Chemical Bonding(12th) flash cards

1. 1.What is a chemical bond, and what fundamentally drives its formation?

   A chemical bond is the attractive force holding atoms together in a molecule or ion. Bonds form because the resulting aggregate has lower potential energy (greater stability) than the separated atoms; atoms tend toward a stable (usually noble-gas) electronic configuration.

   Hint: Energy minimization

2. 2.State the octet rule (Kössel–Lewis).

   Atoms combine by gaining, losing, or sharing electrons so as to attain a stable outer shell of 8 electrons (octet), resembling the nearest noble gas. Hydrogen aims for a duplet (2 electrons).

   Hint: 8 in the valence shell

3. 3.How does an ionic (electrovalent) bond form?

   By complete transfer of one or more electrons from an electropositive atom (low IE) to an electronegative atom (high electron affinity), producing oppositely charged ions held by electrostatic attraction. Example: $\text{Na}^+\text{Cl}^-$.

   Hint: Transfer, not sharing

4. 4.What atomic-property conditions favor formation of a stable ionic bond?

   (1) Low ionization enthalpy of the metal, (2) high (negative) electron gain enthalpy of the non-metal, (3) high lattice enthalpy of the resulting solid. Large electronegativity difference (roughly $>1.7$) also favors ionic character.

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

5. 5.Define lattice enthalpy of an ionic solid.

   The energy released when one mole of a solid ionic compound is formed from its constituent gaseous ions (or the energy required to separate one mole of solid into gaseous ions). E.g. $\text{Na}^+(g)+\text{Cl}^-(g)\rightarrow \text{NaCl}(s)$; larger magnitude means a more stable crystal.

   Hint: Gaseous ions → solid

6. 6.How do ionic charge and internuclear distance affect lattice energy?

   Lattice energy $U \propto \dfrac{q_+ q_-}{r_+ + r_-}$. It increases with larger ionic charges and decreases with larger interionic distance (bigger ions). Thus $\text{MgO} > \text{NaF}$ and $\text{LiF} > \text{CsF}$.

   Hint: Higher charge, smaller ions ⇒ bigger U

7. 7.State the Born–Haber cycle and its purpose.

   It is a thermochemical cycle (application of Hess's law) relating lattice enthalpy to measurable quantities: sublimation, ionization, bond dissociation, electron gain, and formation enthalpies. Used to calculate lattice enthalpy indirectly since it cannot be measured directly.

   Hint: Hess's law for ionic solids

8. 8.Write the Born–Haber terms whose sum equals $\Delta_f H$ for $\text{NaCl}$.

   $\Delta_f H = \Delta_{sub}H(\text{Na}) + IE(\text{Na}) + \tfrac{1}{2}\Delta_{diss}H(\text{Cl}_2) + \Delta_{eg}H(\text{Cl}) + U_{lattice}$, where $U_{lattice}$ (lattice enthalpy of formation) is negative.

   Hint: Sublimation + IE + ½ bond + EA + lattice

9. 9.List the general properties of ionic compounds.

   High melting/boiling points, hard but brittle crystalline solids, conduct electricity in molten/aqueous state (not solid), generally soluble in polar solvents (water), insoluble in non-polar solvents, non-directional bonding.

   Hint: Hard, high m.p., conduct when molten

10. 10.Why don't ionic solids conduct electricity in the solid state but do when molten?

    In the solid, ions are locked in fixed lattice positions and cannot migrate. On melting (or dissolving), ions become free to move and carry charge, so the melt/solution conducts.

    Hint: Mobility of ions

11. 11.How does a covalent bond form (Lewis picture)?

    By mutual sharing of one or more electron pairs between atoms, each shared pair counting toward the octet of both atoms. Sharing occurs between atoms of similar (usually high) electronegativity. Example: $\text{Cl}-\text{Cl}$.

    Hint: Shared pairs

12. 12.Distinguish sigma ($\sigma$) and pi ($\pi$) bonds.

    A $\sigma$ bond forms by head-on (axial) overlap of orbitals along the internuclear axis (s–s, s–p, p–p, or hybrid); it is cylindrically symmetric and stronger. A $\pi$ bond forms by sideways (lateral) overlap of parallel p orbitals; weaker and it has a nodal plane through the axis.

    Hint: Head-on vs sideways

13. 13.In a triple bond, how many sigma and pi bonds are present?

    One $\sigma$ and two $\pi$ bonds. A single bond = 1 $\sigma$; a double bond = 1 $\sigma$ + 1 $\pi$; a triple bond = 1 $\sigma$ + 2 $\pi$. The first bond between two atoms is always a $\sigma$ bond.

    Hint: 1σ + 2π

14. 14.What is a coordinate (dative) bond?

    A covalent bond in which both shared electrons come from the same atom (the donor). Once formed, it is indistinguishable from a normal covalent bond. Examples: $\text{NH}_4^+$ (N→H), $\text{H}_3\text{O}^+$, $\text{O}_3$, adducts like $\text{H}_3\text{N}\rightarrow\text{BF}_3$.

    Hint: Both electrons from one atom

15. 15.How do you draw a Lewis (electron-dot) structure? Give the steps.

    (1) Count total valence electrons (add for anion charge, subtract for cation). (2) Choose the least electronegative atom (except H) as central. (3) Place single bonds; distribute remaining electrons as lone pairs to complete octets on terminal atoms first. (4) Form multiple bonds if the central atom lacks an octet. (5) Verify with formal charges.

    Hint: Count, skeleton, distribute, multiple bonds, check

16. 16.Define formal charge and give its formula.

    Formal charge (FC) on an atom $= (\text{valence electrons}) - (\text{lone-pair electrons}) - \tfrac{1}{2}(\text{bonding electrons})$. It shows the hypothetical charge assuming equal sharing, helping choose the best Lewis structure.

    Hint: $V - L - \tfrac{1}{2}B$

17. 17.Which Lewis structure is preferred based on formal charges?

    The structure with formal charges closest to zero, and (when charges exist) with the negative formal charge on the more electronegative atom. The sum of formal charges must equal the overall charge of the species.

    Hint: Minimize FC; negative on electronegative atom

18. 18.Give three important exceptions to the octet rule.

    (1) Incomplete octet (electron-deficient): $\text{BeCl}_2$ (4e), $\text{BF}_3$ (6e), $\text{AlCl}_3$. (2) Odd-electron molecules: $\text{NO}$, $\text{NO}_2$, $\text{ClO}_2$. (3) Expanded octet: $\text{PCl}_5$ (10e), $\text{SF}_6$ (12e), $\text{IF}_7$ (14e) using d-orbitals.

    Hint: Deficient, odd-electron, expanded

19. 19.Why can period-3 (and beyond) elements expand their octet but period-2 cannot?

    Elements from period 3 onward have accessible, energetically available vacant d-orbitals in their valence shell to accommodate more than 8 electrons. Period-2 elements (C, N, O, F) have only 2s and 2p orbitals (max 8), so they cannot exceed the octet.

    Hint: Availability of d-orbitals

20. 20.State the postulates of VSEPR theory.

    (1) Molecular shape depends on the number of electron pairs (bonding + lone) around the central atom. (2) Electron pairs arrange to minimize repulsion (maximum separation). (3) Repulsion order: lp–lp > lp–bp > bp–bp. (4) Lone pairs and multiple bonds distort ideal angles.

    Hint: Pairs spread out to minimize repulsion

21. 21.Give the repulsion order between lone pairs (lp) and bond pairs (bp).

    $\text{lp–lp} > \text{lp–bp} > \text{bp–bp}$. Lone pairs occupy more angular space (held by one nucleus), so increasing lone pairs on the central atom reduces the bond angle.

    Hint: Lone pairs push hardest

22. 22.List the ideal geometries for 2, 3, 4, 5, and 6 electron domains.

    2 → linear ($180°$); 3 → trigonal planar ($120°$); 4 → tetrahedral ($109.5°$); 5 → trigonal bipyramidal ($90°,120°,180°$); 6 → octahedral ($90°,180°$).

    Hint: Linear, trig planar, tetra, TBP, octa

23. 23.Explain the bond-angle trend $\text{CH}_4 > \text{NH}_3 > \text{H}_2\text{O}$.

    All have four electron domains (tetrahedral). $\text{CH}_4$ has 0 lone pairs ($109.5°$); $\text{NH}_3$ has 1 lp ($107°$); $\text{H}_2\text{O}$ has 2 lp ($104.5°$). More lone pairs → greater lp–bp repulsion → smaller bond angle.

    Hint: Increasing lone pairs shrink the angle

24. 24.Why is the bond angle in $\text{NH}_3$ ($107°$) larger than in $\text{NF}_3$ ($102°$)?

    In $\text{NH}_3$, N is more electronegative than H, so bonding electrons lie closer to N, increasing bp–bp repulsion and widening the angle. In $\text{NF}_3$, F is more electronegative and pulls bonding electrons away from N, reducing bp–bp repulsion and narrowing the angle.

    Hint: Where the bonding pair sits

25. 25.Define hybridization.

    The intermixing of atomic orbitals of similar energy to form an equal number of new, equivalent hybrid orbitals with definite geometry, better suited for bonding. It explains observed shapes and equal bond strengths (e.g. all four C–H bonds in $\text{CH}_4$ are identical).

    Hint: Mixing orbitals of similar energy

26. 26.Match hybridization to geometry: $sp$, $sp^2$, $sp^3$, $sp^3d$, $sp^3d^2$, $sp^3d^3$.

    $sp$ → linear ($180°$); $sp^2$ → trigonal planar ($120°$); $sp^3$ → tetrahedral ($109.5°$); $sp^3d$ → trigonal bipyramidal; $sp^3d^2$ → octahedral; $sp^3d^3$ → pentagonal bipyramidal.

    Hint: Steric number 2→7

27. 27.Give the steric-number formula used to predict hybridization.

    Steric number (SN) $=$ (number of $\sigma$-bonded atoms) $+$ (number of lone pairs on central atom). SN $=2\Rightarrow sp$, $3\Rightarrow sp^2$, $4\Rightarrow sp^3$, $5\Rightarrow sp^3d$, $6\Rightarrow sp^3d^2$, $7\Rightarrow sp^3d^3$.

    Hint: σ atoms + lone pairs

28. 28.Give the shortcut formula for hybridization/lone pairs of a central atom.

    $H = \tfrac{1}{2}\big[V + M - C + A\big]$, where $V=$ valence electrons of central atom, $M=$ monovalent atoms attached, $C=$ cation charge, $A=$ anion charge. $H=2,3,4,5,6,7$ gives $sp, sp^2, sp^3, sp^3d, sp^3d^2, sp^3d^3$.

    Hint: ½[V + monovalent − cation + anion]

29. 29.Why do $\pi$ bonds not affect hybridization/geometry in VSEPR?

    Only $\sigma$ bonds and lone pairs determine the hybridization and the arrangement of electron domains. A multiple bond counts as one domain; the extra $\pi$ electrons lie in unhybridized p orbitals and only fine-tune bond angles.

    Hint: Count a double/triple bond as one domain

30. 30.State the main ideas of Valence Bond Theory (VBT).

    A covalent bond forms by overlap of half-filled atomic orbitals of two atoms with opposite spins; greater overlap → stronger bond. Electrons remain largely localized on the parent atoms. It explains directional bonding, hybridization, and bond strength but not magnetic behavior of $\text{O}_2$.

    Hint: Overlap of half-filled orbitals, opposite spins

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