The orbital angular momentum for a $d -$ electron is:

\sqrt{6} \frac{h}{2 π}

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\sqrt{2} h

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\frac{h}{2 π}

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2 h

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Solution

The correct option is D $\sqrt{6} \frac{h}{2 π}$
For a $d -$ electron, $l = 2$
$∴$ orbital angular momentum $= \sqrt{l (l + 1)} \frac{h}{2 π} = \sqrt{6} \frac{h}{2 π}$

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