A) \[{{E}_{1}}\]
B) \[Zn|ZnS{{O}_{4}}\,(0.01M)||CuS{{O}_{4}}(1.0M)|Cu\]
C) \[ZnS{{O}_{4}}\]
D) All of these
Correct Answer: D
Solution :
The compounds having metal ions with unpaired electrons, are coloured. Thus, in \[{{H}_{2}}/Pd-BaS{{O}_{4}}\] \[LiAl{{H}_{4}}/ether\] (One unpaired electron) in \[NaB{{H}_{4}}/ether/{{H}_{3}}{{O}^{+}}\] \[C{{H}_{3}}N{{H}_{2}}\] (Five unpaired electrons) in \[{{(C{{H}_{3}})}_{2}}CHN{{H}_{2}}\] \[C{{H}_{3}}-NH-C{{H}_{3}}\] (Three unpaired electrons) Hence, all compounds are coloured.
You need to login to perform this action.
You will be redirected in
3 sec
