Chemical Bonding and Molecular Structure MCQs

Large difference in electronegativity between Cs (≈0.79) and F (≈3.98) gives ΔEN ≈3.2, well above the ionic threshold (~2.0).

Cs readily loses one electron to form Cs⁺, while F gains one electron to form F⁻, producing a lattice of oppositely charged ions.

The resulting strong electrostatic attraction makes the solid CsF predominantly ionic.

Hence the correct answer is CsF.

Count valence electrons: C has 4, O has 6 → total 10. In the triple‑bond structure C is bonded to O with three bonds and has one lone pair, so C owns 5 electrons (3 from bonds + 2 from its lone pair). Formal charge = valence – (owned electrons) = 4 – 5 = –1. Hence the formal charge on carbon is –1.

Sulfur has 6 valence electrons; with four F atoms it forms four S–F bonds and retains one lone pair → 5 electron domains.

According to VSEPR, five domains adopt a trigonal‑bipyramidal arrangement; the lone pair occupies an equatorial position to minimize repulsion.

The resulting shape after ignoring the lone pair is a distorted tetrahedron known as a see‑saw geometry.

Thus the molecular geometry of SF₄ is see‑saw.

Phosphorus in PCl₅ forms five σ‑bonds, requiring five hybrid orbitals.

The central atom uses its 3s, 3p (three) and one 3d orbital to generate five equivalent hybrids.

Five hybrid orbitals correspond to sp³d hybridization.

Thus the hybridization of phosphorus in PCl₅ is sp3d.

O₂ molecular orbital diagram places 12 valence electrons in σ2s, σ*2s, σ2p, π2p, π*2p and σ*2p orbitals.

After filling σ2s, σ*2s, σ2p and the two degenerate π2p orbitals, four electrons remain for the π*2p set.

According to Hund’s rule these four electrons occupy the two π* antibonding orbitals as two unpaired electrons.

Thus O₂ shows paramagnetism due to two unpaired electrons in π* orbitals.

Count valence electrons of O (6) and assign electrons in the Lewis structure O–O=O with one double bond and one single bond; the central O has 2 bonds (one double, one single) and 2 lone pairs → assigned electrons = 2 (from single bond) + 4 (from double bond) + 4 (lone pairs) = 10. Formal charge = valence – assigned = 6 – 10 = +1. Hence the central oxygen carries a +1 charge.

PCl5 has five electron domains around phosphorus → AX5.

According to VSEPR, five regions adopt a trigonal‑bipyramidal arrangement to minimize repulsion.

The three equatorial bonds lie at 120° and the two axial bonds are 180° apart.

Hence the shape is trigonal bipyramidal.

Carbon in ethene forms three sigma bonds (two C–H and one C=C).

Each sigma bond requires one hybrid orbital, so three hybrids are needed.

The hybridisation that gives three equivalent orbitals is sp² (one s + two p).

Thus the central carbon in ethene is sp².

Add the electrons of two He atoms: 2 + 2 = 4 electrons fill the σ1s bonding and σ1s* antibonding MOs (2 each), giving bond order = (2 − 2)/2 = 0.

A zero bond order means the molecule is not bound, so He₂ is not a stable species.

Hence He₂ does NOT exist according to molecular orbital theory. He2.

Hydrogen bonding requires a highly electronegative atom attached to hydrogen, creating a strong dipole.

Fluorine, oxygen and nitrogen are the most electronegative elements that can bear lone‑pair electrons, allowing the H‑X bond to act as a hydrogen‑bond donor and the lone pair to accept.

Thus hydrogen bonding occurs when hydrogen is bonded to F, O or N. Fluorine, oxygen, nitrogen.

Small highly charged cations polarize the electron cloud of the anion, and large, easily polarizable anions enhance this effect.

Greater polarization distorts the ionic electron distribution toward sharing, increasing covalent character.

Thus a small cation together with a large anion leads to the most covalent character.

Formal charge = valence electrons – ( lone‑pair electrons + ½ bonding electrons).

Nitrogen has 5 valence electrons, no lone pairs in NH₄⁺, and forms four N–H bonds (8 bonding electrons).

FC = 5 – (0 + ½·8) = 5 – 4 = +1.

Thus the formal charge on nitrogen in NH₄⁺ is +1.

Xe has 8 valence electrons; each F contributes one electron for a Xe–F bond, giving 4 bonding pairs.

The remaining 2 electrons form one lone pair on Xe, which occupies one position of an octahedral electron‑pair geometry.

Two lone‑pair positions are opposite each other, leaving the four F atoms in the same plane → square planar molecular shape.

Thus the shape of XeF₄ is square planar.

Each carbon in ethyne is sp‑hybridised, giving one σ bond to the other carbon and one σ bond to hydrogen → 2 C–H σ bonds + 1 C≡C σ bond = 3 σ bonds total.

The remaining two unhybridised p orbitals on each carbon overlap side‑by‑side forming two π bonds in the triple bond.

Thus ethyne contains three sigma bonds and two pi bonds, which matches the answer 3 sigma and 2 pi.

N₂ has 10 valence electrons: 5 from each N. In MO diagram, σ2s, σ*2s, σ2p, π2p (two degenerate) are filled; σ*2p remains empty.

Bond order = (bonding electrons – antibonding electrons)/2 = (8 – 2)/2 = 3.

Thus the bond order of N₂ is 3.

Lattice enthalpy ∝ (z⁺·z⁻)/r where z are ionic charges and r is the inter‑ionic distance.

Higher charges increase the electrostatic attraction, while a smaller ionic radius reduces r, both raising the energy required to separate the lattice.

Thus the greatest lattice enthalpy occurs for ions that are both small and carry high charges.

Count valence electrons of S (group 6) = 6. In the drawn structure S is bonded to four O atoms, two with double bonds (4 electrons each) and two with single bonds (2 electrons each).

Assign electrons to S: half of each bond → 2 from each double bond (total 4) and 1 from each single bond (total 2), giving 6 electrons around S.

Formal charge = valence electrons – (non‑bonding electrons + ½ bonding electrons) = 6 – (0 + 6) = 0.

Thus the formal charge on sulfur is 0.

Cl has five electron pairs (3 bonding + 2 lone) → AX₃E₂ electron‑pair geometry is trigonal bipyramidal.

The two lone pairs occupy the equatorial positions to minimize repulsion, leaving three bonding pairs in the axial and one equatorial position.

Only three atoms remain, arranged as two axial and one equatorial bond, giving a T‑shaped molecular geometry.

Hence the shape of ClF₃ is T-shaped.

BeCl2 has a linear geometry; the Be atom forms two sigma bonds and has no lone pairs, requiring two sp hybrid orbitals.

The remaining two unhybridized p orbitals on Be hold the empty p‑π system for bonding with Cl.

Thus the central atom Be is sp hybridized, matching the given answer.

An electron placed in an antibonding orbital occupies a higher‑energy level that opposes the bonding interaction.

Bond order = (number of bonding electrons – number of antibonding electrons)/2, so adding an antibonding electron reduces this value.

A lower bond order means a weaker, less stable bond, thus the molecule’s stability decreases.