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Chemical Bonding - Exceptions of Octet Rule-advanced

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Octet Rule, based on the electronic theory of chemical bonding, states that atoms tend to combine in such a way that they each have eight electrons in their valence shells, giving them the same electronic configuration as a noble gas (ns2np6ns^2 np^6).

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Incomplete Octet: In some compounds, the number of electrons surrounding the central atom is less than eight. This is common in elements with fewer than four valence electrons, such as LiLi, BeBe, and BB. Examples include LiClLiCl, BeH2BeH_2, and BCl3BCl_3.

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Odd-Electron Molecules: In molecules with an odd number of electrons, like nitric oxide (NONO) and nitrogen dioxide (NO2NO_2), the octet rule cannot be satisfied for all atoms.

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The Expanded Octet: Elements in and beyond the third period of the periodic table have 3d3d orbitals available for bonding. In compounds like PCl5PCl_5, SF6SF_6, and H2SO4H_2SO_4, the central atom has more than eight valence electrons.

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Noble Gas Compounds: Although the octet rule is based on the chemical inertness of noble gases, some noble gases like Xenon (XeXe) and Krypton (KrKr) react with oxygen and fluorine to form compounds like XeF2XeF_2, KrF2KrF_2, and XeOF2XeOF_2.

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Formal Charge: It is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure. It helps in determining the most stable structure when multiple Lewis structures are possible.

📐Formulae

FC=[V]−[L]−12[S]FC = [V] - [L] - \frac{1}{2}[S]

Total Valence Electrons=∑Valence electrons of individual atoms±ionic charge\text{Total Valence Electrons} = \sum \text{Valence electrons of individual atoms} \pm \text{ionic charge}

💡Examples

Problem 1:

Explain why PCl5PCl_5 is considered an exception to the octet rule.

Solution:

In PCl5PCl_5, Phosphorus (PP) is the central atom. The atomic number of PP is 1515, with an electronic configuration of [Ne]3s23p3[Ne] 3s^2 3p^3. It has 55 valence electrons. In PCl5PCl_5, it forms five covalent bonds with five Chlorine atoms.

Explanation:

Since each bond consists of 22 shared electrons, Phosphorus is surrounded by 5×2=105 \times 2 = 10 electrons. This exceeds the 88 electrons required by the octet rule, making it an 'Expanded Octet' exception.

Problem 2:

Calculate the formal charge on the Boron atom in BF3BF_3.

Solution:

For Boron (BB) in BF3BF_3:

  1. Valence electrons (VV) = 33
  2. Lone pair electrons (LL) = 00
  3. Shared electrons (SS) = 66 (from 33 single bonds) FC=3−0−12(6)=3−3=0FC = 3 - 0 - \frac{1}{2}(6) = 3 - 3 = 0

Explanation:

Even though the formal charge is 00, Boron only has 66 electrons in its valence shell in BF3BF_3, making it an incomplete octet molecule.

Problem 3:

Identify the number of electrons around the central atom in Sulphur Hexafluoride (SF6SF_6).

Solution:

Sulphur (SS) has 66 valence electrons ([Ne]3s23p4[Ne] 3s^2 3p^4). In SF6SF_6, it forms 66 single bonds with 66 Fluorine atoms.

Explanation:

Total electrons around Sulphur = 6 bonds×2 electrons/bond=12 electrons6 \text{ bonds} \times 2 \text{ electrons/bond} = 12 \text{ electrons}. This is an example of an expanded octet because 12>812 > 8.