A) \[n=3,l=2\]
B) \[n=3,l=0\]
C) \[n=2,l=1\]
D) \[n=4,l=0\]
Correct Answer: C
Solution :
Lower the \[3NaCNS+FeC{{l}_{3}}\xrightarrow{{}}\underset{\begin{smallmatrix} ferric \\ sulphocyanide \\ (blood\,red) \end{smallmatrix}}{\mathop{Fe{{(CNS)}_{3}}}}\,+3NaCl\] of an electron, lower will be its energy. If for any two electrons \[N{{a}_{2}}{{B}_{4}}{{O}_{7}}\xrightarrow{{{740}^{o}}C}2\underbrace{2\underset{\begin{smallmatrix} sodium \\ metaborate \end{smallmatrix}}{\mathop{NaB{{O}_{2}}}}\,+\underset{\begin{smallmatrix} boric \\ anhydride \end{smallmatrix}}{\mathop{{{B}_{2}}{{O}_{3}}}}\,}_{glassy\,\,bead}\] is same, the electron with lower value of n, has lower energy. Hence, the correct order of energy is \[W=\frac{EN\,V}{1000}\] \[N=\frac{1.575\times 1000}{63\times 250}\] \[\because \] \[{{H}_{2}}{{C}_{2}}{{O}_{4}}.2{{H}_{2}}O=\frac{126}{2}=63]\] Electron with \[=0.1N\] and \[{{H}_{2}}{{C}_{2}}{{O}_{4}}.2{{H}_{2}}O\] has least energy.
You need to login to perform this action.
You will be redirected in
3 sec
