Co Ordination Compounds Question 116
Question: The d-electron configurations of $ C{r^{2+}},M{n^{2+}},F{e^{2+}} $ and $ C{o^{2+}} $ are $ d^{4},d^{5},d^{6} $ and $ d^{7} $ respectively. Which one of the following will exhibit the lowest paramagnetic behaviour? (Atomic no. Cr = 24, Mn = 25, Fe= 26, Co = 27).
Options:
A) $ {{[Co{{(H_2O)}_6}]}^{2+}} $
B) $ {{[Cr{{(H_2O)}_6}]}^{2+}} $
C) $ {{[Mn{{(H_2O)}_6}]}^{2+}} $
D) $ {{[Fe{{(H_2O)}_6}]}^{2+}} $
Show Answer
Answer:
Correct Answer: A
Solution:
| Electronic configuration | No. of unpaired electrons |
|---|---|
| $ C{o^{2+}} $ | $ d^{7} $ |
| $ C{r^{2+}} $ | $ d^{4} $ |
| $ M{n^{2+}} $ | $ d^{5} $ |
| $ F{e^{2+}} $ | $ d^{6} $ |
$ \therefore $ Since $ C{o^{2+}} $ has lowest no.
of unpaired electrons hence lowest paramagnetic behaviour is shown by $ {{[Co{{(H_2O)}_6}]}^{2+}} $
BETA
