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

Which quantum number determines the shape of an electron's orbital, and what value corresponds to a d-orbital?

Answer: Angular momentum quantum number (l), l = 2

The angular momentum quantum number (l) defines the shape of an orbital. l = 0 corresponds to s-orbitals, l = 1 to p-orbitals, and l = 2 to d-orbitals. The principal quantum number (n) defines the energy level/shell, not the shape.

An element has the electron configuration [Kr] 4d¹⁰ 5s² 5p⁴. What is the total number of unpaired electrons in its ground state?

Answer: 2

The 5p⁴ subshell has 3 orbitals. By Hund's rule, electrons fill each orbital singly before pairing. With 4 electrons in 3 orbitals, two orbitals get 1 electron each and one gets 2 (paired). This results in exactly 2 unpaired electrons. The 4d¹⁰ and 5s² subshells are fully paired.

Which of the following correctly explains why the second ionization energy of sodium (Na) is dramatically higher than its first ionization energy?

Answer: The second electron is removed from a fully filled, stable noble gas core configuration

After Na loses its first valence electron (from 3s¹), the resulting Na⁺ ion has the electron configuration of neon ([He] 2s² 2p⁶) — a stable noble gas core. Removing the second electron requires breaking into this stable, fully-filled configuration, which requires a vastly greater amount of energy. The atomic radius actually decreases after ionization, and shielding decreases (not increases) with fewer electrons.

A molecule of SF₄ has 4 bonding pairs and 1 lone pair on the central sulfur atom. What is the correct molecular geometry (not electron geometry)?

Answer: See-saw (seesaw)

SF₄ has 5 electron domains around sulfur (4 bonding + 1 lone pair), giving it a trigonal bipyramidal electron geometry. However, molecular geometry describes only the positions of atoms, not lone pairs. With the lone pair occupying an equatorial position, the four fluorine atoms adopt a see-saw shape. Tetrahedral is 4 domains (no lone pairs); square planar requires 4 bonding + 2 lone pairs.

In a Born-Haber cycle for the formation of MgCl₂, which step accounts for the largest energy input (most endothermic)?

Answer: Second ionization energy of magnesium

The second ionization energy of magnesium (removing an electron from Mg⁺ to form Mg²⁺) is the largest single energy-consuming step. Mg⁺ has a [Ne] 3s¹ configuration, and removing that remaining valence electron is far more difficult than removing the first (from neutral Mg with [Ne] 3s²). The lattice energy released offsets all endothermic steps, but among the inputs, IE₂ of Mg (~1451 kJ/mol) exceeds IE₁ (~738 kJ/mol), sublimation (~148 kJ/mol), and Cl₂ bond dissociation (~242 kJ/mol).

Which type of intermolecular force is primarily responsible for the unusually high surface tension and boiling point of water compared to H₂S, despite H₂S having a larger molar mass?

Answer: Hydrogen bonding

Water (H₂O) exhibits hydrogen bonding because oxygen is highly electronegative and has lone pairs that attract the δ⁺ hydrogen atoms of adjacent molecules. This uniquely strong intermolecular force requires significantly more energy to overcome than the dipole-dipole and London dispersion forces present in H₂S. Although H₂S has a higher molar mass (larger London forces), water's hydrogen bonds dominate, giving it a much higher boiling point (100°C vs. −60°C) and surface tension. Ion-dipole forces involve actual ions, which are not present here.