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NEET CHEMISTRYMedium

Given the following five species: NH3,AlCl3,BeCl2,CCl4,PCl5NH_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:

A

One

B

Three

C

Two

D

Four

Step-by-Step Solution

  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:
  • NH3NH_3: Nitrogen has 5 valence electrons. It forms 3 sigma bonds with H and has 1 lone pair. Total electrons = 3×23 \times 2 (bonding) + 2 (lone pair) = 8 electrons. (Follows octet rule) .
  • AlCl3AlCl_3: Aluminum has 3 valence electrons. It forms 3 sigma bonds with Cl. Total electrons = 3×23 \times 2 = 6 electrons. (Incomplete octet, electron deficient) .
  • BeCl2BeCl_2: Beryllium has 2 valence electrons. It forms 2 sigma bonds with Cl. Total electrons = 2×22 \times 2 = 4 electrons. (Incomplete octet, electron deficient) .
  • CCl4CCl_4: Carbon has 4 valence electrons. It forms 4 sigma bonds with Cl. Total electrons = 4×24 \times 2 = 8 electrons. (Follows octet rule) .
  • PCl5PCl_5: Phosphorus has 5 valence electrons. It forms 5 sigma bonds with Cl. Total electrons = 5×25 \times 2 = 10 electrons. (Expanded octet/Hypervalent) .
  1. Count exceptions: The species that do not have 8 electrons are AlCl3AlCl_3 (6), BeCl2BeCl_2 (4), and PCl5PCl_5 (10).
  2. Conclusion: There are 3 such species.
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