CBSE Class 11 Chemistry Chapter 4: Chemical Bonding and Molecular Structure NCERT Solutions

Solution: A chemical bond is fundamentally an attractive force that binds together the constituent particles, such as atoms or ions, within a chemical species, ensuring its stability. The formation of these bonds is driven by the inherent tendency of systems to reach a state of lower energy and greater stability. This stability is often achieved when atoms attain the electron configuration of the nearest noble gases, characterized by fully filled outermost electron shells. Elements with incomplete valence shells are generally unstable and reactive. To achieve stability, atoms can interact by either sharing electrons, forming a covalent bond, or by transferring one or more electrons from one atom to another, resulting in an ionic bond.

Solution: Lewis dot symbols represent the valence electrons of an atom. The number of dots around the element's symbol corresponds to its valence electrons.

Magnesium (Mg): Mg is in Group 2, so it has 2 valence electrons. Lewis symbol: Mg with two dots.

Sodium (Na): Na is in Group 1, so it has 1 valence electron. Lewis symbol: Na with one dot.

Boron (B): B is in Group 13, so it has 3 valence electrons. Lewis symbol: B with three dots.

Oxygen (O): O is in Group 16, so it has 6 valence electrons. Lewis symbol: O with six dots.

Nitrogen (N): N is in Group 15, so it has 5 valence electrons. Lewis symbol: N with five dots.

Bromine (Br): Br is in Group 17, so it has 7 valence electrons. Lewis symbol: Br with seven dots. Mg Na ⋅

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\bullet B \bullet\\

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\bullet O ⋅ \\

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\bullet N \\

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\end{array}

\begin{array}{c}

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\bullet Br ⋅ \\

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Solution: Lewis symbols represent valence electrons. For ions, the charge indicates the number of electrons gained or lost.

1. Sulphur (S) and S²⁻ ion:

   Sulphur (S) is in Group 16 and has 6 valence electrons. Its Lewis symbol is S with six dots.

   The S²⁻ ion has gained two electrons to achieve a stable octet. Thus, it has 6 + 2 = 8 valence electrons. Its Lewis symbol is [S with eight dots]²⁻.

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   \bullet ⋅ \\

   \bullet S ⋅ \\

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   \bullet S ⋅ \\

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   \end{array} \right]^{2-}

2. Aluminium (Al) and Al³⁺ ion:

   Aluminium (Al) is in Group 13 and has 3 valence electrons. Its Lewis symbol is Al with three dots.

   The Al³⁺ ion has lost its three valence electrons. Thus, it has 0 valence electrons shown around Al. Its Lewis symbol is [Al]³⁺.

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   \bullet Al \\

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   \end{array} and \begin{bmatrix} Al \end{bmatrix}^{3+}

3. Hydrogen (H) and H⁻ ion:

   Hydrogen (H) has 1 valence electron. Its Lewis symbol is H with one dot.

   The H⁻ ion has gained one electron to achieve the stable configuration of Helium (2 electrons). Thus, it has 1 + 1 = 2 electrons. Its Lewis symbol is [H with two dots]⁻. ⋅ H and

   [H:]^{-}

Solution: Lewis structures represent the bonding and lone pair electrons in a molecule or ion.

1. Hydrogen Sulphide (H₂S):

• Total valence electrons: 2(from H) + 6(from S) = 8
• Sulphur is the central atom. Connect H atoms to S with single bonds.
• Structure: H–S–H
• Distribute remaining electrons (8 - 4 = 4) as lone pairs on S.
• Lewis Structure: $$\backslash begin\{array\}\{c\}\; \backslash bullet\; \cdot \; \backslash \backslash \; H\; -\; S\; -\; H\; \backslash \backslash \; \backslash bullet\; \cdot \; \backslash end\{array\}$$

2. Silicon Tetrachloride (SiCl₄):

• Total valence electrons: 4(from Si) + 4(from Cl) = 20
• Silicon is the central atom. Connect Cl atoms to Si with single bonds.
• Structure: Cl–Si–Cl with two Cl atoms above and below.
• Distribute remaining electrons (20 - 8 = 12) as lone pairs on Cl atoms (3 lone pairs each).
• Lewis Structure: (Simplified representation showing bonds and lone pairs on Cl)

3. Beryllium Fluoride (BeF₂):

• Total valence electrons: 2(from Be) + 2(from F) = 10
• Beryllium is the central atom. Connect F atoms to Be with single bonds.
• Structure: F–Be–F
• Distribute remaining electrons (10 - 4 = 6) as lone pairs on F atoms (3 lone pairs each).
• Lewis Structure: $$F\; -\; Be\; -\; F$$ (with 3 lone pairs on each F)

4. Carbonate Ion (CO₃²⁻):

• Total valence electrons: 4(from C) + 3(from O) + 2(charge) = 18
• Carbon is the central atom. Connect O atoms to C with single bonds initially.
• Structure: O–C–O with one O above.
• To satisfy octets, form a double bond between C and one O.
• Distribute remaining electrons (18 - 8 = 10, after forming one double bond and two single bonds) as lone pairs on O atoms. The double-bonded O gets 2 lone pairs, single-bonded Os get 3 lone pairs each.
• Lewis Structure: (Resonance structures exist)

5. Formic Acid (HCOOH):

• Total valence electrons: 1(from H) + 4(from C) + 2(from O) + 1(from OH) = 8
• Structure: H is bonded to C, C is bonded to one O with a double bond, and to another O with a single bond. The second O is bonded to H.
• Distribute remaining electrons as lone pairs on the double-bonded O.
• Lewis Structure: $$\backslash begin\{array\}\{c\}\; O\; \backslash \backslash \; ||\; \backslash \backslash \; H\; -\; C\; -\; OH\; \backslash end\{array\}$$