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Chemical Bonding flash cards

Master Chemical Bonding through 106 NEET-level recall cards, systematically structured one idea at a time. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.

Chemical Bonding, question and answer

30 of this chapter's 106 cards, laid out open so you can read straight through. The remaining 76 are in the interactive deck, where the answer stays hidden until you commit to one.

  1. 1.What is a chemical bond?

    An attractive force that holds together the constituent particles (atoms, ions or molecules) in a chemical species. Atoms bond to achieve a lower energy, more stable configuration.

    Hint: Force holding particles together

  2. 2.State the octet rule.

    Atoms tend to gain, lose or share electrons so as to acquire a stable outer shell of 8 electrons (the noble-gas configuration ns2np6ns^2np^6).

    Hint: 8 electrons in valence shell

  3. 3.What is the Kossel–Lewis approach to bonding?

    Atoms attain the stable noble-gas octet either by transferring electrons (ionic/electrovalent bond) or by sharing electrons (covalent bond), with valence electrons represented as Lewis dot symbols.

    Hint: Transfer vs sharing to get octet

  4. 4.What is a Lewis dot symbol?

    A representation of an atom's valence electrons as dots placed around its chemical symbol. E.g. nitrogen has 5 valence electrons, so N\text{N} is drawn with 5 dots.

    Hint: Dots = valence electrons

  5. 5.How is an ionic (electrovalent) bond formed?

    By the complete transfer of one or more electrons from a metal (low ionization enthalpy) to a non-metal (high electron gain enthalpy), producing oppositely charged ions held by electrostatic attraction.

    Hint: Electron transfer, metal to non-metal

  6. 6.How is a covalent bond formed?

    By the mutual sharing of one or more electron pairs between two atoms, each contributing electrons so that both attain a stable octet.

    Hint: Shared electron pairs

  7. 7.Distinguish single, double and triple covalent bonds.

    A single bond shares 1 electron pair, a double bond shares 2 pairs, and a triple bond shares 3 pairs. Bond strength and bond order increase; bond length decreases from single to triple.

    Hint: 1, 2 or 3 shared pairs

  8. 8.What favours the formation of an ionic bond?

    Low ionization enthalpy of the metal, high (negative) electron gain enthalpy of the non-metal, and a high lattice enthalpy of the resulting crystal.

    Hint: Easy to lose, easy to gain, stable lattice

  9. 9.What is lattice enthalpy of an ionic solid?

    The energy released when one mole of an ionic solid is formed from its constituent gaseous ions (or, equivalently, the energy required to separate one mole of the solid into gaseous ions). Higher lattice enthalpy means a more stable ionic solid.

    Hint: Energy of forming crystal from gaseous ions

  10. 10.How does lattice enthalpy depend on ionic charge and size?

    Lattice enthalpy increases with higher ionic charges and decreases with larger ionic radii, since electrostatic attraction q1q2r\propto \dfrac{q_1 q_2}{r}.

    Hint: More charge, smaller size = stronger

  11. 11.What is the octet rule limitation of the incomplete octet?

    Some stable molecules have a central atom with fewer than 8 electrons, e.g. LiCl\text{LiCl}, BeH2\text{BeH}_2 (4 e^- on Be) and BF3\text{BF}_3 (6 e^- on B).

    Hint: Less than 8: Be, B compounds

  12. 12.What is the expanded octet exception to the octet rule?

    Elements of period 3 and beyond can use vacant dd-orbitals to accommodate more than 8 electrons, e.g. PF5\text{PF}_5 (10 e^-), SF6\text{SF}_6 (12 e^-), H2SO4\text{H}_2\text{SO}_4.

    Hint: More than 8: d-orbitals available

  13. 13.Give an example of an odd-electron molecule that violates the octet rule.

    Molecules with an odd number of electrons such as NO\text{NO} (nitric oxide) and NO2\text{NO}_2 (nitrogen dioxide) cannot satisfy the octet rule for all atoms.

    Hint: NO, NO2 — odd electrons

  14. 14.Name three other limitations of the octet rule.

    It does not explain the shapes of molecules, does not account for the relative stability/energy of molecules, and does not apply to noble gas compounds like XeF2\text{XeF}_2 or many transition-metal compounds.

    Hint: Shape, stability, noble gases

  15. 15.How do you draw a Lewis structure (steps)?

    (1) Sum all valence electrons (add for negative charge, subtract for positive). (2) Choose the least electronegative atom as central. (3) Connect atoms with single bonds. (4) Complete octets of terminal atoms, then the central atom, using lone pairs. (5) Form multiple bonds if the central atom lacks an octet.

    Hint: Count e-, central atom, bonds, octets

  16. 16.Write the Lewis structure electron count for CO2\text{CO}_2.

    Total valence electrons =4+2(6)=16= 4 + 2(6) = 16. Structure is O=C=O\text{O}=\text{C}=\text{O} with two C=O double bonds; each O has 2 lone pairs, C has none, all atoms have octets.

    Hint: 16 electrons, two double bonds

  17. 17.What is the formula for formal charge?

    Formal charge=(valence electrons)(non-bonding electrons)12(bonding electrons)\text{Formal charge} = (\text{valence electrons}) - (\text{non-bonding electrons}) - \tfrac{1}{2}(\text{bonding electrons}).

    Hint: V − lone − half bonding

  18. 18.Why are formal charges useful?

    They help select the most plausible Lewis structure: the preferred structure is usually the one with formal charges closest to zero and with any negative formal charge on the more electronegative atom.

    Hint: Pick most stable resonance/Lewis form

  19. 19.Calculate the formal charge on the central O in ozone (with one single and one double bond).

    Central O: valence =6=6, lone pair electrons =2=2, bonding electrons =6=6 (one single + one double). Formal charge =623=+1= 6 - 2 - 3 = +1.

    Hint: 6 − 2 − 3

  20. 20.What is bond length?

    The equilibrium distance between the nuclei of two bonded atoms in a molecule, usually measured in picometres (pm) or angstroms.

    Hint: Internuclear distance

  21. 21.How is bond length related to bond order?

    As bond order increases, bond length decreases (and bond strength increases). E.g. C–C (154 pm) > C=C (134 pm) > C≡C (120 pm).

    Hint: Higher order = shorter

  22. 22.What is covalent radius?

    One-half of the distance between the nuclei of two identical atoms bonded by a single covalent bond (e.g. half the internuclear distance in Cl2\text{Cl}_2).

    Hint: Half of single-bond distance

  23. 23.Define bond enthalpy.

    The amount of energy required to break one mole of bonds of a particular type in gaseous molecules, expressed in kJ mol1\text{kJ mol}^{-1}. Larger bond enthalpy indicates a stronger bond.

    Hint: Energy to break 1 mole of bonds

  24. 24.How does bond enthalpy vary for single, double and triple bonds?

    It increases with bond multiplicity: e.g. C–C (347347), C=C (611611), C≡C (837837 kJ mol1\text{kJ mol}^{-1}). More shared pairs means a stronger, harder-to-break bond.

    Hint: Triple > double > single

  25. 25.Define bond order.

    The number of bonds (shared electron pairs) between two atoms. In VBT it is the number of shared pairs; in MOT, Bond order=12(NbNa)\text{Bond order} = \tfrac{1}{2}(N_b - N_a).

    Hint: Number of bonds between atoms

  26. 26.What is bond angle?

    The angle between the two bonds (orbitals containing bonding electron pairs) around the central atom, e.g. the H–O–H angle in water is 104.5104.5^\circ.

    Hint: Angle between two bonds at central atom

  27. 27.What is resonance?

    When a single Lewis structure cannot describe a molecule adequately, the actual structure is represented as a resonance hybrid of two or more canonical (contributing) structures that differ only in electron positions.

    Hint: Hybrid of several canonical forms

  28. 28.Are resonance structures real, distinct forms of a molecule?

    No. The individual canonical forms do not exist; the molecule has a single real structure (the resonance hybrid) that is more stable than any contributing form. Resonance energy is the extra stability gained.

    Hint: Only the hybrid is real

  29. 29.Show the resonance in the carbonate ion CO32\text{CO}_3^{2-}.

    The double bond is delocalised over all three C–O bonds, giving three equivalent resonance structures. All three C–O bonds are identical, with a bond order of 1.331.33.

    Hint: Three equivalent forms, bond order 1.33

  30. 30.What is resonance energy?

    The difference in energy between the actual molecule (resonance hybrid) and the most stable single canonical structure. Greater resonance energy means greater stability of the hybrid.

    Hint: Extra stability from delocalisation

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