Problem 145
Question
The number of unpaired electrons in the complex ion \(\left[\mathrm{CoF}_{6}\right]^{3-}\) is ____ .
Step-by-Step Solution
Verified Answer
There are four unpaired electrons in \([\text{CoF}_6]^{3-}\).
1Step 1: Determine the Oxidation State of Cobalt
In the complex ion \([\mathrm{CoF}_6]^{3-}\), fluorine has a -1 oxidation state. Let the oxidation state of cobalt be \(x\). The equation becomes: \[x + 6(-1) = -3\]. Solving this gives \(x = +3\), so cobalt is in the +3 oxidation state.
2Step 2: Determine the Electron Configuration of \( \text{Co}^{3+} \)
Cobalt in its neutral state \(\text{Co}\) has an atomic number of 27, with the electron configuration \([\text{Ar}] 3d^7 4s^2 \). In the +3 oxidation state, it loses three electrons, resulting in the configuration \([\text{Ar}] 3d^6 \).
3Step 3: Assess the Ligand Field Strength
Fluoride (\(\text{F}^-\)) is a weak field ligand, meaning it does not cause electron pairing in the 3d orbitals. This means \(\text{Co}^{3+}\) in \([\text{CoF}_6]^{3-}\) will exhibit a high-spin configuration.
4Step 4: Determine the Number of Unpaired Electrons
With a high-spin configuration, the \(3d^6\) has the distribution \(\uparrow\downarrow \uparrow \uparrow \uparrow \uparrow\). The orbital filling for high-spin \(\text{Co}^{3+}\) in \([\text{CoF}_6]^{3-}\) results in four unpaired electrons in the \(3d\) orbitals.
Key Concepts
Oxidation StateElectron ConfigurationHigh-Spin ComplexUnpaired Electrons
Oxidation State
When working with transition metal complexes, one of the first steps is to determine the oxidation state. This is crucial because it affects the electron configuration and chemical behavior. In the complex ion \([\mathrm{CoF}_6]^{3-}\), each fluorine contributes a -1 oxidation state. Since there are six fluorine atoms, their combined oxidation state is -6. The complex has an overall charge of -3. By letting the oxidation state of cobalt be \(x\), we set up the equation: \[x + 6(-1) = -3\]. Solving this, we find that \(x = +3\), which tells us that cobalt is in the +3 oxidation state. Understanding oxidation states helps in predicting how many electrons are involved in bonding and thus the resulting magnetic properties of the compound.
Electron Configuration
The electron configuration provides insights into how electrons are distributed among the orbitals of an atom. In its neutral form, cobalt has an atomic number of 27, resulting in an electron configuration of \([\text{Ar}] 3d^7 4s^2\). When cobalt forms a +3 oxidation state as in \(\text{Co}^{3+}\), it loses three electrons, typically from the outermost orbitals first. Therefore, the new configuration becomes \([\text{Ar}] 3d^6\). This change in electron distribution significantly influences the physical and chemical properties of the complex, as fewer electrons are available for bonding and other interactions.
High-Spin Complex
Some ligands, like fluoride in \([\mathrm{CoF}_6]^{3-}\), are considered weak field ligands. This means they are unable to strongly pair up electrons within the d-orbitals. Consequently, in the presence of weak field ligands, the complex tends to adopt a high-spin configuration. In high-spin complexes, the electrons occupy the orbitals singly first, following Hund's rule, before any pairing occurs. This results in the maximum number of unpaired electrons and increased magnetic properties. For \(\text{Co}^{3+}\) with configuration \([\text{Ar}] 3d^6\), the high-spin configuration distributes electrons as: \(\uparrow\downarrow \uparrow \uparrow \uparrow \uparrow\), leading to four unpaired electrons.
Unpaired Electrons
Unpaired electrons are electrons that remain alone in an orbital without another electron to pair with it. These unpaired electrons are the reason for the magnetic properties of a compound. In the \([\mathrm{CoF}_6]^{3-}\) complex, the presence of a high-spin configuration results in four unpaired electrons within the d-orbitals of cobalt. Generally, the more unpaired electrons there are, the stronger the magnetic attraction a compound will have to a magnetic field. It's these magnetic characteristics that make understanding unpaired electrons important in coordination chemistry, as they directly affect how a substance behaves in magnetic environments.
Other exercises in this chapter
Problem 141
Amongst the following, the total number of species which are diamagnetic is ______ . \(\mathrm{K}_{4}[\mathrm{Fe}(\mathrm{CN})], \mathrm{K}_{3}\left[\mathrm{Cr}
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When excess of \(\mathrm{KCN}\) is added to aqueous solution of copper sulphate a co-ordination compound \(\mathrm{K}_{\mathrm{x}}\left[\mathrm{Cu}(\mathrm{CN})
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In \(\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}\), the isomerism shown is (a) optical (b) ionization (c) geometrical (d) ligand
View solution Problem 148
A similarity between optical and geometrical isomerism is that (a) if in a compound one is present then the other will also be present (b) each gives equal numb
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