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How do we write spin multiplicity for Mn(2+), Mn(7+)?
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Monehin Oluwatoyin
How do we write spin multiplicity for Mn(2+), Mn(7+)?
Spin multiplicity = $2S+1$, where $S$ is the total spin angular momentum. Now $S = \frac{n}{2}$ where $n$ represents total number of unpaired electrons. So now we can write spin multiplicity = $n+1$.
Now coming to your question,
$\ce{Mn}$ has 5 unpaired electrons in it. Therefore, its spin multiplicity $(S) = 5+1 = 6$.
Similarly, $\ce{Mn^{2+}}$ has 5 unpaired elections in it. Therefore, its spin multiplicity $(S) = 5+1=6$.
$\ce{Mn^{7+}}$ has zero number of unpaired electrons. Therefore, its spin multiplicity $(S) = 0+1=1$.
Spin multiplicity = $2S+1$, where $S$ is the total spin angular momentum. Now $S = \frac{n}{2}$ where $n$ represents total number of unpaired electrons. So now we can write spin multiplicity = $n+1$.
Now coming to your question,
$\ce{Mn}$ has 5 unpaired electrons in it. Therefore, its spin multiplicity $(S) = 5+1 = 6$.
Similarly, $\ce{Mn^{2+}}$ has 5 unpaired elections in it. Therefore, its spin multiplicity $(S) = 5+1=6$.
$\ce{Mn^{7+}}$ has zero number of unpaired electrons. Therefore, its spin multiplicity $(S) = 0+1=1$.
Spin multiplicity = $2S+1$, where $S$ is the total spin angular momentum. Now $S = \frac{n}{2}$ where $n$ represents total number of unpaired electrons. So now we can write spin multiplicity = $n+1$.
Now coming to your question,
Spin multiplicity = $2S+1$, where $S$ is the total spin angular momentum. Now $S = \frac{n}{2}$ where $n$ represents total number of unpaired electrons. So now we can write spin multiplicity = $n+1$.
Now coming to your question,
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