The correct statement(s) about Cr^{2+} and Mn^{3+} is (are)
[Atomic numbers of Cr = 24 and Mn = 25]

Q: Why is $\text{Cr}^{3+}$ considered stable even though it's not a $\text{d}^5$ configuration?

In an aqueous medium, the five degenerate d-orbitals split into two sets: a lower energy $\text{t}_{2\text{g}}$ set (3 orbitals) and a higher energy $\text{e}_{\text{g}}$ set (2 orbitals). A $\text{d}^3$ configuration perfectly half-fills the $\text{t}_{2\text{g}}$ level, providing significant crystal field stabilization energy (CFSE).

Q: Why doesn't $\text{Mn}^{3+}$ act as a reducing agent to become $\text{Mn}^{4+}$?

To become $\text{Mn}^{4+}$, it would need to lose another electron, reaching a $\text{d}^3$ state. While $\text{d}^3$ is stable in water, the 4th ionization enthalpy of Manganese is extremely high. Gaining an electron to achieve the highly stable, exactly half-filled $\text{d}^5$ state ($\text{Mn}^{2+}$) is energetically much more favorable.

Q: What exactly does a reducing agent do?

A reducing agent reduces another substance. In the process, it gets oxidized itself, which means it loses electrons. So, if an ion is a good reducing agent, it wants to give away an electron to reach a more stable state.

Q: Are all $\text{d}^4$ transition metal ions unstable?

Yes, $\text{d}^4$ configurations are generally less stable than their neighboring $\text{d}^3$ or $\text{d}^5$ states. They act as reactive intermediates, strongly driven to either lose an electron (like $\text{Cr}^{2+}$) or gain an electron (like $\text{Mn}^{3+}$) to achieve a more stable electronic arrangement.

Q: Why is Chromium's ground state $3\text{d}^5 4\text{s}^1$ instead of $3\text{d}^4 4\text{s}^2$?

The $3\text{d}^5 4\text{s}^1$ configuration allows Chromium to have a perfectly half-filled 3d subshell. Half-filled subshells have higher exchange energy and a more symmetrical electron distribution, making them significantly more stable than a $3\text{d}^4$ configuration.

Question

The correct statement(s) about $	ext{Cr}^{2+}$ and $	ext{Mn}^{3+}$ is (are)
[Atomic numbers of $	ext{Cr} = 24$ and $	ext{Mn} = 25$]

Options:

(A) $	ext{Cr}^{2+}$ is a reducing agent

(B) $	ext{Mn}^{3+}$ is an oxidizing agent

(C) Both $	ext{Cr}^{2+}$ and $	ext{Mn}^{3+}$ exhibit $	ext{d}^4$ electronic configuration

(D) When $	ext{Cr}^{2+}$ is used as a reducing agent, the chromium ion attains $	ext{d}^5$ electronic configuration

Sigma Percentile · Accepted Answer

The behavior of transition metal ions in chemical reactions is a fascinating dance of electrons, driven entirely by the quest for stability. In this problem, we are presented with two ions: $	ext{Cr}^{2+}$ and $	ext{Mn}^{3+}$. At first glance, they might seem quite similar, but their chemical personalities are completely opposite. Let's dive into the quantum mechanics that dictate their behavior.

### The Setup: Electronic Configurations

Before we can understand how these ions react, we must know their starting positions. We begin with the ground state electronic configurations of the neutral atoms.

Chromium ($	ext{Z} = 24$) is a famous exception to the Aufbau principle. Instead of the expected $3	ext{d}^4 4	ext{s}^2$, one electron from the $4	ext{s}$ orbital is promoted to the $3	ext{d}$ orbital. This gives Chromium a configuration of $[	ext{Ar}] 3	ext{d}^5 4	ext{s}^1$. Why? Because a half-filled d-subshell ($3	ext{d}^5$) possesses a highly symmetrical electron distribution and maximum exchange energy, making it exceptionally stable.

Manganese ($	ext{Z} = 25$), on the other hand, follows the standard rules. Its configuration is $[	ext{Ar}] 3	ext{d}^5 4	ext{s}^2$.

Now, let's form the ions mentioned in the question. To create $	ext{Cr}^{2+}$, we remove two electrons from neutral Chromium. We take one from the $4	ext{s}$ orbital and one from the $3	ext{d}$ orbital, leaving us with:
$$ 	ext{Cr}^{2+}: [	ext{Ar}] 3	ext{d}^4 $$

To create $	ext{Mn}^{3+}$, we remove three electrons from neutral Manganese. We take two from the $4	ext{s}$ orbital and one from the $3	ext{d}$ orbital, resulting in:
$$ 	ext{Mn}^{3+}: [	ext{Ar}] 3	ext{d}^4 $$

**Both $	ext{Cr}^{2+}$ and $	ext{Mn}^{3+}$ exhibit a $	ext{d}^4$ electronic configuration.** This confirms that Option (C) is absolutely correct.

### The $	ext{d}^4$ Dilemma

A $	ext{d}^4$ configuration is inherently restless. It sits awkwardly between two highly stable states: the $	ext{d}^3$ state (which is very stable in aqueous solutions) and the $	ext{d}^5$ state (which is stable everywhere due to being exactly half-filled). Because of this, $	ext{d}^4$ ions are highly reactive. They desperately want to either lose an electron to become $	ext{d}^3$ or gain an electron to become $	ext{d}^5$.

### Chromium's Path: The Reducing Agent

Let's look at $	ext{Cr}^{2+}$. If it acts as a reducing agent, it must reduce another species, which means it must oxidize itself. Oxidation is the loss of electrons.

$$ 	ext{Cr}^{2+} ightarrow 	ext{Cr}^{3+} + 	ext{e}^- $$

In terms of orbitals, this transition is:
$$ 3	ext{d}^4 ightarrow 3	ext{d}^3 $$

Why is this favorable? In an aqueous medium, the water molecules act as ligands and surround the central metal ion. This creates an octahedral crystal field, which splits the five degenerate d-orbitals into two distinct energy levels: a lower energy set of three orbitals called $	ext{t}_{2	ext{g}}$, and a higher energy set of two orbitals called $	ext{e}_{	ext{g}}$.

A $	ext{d}^3$ configuration perfectly half-fills the lower $	ext{t}_{2	ext{g}}$ level (one electron in each of the three orbitals). This arrangement provides a massive amount of Crystal Field Stabilization Energy (CFSE). Because $	ext{Cr}^{3+}$ is so incredibly stable in water, $	ext{Cr}^{2+}$ is highly motivated to throw away an electron to reach that state.

Therefore, **$	ext{Cr}^{2+}$ is a strong reducing agent.** This makes Option (A) correct. Furthermore, since it attains a $	ext{d}^3$ configuration (not $	ext{d}^5$), Option (D) is incorrect.

### Manganese's Path: The Oxidizing Agent

Now, let's examine $	ext{Mn}^{3+}$. It also starts with a $	ext{d}^4$ configuration. However, its path to stability is different. If it acts as an oxidizing agent, it must oxidize another species, meaning it must reduce itself. Reduction is the gain of electrons.

$$ 	ext{Mn}^{3+} + 	ext{e}^- ightarrow 	ext{Mn}^{2+} $$

In terms of orbitals, this transition is:
$$ 3	ext{d}^4 ightarrow 3	ext{d}^5 $$

As we discussed earlier, a $	ext{d}^5$ configuration is exactly half-filled. This state is universally stable due to maximum exchange energy. The drive for $	ext{Mn}^{3+}$ to grab an electron and achieve this $	ext{d}^5$ nirvana is immense.

Therefore, **$	ext{Mn}^{3+}$ is a strong oxidizing agent.** This makes Option (B) correct.

### The Final Verdict

By understanding the underlying quantum mechanics and the drive for orbital stability, we can easily predict the chemical behavior of these transition metal ions. $	ext{Cr}^{2+}$ sheds an electron to find stability in the split crystal field ($	ext{t}_{2	ext{g}}^3$), while $	ext{Mn}^{3+}$ snatches an electron to achieve a perfectly half-filled subshell ($	ext{d}^5$).

The correct statements are (A), (B), and (C).

Animated Solution for Chemistry - d and f-Block Elements: The correct statement(s) about $Cr^{2 +}$ and $Mn^{3 +}$ is (are) [Atomic numbers of $Cr = 24$ and $Mn = 25$ ]

Select Answer:

Visualized Solution

$Electronic Configuration of Cr and Mn$

$Configuration of Cr^{2 +} and Mn^{3 +}$

$Cr^{2 +} as a Reducing Agent$

$Mn^{3 +} as an Oxidizing Agent$

$Conclusion$

The Sigma Insight: d-block Elements

The Setup

The $d^{4}$ Dilemma

Chromium's Path

Manganese's Path

The Final Verdict

Similar Questions

The correct order of following $3 d$ -metal oxides, according to their oxidation numbers is (A) $CrO_{3}$ (B) $Fe_{2} O_{3}$ (C) $MnO_{2}$ (D) $V_{2} O_{5}$ (E) $Cu_{2} O$

In which of the following pairs, the outer most electronic configuration will be the same?

In context with the transition elements, which of the following statements is incorrect?

The incorrect statement(s) among (1)-(3) is (are) 1. W(VI) is more stable than Cr(VI). 2. in the presence of HCl, permanganate titrations provide satisfactory results. 3. some lanthanoid oxides can be used as phosphors.

Fusion of $MnO_{2}$ with $KOH$ in presence of $O_{2}$ produces a salt $W$ . Alkaline solution of $W$ upon eletrolytic oxidation yields another salt $X$ . The manganese containing ions present in $W$ and $X$ , respectively, are $Y$ and $Z$ . Correct statement(s) is (are)

The incorrect statement is :

The correct order of $E_{M^{2 +} / M}^{\circ}$ values with negative sign for the four successive elements Cr, Mn, Fe and Co is

The sum of the total number of bonds between chromium and oxygen atoms in chromate and dichromate ions is .........

The electrode potential of $M^{2 +} / M$ of 3d-series elements shows positive value for

In the structure of the dichromate ion, there is a

Animated Solution for Chemistry - d and f-Block Elements: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The correct statement(s) about Cr2+ and Mn3+ is (are) [Atomic numbers of Cr=24 and Mn=25]