A complex formulated as $\ce{Cs_x[NbCl6]}$ has spin-only magnetic moment $\mu_{s.o}$ of approximately 2 B.M. From this value calculate how many unpaired electrons are present.

Spin-only magnetic moment is related to the number of unpaired electrons by the formula

$$\mu_{s.o}= \sqrt{n(n+1)}$$ where $n$ is the number of unpaired electrons.

I let $\sqrt{n(n+1} = 2$ and solved for $n$. I got two values one $+$ and one $-$. Do I just take the positive value and round it to the nearest whole number. In this case I got $n_+= \sqrt{5} - 1 \approx 1.2$.

Does this mean there is $1$ unpaired electron?


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