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Atomic numbers of \(\mathrm{Cr}\) and \(\mathrm{Fe}\) are respectively 25 and 26\. Which of the following is paramagnetic due to the spin of the electron? (a) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) (b) \(\left[\mathrm{Fe}(\mathrm{CO})_{5}\right]\) (c) \(\left[\mathrm{Cr}(\mathrm{CO})_{6}\right]\) (d) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4}\)

Short Answer

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(a) \([\mathrm{Cr}(\mathrm{NH}_3)_6]^{3+}\) is paramagnetic.

Step by step solution

01

Determine Electron Configuration

First, determine the electron configurations for Cr (Chromium) and Fe (Iron). Chromium (atomic number 25) has the electron configuration \[ \text{Cr}: [\text{Ar}] 3d^5 4s^1 \]Iron (atomic number 26) has the electron configuration \[ \text{Fe}: [\text{Ar}] 3d^6 4s^2 \]
02

Analyze Coordination Compounds

Review the coordination compounds to determine their electron configurations in a complex. In octahedral complexes, ligands split the 3d orbitals into two sets with different energies: \(t_{2g}\) and \(e_g\). Ligands like NH3 and CN^- are strong field ligands, while CO is very strong field and can cause low spin.
03

Examine Each Complex for Paramagnetic Property

(a) \( [\text{Cr}(\text{NH}_3)_6]^{3+} \): - \([\text{Cr}^{3+}] = [\text{Ar}] 3d^3 \), three unpaired electrons make it paramagnetic.(b) \( [\text{Fe}(\text{CO})_5] \):- Carbon monoxide is a strong field ligand; Fe in this complex has no unpaired electrons, not paramagnetic.(c) \( [\text{Cr}(\text{CO})_6] \):- Here, \(\text{Cr}^0\) state, CO ligand causes pairing of electrons, no unpaired electrons, not paramagnetic.(d) \( [\text{Fe}(\text{CN})_6]^{4-} \): - \([\text{Fe}^{2+}] = [\text{Ar}] 3d^6 \), CN^- causes all electrons to pair, not paramagnetic.
04

Identify Paramagnetic Species

Check for unpaired electrons. The complex \([\text{Cr}(\text{NH}_3)_6]^{3+}\) has unpaired electrons due to Cr's \(3d^3\) electron configuration, making it paramagnetic.

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Key Concepts

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

Electron Configuration
The electron configuration of an element describes the distribution of its electrons in different orbitals around the nucleus. Understanding electron configurations allows us to predict the behavior of atoms in different chemical contexts. For elements like Chromium (Cr) and Iron (Fe), each has a unique electron configuration based on their atomic numbers. Chromium, with an atomic number of 25, has the configuration \( [\text{Ar}] 3d^5 4s^1 \). This means that after filling the first 18 electrons up to Argon, five electrons occupy the 3d subshell and one electron in the 4s subshell.
Iron, with an atomic number of 26, has the configuration \( [\text{Ar}] 3d^6 4s^2 \). Here, six electrons are in the 3d subshell, and two electrons fill the 4s subshell post-Argon. By analyzing these configurations, we can determine where electrons are likely unpaired, which relates directly to magnetic properties.
When atoms form coordination compounds, the electron configuration may change depending on the ligands' influence.
Coordination Compounds
Coordination compounds are complex structures consisting of a central metal atom bonded to surrounding molecules or ions known as ligands. These compounds play a significant role in fields like chemistry and biochemistry due to their diverse properties and functions.
A key concept in studying coordination compounds is the geometrical arrangement, such as octahedral or tetrahedral structures, formed by the ligands around the metal center. In an octahedral complex, the ligands can split the d orbitals into two distinct energy levels, known as \( t_{2g} \) and \( e_g \).
Ligands significantly affect the metal center's electron configuration. Strong field ligands, like CN^- and NH3, can cause considerable splitting of energy levels in d orbitals, leading to differences in electron pairing. Understanding these properties helps us predict the compound's magnetic behavior and electronic properties.
Strong Field Ligands
Strong field ligands are ligands that produce a large splitting between the \( t_{2g} \) and \( e_g \) orbitals in a coordination complex. This splitting influences whether the compound will have high-spin or low-spin configurations.
In contexts where the ligand field is strong, such as with CO and CN^-, the large energy difference usually results in paired electrons, leading to low-spin states. This occurs because the energy required to pair up electrons is less than the energy gap between the \( t_{2g} \) and \( e_g \) orbitals under strong fields. This plays a crucial role in determining the magnetic properties of coordination compounds.
  • NH3, often seen as an intermediate or strong field ligand, can lead to paramagnetism in cases like \( [\text{Cr}(\text{NH}_3)_6]^{3+} \). Here, the splitting allows some unpaired electrons despite the field's relative strength.
  • Other ligands like CO cause more significant splitting, resulting in paired electrons and non-paramagnetic properties.
Unpaired Electrons
Unpaired electrons are electrons that are alone in an orbital and not paired with another electron. This is crucial for determining the magnetic properties of a substance. When electrons are unpaired, they contribute to a material's paramagnetic nature, making it attracted to magnetic fields.
For example, in \( [\text{Cr}(\text{NH}_3)_6]^{3+} \), chromium maintains a \( 3d^3 \) configuration with three unpaired electrons, rendering the complex paramagnetic. The absence of unpaired electrons results in diamagnetic behavior, where the material is slightly repelled by magnetic fields.
  • To identify unpaired electrons, examine the electron configuration of the metal center in the complex.
  • Consider the ligand field's influence on electron pairing and orbital splitting.
  • The presence or absence of unpaired electrons not only impacts magnetism but also affects the compound's color and reactivity.
Ultimately, recognizing these properties in coordination complexes is essential for predicting and understanding their chemical behavior.

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Most popular questions from this chapter

The number of chloride ions produced by the complex tetraamminechloroplatinum(IV) chloride in an aqueous solution is (a) 1 (b) 2 (c) 3 (d) 4

Identify the coordination compounds (a) \(\mathrm{FeSO}_{4} \cdot\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{Fe}(\mathrm{CN})_{2} \cdot 4 \mathrm{KCN}\) (c) \(\mathrm{KCl} \cdot \mathrm{MgCl}_{2} .6 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{CuSO}_{4} .4 \mathrm{NH}_{3}\)

For a complex \(\mathrm{MA}_{3} \mathrm{~B}_{3}\) possessing a trigonal prismatic geometry, the number of possible isomers are (a) 3 (b) 4 (c) 5 (d) 6

The complex ion which has no d electrons in the central metal atom is (Atomic number \(\mathrm{Cr}=24\), \(\mathrm{Mn}=25, \mathrm{Fe}=26, \mathrm{Co}=27)\) (a) \(\left[\mathrm{MnO}_{4}\right]\) (b) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) (c) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3}\) (d) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\)

$$ \begin{aligned} &\text { Match the following }\\\ &\begin{array}{ll} \hline \text { Column-I } & \text { Column-II } \\ \hline \text { (a) }\left[\mathrm{MnCl}_{6}\right]^{3-} & \text { (p) One unpaired electron } \\ \text { (b) }\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} & \text { (q) } \mathrm{d}^{2} \mathrm{sp}^{3} \\ \text { (c) }\left[\mathrm{CoF}_{6}\right]^{3-} & \text { (r) } \mathrm{sp}^{3} \mathrm{~d}^{2} \\ \text { (d) }\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2-} & \text { (s) Four unpaired electrons } \\ & \text { (t) Three unpaired electrons } \\ \hline \end{array} \end{aligned} $$

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