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Chemical Bonding - Octet Rule-advanced

Grade 9CBSE

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

🔑Concepts

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The Octet Rule states that atoms of elements combine to form molecules in such a way that each atom has eight electrons in its valence shell, achieving the stable electronic configuration of a noble gas.

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The Duet Rule is a specific case for elements like Hydrogen (HH), Lithium (LiLi), and Beryllium (BeBe), which strive to achieve two electrons in their outermost shell, resembling the Helium (HeHe) configuration.

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Valency is the combining capacity of an atom, determined by the number of electrons it needs to lose, gain, or share to complete its octet. For atoms with vv valence electrons, if v≤4v \le 4, valency is vv; if v>4v > 4, valency is 8−v8 - v.

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Lewis Dot Structures are diagrams that show the bonding between atoms of a molecule and the lone pairs of electrons that may exist. Valence electrons are represented as dots around the chemical symbol.

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Ionic (Electrovalent) Bonding occurs when one atom transfers electrons to another. For example, NaNa (2,8,12, 8, 1) loses an electron to ClCl (2,8,72, 8, 7) to form Na+Na^+ and Cl−Cl^-, both achieving the octet configuration.

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Covalent Bonding involves the sharing of electron pairs between atoms. In Cl2Cl_2, each Chlorine atom shares one electron to complete their respective octets.

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Advanced Exceptions - Incomplete Octet: In some compounds like BCl3BCl_3 or BeH2BeH_2, the central atom has fewer than 8 electrons (e.g., Boron in BCl3BCl_3 has only 6 electrons).

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Advanced Exceptions - Expanded Octet: Elements in the 3rd period and beyond (like PP, SS, ClCl) can accommodate more than 8 electrons due to the availability of dd-orbitals. Examples include PCl5PCl_5 (10 electrons) and SF6SF_6 (12 electrons).

📐Formulae

Valency=8−number of valence electrons (if >4)\text{Valency} = 8 - \text{number of valence electrons (if } > 4)

Formal Charge=[Total valence e−]−[Non-bonding e−]−12[Bonding e−]\text{Formal Charge} = [\text{Total valence } e^-] - [\text{Non-bonding } e^-] - \frac{1}{2}[\text{Bonding } e^-]

Electronic Configuration of Noble Gases=ns2np6 (except He)\text{Electronic Configuration of Noble Gases} = ns^2 np^6 \text{ (except He)}

💡Examples

Problem 1:

Show the formation of Magnesium Chloride (MgCl2MgCl_2) using the Octet Rule and electron transfer.

Solution:

Mg→Mg2++2e−Mg \rightarrow Mg^{2+} + 2e^- (Electronic config: 2,8,2→2,82, 8, 2 \rightarrow 2, 8) 2Cl+2e−→2Cl−2Cl + 2e^- \rightarrow 2Cl^- (Electronic config: 2,8,7→2,8,82, 8, 7 \rightarrow 2, 8, 8) Combined: Mg2++2[Cl]−→MgCl2Mg^{2+} + 2[Cl]^- \rightarrow MgCl_2

Explanation:

Magnesium has 2 valence electrons. It loses them to achieve a stable octet (2,82, 8). Each of the two Chlorine atoms accepts one electron to complete its own octet (2,8,82, 8, 8). The electrostatic attraction between Mg2+Mg^{2+} and Cl−Cl^- forms the ionic bond.

Problem 2:

Explain the covalent bonding in Phosphorus Pentachloride (PCl5PCl_5) and identify if it follows the Octet Rule.

Solution:

In PCl5PCl_5, Phosphorus (PP) is the central atom. It has 5 valence electrons (3s23p33s^2 3p^3). It forms 5 covalent bonds with 5 Chlorine atoms.

Explanation:

Each P−ClP-Cl bond involves sharing one pair of electrons. Consequently, Phosphorus is surrounded by 5×2=105 \times 2 = 10 electrons. This is an example of an Expanded Octet, which is an exception to the standard Octet Rule, made possible because Phosphorus has vacant 3d3d orbitals.

Problem 3:

Calculate the valency of Oxygen (Z=8Z=8) and Nitrogen (Z=7Z=7) using the Octet Rule.

Solution:

Oxygen: K=2,L=6K=2, L=6. Valency =8−6=2= 8 - 6 = 2. Nitrogen: K=2,L=5K=2, L=5. Valency =8−5=3= 8 - 5 = 3.

Explanation:

Oxygen needs to gain 2 electrons to reach 8, and Nitrogen needs to gain 3 electrons to reach 8 in their respective valence shells.