Given the following five species: $NH_3, AlCl_3, BeCl_2, CCl_4, PCl_5$. The total number of these species that

Question

Given the following five species: $NH_3, AlCl_3, BeCl_2, CCl_4, PCl_5$. The total number of these species that do not have eight electrons around the central atom in its/their outermost shell, is:

Accepted Answer

1. **Analyze the Octet Rule:** The octet rule states that atoms tend to combine in such a way that they each have eight electrons in their valence shells. However, there are exceptions.
2. **Analyze each species:**
   - **$NH_3$:** Nitrogen has 5 valence electrons. It forms 3 sigma bonds with H and has 1 lone pair. Total electrons = $3 	imes 2$ (bonding) + 2 (lone pair) = **8 electrons**. (Follows octet rule) .
   - **$AlCl_3$:** Aluminum has 3 valence electrons. It forms 3 sigma bonds with Cl. Total electrons = $3 	imes 2$ = **6 electrons**. (Incomplete octet, electron deficient) .
   - **$BeCl_2$:** Beryllium has 2 valence electrons. It forms 2 sigma bonds with Cl. Total electrons = $2 	imes 2$ = **4 electrons**. (Incomplete octet, electron deficient) .
   - **$CCl_4$:** Carbon has 4 valence electrons. It forms 4 sigma bonds with Cl. Total electrons = $4 	imes 2$ = **8 electrons**. (Follows octet rule) .
   - **$PCl_5$:** Phosphorus has 5 valence electrons. It forms 5 sigma bonds with Cl. Total electrons = $5 	imes 2$ = **10 electrons**. (Expanded octet/Hypervalent) .
3. **Count exceptions:** The species that do *not* have 8 electrons are $AlCl_3$ (6), $BeCl_2$ (4), and $PCl_5$ (10).
4. **Conclusion:** There are **3** such species.

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