Consider the following statements.I. Eu2+ has f7 configuration and it is a strong reducing agent.II. Yb2+ has f14 configuration and it is a reductant.III.TbIV has half-filled f-orbitals and is an oxidant.

The Electronic configuration of Europium is 4f76s2 and Ytterbium is 4f146s2.Eu and Yb in +2 oxidation states acquires half filled (4f7) and full filled (4f14) configuration but still they oxidized to their common +3 state and thus acts as reductant.Tb in +4 state has 4f7 half filled electronic configuration but it readily get reduced to common state +3 and acts as Oxidant.

Consider the following statements.I. Eu2+ has f7 configuration and it is a strong reducing agent.II. Yb2+ has f14 configuration and it is a reductant.III.TbIV has half-filled f-orbitals and is an oxidant.

The Electronic configuration of Europium is 4f76s2 and Ytterbium is 4f146s2.Eu and Yb in +2 oxidation states acquires half filled (4f7) and full filled (4f14) configuration but still they oxidized to their common +3 state and thus acts as reductant.Tb in +4 state has 4f7 half filled electronic configuration but it readily get reduced to common state +3 and acts as Oxidant.

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Consider the following statements.
I. Eu²⁺ has f⁷ configuration and it is a strong reducing agent.
II. Yb²⁺ has f¹⁴ configuration and it is a reductant.
III.Tb^IV has half-filled f-orbitals and is an oxidant.

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Only II and III

I, II and III

Only I and II

Only I

answer is D.

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Detailed Solution

The Electronic configuration of Europium is 4f⁷6s² and Ytterbium is 4f¹⁴6s².Eu and Yb in +2 oxidation states acquires half filled (4f⁷) and full filled (4f¹⁴) configuration but still they oxidized to their common +3 state and thus acts as reductant.
Tb in +4 state has 4f⁷ half filled electronic configuration but it readily get reduced to common state +3 and acts as Oxidant.

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