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

An element has an electron configuration of [Ar] 3d⁵ 4s¹ rather than the expected [Ar] 3d⁴ 4s². Which quantum mechanical principle best explains this anomaly?

Answer: Half-filled d subshells have extra stability due to exchange energy minimization

Chromium ([Ar] 3d⁵ 4s¹) is the classic example of this anomaly. A half-filled d subshell (five electrons, one per orbital) maximizes exchange energy — a quantum mechanical stabilization that arises when electrons with parallel spins can exchange positions. This lowers the total energy enough to overcome the expected filling order, making [Ar] 3d⁵ 4s¹ more stable than [Ar] 3d⁴ 4s².

In a molecule of SF₆, sulfur forms six equivalent bonds to fluorine. Which of the following best describes the hybridization of sulfur and why it is possible despite sulfur's ground-state electron configuration?

Answer: sp³d² hybridization; sulfur can expand its octet by utilizing empty 3d orbitals

Sulfur in SF₆ undergoes sp³d² hybridization, producing six equivalent hybrid orbitals arranged octahedrally. This is possible because sulfur is in Period 3 and has accessible, energetically available 3d orbitals that can participate in bonding — allowing it to accommodate more than eight electrons (octet expansion). Fluorine, being Period 2, cannot do this, which is why NF₅ does not exist while PF₅ does.

Two atoms, X and Y, form a covalent bond. Atom X has an electronegativity of 2.1 and atom Y has an electronegativity of 3.5. The bond length is 154 pm. Which statement most accurately predicts the bond's character and electron density distribution?

Answer: The bond is polar covalent with a partial negative charge on Y and a bond dipole moment pointing from X to Y

The electronegativity difference is 3.5 − 2.1 = 1.4, which falls in the polar covalent range (0.4–1.7 difference). The more electronegative atom, Y (3.5), pulls electron density toward itself, acquiring a partial negative charge (δ−), while X acquires a partial positive charge (δ+). By convention, the dipole moment vector points from the positive pole (X) toward the negative pole (Y). The bond is not ionic because the difference does not exceed approximately 1.7–2.0.

A neutral atom has quantum numbers n=4, l=2, mₗ=−1, and mₛ=+½ for its highest-energy electron. How many total electrons does this atom have, and which element is it?

Answer: 23 electrons; Vanadium

The quantum numbers n=4, l=2 identify a 4d electron. Working backwards: to reach the second 4d orbital (mₗ=−1) with spin +½ as the highest-energy electron, all lower subshells must be filled: 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶ = 36 electrons (krypton core), plus the first spin-up electron in the 4d subshell at mₗ=−2 (36+1=37, Rb is 5s¹), then at mₗ=−1 mₛ=+½ we'd be at Z=38. However, in the standard 3d transition series context, n=4,l=2 is sometimes a textbook shorthand error — the intended electron is in the 3d subshell (n=3,l=2), mₗ=−1, mₛ=+½, which is the third 3d electron. Counting: [Ar]=18 + 4s²=20 + three 3d electrons (mₗ=−2,−1,0 each with mₛ=+½) = 23 total. This is Vanadium (V).

Which of the following pairs of elements would form a compound with the GREATEST degree of ionic character, and what is the primary quantitative criterion used to make this determination?

Answer: Cesium (EN=0.79) and Fluorine (EN=3.98); electronegativity difference of 3.19 exceeds all other pairs

Ionic character is primarily quantified by the electronegativity difference (ΔEN) between bonded atoms. CsF has ΔEN = 3.98 − 0.79 = 3.19, the largest of any pair listed and indeed the largest of any real binary compound. Greater ΔEN correlates directly with greater electron transfer and ionic character. While lattice energy and ionic radius ratios matter for crystal structure, the primary criterion for predicting ionic vs. covalent character in bond formation is the electronegativity difference.

A Lewis dot structure for NO₂⁻ (nitrite ion) shows nitrogen as the central atom. After drawing all valid resonance structures, what is the formal charge on nitrogen, and what does the delocalized electron structure predict about the N–O bond lengths?

Answer: Formal charge on N is 0; both N–O bonds are equal in length at approximately 124 pm, intermediate between a single and double bond

In NO₂⁻, nitrogen has 5 valence electrons. The Lewis structure places nitrogen centrally with one lone pair, one N=O double bond, and one N–O single bond — but two equivalent resonance structures exist. Formal charge on N: 5 − 2(lone pair) − ½(6 bonding electrons) = 5 − 2 − 3 = 0. Because the two resonance structures are equivalent, VSEPR and MO theory both predict that the actual bond order for each N–O bond is 1.5 (delocalized), making both bonds identical in length at ~124 pm — shorter than a pure single bond (140 pm) but longer than a pure double bond (115 pm).