For a $d$ electron, the orbital angular momentum is :

\sqrt{6} \frac{h}{2 π}

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\sqrt{2} \frac{h}{2 π}

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\sqrt{6} \frac{h}{4 π}

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\frac{h}{2 π}

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Solution

The correct option is A $\sqrt{6} \frac{h}{2 π}$
Orbital angular momentum $(l) = \sqrt{l (l + 1)} \frac{h}{2 π}$ $= \sqrt{6} \frac{h}{2 π}$ ( $l = 2$ for d-orbital)

Hence, option A is correct.

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