Chemistry - CHEMICAL BONDING Question with Solution | TestHub

$\bar{\mathrm{\mu }}=\mathrm{e}.\mathrm{d}$

$\text{e}\text{=}\text{1}\text{.167e}\left(\frac{\text{1}\text{.602}\text{×}{\text{10}}^{-\text{19}}\text{C}}{{\text{1e}}^{-}}\right)$

$=2.675\times {10}^{-20}\text{C}$

$\text{d}=154.6\text{pm}=154.6\times {10}^{-12}\text{m}$

So$\overline{\mu }=(2.675\times {10}^{-20}\text{C})(154.6\times {10}^{-12})$

$=4.136\times {10}^{-30}\text{Cm}$

$=\frac{4.136\times {10}^{-30}}{3.34\times {10}^{-30}}=1.24\text{D}$