To solve this problem, we need to determine the angular momentum of an electron in a $d$-orbital.$\\$The angular momentum of an electron in an atomic orbital is given by the formula:$\\ L = \sqrt{l(l+1)} \cdot \frac{h}{2\pi} \\$where:$\\$• $L$ is the angular momentum,$\\$• $l$ is the azimuthal quantum number (also known as the angular momentum quantum number),$\\$• $h$ is Planck's constant.$\\$For a $d$-orbital, the azimuthal quantum number $l$ is $2. \\$Substitute $l = 2$ into the formula:$\\ L = \sqrt{2(2+1)} \cdot \frac{h}{2\pi} \\ L = \sqrt{6} \cdot \frac{h}{2\pi} \\$Since the options are given in terms of $h,$ we can express $L$ as:$\\ L = \sqrt{6} \cdot \frac{h}{2\pi} \cdot 2\pi \\ L = \sqrt{6} \cdot h \\$Therefore, the angular momentum of an electron in a $d$-orbital is $\sqrt{6} h. \\$This corresponds to Option 4.