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(C) The magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n+2)} \, B.M.$,where $n$ is the number of unpaired electrons.Given $\mu = 1.73

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$3$

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(C) The magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n+2)} \, B.M.$,where $n$ is the number of unpaired electrons.Given $\mu = 1.73 \, B.M.$,we have $\sqrt{n(n+2)} = 1.73$.Squaring both sides,$n(n+2) = 3$,which simplifies to $n^2 + 2n - 3 = 0$.Solving the quadratic equation,$(n+3)(n-1) = 0$,so $n = 1$.Titanium $(Ti)$ has an atomic number of $22$ and an electronic configuration of $[Ar] 3d^2 4s^2$.For $n = 1$ (one unpaired electron),the configuration must be $3d^1$,which corresponds to the $Ti^{3+}$ ion (oxidation state $+3$).

If a titanium compound exhibits a magnetic moment of $1.73 \, B.M.$,what is the oxidation state of $Ti$ in the compound?

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