A) Two
B) Four
C) Three
D) Five
Correct Answer: D
Solution :
\[M{{n}^{25}}\xrightarrow{{}}3{{d}^{5}}+4{{s}^{2}}\] \[M{{n}^{2+}}\xrightarrow{{}}3{{d}^{5}}\] In presence of weak ligand field, there will be no pairing of electrons. So it will form a high spin complex. i.e. the number of unpaired electrons = 5.You need to login to perform this action.
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