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NEET CHEMISTRYStructure of AtomMedium

Question

Which one of the elements with the following outer orbital configurations may exhibit the largest number of oxidation states?

A

3d^3, 4s^2

B

3d^5, 4s^1

C

3d^5, 4s^2

D

3d^2, 4s^2

Step-by-Step Solution

The variety of oxidation states in transition elements arises from the participation of both the (n−1)d(n-1)d and nsns electrons in bonding. The element with the maximum number of unpaired electrons in the d-subshell plus electrons in the s-subshell will exhibit the largest number of oxidation states.

  1. Analyze Configurations: 3d34s23d^3 4s^2 (Vanadium): 3+2=53+2=5 valence electrons. Max O.S. is +5. 3d54s13d^5 4s^1 (Chromium): 5+1=65+1=6 valence electrons. Max O.S. is +6. 3d54s23d^5 4s^2 (Manganese): 5+2=75+2=7 valence electrons. Max O.S. is +7. 3d24s23d^2 4s^2 (Titanium): 2+2=42+2=4 valence electrons. Max O.S. is +4.

  2. Conclusion: Manganese (3d54s23d^5 4s^2) can utilize all 7 valence electrons, showing oxidation states from +2 to +7, which is the maximum number in the 3d series.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Structure of Atom. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYStructure of Atomelementsfollowingorbitalconfigurationsexhibit

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