To draw the Lewis structure for beryllium hydride (BeH₂), we start by determining the total number of valence electrons. Beryllium (Be) has 2 valence electrons, and each hydrogen (H) has 1 valence electron. Since there are two hydrogens, we have:
Total valence electrons = 2 (from Be) + 1×2 (from H) = 4 valence electrons.
Next, we place the beryllium atom in the center since it is less electronegative than hydrogen. We can then surround it with the two hydrogen atoms:
Be
/
\
H H
Now, we form bonds between the beryllium and the hydrogen atoms. Each H atom shares its one valence electron with Be, forming two single covalent bonds:
H:Be:H
Thus, the Lewis structure for BeH₂ looks like this:
H — Be — H
Now, let’s evaluate whether beryllium satisfies the octet rule. The octet rule states that atoms tend to form bonds in such a way that they each have eight electrons in their valence shell, resembling the electron configuration of noble gases. However, beryllium is an exception to this rule.
In the case of BeH₂, beryllium actually has only 4 electrons around it (two from each bond with hydrogen). This means that beryllium does not complete an octet. Instead, it is stable with just four electrons.
Conclusion: The Lewis structure of BeH₂ shows that beryllium does not satisfy the octet rule as it only has four electrons in its valence shell. Beryllium is somewhat unique in its bonding behavior, as it prefers to form two bonds and does not require eight electrons to be stable.