Problem 8
Question
Which of the following complex species does not involve inner orbital hybridization? (a) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) (b) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) (c) \(\left[\mathrm{CoF}_{6}\right]^{3-}\) (d) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3}\)
Step-by-Step Solution
Verified Answer
\\([\mathrm{CoF}_{6}]^{3-}\\) does not involve inner orbital hybridization.
1Step 1: Understand Inner and Outer Orbital Hybridization
Inner orbital or low-spin complexes involve the pairing of electrons in the t2g orbitals, usually resulting in d2sp3 hybridization. Meanwhile, outer orbital or high-spin complexes use outer 3d orbitals, resulting in sp3d2 hybridization. Identifying the presence of strong or weak ligands can assist in determining the type.
2Step 2: Analyze \\([\mathrm{Co}(\mathrm{NH}_{3})_{6}]^{3+}\\)
Cobalt in \([\mathrm{Co}(\mathrm{NH}_{3})_{6}]^{3+}\) is in the +3 oxidation state, with an electronic configuration of \([\mathrm{Ar}] 3d^6\). Ammonia (NH3) is a strong field ligand, causing pairing of electrons, leading to inner d2sp3 hybridization.
3Step 3: Analyze \\([\mathrm{Cr}(\mathrm{NH}_{3})_{6}]^{3+}\\)
Chromium in \([\mathrm{Cr}(\mathrm{NH}_{3})_{6}]^{3+}\) is also with +3 oxidation, having an electronic configuration of \([\mathrm{Ar}] 3d^3\). NH3 being a strong field ligand pairs electrons for inner orbital hybridization d2sp3.
4Step 4: Analyze \\([\mathrm{CoF}_{6}]^{3-}\\)
Cobalt in \([\mathrm{CoF}_{6}]^{3-}\) with +3 oxidation has \([\mathrm{Ar}] 3d^6\) configuration. Fluoride (F) is a weak field ligand causing no electron pairing, leading to outer orbital sp3d2 hybridization.
5Step 5: Analyze \\([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\\)
Iron in \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\), with +3 oxidation, has an electronic configuration of \([\mathrm{Ar}] 3d^5\). Cyanide (CN-) is a strong field ligand, resulting in electron pairing and inner d2sp3 hybridization.
6Step 6: Identify the Exception
Among the given complexes, \([\mathrm{CoF}_{6}]^{3-}\) does not involve inner orbital hybridization but rather outer orbital hybridization because of the weak field ligand fluoride.
Key Concepts
Inner Orbital HybridizationOuter Orbital HybridizationField Strength of LigandsElectronic ConfigurationHigh-spin and Low-spin Complexes
Inner Orbital Hybridization
Inner orbital hybridization refers to a specific mixing of atomic orbitals that often occurs in transition metal complexes when strong field ligands are present. In this type of hybridization, electrons are paired up in the metal's inner 3d orbitals (particularly the t2g set) before going out to hybridize. This leads to the formation of d2sp3 hybrid orbitals. Typically, this results in low-spin configurations, as electrons are tightly paired.
- Occurs with strong field ligands, like cyanide (CN-) or ammonia (NH3).
- Forms d2sp3 hybrid orbitals.
- Resulting complexes are usually low-spin.
Outer Orbital Hybridization
Outer orbital hybridization is seen in transition metal complexes when weak field ligands are around. In contrast to inner orbital hybridization, this type uses the metal's outer or higher energy d orbitals (like 4d), leading to sp3d2 hybridization. Because the electrons do not pair in inner t2g orbitals, they remain unpaired, resulting in high-spin configurations.
- Facilitated by weak field ligands, such as fluoride (F-).
- Leads to sp3d2 hybrid orbitals.
- Complexes usually end up in high-spin states.
Field Strength of Ligands
Ligands in coordination chemistry can be classified by their ability to influence the distribution of electrons in the metal atom or ion, a property known as "field strength." The field strength of a ligand determines whether a complex will exhibit inner orbital or outer orbital hybridization.
- Strong field ligands: They cause pairing of electrons (e.g., CN-, NH3).
- Weak field ligands: They do not cause electron pairing (e.g., F-).
Electronic Configuration
The electronic configuration of the central metal atom/ion in a complex provides a blueprint for determining its chemical behavior and properties. It indicates how electrons are arranged around the nucleus, commonly represented using notations based on the Aufbau principle. For transition metals, the 3d and 4s orbitals are particularly important.
) depending on the ligands around it, the electrons may undergo pairing (inner orbital formation) or remain unpaired (outer orbital formation). Identifying configurations helps predict whether the complex is paramagnetic or diamagnetic, among other characteristics.
- Influences the hybridization type within the complex.
- Determines magnetic and optical properties.
) depending on the ligands around it, the electrons may undergo pairing (inner orbital formation) or remain unpaired (outer orbital formation). Identifying configurations helps predict whether the complex is paramagnetic or diamagnetic, among other characteristics.
High-spin and Low-spin Complexes
In coordination chemistry, the concepts of high-spin and low-spin complexes arise from the interactions between the metal ion and the ligands. These interactions affect the distribution of electrons among the d-orbitals.
- High-spin complexes: Occur with weak field ligands leading to minimal electron pairing and maximum unpaired electrons.
- Low-spin complexes: Result from strong field ligands, causing electron pairing and fewer unpaired electrons.
- Magnetism: High-spin complexes are typically paramagnetic, while low-spin are diamagnetic or less paramagnetic.
- Color: The difference in d-electron arrangements affects light absorption, influencing the color of the complex.
Other exercises in this chapter
Problem 6
In the spectrochemical series, the magnitude of the crystal field splitting is maximum for which ion? (a) \(\mathrm{F}^{-}\) (b) \(\mathrm{NO}_{2}^{-}\) (c) \(\
View solution Problem 7
The correct name of the compound \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]\left(\mathrm{NO}_{3}\right)_{2}\) according to system is (a) tetraam
View solution Problem 9
The complex \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4}\) will give a white precipitate with (a) \(\mathrm{PbCl}_{2}\
View solution Problem 10
Pentaamminenitrocobalt (III) cation possesses the property of (a) physiosorption (b) chirality (c) linkage isomerism (d) versatile reducing agent
View solution