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The correct order of E(M^(2+)//M)^(@) Va...

The correct order of `E_(M^(2+)//M)^(@)` Values with negative sign for the four successive elements `Cr, Mn, Fe` and `Co` is:

A

Mn > Cr > Fe > Co

B

Cr > Fe > Mn > Co

C

Fe > Mn > Cr > Co

D

Cr > Mn > Fe > Co

Text Solution

AI Generated Solution

The correct Answer is:
To determine the correct order of \( E^{\circ}_{M^{2+}/M} \) values with a negative sign for the elements Chromium (Cr), Manganese (Mn), Iron (Fe), and Cobalt (Co), we can follow these steps: ### Step 1: Identify the Electronic Configurations 1. **Chromium (Cr)**: Atomic number = 24 - Electronic configuration: \( [Ar] 3d^5 4s^1 \) 2. **Manganese (Mn)**: Atomic number = 25 - Electronic configuration: \( [Ar] 3d^5 4s^2 \) 3. **Iron (Fe)**: Atomic number = 26 - Electronic configuration: \( [Ar] 3d^6 4s^2 \) 4. **Cobalt (Co)**: Atomic number = 27 - Electronic configuration: \( [Ar] 3d^7 4s^2 \) ### Step 2: Analyze the Stability of the Oxidation States - When these metals lose two electrons to form \( M^{2+} \), we need to consider their electronic configurations after losing the electrons. 1. **For Manganese (Mn)**: - Loses 2 electrons: \( 3d^5 4s^2 \) → \( 3d^5 \) (half-filled, stable configuration) 2. **For Chromium (Cr)**: - Loses 2 electrons: \( 3d^5 4s^1 \) → \( 3d^4 4s^0 \) (less stable than Mn) 3. **For Iron (Fe)**: - Loses 2 electrons: \( 3d^6 4s^2 \) → \( 3d^6 4s^0 \) (stable but not as favorable as Mn) 4. **For Cobalt (Co)**: - Loses 2 electrons: \( 3d^7 4s^2 \) → \( 3d^7 4s^0 \) (least favorable) ### Step 3: Determine the Order of Ease of Losing Electrons - The ease of losing electrons generally decreases as we move from left to right in the periodic table due to increasing nuclear charge and decreasing atomic size. - Therefore, the order of ease of losing two electrons is: 1. **Manganese (Mn)** - most favorable due to half-filled configuration. 2. **Chromium (Cr)** - larger atomic size allows easier loss than Fe and Co. 3. **Iron (Fe)** - less favorable than Cr but more than Co. 4. **Cobalt (Co)** - least favorable. ### Step 4: Write the Final Order Thus, the correct order of \( E^{\circ}_{M^{2+}/M} \) values with a negative sign is: - **Mn > Cr > Fe > Co** ### Conclusion The correct order is: **Manganese (Mn), Chromium (Cr), Iron (Fe), Cobalt (Co)**.
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  • The correct order of the number of unpaired electrons in the ions Cu^(2+), Ni^(2+), Fe^(3+) , and Cr^(3+) is

    A
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    B
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    C
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