Problem 131 - A Complete Resource Book in Chemistry for JEE Main

Question

Match the following (a) \(\left[\mathrm{MnCl}_{6}\right]^{3-}\) (p) One unpaired electron (b) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) (q) \(\mathrm{d}^{2} \mathrm{sp}^{3}\) (c) \(\left[\mathrm{CoF}_{6}\right]^{3-}\) (r) \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) (d) \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{2-}\) (s) Four unpaired electrons (t) Three unpaired electrons

Step-by-Step Solution

Answer

(a) s, (b) q, (c) t, (d) r

Step 1: Understanding Complexes and Notations

Each complex ion is represented with a central metal atom surrounded by ligands within bracket notation. Let’s identify the oxidation state and electron configuration. \(\left[\mathrm{MnCl}_{6}\right]^{3-}\) features Mn, \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) has Fe, \(\left[\mathrm{CoF}_{6}\right]^{3-}\) includes Co, and \(\left[\mathrm{Fe}(\mathrm{H}_{2}\mathrm{O})_{6}\right]^{2-}\) also contains Fe.

Step 2: Calculation of Oxidation States

Determine the oxidation state of the metals by considering the charges provided by ligands: Cl is -1, CN is -1, F is -1, and \(\mathrm{H}_2\mathrm{O}\) is neutral. For \(\left[\mathrm{MnCl}_{6}\right]^{3-}\), Mn is \(+3\); \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\), Fe is \(+3\); \(\left[\mathrm{CoF}_{6}\right]^{3-}\), Co is \(+3\); and \(\left[\mathrm{Fe}(\mathrm{H}_{2}\mathrm{O})_{6}\right]^{2-}\), Fe is \(+2\).

Step 3: Determine Electron Configurations

Find the electron configuration based on oxidation state. Mn\(^{3+}\): \([\mathrm{Ar}]\: 3d^4\), Fe\(^{3+}\): \([\mathrm{Ar}]\: 3d^5\), Co\(^{3+}\): \([\mathrm{Ar}]\: 3d^6\), Fe\(^{2+}\): \([\mathrm{Ar}]\: 3d^6\).

Step 4: Identify Spin States

Determine the number of unpaired electrons. Mn\(^{3+}\) with \([\mathrm{Ar}]\: 3d^4\) has 4 unpaired. Fe\(^{3+}\) with \([\mathrm{Ar}]\: 3d^5\) has 5 unpaired. Co\(^{3+}\) with \([\mathrm{Ar}]\: 3d^6\) when using strong field ligands lacks seen unpaired and low spin would have 0, but in given pairs it should be high spin case due to weak field hence 3 unpaired. Fe\(^{2+}\) with \([\mathrm{Ar}]\: 3d^6\) historically would normally be low spin, depends on water's stronger field, therefore 4 unpaired is preferred if high spin is considered due to water.

Step 5: Assign Hybridization and Match

For \([\mathrm{MnCl}_6]^{3-}\), with four unpaired electrons matches (s), follows \(\text{sp}^3\text{d}^2\); initially it'd be high spin and not low due to Cl. \([\mathrm{Fe}(\mathrm{CN})_6]^{3-}\), three unpaired due CN as stronger ties \text{d}^2\text{sp}^3, is incorrect expected for others thus defaulted to none. \([\mathrm{CoF}_6]^{3-}\), 3 unpaired as gave aligns with Co indicating \(\text{sp}^3\text{d}^2\). \([\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{2-}\), default unpaired expectations cause doubts but almost general to initial default to actual high spin transition especially in thought exercise influences.

Key Concepts

Complex Ions

Complex ions are fascinating structures in coordination chemistry where a central metal ion is surrounded by molecules or ions known as ligands. These ligands form coordinate bonds with the metal, leading to the creation of a complex structure.

For example, in the complex ion

• \([MnCl_{6}]^{3-}\) has manganese (Mn) as the central metal ion surrounded by six chloride ions (Cl).
• \([Fe(CN)_{6}]^{3-}\) features iron (Fe) surrounded by six cyanide ions (CN).
• \([CoF_{6}]^{3-}\) consists of cobalt (Co) with six fluoride ions (F).
• \([Fe(H_{2}O)_{6}]^{2-}\) involves iron (Fe) coordinated with six water molecules.

These complex ions are crucial in understanding the behavior of compounds, their color, magnetic properties, and reactivity. The particular combination of a central ion and its ligands defines the entire geometry and the electronic characteristics of the complex.

Oxidation States

The concept of oxidation state is central to understanding how electrons are arranged within complex ions. It refers to the charge of the central metal atom if we assume that all ligands are removed along with their electrons.

To determine the oxidation state:

• Identify the charge contributed by the ligands. For instance, Cl, CN, and F ions each contribute a charge of -1 while water (H₂O) is neutral.
• Subtract the total charge of the ligands from the overall charge of the complex ion to find the oxidation state of the metal.

For example, in

• \([MnCl_{6}]^{3-}\), manganese has an oxidation state of +3.
• \([Fe(CN)_{6}]^{3-}\), iron has an oxidation state of +3.
• \([CoF_{6}]^{3-}\), cobalt's oxidation state is +3.
• \([Fe(H_{2}O)_{6}]^{2-}\), iron holds a +2 oxidation state.

Understanding these oxidation states helps us in predicting the reactivity and stability of the complex ions.

Electron Configuration

Electron configuration describes how electrons are distributed among the various orbitals in an atom. When dealing with complex ions, it is essential to identify the electron configuration of the metal ion after accounting for its oxidation state.

For a better grasp, consider these configurations:

• Manganese (Mn) in the oxidation state of +3 has the configuration of \([Ar] 3d^{4}\).
• Iron (Fe) at +3 oxidation is configured as \([Ar] 3d^{5}\).
• Cobalt (Co) also at +3, deviates to \([Ar] 3d^{6}\).
• Iron (Fe) once again, but at +2, remains at \([Ar] 3d^{6}\).

Recognizing the electron configuration allows us to determine factors like the number of unpaired electrons, which influence the magnetic properties of these complexes.

Hybridization

Hybridization is the concept where atomic orbitals mix to create new hybrid orbitals, playing a vital role in the geometry and bonding properties of complex ions.

In these complexes, the arrangement of ligands forces the central atom to adjust its orbitals for optimal bonding:

• For \([MnCl_{6}]^{3-}\), with four unpaired electrons, hybridization follows the \( ext{sp}^{3} ext{d}^{2}\) pattern, suggesting an octahedral arrangement.
• \([Fe(CN)_{6}]^{3-}\) shows\(d^{2}sp^{3}\), indicating a weaker need for hybridization due to CN being a strong field ligand.
• \([CoF_{6}]^{3-}\) aligns to \( ext{sp}^{3} ext{d}^{2}\), consistent for fluoride ions indicative of high spin systems.
• \([Fe(H_{2}O)_{6}]^{2-}\) also displays an octahedral geometry usually, with electron pair repulsions tailoring its magnetic nature.

Hybridization helps in predicting the spatial arrangement of the ligands around the metal ion, playing a critical role in determining the chemical and physical properties of the complex ion.