A) \[3B{{r}_{2}}(l)+3{{H}_{2}}O(l)\]
B) \[(B{{r}_{2}})\]
C) \[\frac{d[B{{r}_{2}}]}{dt}=-\frac{3}{5}\frac{d[B{{r}^{-}}]}{dt}\]
D) \[\frac{d[B{{r}_{2}}]}{dt}=-\frac{5}{3}\frac{d[B{{r}^{-}}]}{dt}\]
Correct Answer: C
Solution :
Key Idea For an electron, n maybe 0, 1, 2 ... and \[=\frac{\sigma }{{{\varepsilon }_{0}}}(a-b+c)=\frac{\sigma }{{{\varepsilon }_{0}}}(2a)\] to n -1 and m \[m=-l\] to \[(\because c=a+b)\](including 0) and \[{{V}_{B}}=\frac{1}{4\pi {{\varepsilon }_{0}}}-\frac{4\pi {{a}^{2}}}{a}-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\sigma 4\pi {{b}^{2}}}{b}\] Hence; if \[+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{\sigma 4\pi {{c}^{2}}}{c}\]\[=\frac{\sigma }{{{\varepsilon }_{0}}}\left( \frac{{{a}^{2}}}{c}-b+c \right)\frac{\sigma }{{{\varepsilon }_{0}}}(2a)\,(\because c=a+b)\]to \[{{V}_{C}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{\sigma 4\pi {{a}^{2}}}{c}-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\sigma 4\pi {{b}^{2}}}{c}\] \[+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{\sigma 4\pi {{c}^{2}}}{c}\] \[=\frac{\sigma }{{{\varepsilon }_{0}}}\left( \frac{{{a}^{2}}}{c}-\frac{{{b}^{2}}}{c}+c \right)=\frac{\sigma }{{{\varepsilon }_{0}}}(2a)\,(\because c=a+b)\]to\[{{V}_{A}}={{V}_{C}}\ne {{V}_{B}}\] \[\therefore \] \[{{v}_{S}}=v+{{v}_{B}}\] Therefore, option (c) is not a permissible set of quantum numbers.
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