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B.Sc Chemistry Chemical Bonding — practice questions

Explore 54 free Chemical Bonding questions in B.Sc Chemistry Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Chemical Bonding questions and exercises with answers

Showing questions 1–30 of 54.

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 →An ELECTROVALENT (ionic) bond is formed when:Read the answer and worked solution →A COVALENT bond is formed by:Read the answer and worked solution →The octet rule states atoms tend to:Read the answer and worked solution →VSEPR stands for:Read the answer and worked solution →The shape of methane (CH₄) is:Read the answer and worked solution →Water (H₂O) molecular shape is:Read the answer and worked solution →Bond order in N₂ molecule is:Read the answer and worked solution →Dipole moment of CO₂ is:Read the answer and worked solution →In NH₃, hybridization of N is:Read the answer and worked solution →BF₃ has shape:Read the answer and worked solution →In SF₆, central S has hybridization:Read the answer and worked solution →The order of bond strength: single < double < triple. Bond LENGTHS go in:Read the answer and worked solution →Bond angle in NH₃ compared to H₂O:Read the answer and worked solution →Molecule with linear shape (besides CO₂):Read the answer and worked solution →Number of σ and π bonds in C₂H₄ (ethene):Read the answer and worked solution →Bond order from MOT (molecular orbital theory):Read the answer and worked solution →Among NaCl, HCl, H₂O — order of bond polarity:Read the answer and worked solution →Lattice energy of an ionic crystal depends on:Read the answer and worked solution →Hybridization in PCl₅ (gaseous phosphorus pentachloride):Read the answer and worked solution →XeF₄ has shape:Read the answer and worked solution →In benzene C₆H₆, each C is:Read the answer and worked solution →Number of π bonds in HCN:Read the answer and worked solution →Among HF, HCl, HBr, HI: order of bond ENTHALPY (strength):Read the answer and worked solution →Effective nuclear charge (Z_eff) felt by an electron in 2p orbital of nitrogen, given Z = 7, screening constant σ ≈ 3.25:Read the answer and worked solution →Among NaCl, NaBr, NaI: order of melting points: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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