- Basic Math and Science Atomic Structure and Bonding Questions and Answers Flashcards

Which of the following best explains why the second ionization energy of magnesium (Mg) is significantly higher than its first ionization energy, but the second ionization energy of sodium (Na) is dramatically higher than that same ratio?

Answer: After sodium loses one electron it achieves a noble gas configuration, so removing the second electron disrupts a fully filled n=2 shell

Na⁺ has the electron configuration of neon ([He] 2s² 2p⁶), a complete and highly stable noble gas configuration. Removing a second electron means breaking into this stable core, requiring an enormous amount of energy — far more than the jump seen between IE1 and IE2 for Mg, where both electrons being removed come from the same 3s subshell.

In a molecule with sp² hybridization, the unhybridized p orbital on each carbon participates in π bonding. Which statement correctly describes the geometry and bond angles of such a molecule?

Answer: Trigonal planar geometry with bond angles of exactly 120°, with the π bond perpendicular to the molecular plane

sp² hybridization produces three hybrid orbitals arranged in a trigonal planar geometry with 120° bond angles. The remaining unhybridized p orbital on each carbon is perpendicular to this plane and overlaps side-by-side with adjacent p orbitals to form the π bond, which lies above and below the molecular plane.

An element has the electron configuration [Xe] 4f¹⁴ 5d⁶ 6s². Which of the following statements about this element is MOST accurate?

Answer: It is a 5d transition metal (osmium), and the filled 4f subshell is a result of lanthanide contraction affecting its properties

The configuration [Xe] 4f¹⁴ 5d⁶ 6s² corresponds to osmium (Os), a 5d transition metal in period 6. The completely filled 4f¹⁴ subshell results from the lanthanide series filling before the 5d series. Lanthanide contraction (the poor shielding of f electrons) causes 5d and 6s orbitals to be smaller and more tightly held than expected, influencing osmium's chemical behavior.

Two atoms form a bond with a very high lattice energy. One atom has an electronegativity of 0.9 and the other has an electronegativity of 4.0. Which bond property combination is MOST consistent with this description?

Answer: Ionic bond with large charge magnitudes and small ionic radii contributing to high lattice energy

The electronegativity difference is 3.1 (4.0 − 0.9), which greatly exceeds the 1.7 threshold for ionic character. Ionic bonds with large ion charges (e.g., Mg²⁺ and O²⁻) and small ionic radii produce very high lattice energies because lattice energy is proportional to (charge product)/(sum of radii). The combination of large electronegativity difference, high charges, and small ions maximizes the Coulombic attraction captured in the Born-Landé equation.

Which of the following pairs of molecules are isoelectronic AND have the same molecular geometry?

Answer: N₂O and CO₂

N₂O (nitrous oxide) and CO₂ are isoelectronic: both have 16 valence electrons total and consist of three atoms in the pattern (X=Y=Z), giving a linear geometry. CO₂ is O=C=O (16 e⁻), and N₂O is N≡N⁺–O⁻ / N=N=O (16 e⁻). NO₂ has an odd number of electrons and is bent; SO₂ has a lone pair on sulfur making it bent; BF₃ and NH₃ differ in geometry (trigonal planar vs. trigonal pyramidal).

When solving the quadratic equation 3x² − 5x − 2 = 0 using the quadratic formula, what are the correct roots, and which mathematical principle explains why two real solutions exist?

Answer: x = 2 and x = −1/3; the discriminant (b² − 4ac) is positive, guaranteeing two distinct real roots

Using the quadratic formula: x = [5 ± √(25 + 24)] / 6 = [5 ± √49] / 6 = [5 ± 7] / 6. This gives x = 12/6 = 2 and x = −2/6 = −1/3. The discriminant b² − 4ac = 25 − 4(3)(−2) = 25 + 24 = 49 > 0, confirming two distinct real roots. This can be verified by factoring: (3x + 1)(x − 2) = 0.