Which of the following lanthanoids show +2 oxidation state besides the characteristic oxidation state +3 of lanthanoids?
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Eu
Yb
Explanation:
Cerium (Z = 57) ⇒ Electronic configuration $=[\text{Xe}]4\text{f}^55\text{d}^06\text{s}^2$
Oxidation state of Ce +3, +4.
Europium (Z = 63) ⇒ Electronic configuration $=[\text{Xe}]4\text{f}^75\text{d}^06\text{s}^2$
Oxidation state of Eu = +2, +3.
Ytterbium (Z = 70) ⇒ Electronic configuration $=[\text{Xe}]4\text{f}^{14}5\text{d}^06\text{s}^2$
Oxidation state of Yb = +2, +3.
Holmium (Z = 67) ⇒ Electronic configuration $=[\text{Xe}]4\text{f}^{11}5\text{d}^06\text{s}^2$
Oxidation state of Ho = +3.
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$\ce{KMnO_4}$ acts as an oxidising agent in acidic medium. The number of moles of $\ce{KMnO_4}$ that will be needed to react with one mole of sulphide ions in acidic solution is