For the four successive transition elements (Cr, Mn, Fe and Co), the stability of +2 oxidation state will be there in which of the following order?(At. nos. Cr = 24, Mn = 25, Fe = 26, Co = 27)a)Mn Fe Cr Cob)Fe Mn Co Crc)Co Mn Fe Crd)Cr Mn Co FeCorrect answer is option 'A'. Can you explain this answer?

Question

For the four successive transition elements (Cr, Mn, Fe and Co), the stability of +2 oxidation state will be there in which of the following order?(At. nos. Cr = 24, Mn = 25, Fe = 26, Co = 27)a)Mn > Fe > Cr > Cob)Fe > Mn > Co > Crc)Co > Mn > Fe > Crd)Cr > Mn > Co > FeCorrect answer is option 'A'. Can you explain this answer?

Answer

The stability of 2 oxidation state in the four successive transition elements (Cr, Mn, Fe, and Co) can be explained based on the following factors:

1. Electronic configuration: The electronic configuration of these elements is as follows:

- Chromium (Cr): [Ar] 3d5 4s1
- Manganese (Mn): [Ar] 3d5 4s2
- Iron (Fe): [Ar] 3d6 4s2
- Cobalt (Co): [Ar] 3d7 4s2

2. Screening effect: The screening effect of the 3d electrons on the 4s electrons decreases as we move from left to right across the period. This is due to the increase in nuclear charge which attracts the 4s electrons more towards the nucleus.

3. Ionization energy: The ionization energy of the 4s electrons also increases as we move from left to right across the period. This is due to the increase in nuclear charge which makes it harder to remove an electron from the outermost shell.

Based on the above factors, the stability of 2 oxidation state in the four successive transition elements (Cr, Mn, Fe, and Co) can be explained as follows:

- Chromium (Cr): In Cr, the 3d5 electrons have a strong screening effect on the 4s electron, making it easier to remove. Also, the ionization energy of the 4s electron is relatively low. Therefore, Cr has a stable 2 oxidation state.
- Manganese (Mn): In Mn, the 3d5 electrons still have a strong screening effect on the 4s electron, but the ionization energy of the 4s electron is higher than in Cr. Therefore, Mn has a less stable 2 oxidation state compared to Cr.
- Iron (Fe): In Fe, the 3d6 electrons have a weaker screening effect on the 4s electron, and the ionization energy of the 4s electron is higher than in Mn. Therefore, Fe has a less stable 2 oxidation state compared to Mn and Cr.
- Cobalt (Co): In Co, the 3d7 electrons have an even weaker screening effect on the 4s electron, and the ionization energy of the 4s electron is the highest among the four elements. Therefore, Co has the least stable 2 oxidation state among the four elements.

Therefore, the stability of 2 oxidation state in the four successive transition elements (Cr, Mn, Fe, and Co) follows the order: Mn > Fe > Cr > Co.

Answer

The order can be explained using the idea of spin correlation. Spin correlation refers to lowering of energy for like (parallel) spins. Spin correlation leading to decrease in repulsion for electrons of like spins than for electrons of different spins is called exchange energy.

Spin correlation and its exchange energy gives an electronic configuration a special stability which is greatest for half-filled electronic configurations.

Mn²⁺(d⁵) gets stabilisation due to half-filled configuration.In Fe²⁺(d⁶) the placing of one extra electron in a subshell destabilises. Placing of 2 electrons in Co²⁺(d⁷) destabilisesit more.Cr²⁺(d⁴) has one vacant subshell. Fe²⁺ gets more Stabilisation compared to Cr²⁺ through exchange energy. So the order is as follows:

Mn > Fe > Cr > Co

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