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प्रश्न
EΘ of Cu is + 0.34V while that of Zn is – 0.76V. Explain.
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उत्तर
High ionisation enthalpy to change \[\ce{Cu(s)}\] to \[\ce{Cu^2+}\] is not balanced by hydration enthalpy. Therefore, it exhibits a positive EΘ value. However, \[\ce{Zn}\] exhibits a lower value of ionization enthalpy because a stable 3d10 configuration is attained after losing two electrons. The hydration energy for \[\ce{Zn^2+}\] is comparable to that of \[\ce{Cu^2+}\]. Therefore, E° for Zn is negative.
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संबंधित प्रश्न
Which of the following cations are coloured in aqueous solutions and why ?
Sc3+, V3+, Ti4+, Mn2+ (At. Nos. Sc = 21, V = 23, Ti = 22, Mn = 25)
Complete the following chemical equations:
`(i) Cr_2O_7^(2-)+6Fe^(2+)+14H^+ ->`
`(ii) 2CrO_4^(2-)+2H^+ ->`
`(iii) 2MnO_4^-+5C_2O_4^(2-)+16H^+ ->`
The elements of 3d transition series are given as: Sc Ti V Cr Mn Fe Co
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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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(i) \[\ce{Co^{2+}}\]
(ii) \[\ce{Cr^{2+}}\]
(iii) \[\ce{Mn^{2+}}\]
(iv) \[\ce{Cr^{3+}}\]
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Reason (R): Greater number of unpaired electrons in transition metals results in weak metallic bonding.
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Atomic sizes
