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प्रश्न
Why are interstitial compounds well known for transition metals?
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उत्तर
Interstitial compounds do not correspond to any normal oxidation state of the metal. Because of their composition, these compounds are referred to as interstitial compounds. The principal physical and chemical characteristics of these compounds are as follows:
- They have high melting points, higher than those of pure metals.
- They are very hard; some borides approach diamond in hardness.
- They retain metallic conductivity.
- They are chemically inert.
संबंधित प्रश्न
Why do interstitial compounds have higher melting points than corresponding pure metals?
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E° value for the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+.
In what way is the electronic configuration of the transition elements different from that of the non-transition elements?
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Read the passage given below and answer the following question:
The transition metals when exposed to oxygen at low and intermediate temperatures form thin, protective oxide films of up to some thousands of Angstroms in thickness. Transition metal oxides lie between the extremes of ionic and covalent binary compounds formed by elements from the left or right side of the periodic table. They range from metallic to semiconducting and deviate by both large and small degrees from stoichiometry. Since electron bonding levels are involved, the cations exist in various valence states and hence give rise to a large number of oxides. The crystal structures are often classified by considering a cubic or hexagonal close-packed lattice of one set of ions with the other set of ions filling the octahedral or tetrahedral interstices. The actual oxide structures, however, generally show departures from such regular arrays due in part to distortions caused by packing of ions of different size and to ligand field effects. These distortions depend not only on the number of d-electrons but also on the valence and the position of the transition metal in a period or group.
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Reason: Oxides of transition metals are always stoichiometric.
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\[\ce{3MnO^{2-}4 + 4H^+ -> \underline{}\underline{}\underline{}\underline{} + \underline{}\underline{}\underline{}\underline{} + 2H2O}\]
