Purpose: Self-check electron configuration, Lewis, VSEPR, hybridization, and IMFs.
Q1
Ground-state electron configuration of Fe (Z=26).
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[Ar] 4s² 3d⁶ (or 1s²2s²2p⁶3s²3p⁶4s²3d⁶).
Q2
Configuration of Cu — note the anomaly.
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[Ar] 4s¹ 3d¹⁰ — half/fully filled d-subshell stability.
Q3
How many unpaired electrons in N (Z=7)?
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3 — 2p³ has one e⁻ in each of three p orbitals (Hund's rule).
Q4
VSEPR geometry of NH₃?
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AX₃E (3 bond pairs + 1 lone pair) → trigonal pyramidal; bond angle ~107°.
Q5
Geometry of CO₂.
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Linear, AX₂; bond angle 180°.
Q6
Geometry of SF₆.
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Octahedral, AX₆; 90° angles.
Q7
Polar or nonpolar: CCl₄?
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Nonpolar — symmetric tetrahedral; bond dipoles cancel.
Q8
Hybridization at the central C in CH₄, C₂H₄, C₂H₂.
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CH₄ → sp³; C₂H₄ → sp²; C₂H₂ → sp.
Q9
Hybridization of C in CO₃²⁻.
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sp² — trigonal planar with 3 σ-bonds; π electrons delocalized.
Q10
Strongest IMF in liquid water.
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Hydrogen bonding (O–H...O).
Q11
Rank b.p.: CH₄, NH₃, H₂O, HF.
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CH₄ (−161) < NH₃ (−33) < HF (20) < H₂O (100). H₂O highest due to two H-bond donors and two acceptors per molecule.
Q12
Why is iodine (I₂) a solid at room temperature while F₂ is a gas?
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Larger I₂ has more electrons / greater polarizability → stronger London dispersion forces.