Problem 67 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following are diamagnetic? (1) \(\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) (2) \(\mathrm{K}_{3}\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]\) (3) \(\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\) (4) \(\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]\) Select the correct answer using the codes given below: (a) \(\mathrm{l}\) and 2 (b) 2,3 and 4 (c) 1,3 and 4 (d) 2 and 4

Step-by-Step Solution

Answer

Diamagnetic: 1, 3, and 4. Correct answer: (c) 1, 3, and 4.

Step 1: Understand Diamagnetic vs Paramagnetic

Diamagnetic substances have all their electrons paired, meaning no unpaired electrons are present. Paramagnetic substances contain at least one unpaired electron.

Step 2: Determine Oxidation States

Calculate the oxidation state of the metal in each complex: 1. For K₄[Fe(CN)₆], the oxidation state of Fe is +2. 2. For K₃[Cr(CN)₆], the oxidation state of Cr is +3. 3. For K₃[Co(CN)₆], the oxidation state of Co is +3. 4. For K₂[Ni(CN)₄], the oxidation state of Ni is +2.

Step 3: Write Electronic Configurations

Write down the electronic configuration of each metal ion in its oxidation state: 1. Fe²⁺: [Ar] 3d⁶ 2. Cr³⁺: [Ar] 3d³ 3. Co³⁺: [Ar] 3d⁶ 4. Ni²⁺: [Ar] 3d⁸

Step 4: Analyze Presence of Unpaired Electrons

Check each electron configuration for unpaired electrons: 1. Fe²⁺: [Ar] 3d⁶ - can be either low-spin or high-spin but K₄[Fe(CN)₆] is a low-spin complex with no unpaired electrons. This makes it diamagnetic. 2. Cr³⁺: [Ar] 3d³ - has 3 unpaired electrons, making it paramagnetic. 3. Co³⁺: [Ar] 3d⁶ - in the low-spin configuration, this is diamagnetic as all electrons are paired. 4. Ni²⁺: [Ar] 3d⁸ - has 2 unpaired electrons in general but K₂[Ni(CN)₄] is a square planar complex and is diamagnetic due to all paired electrons.

Key Concepts

Oxidation States

Oxidation states are a crucial concept in chemistry, helping us determine how many electrons an atom has gained or lost when forming a chemical compound. The oxidation state is represented by a number, which is often positive or negative. For example, in the case of \(\mathrm{K}_{4}[\mathrm{Fe}(\mathrm{CN})_{6}]\), the iron (Fe) has an oxidation state of +2. This is determined by considering the charges of all components: potassium (K) is +1 and cyanide (CN) is -1. Calculations must ensure the sum of the oxidation states equals the overall charge on the complex.

Knowing the oxidation states helps in understanding how electrons are distributed in a compound, influencing properties like magnetism and reactivity. It's especially important in coordination compounds, where metals can exhibit different behaviors based on their oxidation states, impacting their magnetic properties. Knowing the oxidation state is the first step in finding out whether a complex is paramagnetic or diamagnetic.

Electron Configuration

Electron configuration refers to the arrangement of electrons in an atom or ion. Each element has a unique electron configuration which determines its chemical behavior. In complexes, especially involving transition metals, electron configurations decide many features, including magnetic properties.

For transition metals, their d electrons play a significant role. For instance:

• For \(\mathrm{Fe^{2+}}\), the electron configuration is \([\mathrm{Ar}]\ 3d^6\). This means it has six electrons in its d orbitals.
• For \(\mathrm{Cr^{3+}}\), it is \([\mathrm{Ar}]\ 3d^3\).
• For \(\mathrm{Co^{3+}}\), it is \([\mathrm{Ar}]\ 3d^6\).
• For \(\mathrm{Ni^{2+}}\), it is \([\mathrm{Ar}]\ 3d^8\).

The configuration can indicate whether the electrons are all paired or if any are unpaired, which is pivotal for identifying materials as diamagnetic or paramagnetic. Paired electrons lead to diamagnetism, while unpaired electrons result in paramagnetism.

Understanding how to write and interpret these configurations is vital to predict how a metal will behave in different complexes, such as low-spin or high-spin scenarios.

Low-Spin and High-Spin Complexes

In coordination chemistry, low-spin and high-spin complexes are types of splitting observed in the d orbitals of transition metals, significantly affecting their magnetic properties. In low-spin complexes, electrons tend to pair up in lower energy orbitals before they occupy higher ones, often resulting in no unpaired electrons, which makes the compound diamagnetic.

For example, \(\mathrm{K}_{4}[\mathrm{Fe}(\mathrm{CN})_{6}]\) undergoes a low-spin arrangement due to the strong field ligand, cyanide (CN), causing no unpaired electrons and the complex being diamagnetic.

• Low-spin complexes: Typically formed when the ligand field is strong (like \(\mathrm{CN^-}\)), promoting electron pairing.
• High-spin complexes: Formed when the ligand field is weak, leading to unpaired electrons spread out over available orbitals.

In contrast, high-spin complexes involve increased splitting energy allowing more unpaired electrons, thus being paramagnetic. Understanding whether a complex forms low-spin or high-spin configurations is crucial in predicting the electronic and magnetic behavior of metal complexes, aiding in studying and designing materials for various applications.