A) 4 \[\to \] 2
B) 2 \[\to \] 4
C) 3 \[\to \] 6
D) 6 \[\to \] 2
Correct Answer: C
Solution :
[c] \[{{(H{{e}^{+}})}_{2\to }}=(L{{i}^{2+}}){{n}_{4\to {{n}_{3}}}}\] \[\therefore \frac{{{Z}_{1}}}{{{Z}_{2}}}=\frac{{{n}_{2}}}{{{n}_{4}}}=\frac{{{n}_{1}}}{{{n}_{3}}}\] or \[\frac{2}{3}=\frac{2}{{{n}_{4}}}=\frac{4}{{{n}_{3}}}\] \[\therefore {{n}_{4}}=3\] and \[{{n}_{3}}=6\] \[\therefore \] transition in \[L{{i}^{2+}}\]ion=3\[\to \]6
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