GATE Life Sciences Structure and Bonding Syllabus

Structure and Bonding syllabus — full chapter and topic list

Expand any chapter to see its topics and sub-topics. This is the whole examinable outline for Structure and Bonding in GATE Life Sciences, not a summary of it.

1. 1

   Ionic and Covalent Bonding

   6 topics

   • MO and VB Approaches for Diatomic Molecules
   • VSEPR Theory and Shape of Molecules
   • Hybridization
   • Resonance
   • Dipole Moment
   • Structure Parameters such as Bond Length, Bond Angle, and Bond Energy
2. 2

   Hydrogen Bonding and Van der Waals Interactions

   2 topics

   • Hydrogen Bonding
   • Van der Waals Interactions
3. 3

   Ionic Solids

   2 topics

   • Ionic Radii
   • Lattice Energy (Born-Haber Cycle)
4. 4

   HSAB Principle

   overview

   Examined as a single unit within Structure and Bonding — no further topic split in the official outline.

Structure and Bonding flashcards for GATE Life Sciences

22 of 50 cards from the Structure and Bonding deck — real questions with worked answers.

1. In Molecular Orbital (MO) theory, how do atomic orbitals combine to form molecular orbitals?

   Atomic orbitals combine by Linear Combination of Atomic Orbitals (LCAO). $N$ atomic orbitals give $N$ molecular orbitals: in-phase (additive) combination forms a lower-energy bonding MO, and out-of-phase (subtractive) combination forms a higher-energy antibonding MO.

2. What is the formula for bond order in MO theory, and what does it indicate?

   $$\text{Bond Order} = \frac{1}{2}\left(N_b - N_a\right)$$ where $N_b$ = electrons in bonding MOs and $N_a$ = electrons in antibonding MOs. A higher bond order means a stronger, shorter bond; a bond order of $0$ means the molecule does not exist.

3. What is the MO electron configuration and bond order of $\ce{O2}$, and why is it paramagnetic?

   $\ce{O2}$: $\sigma_{2s}^{2}\,\sigma_{2s}^{*2}\,\sigma_{2p_z}^{2}\,\pi_{2p}^{4}\,\pi_{2p}^{*2}$. Bond order $=\frac{1}{2}(8-4)=2$. It is paramagnetic because the two $\pi^{*}$ electrons occupy degenerate orbitals singly with parallel spins.

4. For diatomics up to $\ce{N2}$, how does $s$–$p$ mixing alter the MO energy ordering?

   For $\ce{Li2}$ through $\ce{N2}$, $s$–$p$ mixing raises $\sigma_{2p_z}$ above the $\pi_{2p}$ orbitals, giving the order $\pi_{2p} < \sigma_{2p_z}$. For $\ce{O2}$, $\ce{F2}$ the mixing is small, so $\sigma_{2p_z} < \pi_{2p}$.

5. How does Valence Bond (VB) theory describe a covalent bond?

   VB theory describes a covalent bond as the overlap of two singly-occupied atomic orbitals on adjacent atoms, pairing electrons of opposite spin in the region between the nuclei. Bonds remain localized between the two bonded atoms.

6. What is the key conceptual difference between MO theory and VB theory?

   MO theory treats electrons as delocalized over the whole molecule in molecular orbitals, while VB theory keeps electrons localized in bonds formed by overlapping atomic orbitals (with resonance/hybridization added to fix shortcomings).

7. Distinguish a $\sigma$ bond from a $\pi$ bond in terms of orbital overlap.

   A $\sigma$ bond forms by head-on (axial) overlap of orbitals along the internuclear axis and is cylindrically symmetric. A $\pi$ bond forms by sideways (lateral) overlap of $p$ orbitals above and below the axis, with a nodal plane containing the axis; $\sigma$ bonds are stronger than $\pi$ bonds.

8. State the central postulate of VSEPR theory.

   Electron pairs (bonding and lone pairs) in the valence shell of the central atom arrange themselves as far apart as possible to minimize mutual repulsion, thereby determining the molecular geometry.

9. Rank the strength of electron-pair repulsions in VSEPR theory.

   $$\text{lone pair–lone pair} > \text{lone pair–bond pair} > \text{bond pair–bond pair}$$ Lone pairs occupy more angular space and compress adjacent bond angles.

10. What is the steric number (electron domain count), and how is it calculated?

    The steric number = (number of atoms bonded to the central atom) + (number of lone pairs on the central atom). It determines the electron-pair geometry. A double or triple bond counts as one domain.

11. Give the electron-pair geometry and ideal bond angle for steric numbers 2 through 6.

    SN 2: linear, $180^{\circ}$. SN 3: trigonal planar, $120^{\circ}$. SN 4: tetrahedral, $109.5^{\circ}$. SN 5: trigonal bipyramidal, $90^{\circ}/120^{\circ}$. SN 6: octahedral, $90^{\circ}$.

12. Predict the shape and bond angle of $\ce{H2O}$ using VSEPR.

    $\ce{H2O}$ has steric number 4 (2 bond pairs + 2 lone pairs), so it is bent/angular. Lone-pair repulsion compresses the angle from $109.5^{\circ}$ to about $104.5^{\circ}$.

13. Predict the shape and bond angle of $\ce{NH3}$ using VSEPR.

    $\ce{NH3}$ has steric number 4 (3 bond pairs + 1 lone pair), giving a trigonal pyramidal shape with a bond angle of about $107^{\circ}$ (reduced from $109.5^{\circ}$ by the lone pair).

14. What geometry does $\ce{SF6}$ adopt and why?

    $\ce{SF6}$ has steric number 6 (6 bond pairs, no lone pairs), giving a regular octahedral geometry with all $\ce{F-S-F}$ angles equal to $90^{\circ}$ (and $180^{\circ}$ trans).

15. In a trigonal bipyramidal arrangement, where do lone pairs preferentially go, and why?

    Lone pairs occupy equatorial positions because there they experience only two $90^{\circ}$ repulsions (versus three in axial positions), minimizing repulsion. This gives shapes like seesaw ($\ce{SF4}$), T-shaped ($\ce{ClF3}$), and linear ($\ce{XeF2}$).

16. What is hybridization?

    Hybridization is the mixing of atomic orbitals of similar energy on the same atom to form an equal number of new, equivalent hybrid orbitals oriented to give the observed molecular geometry and to maximize bonding overlap.

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

    $sp$: linear ($180^{\circ}$); $sp^2$: trigonal planar ($120^{\circ}$); $sp^3$: tetrahedral ($109.5^{\circ}$); $sp^3d$: trigonal bipyramidal; $sp^3d^2$: octahedral.

18. How do you determine the hybridization of a central atom from its steric number?

    Steric number = hybridization: SN 2 $\to sp$, SN 3 $\to sp^2$, SN 4 $\to sp^3$, SN 5 $\to sp^3d$, SN 6 $\to sp^3d^2$. (Steric number counts bonded atoms plus lone pairs.)

19. What is the hybridization and shape of the carbon atoms in ethyne, $\ce{C2H2}$?

    Each carbon is $sp$ hybridized, giving a linear geometry ($180^{\circ}$). The triple bond consists of one $\sigma$ bond (from $sp$ overlap) and two $\pi$ bonds (from unhybridized $p$ orbitals).

20. How does increasing $s$-character in a hybrid orbital affect bond angle and electronegativity?

    More $s$-character gives larger bond angles (closer to the orbital's geometry) and higher effective electronegativity, because $s$ orbitals are closer to the nucleus. Order of $s$-character: $sp\,(50\%) > sp^2\,(33\%) > sp^3\,(25\%)$.

21. What is resonance?

    Resonance is the representation of a molecule whose true bonding cannot be shown by a single Lewis structure as a weighted combination (hybrid) of two or more valid contributing structures that differ only in the placement of electrons, not atoms.

22. How does resonance affect molecular stability and bond lengths?

    Resonance delocalizes electrons, lowering the energy and stabilizing the molecule (resonance/delocalization energy). It also equalizes bond lengths; e.g., all six $\ce{C-C}$ bonds in benzene are identical, intermediate between single and double bonds.

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