Answer
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Hint:
They are both elements of the first transition series with their outermost configuration Cu - $\left[ Ar \right]3{{d}^{10}}4{{s}^{1}}$ , $C{{u}^{+2}}$ $\left[ Ar \right]3{{d}^{9}}$ and of Zinc is $\left[ Ar \right]3{{d}^{10}}4{{s}^{2}}$ , $Z{{n}^{+2}}$ $\left[ Ar \right]3{{d}^{10}}$.
Because of the difference in reactivity in Cu and Zn they show coloured and colourless solutions respectively.
Complete step by step answer:
- We already know that transition elements i.e. the d – block elements are those whose atoms in ground state or in one of the common oxidation states have partly filled d – subshell.
- Cu do not have any partly filled d – subshell in their ground state but have partially filled d- subshell in their most common state, therefore we consider it as a transition element.
- But Zn is not considered as a transition element because it has completely filled d – subshell in ground as well as the common oxidation state which is +2.
- Because of the partly filled d – orbital we observe that most transition metals ions and compounds are coloured in solid or in solution state.
- This is because the unpaired \[{{e}^{-}}\] undergo d –d transition and absorb and emit light in visible range.
- But as Zn has completely filled- orbital its salt is white.
- It is seen that Cu shows positive reduction potential due its low hydration energy which does not compensate for high enthalpy of atomisation.
- While the reduction potential of Zn is more negative for +2 states due to their stable configuration.
- As their reactivity is related to their reduction potential and electronic configuration.
- Thus, we can conclude that as zinc is more reactive than copper and displaces copper from copper sulphate solution, the aqueous solution of copper sulphate is blue in colour and that of zinc sulphate is colourless.
Thus, the answer is option D. blue and colourless respectively.
Note:
-$C{{u}^{+2}}$ appears blue by absorption of red wavelength, while Zinc sulphate is colourless because of completely filled d – orbitals.
-Also because of the negative reduction potential of $Z{{n}^{+2}}$ it is more reactive than copper.
They are both elements of the first transition series with their outermost configuration Cu - $\left[ Ar \right]3{{d}^{10}}4{{s}^{1}}$ , $C{{u}^{+2}}$ $\left[ Ar \right]3{{d}^{9}}$ and of Zinc is $\left[ Ar \right]3{{d}^{10}}4{{s}^{2}}$ , $Z{{n}^{+2}}$ $\left[ Ar \right]3{{d}^{10}}$.
Because of the difference in reactivity in Cu and Zn they show coloured and colourless solutions respectively.
Complete step by step answer:
- We already know that transition elements i.e. the d – block elements are those whose atoms in ground state or in one of the common oxidation states have partly filled d – subshell.
- Cu do not have any partly filled d – subshell in their ground state but have partially filled d- subshell in their most common state, therefore we consider it as a transition element.
- But Zn is not considered as a transition element because it has completely filled d – subshell in ground as well as the common oxidation state which is +2.
- Because of the partly filled d – orbital we observe that most transition metals ions and compounds are coloured in solid or in solution state.
- This is because the unpaired \[{{e}^{-}}\] undergo d –d transition and absorb and emit light in visible range.
- But as Zn has completely filled- orbital its salt is white.
- It is seen that Cu shows positive reduction potential due its low hydration energy which does not compensate for high enthalpy of atomisation.
- While the reduction potential of Zn is more negative for +2 states due to their stable configuration.
- As their reactivity is related to their reduction potential and electronic configuration.
- Thus, we can conclude that as zinc is more reactive than copper and displaces copper from copper sulphate solution, the aqueous solution of copper sulphate is blue in colour and that of zinc sulphate is colourless.
Thus, the answer is option D. blue and colourless respectively.
Note:
-$C{{u}^{+2}}$ appears blue by absorption of red wavelength, while Zinc sulphate is colourless because of completely filled d – orbitals.
-Also because of the negative reduction potential of $Z{{n}^{+2}}$ it is more reactive than copper.
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