Atomic Structure and Bonding Flashcards

An element has the electron configuration [Kr] 4d¹⁰ 5s² 5p⁴. Which of the following correctly identifies this element and its most stable ion?

Answer: Tellurium (Te), forming Te²⁻

The configuration [Kr] 4d¹⁰ 5s² 5p⁴ places this element in period 5, group 16 — that is Tellurium (Te). With 6 valence electrons (5s²5p⁴), Te most readily gains 2 electrons to achieve the stable [Kr] 4d¹⁰ 5s² 5p⁶ noble-gas configuration, forming the Te²⁻ ion.

Which pair of atoms would form the bond with the GREATEST degree of ionic character based on electronegativity difference?

Answer: Cs and F

Ionic character increases with electronegativity difference (ΔEN). Cs has an EN of ~0.79 and F has an EN of 3.98, giving ΔEN ≈ 3.19 — the largest among these pairs. Li–Cl ≈ 2.0, Ba–O ≈ 2.5, and K–Br ≈ 1.96. The Cs–F bond therefore has the greatest ionic character.

A neutral atom of element X has 3 complete electron shells and its outermost subshell is exactly half-filled with unpaired electrons. Which of the following is element X?

Answer: Phosphorus (P)

Three complete shells means the valence electrons are in n=3. A half-filled outermost subshell with all electrons unpaired is the hallmark of a p³ configuration (one electron in each of the three p orbitals). The n=3 element with configuration [Ne] 3s² 3p³ is Phosphorus (P). Sulfur has 3p⁴ (not half-filled). Aluminum has 3p¹. Arsenic is in period 4, not period 3.

In a molecule of SF₄, the central sulfur atom obeys an expanded octet. What is the electron geometry (including lone pairs) and the molecular geometry (shape) of SF₄?

Answer: Trigonal bipyramidal electron geometry; see-saw molecular geometry

Sulfur in SF₄ has 4 bonding pairs and 1 lone pair, totaling 5 electron domains. Five electron domains produce a trigonal bipyramidal electron geometry. The lone pair occupies an equatorial position (minimizing repulsion), pushing the four F atoms into a see-saw (or seesaw/butterfly) molecular shape.

The first four successive ionization energies of an unknown element are approximately 738, 1,451, 7,733, and 10,540 kJ/mol. To which group in the periodic table does this element most likely belong?

Answer: Group 2

There is a dramatic jump between the 2nd ionization energy (1,451 kJ/mol) and the 3rd (7,733 kJ/mol) — a nearly fivefold increase. This large jump occurs when the next electron to be removed comes from a complete, lower-energy shell (a noble-gas core). This means the element has 2 valence electrons and belongs to Group 2 (alkaline earth metals). The actual values closely match those of Magnesium.

Which of the following correctly ranks the bond lengths from SHORTEST to LONGEST for the carbon–carbon bonds found in the species ethane (C₂H₆), ethene (C₂H₄), ethyne (C₂H₂), and benzene (C₆H₆)?

Answer: C₂H₂ < C₆H₆ < C₂H₄ < C₂H₆

Bond length decreases as bond order increases because more shared electrons pull the nuclei closer together. C₂H₂ has a C≡C triple bond (bond order 3, ~120 pm), benzene C₆H₆ has delocalized bonds with bond order 1.5 (~140 pm), C₂H₄ has a C=C double bond (bond order 2, ~134 pm), and C₂H₆ has a C–C single bond (bond order 1, ~154 pm). Correct order shortest to longest: C₂H₂ < C₆H₆ < C₂H₄ < C₂H₆. Note that benzene's delocalized bond order (1.5) makes it shorter than a pure single bond but longer than a double bond.