What may be the stable oxidation state of the transition element with the following d- electron configurations in the ground state of their atoms: , , and ?
The stable oxidation states for the given d-electron configurations are:
3d³ configuration: The stable oxidation state is +5. This corresponds to Vanadium (V), where losing all five valence electrons (two from 4s and three from 3d) gives a stable d⁰ configuration (noble gas core).
3d⁵ configuration: The stable oxidation state is +2. This corresponds to Manganese (Mn), where losing two 4s electrons gives a stable half-filled d⁵ configuration. The +7 oxidation state is also possible, giving d⁰ configuration.
3d⁸ configuration: The stable oxidation state is +2. This corresponds to Nickel (Ni), where losing two 4s electrons is the common and stable oxidation state.
3d⁴ configuration: The stable oxidation state is +6. This corresponds to Chromium (Cr), where losing all six valence electrons gives a stable d⁰ configuration.
Explanation
The question tests understanding of how d-electron configurations relate to stable oxidation states. Transition elements achieve stability through d⁰ (noble gas configuration), d⁵ (half-filled), or d¹⁰ (fully-filled) configurations. The textbook context shows that oxidation states vary due to incomplete filling of d orbitals. For each configuration, students must identify the element and determine which oxidation state yields a stable electronic arrangement.
Solution Steps
Step 1: Identify the element for each d-configuration (3d³=V, 3d⁵=Mn, 3d⁸=Ni, 3d⁴=Cr)
Step 2: Determine valence electrons (4s² + 3dⁿ)
Step 3: Apply stability rules - d⁰, d⁵, d¹⁰ configurations are stable
Step 4: Calculate oxidation state needed to achieve stable configuration