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NCERT Exemplar · Q9

Q.The largest oxidation number exhibited by an element depends on its outer electronic configuration. With which of the following outer electronic configurations the element will exhibit largest oxidation number?

(i) 3d^1 4s^2
(ii) 3d^3 4s^2
(iii) 3d^5 4s^1
(iv) 3d^5 4s^2
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The maximum oxidation state an element can exhibit equals the total number of valence electrons available for bonding. Configuration 3d⁵ 4s² offers seven valence electrons, the most among the options, so the answer is (iv).

Why valence electrons determine maximum oxidation state

An element's highest oxidation number reflects the complete removal or sharing of all its valence electrons in chemical bonding. For transition metals, both the outermost ss electrons and the penultimate dd electrons participate in bonding, so we count them together.

The key insight: an atom can lose or share electrons from its valence shell until that shell is emptied (or until it reaches a particularly stable configuration). The more valence electrons available, the higher the potential oxidation state.

Step-by-step comparison

Let's count the total valence electrons (nd+(n+1)sn\text{d} + (n+1)\text{s}) for each configuration:

  1. Option (i): 3d1 4s23\text{d}^1 \, 4\text{s}^2

    Total valence electrons = 1+2=31 + 2 = 3

    Maximum oxidation state = +3+3

    (Example: Sc in ScF₃)

  2. Option (ii): 3d3 4s23\text{d}^3 \, 4\text{s}^2

    Total valence electrons = 3+2=53 + 2 = 5

    Maximum oxidation state = +5+5

    (Example: V in V₂O₅)

  3. Option (iii): 3d5 4s13\text{d}^5 \, 4\text{s}^1

    Total valence electrons = 5+1=65 + 1 = 6

    Maximum oxidation state = +6+6

    (Example: Cr in CrO₃ or K₂Cr₂O₇)

  4. Option (iv): 3d5 4s23\text{d}^5 \, 4\text{s}^2

    Total valence electrons = 5+2=75 + 2 = 7

    Maximum oxidation state = +7+7

    (Example: Mn in KMnO₄, where Mn is in the +7+7 state) …

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