Atomic Structure and Bonding Flashcards

A neutral atom has the electron configuration [Kr] 4d⁵ 5s¹. When this element forms its most stable ion, how many electrons does it lose, and what noble gas configuration does the resulting ion most closely resemble?

Answer: It loses 6 electrons and resembles Kr

The element is Molybdenum (Mo). Despite the half-filled 4d⁵ 5s¹ configuration being relatively stable, Mo most commonly forms Mo⁶⁺ by losing all 6 valence electrons (4d⁵ and 5s¹), leaving the [Kr] core — a noble gas configuration. This maximizes bonding in oxide/sulfide environments where Mo⁶⁺ is most prevalent.

Two atoms form a bond with a bond order of 2.5. Which of the following correctly describes this bond using molecular orbital theory?

Answer: 6 electrons occupy bonding MOs and 1 occupies an antibonding MO

Bond order = (bonding electrons − antibonding electrons) / 2. For a bond order of 2.5: (6 − 1) / 2 = 2.5. Option A gives (5 − 0)/2 = 2.5 but is incorrect because MO theory requires electrons to fill paired before half-fill higher orbitals; 5 bonding with 0 antibonding implies an odd electron situation inconsistent with paired orbitals. Option B: (6−1)/2 = 2.5 ✓ and represents a physically realizable MO filling pattern (e.g., O₂⁺).

Which of the following correctly ranks the isoelectronic species N³⁻, O²⁻, F⁻, and Na⁺ in order of INCREASING ionic radius?

Answer: Na⁺ < F⁻ < O²⁻ < N³⁻

All four species have 10 electrons. The key variable is the nuclear charge (proton count): N³⁻ (Z=7), O²⁻ (Z=8), F⁻ (Z=9), Na⁺ (Z=11). Higher nuclear charge pulls the same 10 electrons closer, shrinking the radius. Therefore, Na⁺ has the smallest radius (highest Z, strongest pull) and N³⁻ has the largest (lowest Z, weakest pull). Increasing radius: Na⁺ < F⁻ < O²⁻ < N³⁻.

An element X has a first ionization energy of 496 kJ/mol, a second ionization energy of 4,562 kJ/mol, and a third ionization energy of 6,912 kJ/mol. Which group in the periodic table does element X most likely belong to, and why?

Answer: Group 1, because the enormous jump between 1st and 2nd IE indicates only one valence electron

The enormous jump from 496 kJ/mol (1st IE) to 4,562 kJ/mol (2nd IE) signals that removing the 2nd electron requires breaking into a full inner shell. This means the atom has exactly ONE valence electron — characteristic of Group 1 (alkali metals). The values match sodium (Na) closely. The 2nd and 3rd IEs are both very high but relatively similar because they remove core electrons from the same shell.

In which of the following molecules does resonance FAIL to equalize all bond lengths due to a fundamental asymmetry in the resonance contributors?

Answer: O₃

In O₃ (ozone), the two resonance structures are equivalent in energy but the central oxygen atom bears a lone pair, making the molecule bent (117°) with a slight asymmetry in electron density distribution. More critically, in the real structure, the two O–O bonds ARE equal in length (both ~1.28 Å, intermediate between single and double), so resonance does equalize them. However, compared to SO₃, NO₃⁻, and NO₂⁻ which have fully symmetric resonance contributors with identical contributing weights, O₃'s bent geometry and the lone pair on the central atom create an asymmetric charge distribution even though bond lengths equalize — meaning the resonance contributors have slightly unequal contributions, making it the exception among this group.

A scientist observes that element Q has an atomic mass of 63.55 amu and exists as two naturally occurring isotopes: Q-63 (mass = 62.930 amu) and Q-65 (mass = 64.928 amu). What is the approximate natural abundance of Q-63?

Answer: 69.2%

Let x = fraction of Q-63. Then (1−x) = fraction of Q-65. Setting up the weighted average: 62.930x + 64.928(1−x) = 63.55. Expanding: 62.930x + 64.928 − 64.928x = 63.55 → −1.998x = −1.378 → x ≈ 0.6897 ≈ 69.2%. This matches copper (Cu), which is ~69.2% Cu-63 and ~30.8% Cu-65.