Problem 128 - A Complete Resource Book in Chemistry for JEE Main

Question

$$ \begin{aligned} &\begin{array}{ll} \text { Match the following } \\ \hline \text { Column-I } & \text { Column-II } \\ \hline \begin{array}{ll} \text { (a) }\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+} & \text { (p) } \mathrm{d}^{2} \mathrm{sp}^{3} \\ \text { (b) }\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{3-} & \text { (q) } \mathrm{sp}^{3} \\ \text { (c) }\left[\mathrm{Co}\left(\mathrm{NO}_{2}\right)_{6}\right]^{3-} & \text { (r) Number of unpaired } \\ &\text { electrons is zero } \end{array} \\ \text { (d) }\left[\mathrm{FeCl}_{4}\right]^{-} \text {(s) Paramagnetic } \\ & \text { (t) Diamangetic } \\ \hline \end{array} \end{aligned} $$

Step-by-Step Solution

Answer

(a) matches (q) and (t); (b) matches (t) and inferred (dsp^2); (c) matches (p) and (r); (d) matches (s) and inferred sp^3.

Step 1: Identify the Hybridization of [Zn(NH3)4]^{2+}

The zinc ion in \([ ext{Zn(NH}_3)_4]^{2+}\) has a +2 oxidation state. Zinc typically forms a d^10 configuration in this state. With four ammonia molecules acting as ligands, and ammonia being a neutral ligand, the complex undergoes \( ext{sp}^3\) hybridization and is diamagnetic due to no unpaired electrons.

Step 2: Determine the Hybridization of [Cu(CN)4]^{3-}

For \([ ext{Cu(CN)}_4]^{3-}\), copper is in the +1 oxidation state, making its electronic configuration d^10. CN^- is a strong field ligand, causing a \( ext{dsp}^2\) hybridization in a square planar arrangement. This complex is also diamagnetic because of no unpaired electrons.

Step 3: Analyze [Co(NO2)6]^{3-}

In \([ ext{Co(NO}_2)_6]^{3-}\), cobalt is in the +3 oxidation state, leading to a d^6 configuration. NO_2^- is a strong field ligand, which results in low spin and pairing of electrons, leading to no unpaired electrons. It is thus diamagnetic, with an \( ext{d}^2 ext{sp}^3\) hybridization.

Step 4: Evaluate [FeCl4]^{-}

In \([ ext{FeCl}_4]^{-}\), iron is in the +3 oxidation state, leaving a d^5 electronic configuration. Chloride is a weak field ligand, which does not cause pairing of electrons, resulting in unpaired electrons and thus the complex is paramagnetic with an \( ext{sp}^3\) hybridization.

Step 5: Match the Above Analyses with Column-II

- \([ ext{Zn(NH}_3)_4]^{2+}\) matches with sp^3 (q) and diamagnetic (t).- \([ ext{Cu(CN)}_4]^{3-}\) matches with square planar, \( ext{dsp}^2\) hybridization (not explicitly mentioned but inferable due to context) and diamagnetic.- \([ ext{Co(NO}_2)_6]^{3-}\) matches with \(d^2sp^3\) (p) and the number of unpaired electrons is zero (r).- \([ ext{FeCl}_4]^{-}\) matches with sp^3 hybridization (not explicitly listed in Column-II but corresponds logically due to its tetrahedral structure) and paramagnetic (s).

Key Concepts

Hybridization

Understanding hybridization is crucial in coordination chemistry, as it explains how atomic orbitals mix to form new hybrid orbitals. This mixing is influenced by the geometry of the complex and the type of ligands present.

Here's a simple breakdown:

• **sp³ Hybridization**: This involves one s orbital and three p orbitals mixing. It's common in tetrahedral complexes like \( \text{[Zn(NH}_3)_4]^{2+} \), where the central metal ion coordinates with four ligands in a tetrahedral shape.
• **dsp² Hybridization**: Involves one d, one s, and two p orbitals. This happens in square planar complexes, such as \( \text{[Cu(CN)}_4]^{3-} \), where a strong-field ligand like CN⁻ induces this specific hybridization.
• **d²sp³ Hybridization**: This results from two d, one s, and three p orbitals combining. It's typical for octahedral complexes, such as \( \text{[Co(NO}_2)_6]^{3-} \).

Hybridization shapes the properties of complexes, influencing bond angles and the overall stability of the coordination complexes.

Paramagnetic

A material or complex is termed paramagnetic if it has one or more unpaired electrons. These unpaired electrons respond to magnetic fields, which can be detected experimentally.

An excellent example from the given complexes is \( \text{[FeCl}_4]^{-} \). Iron in a +3 oxidation state possesses five d electrons (d⁵ configuration).

Since chloride is a weak field ligand, it doesn't cause electron pairing, leading to unpaired electrons:

• This results in a paramagnetic nature, where the presence of unpaired electrons gives rise to a magnetic moment.
• Paramagnetic complexes are attracted to external magnetic fields, a behavior directly opposed to diamagnetic substances.

Understanding the paramagnetic nature helps predict magnetic properties and electron behaviors in coordination complexes.

Diamagnetic

Diamagnetism refers to complexes with all electrons paired. Such compounds do not have any net magnetic moment and are actually slightly repelled by a magnetic field.

For instance, both \( \text{[Zn(NH}_3)_4]^{2+} \) and \( \text{[Cu(CN)}_4]^{3-} \) show diamagnetic properties.

• Zinc's d¹⁰ configuration means all its electrons are paired naturally.
• In the copper complex, CN⁻, being a strong field ligand, pairs all d electrons, eliminating unpaired spins.

Hence, the lack of unpaired electrons in these diamagnetic substances leads to slight repulsion in magnetic fields. This understanding is useful in distinguishing substances based on magnetic susceptibility tests.

Electronic Configuration

Electronic configuration is the distribution of electrons in an atom or molecule's orbitals and is critical in understanding the chemical behavior of coordination complexes.

Let's look at some examples:

• **Zinc in \( \text{[Zn(NH}_3)_4]^{2+} \)**: The electronic configuration is d¹⁰, indicating a fully filled d orbital with no unpaired electrons.
• **Iron in \( \text{[FeCl}_4]^{-} \)**: Here, the electronic configuration is d⁵, leading to five unpaired electrons due to the weak field nature of chloride.
• **Cobalt in \( \text{[Co(NO}_2)_6]^{3-} \)**: Displays a d⁶ configuration, with NO₂⁻ acting as a strong field ligand causing all six d electrons to pair up.
• **Copper in \( \text{[Cu(CN)}_4]^{3-} \)**: Generally holds a d¹⁰ configuration with no unpaired electrons.

Knowing the electronic configuration aids in predicting magnetic properties and the chemical reactivity of different complexes.