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Give a valence bond description of the bonding in each of the following complexes. Include orbital diagrams for the free metal ion and the metal ion in the complex. Tell which hybrid orbitals the metal ion uses and the number of unpaired electrons in each complex. (a) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) (low-spin) (b) \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (high-spin)

Short Answer

Expert verified
[Fe(CN)6]3- has 0 unpaired electrons and uses d2sp3 hybrid orbitals; [Co(H2O)6]2+ has 3 unpaired electrons and uses sp3d2 hybrid orbitals.

Step by step solution

01

Determine the Oxidation State of the Metal

For each complex, we first need to determine the oxidation state of the metal. - In \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\), the oxidation state of Fe can be calculated as follows: Let the oxidation state of Fe be \(x\). Since cyanide (\(\mathrm{CN}^-\)) has a charge of \(-1\), we have: \[ x + 6(-1) = -3 \] Solving for \(x\), we find \(x = +3\), so Fe is in the \(+3\) oxidation state.- In \([\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}\), the oxidation state of Co can be calculated: Let the oxidation state of Co be \(y\). Water is neutral, contributing \(0\): \[ y + 0 = +2 \] Solving for \(y\), we find \(y = +2\), so Co is in the \(+2\) oxidation state.
02

Electron Configuration of the Metal Ion

Now we find the electron configuration for each metal in its oxidized state:- Fe in \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\): The electron configuration of neutral Fe is \([\mathrm{Ar}] 3d^6 4s^2\). In \(\mathrm{Fe}^{3+}\), it has lost 3 electrons, yielding \([\mathrm{Ar}] 3d^5\).- Co in \([\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}\): The electron configuration of neutral Co is \([\mathrm{Ar}] 3d^7 4s^2\). In \(\mathrm{Co}^{2+}\), it has lost 2 electrons, resulting in \([\mathrm{Ar}] 3d^7\).
03

Determine the Ligand Field and Hybridization

Considering the ligand field strength and hybridization:- For \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\) (low-spin complex), CN is a strong field ligand causing pairing in the 3d orbitals. The 3d electrons pair up, resulting in 2 electrons fully occupying 3d orbitals, leading to zero unpaired electrons. The hybridization is \(d^2sp^3\), which uses the 4d, 4s, and three 4p orbitals.- For \([\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}\) (high-spin complex), water is a weak field ligand, allowing more unpaired electrons. The electrons remain unpaired in 3d, resulting in three unpaired electrons. The hybridization is \(sp^3d^2\), using the 4s, three 4p, and two 4d orbitals.
04

Evaluate the Number of Unpaired Electrons

Summarize the information:- In \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\): With 5 electrons in 3d orbitals fully paired, there are 0 unpaired electrons.- In \([\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}\): With electrons fully occupying the 3d orbitals, there are 3 unpaired electrons due to the weak field ligands allowing for high-spin.

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Key Concepts

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

Oxidation State Determination
In coordination chemistry, determining the oxidation state of a metal in a complex is essential. It tells us how many electrons have been removed, which impacts the metal's electronic configuration and magnetic properties.
To find the oxidation state, you sum the known charges of the ligands and the charge of the complex, which allows you to solve for the metal's charge. For example, in \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\) , the cyanide ligand (\(\mathrm{CN^{-}}\)) has a \(-1\) charge. So, for the complex:\[x + 6(-1) = -3\]
where \(x\) is the oxidation state of iron (Fe). Solving, we find \(x = +3\), meaning Fe is in the \(+3\) oxidation state.
  • \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\): Iron (Fe) is in \(+3\) oxidation state.
  • \([\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}\): Water (\(\mathrm{H_2O}\)) is neutral, so cobalt state is solved by \(y+0=+2\), giving Co as \(+2\).
Electron Configuration
Once the oxidation state is known, we can determine the electron configuration of the metal ion in its oxidized form. This is crucial because it impacts the metal's ability to form complexes and its magnetic properties.
Fe has an atomic configuration of \[ [\mathrm{Ar}] 3d^6 4s^2 \]. When it's oxidized to \(\mathrm{Fe^{3+}}\), it loses three electrons, resulting in \[ [\mathrm{Ar}] 3d^5 \].
Similarly, for Co, the neutral configuration is \[ [\mathrm{Ar}] 3d^7 4s^2 \], and losing two electrons to form \(\mathrm{Co^{2+}}\) gives van \[ [\mathrm{Ar}] 3d^7 \].
  • \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\): Iron's electron configuration in \(\mathrm{Fe^{3+}}\) is \[ [\mathrm{Ar}] 3d^5 \].
  • \([\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}\): Cobalt's electron configuration in \(\mathrm{Co^{2+}}\) is \[ [\mathrm{Ar}] 3d^7 \].
Ligand Field Theory
Ligand field theory describes how ligands interact with the metal ion’s d-orbitals, influencing the electronic configuration. It explains complex formation, color, and magnetic properties.
Ligands can be strong or weak field, impacting electron distribution differently. Strong field ligands, like cyanide (\(\mathrm{CN^{-}}\)), can cause electrons to pair up in the metal’s d-orbitals, leading to low-spin complexes, while weak field ligands, like water (\(\mathrm{H_2O}\)), allow unpaired electrons, resulting in high-spin states.
  • \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\): CN is a strong field ligand, resulting in a low-spin configuration with all 3d electrons paired. Zero unpaired electrons.
  • \([\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}\): Water is a weak field ligand, leading to a high-spin configuration with unpaired 3d electrons. Three unpaired electrons.
Hybridization in Coordination Complexes
In coordination chemistry, hybridization plays a key role in determining the shape and bonding nature of complexes. Metal orbitals hybridize to accommodate ligand electron pairs, forming sets of equivalent orbitals.
The type of hybridization depends on the arrangement of the ligands and the d-electron count which is modified by ligand strength. For low-spin and high-spin states, the orbital mixing varies significantly, changing the coordination geometry.
  • \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\): Exhibits \(d^2sp^3\) hybridization, using 4d, 4s, and three 4p orbitals. This leads to an octahedral geometry.
  • \([\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}\): Uses \(sp^3d^2\) hybridization, involving 4s, three 4p, and two 4d orbitals, also resulting in an octahedral complex.
Understanding hybridization helps predict the complex's shape and reactivity, making it a key concept in studying coordination compounds.

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