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PUCET (Punjab) Chemical Bonding and Molecular Structure — practice questions

Explore 97 free Chemical Bonding and Molecular Structure questions in PUCET (Punjab) Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Chemical Bonding and Molecular Structure questions and exercises with answers

Showing questions 1–30 of 97.

What is the shape of a molecule whose central atom is $sp^3$ hybridised and has no lone pairs (e.g. $\mathrm{CH_4}$)?Read the answer and worked solution →![](https://qallery.app/diagrams/v2_bonding_seed_1/img-24.jpeg) What is the hybridisation of the central P atom in $\mathrm{PCl_5}$?Read the answer and worked solution →Arrange the bond angles of $\mathrm{H_2O}$, $\mathrm{NH_3}$ and $\mathrm{CH_4}$ in decreasing order.Read the answer and worked solution →Which of the following molecules has a **non-zero** dipole moment?Read the answer and worked solution →Using molecular orbital theory, the bond order of the $\mathrm{O_2}$ molecule is:Read the answer and worked solution →According to the octet rule, atoms achieve stability by attaining an outer shell configuration of:Read the answer and worked solution →The number of valence electrons in the Lewis dot symbol of nitrogen is:Read the answer and worked solution →In the Lewis structure of CO₃²⁻, the total number of valence electrons to be accounted for is:Read the answer and worked solution →For the NH₄⁺ ion, the total valence electrons available for writing the Lewis structure is:Read the answer and worked solution →In a Lewis structure, the formal charge on an atom is calculated as:Read the answer and worked solution →Which of the following compounds is an example of *incomplete* octet of the central atom?Read the answer and worked solution →Which of the following is most likely to have the greatest covalent character (Fajans' rules)?Read the answer and worked solution →Among H–H, F–F, Cl–Cl, Br–Br, I–I, the bond length is shortest for:Read the answer and worked solution →The average O–H bond enthalpy in water is calculated from the two stepwise dissociation values (502 and 427 kJ mol⁻¹) as:Read the answer and worked solution →Among the following, which molecule has zero net dipole moment?Read the answer and worked solution →One Debye unit (1 D) is equivalent to:Read the answer and worked solution →According to VSEPR theory, the order of repulsion between electron pairs is:Read the answer and worked solution →The shape of a molecule of type AB₄ (no lone pair on A), e.g., CH₄, is:Read the answer and worked solution →The shape of SF₄ is:Read the answer and worked solution →The shape of ClF₃ is:Read the answer and worked solution →XeF₄ adopts which geometry?Read the answer and worked solution →Which of the following overlaps is positive (gives bonding)?Read the answer and worked solution →A sigma (σ) bond is formed by:Read the answer and worked solution →In sp hybridisation, each hybrid orbital has the following s and p character:Read the answer and worked solution →In BeCl₂ (gaseous), the central Be is sp-hybridised. The molecular geometry and Cl–Be–Cl bond angle are:Read the answer and worked solution →The hybridisation of the central atom and shape of NH₃ are:Read the answer and worked solution →Hybridisation of the central atom in PCl₅ is:Read the answer and worked solution →The hybridisation of S in SF₆ is:Read the answer and worked solution →The hybridisation of each carbon in ethyne (C₂H₂) is:Read the answer and worked solution →According to MO theory, the bond order in a molecule is:Read the answer and worked solution →

Worked practice questions

What is the shape of a molecule whose central atom is $sp^3$ hybridised and has no lone pairs (e.g. $\mathrm{CH_4}$)?

  1. Trigonal pyramidal
  2. Tetrahedral
  3. Bent / V-shaped
  4. Square planar
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Correct answer: B. Tetrahedral

Four $sp^3$ hybrid orbitals point to the vertices of a regular tetrahedron, giving a bond angle of $109.5°$. With no lone pairs to compress these angles, the molecular geometry equals the electron-pair geometry. Methane ($\mathrm{CH_4}$) is the textbook example.

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![](https://qallery.app/diagrams/v2_bonding_seed_1/img-24.jpeg) What is the hybridisation of the central P atom in $\mathrm{PCl_5}$?

  1. $sp^2$
  2. $sp^3d^2$
  3. $sp$
  4. $sp^3d$
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Correct answer: D. $sp^3d$

P in $\mathrm{PCl_5}$ has five bonded chlorines and no lone pair. That gives five electron domains. The hybridisation needed is $sp^3d$ (one $s$ + three $p$ + one $d$ orbital = five hybrid orbitals). The molecular geometry is trigonal bipyramidal. $sp^3d^2$ would give six domains and an octahedral geometry, which is what you see in $\mathrm{SF_6}$, not $\mathrm{PCl_5}$.

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Arrange the bond angles of $\mathrm{H_2O}$, $\mathrm{NH_3}$ and $\mathrm{CH_4}$ in decreasing order.

  1. $\mathrm{CH_4} > \mathrm{H_2O} > \mathrm{NH_3}$
  2. $\mathrm{NH_3} > \mathrm{CH_4} > \mathrm{H_2O}$
  3. $\mathrm{H_2O} > \mathrm{NH_3} > \mathrm{CH_4}$
  4. $\mathrm{CH_4} > \mathrm{NH_3} > \mathrm{H_2O}$
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Correct answer: D. $\mathrm{CH_4} > \mathrm{NH_3} > \mathrm{H_2O}$

All three central atoms are $sp^3$ hybridised so the base bond angle is $109.5°$. Lone pairs compress that angle (the lone-pair-to-bond-pair repulsion is stronger than bond-pair-to-bond-pair). Approximate bond angles: - $\mathrm{CH_4}$ (0 lone pairs): $109.5°$ - $\mathrm{NH_3}$ (1 lone pair): $107°$ - $\mathrm{H_2O}$ (2 lone pairs): $104.5°$ More lone pairs means more compression, so the order is $\mathrm{CH_4} > \mathrm{NH_3} > \mathrm{H_2O}$.

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