states is crucial for predicting the behavior of transition metals in complexes.
Unpaired Electrons
Unpaired electrons are a key factor in calculating the magnetic moment of a compound because they produce a magnetic field when they spin. The number of unpaired electrons in a d-orbital influences the magnetic behavior of the entire complex.
For the complexes presented:
  • In \([ ext{MnCl}_4]^{2-}\), 5 unpaired electrons exist owing to Mn’s \(d^5\) state.
  • In \([ ext{CoCl}_4]^{2-}\), you find 3 unpaired electrons from its \(d^7\) configuration.
  • With \([ ext{Fe(CN)}_6]^{4-}\), all electrons are paired, resulting in 0 unpaired electrons.
Unpaired electrons are directly counted using electronic configurations, and thus essential for calculating the spin-only magnetic moment.
Spin-Only Magnetic Moment Formula
The spin-only magnetic moment formula is used to calculate the magnetic properties of a complex based solely on unpaired electron spin, without considering orbital contribution. The formula is given by \( \mu = \sqrt{n(n+2)} \), where \( n \) is the number of unpaired electrons.
This is crucial for transition metal complexes, which often have significant spin contributions to their magnetic moments.
  • For \([ ext{MnCl}_4]^{2-}\) with 5 unpaired electrons, \( \mu = \sqrt{5(5+2)} = 5.92 ext{ B.M.} \).
  • In \([ ext{CoCl}_4]^{2-}\) with 3 unpaired electrons, \( \mu = \sqrt{3(3+2)} = 3.87 ext{ B.M.} \).
  • With no unpaired electrons in \([ ext{Fe(CN)}_6]^{4-}\), \( \mu = 0 ext{ B.M.} \).
By using this formula, we can effectively rank the complexes based on their magnetic moments.
Strong Field Ligands
Strong field ligands have a significant impact on the arrangement of electrons in transition metal complexes. Ligands such as \( ext{CN}^- \) can cause the pairing of electrons within the d-orbitals, leading to fewer unpaired electrons and consequently affecting the magnetic properties.
In the complex \([ ext{Fe(CN)}_6]^{4-}\), the strong field ligand \( ext{CN}^- \) pairs up all the electrons, resulting in no unpaired electrons.
  • This pairing effect decreases the overall magnetic moment.
  • Strong field ligands can therefore transform high-spin configurations to low-spin ones.
Understanding the role of strong field ligands is vital to predicting and comprehending the magnetic behavior of transition metal complexes in chemistry.

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Most popular questions from this chapter

Which one of the following has a square planar geometry? [2007] (a) \(\left[\mathrm{FeCl}_{4}\right]^{2}\) (b) \(\left[\mathrm{NiCl}_{4}\right]^{2}\) (c) \(\left[\mathrm{PtCl}_{4}\right]^{2-}\) (d) \(\left[\mathrm{CoCl}_{4}\right]^{2-}\)

The pair having same magnetic movement is [2016] (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) and \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (b) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{2+}\) and \(\left.\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (c) \(\left[\mathrm{CoCl}_{4}\right]^{2}\) and \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (d) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) and \(\left[\mathrm{CoCl}_{4}\right]^{2}\)

The possible number of co-ordination isomers of \(\mathrm{Pt}(\mathrm{Py})_{4} \mathrm{CuCl}_{4}\) are

For the complex \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) (a) Shape is tetrahedral (b) Hybridization state of central ion is \(\mathrm{sp}^{3}\) (c) Effective atomic number of central ion is 35 (d) Number of unpaired electrons present in the central ion is 1

$$ \begin{aligned} &\text { Match the following }\\\ &\begin{array}{ll} \hline \text { Column-I } & \text { Column-II } \\ \hline \text { (a) }\left[\mathrm{Co}\left(\mathrm{NO}_{2}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\left(\mathrm{NH}_{3}\right)_{2}\right] & \text { (p) } \text { Number of } \\ \mathrm{NO}_{3} & \text { stereoisomers } \\ & =6 \\ \text { (b) }\left[\mathrm{Ni}(\mathrm{en})_{3}\right] \mathrm{Br}_{3} & \text { (q) } \begin{array}{l} \text { Linkage } \\ \text { isomerism } \end{array} \\ \text { (c) }\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}(\mathrm{py})_{3}\right] \mathrm{Br}_{2} & \text { (r) } \begin{array}{l} \text { Ionization } \\ \text { isomerism } \end{array} \\ \text { (d) }\left[\mathrm{Pt}(\mathrm{en})(\mathrm{SCN})_{2}\right]\left(\mathrm{NO}_{3}\right)_{2} & \text { (s) } \begin{array}{l} \text { Optical } \\ \text { isomerism } \end{array} \\ & \text { (t) } \text { Geometrical } \\ & \text { isomerism } \\ \hline \end{array} \end{aligned} $$

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