Problem 132
Question
Match the following (a) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\) (p) Geometrical isomers \(\mathrm{Cl}_{2}\) (b) \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) (q) Paramagnetic (c) \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}\) (r) Diamagnetic (d) \(\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}\) (s) Metal ion with \(+2\) oxidation state (t) \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) hybridization of central metal atom
Step-by-Step Solution
Verified Answer
(a) matches with (p), (b) with (r), (c) with (q), (d) with (s) and (t).
1Step 1: Analyzing compound (a)
Compound (a) is \([\mathrm{Co}(\mathrm{NH}_{3})_{4}(\mathrm{H}_{2}\mathrm{O})_{2}]\). Cobalt is typically in a +3 oxidation state here, resulting in a complex that can exhibit optical or geometrical isomers. However, since optical isomerism isn’t listed, we match it with geometrical isomers. Thus, (a) matches with (p).
2Step 2: Analyzing compound (b)
Compound (b) is \([\mathrm{Pt}(\mathrm{NH}_{3})_{2} \mathrm{Cl}_{2}]\). This is a classic example of a square planar complex known for exhibiting geometrical isomerism but since that pair is taken, we need to look further. Platinum in this complex is in a low oxidation state and the complex is known to be diamagnetic due to complete pairing of electrons in \(\mathrm{d}_{x^2-y^2}\) orbital. This matches with option (r).
3Step 3: Analyzing compound (c)
Compound (c) is \([\mathrm{Co}(\mathrm{H}_{2} \mathrm{O})_{5}\mathrm{Cl}] \mathrm{Cl}\). The central cobalt ion in this case is typically in a +2 or +3 state, which can exhibit paramagnetism due to unpaired electrons. Hence, (c) matches with (q), paramagnetic.
4Step 4: Analyzing compound (d)
Compound (d) is \([\mathrm{Ni}(\mathrm{H}_{2} \mathrm{O})_{6}] \mathrm{Cl}_{2}\). The nickel ion here is in a +2 oxidation state, causing it to match with option (s). Moreover, this complex is known for having \(\mathrm{sp}^{3} \mathrm{d}^{2}\) hybridization. Therefore, it matches with both (s) and (t).
Key Concepts
Geometrical isomerismParamagnetismHybridizationOxidation states
Geometrical isomerism
Geometrical isomerism is a fascinating concept in chemistry which arises due to the differing spatial arrangements of atoms in molecules. It is most often seen in coordination compounds, particularly those with square planar or octahedral geometries.
In square planar complexes, such as \([\mathrm{Pt}(\mathrm{NH}_3)_2 \mathrm{Cl}_2]\), geometrical isomerism occurs due to the different positions substituents can occupy around the central metal atom. Imagine drawing a square, where each corner can be occupied by different atoms or ligands. Depending on where you place these atoms, you can have different isomers called "cis" and "trans" isomers. In a "cis" isomer, similar groups are adjacent, while in the "trans" isomer, they are opposite each other.
This kind of isomerism is crucial because each geometric isomer can have vastly different chemical properties and reactivities. It is noteworthy in drug development where one isomer might be therapeutically effective, and the other not.
In square planar complexes, such as \([\mathrm{Pt}(\mathrm{NH}_3)_2 \mathrm{Cl}_2]\), geometrical isomerism occurs due to the different positions substituents can occupy around the central metal atom. Imagine drawing a square, where each corner can be occupied by different atoms or ligands. Depending on where you place these atoms, you can have different isomers called "cis" and "trans" isomers. In a "cis" isomer, similar groups are adjacent, while in the "trans" isomer, they are opposite each other.
This kind of isomerism is crucial because each geometric isomer can have vastly different chemical properties and reactivities. It is noteworthy in drug development where one isomer might be therapeutically effective, and the other not.
Paramagnetism
Paramagnetism is a property that certain materials or compounds exhibit, which allows them to be attracted by an external magnetic field. This property results from the presence of one or more unpaired electrons in the atom, molecule or compound.
In the context of coordination compounds, such as \([\mathrm{Co}(\mathrm{H}_2 \mathrm{O})_5 \mathrm{Cl}] \mathrm{Cl}\), paramagnetism reveals the nature of the electronic configuration. The presence of unpaired electrons leads to a higher magnetic moment. The more unpaired electrons there are, the stronger the magnetic pull on the compound.
In the context of coordination compounds, such as \([\mathrm{Co}(\mathrm{H}_2 \mathrm{O})_5 \mathrm{Cl}] \mathrm{Cl}\), paramagnetism reveals the nature of the electronic configuration. The presence of unpaired electrons leads to a higher magnetic moment. The more unpaired electrons there are, the stronger the magnetic pull on the compound.
- This property allows scientists and chemists to determine the electronic structure and oxidation state of the metal ions within a complex essentially through magnetic studies.
- The presence or absence of unpaired electrons additionally helps in distinguishing between high spin and low spin complexes.
Hybridization
Hybridization is an essential concept where atomic orbitals mix to form new hybrid orbitals. This process is crucial in explaining the geometry and bonding properties of a coordination complex.
In the case of coordination compounds such as \([\mathrm{Ni}(\mathrm{H}_2 \mathrm{O})_6] \mathrm{Cl}_2\), which exhibits \(\mathrm{sp}^3 \mathrm{d}^2\) hybridization, six hybrid orbitals form to allow the accommodation of ligands. The process allows metal ions to expand their coordination number.
In the case of coordination compounds such as \([\mathrm{Ni}(\mathrm{H}_2 \mathrm{O})_6] \mathrm{Cl}_2\), which exhibits \(\mathrm{sp}^3 \mathrm{d}^2\) hybridization, six hybrid orbitals form to allow the accommodation of ligands. The process allows metal ions to expand their coordination number.
- Different hybridizations, such as \(\mathrm{sp}^3\) or \(\mathrm{d}^2 \mathrm{sp}^3\), lead to different molecular geometries.
- In some cases, it can help predict the magnetic properties of the complex, as certain hybridizations accommodate electron pairings differently.
Oxidation states
Oxidation states play a crucial role in defining the reactivity and stability of coordination complexes. The oxidation state of a metal within a complex influences its color, magnetic behavior, and ligand bonding properties.
Considering \([\mathrm{Ni}(\mathrm{H}_2 \mathrm{O})_6] \mathrm{Cl}_2\), the nickel ion exhibits an oxidation state of \(+2\). This oxidation state is crucial because it determines the number of unpaired electrons and thus the magnetic and optical property of the complex.
Considering \([\mathrm{Ni}(\mathrm{H}_2 \mathrm{O})_6] \mathrm{Cl}_2\), the nickel ion exhibits an oxidation state of \(+2\). This oxidation state is crucial because it determines the number of unpaired electrons and thus the magnetic and optical property of the complex.
- These oxidation states help in balancing redox reactions and serve as an essential tool in identifying the level of electron flow within reactions.
- Furthermore, the oxidation states can guide the feasibility and type of ligand binding that can occur with the metal ion.
Other exercises in this chapter
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