NEET Chemical Bonding and Molecular Structure Previous Year Questions with Solutions

Boron belongs to the second period with electronic configuration 1s² 2s² 2p¹. In the n = 2 principal quantum shell, only s and p subshells exist (one 2s and three 2p orbitals, totaling 4 orbitals). Because there are no vacant d-orbitals in the second energy level, Boron cannot expand its valence shell beyond 8 electrons. Hence, its maximum covalency is strictly limited to four (forming species like BF₄⁻).

• • Statement (1) is completely true.

Beryllium has atomic number 4 with configuration 1s² 2s². Its valence shell is n = 2, which consists of one 2s orbital and three 2p orbitals, providing a total of 1 + 3 = 4 valence orbitals, not three.

• • Statement (2) is false.

Magnesium (Z = 12) and Aluminium (Z = 13) belong to the third period (n = 3). The n = 3 shell contains one 3s, three 3p, and five 3d orbitals (total = 1 + 3 + 5 = 9 valence orbitals). Because vacant 3d orbitals are accessible: - Magnesium can expand its coordination number/covalency up to 6 in complexes like [Mg(H₂O)₆]²⁺. - Aluminium has 9 valence orbitals, not 5, and regularly exhibits a covalency of 6 (e.g., [AlF₆]³⁻).

• • Statements (3) and (4) are both false.

Neon has atomic number Z = 10. A hypothetical homonuclear diatomic neon molecule (Ne₂) contains 2 × 10 = 20 electrons.

• • Core electrons from 1s: 4 electrons.
• • Valence electrons from 2s and 2p: 16 electrons.

For homonuclear diatomics with Z > 7, no significant 2s-2p mixing occurs. The molecular orbitals are ordered by increasing energy: σ_{1s} < σ*_{1s} < σ_{2s} < σ*_{2s} < σ_{2p_z} < (π_{2p_x} = π_{2p_y}) < (π*_{2p_x} = π*_{2p_y}) < σ*_{2p_z}

• • Total electrons accommodated = 2 + 2 + 2 + 2 + 2 + 4 + 4 + 2 = 20 electrons.

The orbital of highest energy that contains electrons in the ground-state electron configuration is defined as the HOMO. In Ne₂, the last two electrons occupy the σ*_{2p_z} (antibonding σ_{2p}) orbital.

• • HOMO is σ*_{2p}.
• • Matches Option (4).

In Molecular Orbital Theory, the bond order is defined as half the difference between the number of bonding electrons (N_b) and antibonding electrons (N_a):

• • A positive bond order (B.O. > 0) indicates a stable molecule.
• • If B.O. = 0 (N_b = N_a), the antibonding destabilization completely cancels the bonding stabilization.
• • The molecule cannot hold together and is unstable or non-existent (e.g., He₂, Be₂).
• • Therefore, Statement I is unequivocally FALSE.

Bond order reflects the multiplicity of the covalent bond between two atoms:

• • For carbon-carbon bonds: C-C (B.O. = 1, length = 154 pm), C=C (B.O. = 2, length = 134 pm), C≡C (B.O. = 3, length = 120 pm).
• • As bond order increases, bond length DECREASES, and bond dissociation enthalpy increases.
• • Statement II states that bond length increases, which is completely FALSE.

Since both statements are false, select Option (2).

• • Matches Option (2) with complete scientific accuracy.

Examine experimental dipole moment values in Debye (D) from NCERT Chapter 4:

• • H₂O has two bond dipoles adding vectorially to two lone pairs in a bent geometry.
• • NH₃ has three bond dipoles reinforcing the single lone pair in a pyramidal geometry.
• • CHCl₃ has opposing C-H and C-Cl bond dipoles, resulting in a net moment of 1.04 D.
• • Order A is completely TRUE.

Determine the number of non-bonding electron pairs on xenon and compare covalent radii:

• • Order B claims XeF₄ > XeO₃ > XeF₂, which contradicts the true count. Order B is FALSE.
• • Covalent bond lengths: N-O single bond (140 pm) > C-H bond (109 pm) > O-H bond (96 pm).
• • Order C claims O-H > C-H > N-O, which is the exact reverse. Order C is FALSE.

Compare bond dissociation enthalpies from NCERT thermodynamics and chemical bonding data:

• • N₂ possesses a triple bond (:N≡N:), O₂ possesses a double bond (O=O), and H₂ has a single bond (H-H).
• • The order N₂ > O₂ > H₂ is completely TRUE.
• • Therefore, A and D only are correct, matching Option (1).

In valence bond theory: Ethane involves sp³-sp³ axial overlap forming one σ-bond. Ethene involves sp²-sp² axial overlap (one σ-bond) and sideways 2p-2p overlap (one π-bond). Ethyne involves sp-sp axial overlap (one σ-bond) and two mutually perpendicular sideways 2p-2p overlaps (two π-bonds).

For homonuclear diatomic molecules with Z ≤ 7, the π2px and π2py molecular orbitals are lower in energy than the σ2pz orbital due to 2s-2p mixing. The 12-electron configuration of C2 is:

The bond order is (8 - 4)/2 = 2. The four valence electrons responsible for the double bond reside entirely in the degenerate π2px and π2py bonding orbitals. Consequently, the double bond in C2 consists entirely of two π-bonds with no σ-bond.

Central atom Nitrogen has 5 valence electrons. It forms 3 single bonds with hydrogen atoms and retains 1 non-bonding lone pair: Steric Number = 3 + 1 = 4 (sp³ hybridization). The electron-pair geometry is tetrahedral, but the molecular shape (positions of nuclei) is Trigonal Pyramidal.

Central atom Bromine has 7 valence electrons. Forming 5 Br-F bonds consumes 5 electrons, leaving 2 electrons (1 lone pair): Steric Number = 5 + 1 = 6 (sp³d² hybridization). Octahedral electron geometry with one lone pair gives a Square Pyramidal shape.

Central atom Xenon has 8 valence electrons. Forming 4 Xe-F bonds leaves 4 non-bonding electrons (2 lone pairs): Steric Number = 4 + 2 = 6 (sp³d² hybridization). To minimize lone pair-lone pair repulsions, the two lone pairs occupy trans axial positions (180° apart), leaving the 4 fluorines in an equatorial plane: Square Planar shape.

Central atom Sulfur has 6 valence electrons, all shared in 6 S-F bonds with 0 lone pairs: Steric Number = 6 + 0 = 6 (sp³d² hybridization). Completely symmetrical regular Octahedral shape.

In the carbonate ion (CO3²⁻), the carbon atom forms one double bond (C=O) and two single bonds (C-O⁻) with formal charges -1 on each single-bonded oxygen. Because all three oxygens are chemically and spatially equivalent, the double bond is delocalized equally among the three oxygen atoms, generating three equivalent canonical Lewis resonance structures:

Both NH3 and NF3 have trigonal pyramidal geometries with one lone pair. In NH3, nitrogen is more electronegative than hydrogen, so N-H bond dipoles reinforce the lone pair dipole: μ = 1.47 D. In NF3, fluorine is more electronegative than nitrogen, so N-F bond dipoles oppose the lone pair dipole, resulting in a tiny net dipole: μ = 0.24 D. Hence μ(NH3) > μ(NF3). Statement 1 is false.

Ozone (O3) is an angular molecule with a central oxygen atom bonded to two terminal oxygens. It possesses only TWO principal canonical resonance structures:

Boron trifluoride (BF3) has sp² hybridization and symmetrical trigonal planar geometry with 120° bond angles. The vector sum of the three identical B-F bond dipole moments is identically zero:

According to the Lewis-Kossel octet rule, main-group atoms tend to gain, lose, or share electrons to attain a stable valence configuration of eight electrons (s² p⁶). Species deviating from 8 valence electrons around the central atom are classified as either electron-deficient (incomplete octet, < 8 e⁻) or hypervalent (expanded octet, > 8 e⁻).

Evaluate the total valence electrons surrounding the central atom in each monomeric Lewis structure:

The species not having eight electrons around their central atom are AlCl₃ (6 e⁻), BeCl₂ (4 e⁻), and PCl₅ (10 e⁻). Thus, exactly 3 species violate the octet rule.

In homonuclear diatomic molecules, atomic orbitals of comparable energy and appropriate symmetry combine to form bonding (σ, π) and antibonding (σ*, π*) molecular orbitals. The internuclear axis is taken as the z-axis, so head-on overlap forms σ 2p_z, and lateral overlaps form degenerate π 2p_x and π 2p_y.

For elements up to nitrogen (Z ≤ 7, total electrons ≤ 14), the 2s and 2p atomic orbital energy levels lie close enough to undergo significant quantum mixing. This interaction raises the energy of the σ 2p_z orbital above the degenerate π 2p_x and π 2p_y orbitals:

Arranging all molecular orbitals in order of increasing energy:

Phosphorus has 5 valence electrons, all bonded to chlorine atoms forming 5 $\sigma$-bonds with zero lone pairs ($sp^3d$ hybridization). The ideal geometry is trigonal bipyramidal:

Sulfur has 6 valence electrons, all forming 6 $\sigma$-bonds with fluorine atoms with zero lone pairs ($sp^3d^2$ hybridization). The regular symmetrical geometry is octahedral:

Bromine has 7 valence electrons: 5 form $\sigma$-bonds with fluorine and 2 form 1 lone pair ($5\text{ bp} + 1\text{ lp} = 6$ electron pairs, $sp^3d^2$ hybridization). According to VSEPR, the lone pair occupies one axial site, giving a square pyramidal shape:

Boron has 3 valence electrons, forming 3 $\sigma$-bonds with fluorine atoms with zero lone pairs ($sp^2$ hybridization). The molecular shape is trigonal planar with bond angles of $120^\circ$:

Boron ($Z = 5$) has the ground-state electronic configuration $1s^2 2s^2 2p^1$. In the excited state, one $2s$ electron is promoted to an empty $2p$ orbital: $1s^2 2s^1 2p_x^1 2p_y^1 2p_z^0$.

The one $2s$ orbital and two $2p$ orbitals mix to form three equivalent $sp^2$ hybrid orbitals directed towards the corners of an equilateral triangle at bond angles of $120^\circ$ (trigonal planar geometry). Each hybrid orbital overlaps with a $2p$ orbital of fluorine to form a $\sigma$-bond.

The central boron atom forms three single $\sigma$-bonds with fluorine and possesses no lone pairs. Each single covalent bond contains 2 shared electrons:

The hybridization is $sp^2$ and the number of electrons around the central atom is 6, corresponding to Option (3).

A molecule is non-polar if the vector sum of all individual bond dipoles and lone-pair moments cancels to zero ($\vec{\mu}_{\text{net}} = \sum \vec{\mu}_i = 0$). This occurs in regular, symmetrical geometries where all ligands attached to the central atom are identical and there are no stereochemically active lone pairs disrupting symmetry.

Antimony ($Sb$) is in Group 15 with 5 valence electrons. In $SbCl_5$, it forms five single covalent bonds with chlorine atoms and possesses zero lone pairs ($sp^3d$ hybridization). Its molecular geometry is a regular trigonal bipyramidal (TBP): - Equatorial Plane: Three $Sb-Cl$ bonds are directed towards the corners of an equilateral triangle at $120^\circ$ bond angles. The vector sum of three equal planar vectors separated by $120^\circ$ is identically zero: $$\vec{\mu}_{\text{eq1}} + \vec{\mu}_{\text{eq2}} + \vec{\mu}_{\text{eq3}} = 0$$ - Axial Axis: Two $Sb-Cl$ bonds lie along a straight line perpendicular to the equatorial plane at $180^\circ$ pointing in opposite directions: $$\vec{\mu}_{\text{ax1}} + \vec{\mu}_{\text{ax2}} = 0$$ Consequently, the overall net dipole moment is zero:

- $POCl_3$: Distorted tetrahedral geometry where the strong $P=O$ dipole cannot be cancelled by the three $P-Cl$ bond dipoles ($\mu = 2.4\text{ D} \neq 0$). - $CH_2O$: Planar geometry, but the strong polar $C=O$ dipole points in the opposite direction of the two weakly polar $C-H$ bonds ($\mu = 2.3\text{ D} \neq 0$). - $NO_2$: Angular/bent molecule (bond angle $\approx 134^\circ$) with an odd electron on nitrogen, giving a permanent dipole moment ($\mu = 0.32\text{ D} \neq 0$).

According to Molecular Orbital Theory (MOT), a homonuclear diatomic molecule exists as a stable chemical entity only if the number of bonding electrons exceeds antibonding electrons ($$N_b > N_a$$), yielding a positive non-zero bond order ($$\text{BO} > 0$$).

Helium ($$Z = 2$$) has configuration $$1s^2$$. For hypothetical $$He_2$$ ($$4\text{ electrons}$$): $$\text{MO Configuration} = (\sigma 1s)^2 (\sigma^* 1s)^2$$ $$\text{Bond Order} = \frac{N_b - N_a}{2} = \frac{2 - 2}{2} = 0$$ Because the antibonding electrons completely cancel the stabilization of the bonding electrons, $$He_2$$ is energetically unstable and does not exist.

- $$Li_2$$ (6 electrons): $$(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 \implies \text{BO} = \frac{4 - 2}{2} = 1$$ (exists in vapor phase). - $$C_2$$ (12 electrons): $$(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x^2 = \pi 2p_y^2) \implies \text{BO} = \frac{8 - 4}{2} = 2$$ (double bond consisting entirely of two pi bonds). - $$O_2$$ (16 electrons): $$\text{BO} = \frac{10 - 6}{2} = 2$$ (paramagnetic with 2 unpaired electrons in $$\pi^* 2p$$).

Only $$He_2$$ has a bond order of zero, confirming that it does not exist.

Dipole moment is a vector quantity (magnitude $$\mu = q \times d$$) pointing from the positive to the negative center of charge. A polyatomic molecule has a net zero dipole moment only when its individual bond dipoles cancel each other symmetrically in space.

- $$\text{BF}_3$$: $$sp^2$$ hybridized, trigonal planar geometry ($$120^\circ$$ bond angles). The vector resultant of any two B-F bonds is equal and opposite to the third B-F bond, so $$\mu_{\text{net}} = 0$$. - $$\text{BeF}_2$$: $$sp$$ hybridized, linear ($$180^\circ$$). The two Be-F bond dipoles oppose and cancel, so $$\mu_{\text{net}} = 0$$. - $$\text{CO}_2$$: $$sp$$ hybridized, linear ($$180^\circ$$). The two C=O bond dipoles cancel, so $$\mu_{\text{net}} = 0$$. - 1,4-dichlorobenzene ($$p$$-dichlorobenzene): The two C-Cl dipole vectors are oriented directly opposite each other at $$180^\circ$$ across the aromatic ring, canceling out completely ($$\mu_{\text{net}} = 0$$).

- $$H_2O$$: $$sp^3$$ hybridized bent geometry with two lone pairs, $$\mu = 1.85\text{ D}$$. - $$NH_3$$: Trigonal pyramidal with one lone pair, $$\mu = 1.47\text{ D}$$. - $$NF_3$$: Trigonal pyramidal, $$\mu = 0.23\text{ D}$$. - $$HF$$: Strongly polar linear diatomic, $$\mu = 1.78\text{ D}$$. - 1,3-dichlorobenzene ($$m$$-dichlorobenzene): The two C-Cl dipoles meet at a $$120^\circ$$ angle, giving a substantial net dipole moment ($$\mu \neq 0$$).

Only the fourth set consists exclusively of non-polar molecules with $$\mu = 0$$.