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Question : 8
Total: 10
Give reasons for the following:
(i) Transition metals form alloys.
(ii)Mn 2 O 3 is basic whereas Mn 2 O 7 is acidic.
(iii)Eu 2 + is a strong reducing agent.
(i) Transition metals form alloys.
(ii)
(iii)
Solution:
(i) Transition metals easily form alloys with other transition metals because they have almost similar size. So they can easily replace each other in the crystal lattice.
(ii) The transition metal oxides in the lower oxidation state of metals are basic in nature and in higher oxidation state they are acidic in nature. The oxidation state ofMn in Mn 2 O 3 is +3 and Mn 2 O 7 has +7. Therefore, Mn 2 O 3 is basic and Mn 2 O 7 is acidic.
(iii) The common oxidation state of lanthanide metals is+ 3 . Eu 2 + is formed by losing the two s electrons, acquires half filled ( 4 f 7 ) configuration. But still, they oxidize to their common +3 state. So, the Eu 2 + loses one electron and is oxidized to Eu 3 + . So Eu 2 + acts as a strong reducing agent.
(ii) The transition metal oxides in the lower oxidation state of metals are basic in nature and in higher oxidation state they are acidic in nature. The oxidation state of
(iii) The common oxidation state of lanthanide metals is
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