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प्रश्न
Write balanced chemical equations for the conversion of `CrO_4^(2-)` to `Cr_2O_7^(2-)` in acidic medium and `Cr_2O_7^(2-)` to `CrO_4^(2-)`
in basic medium.
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उत्तर
`K_2Cr_2O_7 + 2KOH -> 2K_2CrO_4 + H_2O`
`2K_2CrO_4 + 2HCl -> K_2Cr_2O_7 + 2KCl + H_2O`
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संबंधित प्रश्न
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Following are the transition metal ions of 3d series:
Ti4+, V2+, Mn3+, Cr3+
(Atomic numbers: Ti = 22, V = 23, Mn = 25, Cr = 24)
Answer the following:
1) Which ion is most stable in an aqueous solution and why?
2) Which ion is a strong oxidising agent and why?
3) Which ion is colourless and why?
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The transition metals generally form coloured compounds.
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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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| Column I (Property) | Column II (Metal) | |
| (i) | An element which can show +8 oxidation state | (a) \[\ce{Mn}\] |
| (ii) | 3d block element that can show | (b) \[\ce{Cr}\] |
| upto +7 oxidation state | (c) \[\ce{Os}\] | |
| (iii) | 3d block element with highest melting point | (d) \[\ce{Fe}\] |
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\[\ce{A + H^+ -> B + H2O + Na^+}\]
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(Round off to the nearest integer).
[Given: `sqrt3 = 1.73, sqrt2 = 1.41`]
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Reason (R): Greater number of unpaired electrons in transition metals results in weak metallic bonding.
Write the number of unpaired electrons in Cr3+.
(Atomic number of Cr = 24)
