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Question
Account for the following:
Cu+ ion is unstable in aqueous solution.
Solution
Cu2+ is more stable than Cu+ in aqueous medium. Cu2+ has high hydration energy that compensates for the energy required to remove one electron from Cu+ to form Cu2+. So in aqueous medium, Cu+ gives Cu2+ and Cu.
Chemical reaction: \[\ce{2Cu^+_{( aq)} -> Cu^{2+}_{ (aq)} + Cu_{(s)}}\]
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RELATED QUESTIONS
Why do interstitial compounds have higher melting points than corresponding pure metals?
|
`E_((M^(2+)/M)` |
Cr | Mn | Fe | Co | Ni | Cu |
| -0.91 | -1.18 | -0.44 | -0.28 | -0.25 | -0.34 |
From the given data of E0 values, answer the following questions :
(1) Why is `E_(((Cu^(2+))/(Cu)))` value exceptionally positive
(2) Why is `E_(((Mn^(2+))/(Mn)))` value is highly negative as compared to other elements
(3) Which is the stronger reducing agents Cr2+ or Fe2+ ? Give Reason.
Account for the following:
E° value for the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+.
NF3 is possible, but NF5 is not. Why?
Transition metals with lowest melting point is:
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.
In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices on the basis of the above passage.
Assertion: Cations of transition elements occur in various valence states.
Reason: Large number of oxides of transition elements are possible.
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.
In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices on the basis of the above passage.
Assertion: Transition metals form protective oxide films.
Reason: Oxides of transition metals are always stoichiometric.
While filling up of electrons in the atomic orbitals, the 4s orbital is filled before the 3d orbital but reverse happens during the ionisation of the atom. Explain why?
Complete the following reaction and justify that it is a disproportionation reaction:
\[\ce{3MnO^{2-}4 + 4H^+ -> \underline{}\underline{}\underline{}\underline{} + \underline{}\underline{}\underline{}\underline{} + 2H2O}\]
Write the ionic equation for reaction of KI with acidified KMnO4.
