The hydrogen atom is not actually electronegative enough to form bonds to xenon. Were the xenon-hydrogen bond to exist, what would be the structure of XeH4? Double-click any atom and type Xe to change the label.
Draw the molecule by placing atoms on the grid and connecting them with bonds. Show all lone pairs of electrons.
Hydrogen, beryllium, and boron are examples of elements that can have less than an octet of electrons in a covalent compound. Even without an octet, hydrogen achieves the helium configuration, and therefore noble-gas stability, by forming one covalent bond. Beryllium and boron simply do not have enough electrons to form the number of bonds needed to achieve an octet. Instead, Be uses its two valence electrons to form two bonds, and B uses its three valence electrons to form three bonds. None of these elements will have any lone pairs.
Elements in the third row of the periodic table and beyond can have more than an octet of electrons in a covalent compound. This can be explained by the existence of an empty d subshell available to these elements, which allows them to expand their valence to a number greater than eight.
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Our tutors have indicated that to solve this problem you will need to apply the Lewis Dot Structures: Neutral Compounds concept. You can view video lessons to learn Lewis Dot Structures: Neutral Compounds. Or if you need more Lewis Dot Structures: Neutral Compounds practice, you can also practice Lewis Dot Structures: Neutral Compounds practice problems.