Problem 141 - A Complete Resource Book in Chemistry for JEE Main

Question

Amongst the following, the total number of species which are diamagnetic is \(\mathrm{K}_{4}[\mathrm{Fe}(\mathrm{CN})], \mathrm{K}_{3}\left[\mathrm{Cr}(\mathrm{CN})_{6}\right], \mathrm{K}_{3}[\mathrm{Co}(\mathrm{CN})]\) \(\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{CN})_{4}\right],\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}, \mathrm{K}_{2} \mathrm{TiF}_{6}\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\)

Step-by-Step Solution

Answer

Four species are diamagnetic: \([\mathrm{Fe}(\mathrm{CN})_6]^{4-}\), \([\mathrm{Co}(\mathrm{CN})_6]^{3-}\), \([\mathrm{Ni}(\mathrm{CN})_4]^{2-}\), and \([\mathrm{TiF}_6]^{2-}\).

Step 1: Understand Diamagnetism

A substance is considered diamagnetic if all of its electrons are paired. Paramagnetic substances have unpaired electrons. We will check each given species to determine if they have all paired electrons or not.

Step 2: Analyze \(\mathrm{K}_{4}[\mathrm{Fe}(\mathrm{CN})_6]\)

Fe in \(\mathrm{K}_{4}[\mathrm{Fe}(\mathrm{CN})_6]\) is in the oxidation state of +2. The electronic configuration of \(\mathrm{Fe}^{2+}\) is \([\mathrm{Ar}]3d^6\). The strong field ligand CN induces pairing of electrons in the \(3d\) subshell, making \([\mathrm{Fe}(\mathrm{CN})_6]^{4-}\) diamagnetic with all electrons paired.

Step 3: Analyze \(\mathrm{K}_{3}[\mathrm{Cr}(\mathrm{CN})_6]\)

Cr in \(\mathrm{K}_{3}[\mathrm{Cr}(\mathrm{CN})_6]\) is in the oxidation state of +3. The electronic configuration of \(\mathrm{Cr}^{3+}\) is \([\mathrm{Ar}]3d^3\). Even with the strong field ligand CN, \([\mathrm{Cr}(\mathrm{CN})_6]^{3-}\) has unpaired electrons in the \(3d\) subshell, making it paramagnetic.

Step 4: Analyze \(\mathrm{K}_{3}[\mathrm{Co}(\mathrm{CN})_6]\)

Co in \(\mathrm{K}_{3}[\mathrm{Co}(\mathrm{CN})_6]\) is in the oxidation state of +3. The electronic configuration of \(\mathrm{Co}^{3+}\) is \([\mathrm{Ar}]3d^6\). CN, a strong field ligand, causes electron pairing, making \([\mathrm{Co}(\mathrm{CN})_6]^{3-}\) diamagnetic.

Step 5: Analyze \(\mathrm{K}_{2}[\mathrm{Ni}(\mathrm{CN})_4]\)

Ni in \(\mathrm{K}_{2}[\mathrm{Ni}(\mathrm{CN})_4]\) is in the oxidation state of +2. The electronic configuration of \(\mathrm{Ni}^{2+}\) is \([\mathrm{Ar}]3d^8\). CN as a strong field ligand pairs up the electrons, making \([\mathrm{Ni}(\mathrm{CN})_4]^{2-}\) diamagnetic.

Step 6: Analyze \([\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}\)

Co in \([\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}\) is in the oxidation state of +3. The electronic configuration of \(\mathrm{Co}^{3+}\) is \([\mathrm{Ar}]3d^6\). \(\mathrm{NH}_3\) is a weak field ligand, so it does not pair electrons, making \([\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}\) paramagnetic.

Step 7: Analyze \(\mathrm{K}_{2}\mathrm{TiF}_{6}\)

Ti in \(\mathrm{K}_{2}\mathrm{TiF}_{6}\) is in the oxidation state of +4. The electronic configuration of \(\mathrm{Ti}^{4+}\) is \([\mathrm{Ar}]\), which means it has no \(d\) electrons left, making \([\mathrm{TiF}_6]^{2-}\) diamagnetic as all electrons are paired or absent.

Step 8: Analyze \([\mathrm{Pt}(\mathrm{NH}_3)_4]^{2+}\)

Pt in \([\mathrm{Pt}(\mathrm{NH}_3)_4]^{2+}\) is in the oxidation state of +2. The electronic configuration of \(\mathrm{Pt}^{2+}\) is \([\mathrm{Xe}]4f^{14}5d^8\). \(\mathrm{NH}_3\), a weak field ligand, does not pair electrons. However, since the \(5d\) subshell can accommodate all electrons with pairing in this environment, \([\mathrm{Pt}(\mathrm{NH}_3)_4]^{2+}\) is potentially diamagnetic.

Key Concepts

Coordination Compounds

Coordination compounds are fascinating structures where a central metal atom or ion is surrounded by a certain number of ions or molecules, known as ligands. These ligands donate electrons to the metal, forming complex entities called coordination complexes. The nature and arrangement of ligands around the central metal atom significantly affect the properties and stability of these compounds.

A typical example is \( \mathrm{K}_{4}[\mathrm{Fe}(\mathrm{CN})_6] \), where potassium ions surround an iron-cyanide complex. Here, cyanide (CN) acts as a strong field ligand, influencing the electron pairing in iron's \( 3d \) orbitals. Strong field ligands like CN can cause electrons to pair and thereby alter the magnetic properties of the coordination compound. This pairing process can lead to diamagnetism, where all electrons are paired.

Oxidation States

Oxidation states are crucial for understanding the chemistry of coordination compounds. They help to determine the electronic configuration of the metal ion in a compound. For instance, in \( \mathrm{K}_{3}[\mathrm{Cr}(\mathrm{CN})_6] \), chromium is in an oxidation state of +3. This state helps us to deduce the electron configuration as \([\mathrm{Ar}]3d^3\), influencing the compound's reactivity and magnetic properties.

Knowing the oxidation state, we can predict how a metal ion will interact with ligands and affect the compound's magnetic nature. In \( \mathrm{K}_{2}\mathrm{TiF}_6 \), titanium is in the +4 oxidation state, which renders it \([\mathrm{Ar}]\), showing that it has lost all electrons from its \(d\)-orbitals and is diamagnetic.

Electron Configuration

Electron configuration reveals how electrons are distributed within an atom's orbitals. This distribution directly affects the characteristics of coordination compounds. For example, in \( \mathrm{K}_{3}[\mathrm{Co}(\mathrm{CN})_6] \), cobalt \(3+\) has an electron configuration of \([\mathrm{Ar}]3d^6\). Strong field ligands, like CN, cause the \(d\) electrons to pair up, leading to diamagnetism.

Understanding the electron configuration allows us to predict many properties of coordination compounds, including color, magnetic behavior, and bonding nature. For instance, \( \mathrm{K}_{2}[\mathrm{Ni}(\mathrm{CN})_4] \) with nickel in a +2 state has an electron configuration \([\mathrm{Ar}]3d^8\), leading to paired electrons and a diamagnetic nature.

Strong Field Ligands

Strong field ligands, like cyanide (CN) and carbonyl (CO), have a significant influence on the properties of coordination compounds. They cause pairing of \(d\) electrons by creating a large splitting of the \(d\) orbital energies, known as crystal field splitting. In \( \mathrm{K}_{4}[\mathrm{Fe}(\mathrm{CN})_6] \), CN is a strong field ligand that pairs all the \(d\) electrons of iron, creating a diamagnetic complex.

The pairing of electrons results in low spin configurations, where orbitals have paired electrons to the maximum possible extent. This change in electron configuration, driven by strong field ligands, profoundly affects a compound's magnetic properties and its overall stability.

Magnetic Properties in Transition Metals

Magnetic properties in transition metals depend heavily on the presence and arrangement of unpaired electrons in their \(d\) orbitals. A transition metal compound is generally paramagnetic if it has unpaired electrons and diamagnetic if all electrons are paired. In the coordination compound \([\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}\), the presence of unpaired electrons due to the weak field ligand \(\mathrm{NH}_3\) leads it to be paramagnetic.

Conversely, in cases where strong field ligands like \(\mathrm{CN}^-\) are present, as in \([\mathrm{Co}(\mathrm{CN})_6]^{3-}\), they pair electrons resulting in a diamagnetic compound. Understanding these properties helps in predicting the behavior of compounds in magnetic fields, their reactivity, and potential applications in technology.