Problem 76 - A Complete Resource Book in Chemistry for JEE Main

Question

The complex ion which has no d electrons in the central metal atom is (Atomic number \(\mathrm{Cr}=24\), \(\mathrm{Mn}=25, \mathrm{Fe}=26, \mathrm{Co}=27)\) (a) \(\left[\mathrm{MnO}_{4}\right]\) (b) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) (c) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3}\) (d) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\)

Step-by-Step Solution

Answer

The complex ion \([\mathrm{MnO}_{4}]\) has no d electrons in the central metal atom.

Step 1: Identify Central Metal Atoms

First, look at each of the given complex ions and identify the central metal atom in each of them:- \([\mathrm{MnO}_{4}]\): Central atom is Mn.- \([\mathrm{Co}(\mathrm{NH}_{3})_{6}]^{3+}\): Central atom is Co.- \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\): Central atom is Fe.- \([\mathrm{Cr}(\mathrm{H}_{2} \mathrm{O})_{6}]^{3+}\): Central atom is Cr.

Step 2: Determine Oxidation States

Find the oxidation states for each metal in the complex ions:- \([\mathrm{MnO}_{4}]\): Each oxygen is -2, total -8; therefore, Mn is +7 (\(x - 8 = -1\)).- \([\mathrm{Co}(\mathrm{NH}_{3})_{6}]^{3+}\): Each \(\mathrm{NH}_{3}\) is neutral, so Co is +3 (due to the \(3+\) charge).- \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\): Each \(\mathrm{CN}^{-}\) is -1, total -6; thus, Fe is +3 (\(x - 6 = -3\)).- \([\mathrm{Cr}(\mathrm{H}_{2} \mathrm{O})_{6}]^{3+}\): Each \(\mathrm{H}_{2}O\) is neutral, so Cr is +3 (due to the \(3+\) charge).

Step 3: Calculate d-electrons

Calculate the number of d-electrons for each metal based on their oxidation states: - Mn: Atomic number 25, electron configuration [Ar] 3d⁵ 4s²; Mn⁷⁺ is [Ar], 0 d electrons. - Co: Atomic number 27, electron configuration [Ar] 3d⁷ 4s²; Co³⁺ is [Ar] 3d⁶, 6 d electrons. - Fe: Atomic number 26, electron configuration [Ar] 3d⁶ 4s²; Fe³⁺ is [Ar] 3d⁵, 5 d electrons. - Cr: Atomic number 24, electron configuration [Ar] 3d⁵ 4s¹; Cr³⁺ is [Ar] 3d³, 3 d electrons.

Step 4: Determine Final Answer

The complex ion where the metal has no d-electrons is determined from the above calculations. The one with the metal oxidation state that leaves no d-electrons is Mn in \\([\mathrm{MnO}_{4}]\).

Key Concepts

Oxidation States

Oxidation states are vital for understanding the chemical behavior of elements, especially in transition metals. They represent the hypothetical charge of an atom in a molecule, assuming that electrons in all bonds are shared equally. In transition metal complexes, it is crucial to determine the oxidation state to predict how many d-electrons are present.

• For example, in \([\mathrm{MnO}_{4}]\), manganese has an oxidation state of \(+7\). This can be calculated by considering that each oxygen atom has an oxidation state of \(-2\), giving a total of \(-8\) for the four oxygen atoms. Since the overall charge of the permanganate ion is \(-1\), manganese must counterbalance this with a +7 charge (\(x - 8 = -1\)).
• In contrast, in \([\mathrm{Co(\mathrm{NH}_{3})_{6}]^{3+}\), the oxidation state of cobalt is \(+3\). Ammonia is a neutral ligand, contributing no charge.

Calculating oxidation states helps predict the electron configuration of an atom or ion, which is next in our list of core concepts.

Electron Configuration

Electron configuration is the arrangement of electrons in an atom. It is a fundamental concept that describes how electrons fill various energy levels and sub-levels around an atom's nucleus. In transition metals, understanding this configuration is essential as it influences their chemical properties and reactivity. The electron configuration notation follows the principle of filling orbitals in order of increasing energy.

• For manganese (Mn) with an atomic number of 25, the ground state electron configuration is \[ \text{[Ar]} 3d^5 4s^2 \]. When it is in \([\mathrm{MnO}_{4}]\) and Mn is \(+7\), its electron configuration becomes \[ \text{[Ar]} \], meaning it has lost all five \(3d\) and two \(4s\) electrons, resulting in zero d-electrons.
• Cobalt (Co), on the other hand, with atomic number 27, has a ground state electron configuration of \[ \text{[Ar]} 3d^7 4s^2 \]. In the complex \([\mathrm{Co(\mathrm{NH}_{3})_{6}]^{3+}\), it loses three electrons, resulting in a configuration of \[ \text{[Ar]} 3d^6 \], meaning it still retains six d-electrons.

The electron configuration directly impacts the physical and chemical characteristics of transition metals.

Transition Metals

Transition metals are a group of elements found in the d-block of the periodic table, specifically groups 3 through 12. They are characterized by having partially filled d-subshells, which leads to unique properties such as variable oxidation states, formation of colored compounds, and the ability to form multiple types of coordination complexes.

• These metals like chromium, manganese, and iron can exhibit multiple oxidation states. This variability gives them versatility in forming compounds with varying numbers of bonds and geometries.
• The presence of d-electrons allows for various electron transitions, which is why many transition metal compounds are highly colored. This is due to the absorption of specific wavelengths of light as electrons are promoted between d-orbitals.
• Transition metals serve as excellent catalysts due to their ability to lend and receive electrons readily during chemical reactions, enhancing the rate without being consumed themselves.

Understanding transition metals' unique properties is crucial for analyzing their chemical behavior and application in fields such as catalysis, electronic devices, and materials science.