AP Atomic Structure and Periodicity 2

Question 1

An electron in a hydrogen atom transitions from n = 4 to n = 2. What is the wavelength of the emitted photon? (RH = 2.179 × 10⁻¹⁸ J, h = 6.626 × 10⁻³⁴ J·s, c = 3.00 × 10⁸ m/s)

Accepted Answer

486 nm

Answer

ΔE = RH(1/n₁² − 1/n₂²) = 2.179 × 10⁻¹⁸(1/4 − 1/16) = 2.179 × 10⁻¹⁸ × 0.1875 = 4.086 × 10⁻¹⁹ J. λ = hc/E = (6.626 × 10⁻³⁴ × 3.00 × 10⁸)/(4.086 × 10⁻¹⁹) = 486 nm.

Question 2

Which set of quantum numbers is NOT allowed for an electron in an atom?

Accepted Answer

n=3, l=3, ml=+2, ms=+½

Answer

For n = 3, l can be 0, 1, or 2 only (l must be 0 to n−1). l = 3 is not permitted when n = 3, making this set of quantum numbers invalid.

Question 3

The first ionization energies (in kJ/mol) for elements in Period 3 include an unusually high value for Mg (738) compared to Al (577). What best explains why Al has a lower first ionization energy than Mg?

Accepted Answer

Al's outer electron is in a 3p orbital, which is higher in energy and better shielded than Mg's 3s electron

Answer

Al's highest-energy electron occupies a 3p orbital, which has higher energy and experiences more shielding from the 3s electrons than the 3s² configuration of Mg, making it easier to remove.

Question 4

Which of the following correctly lists the ions in order of increasing radius: Na⁺, Mg²⁺, Al³⁺, O²⁻, F⁻?

Accepted Answer

Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻

Answer

All five are isoelectronic (10 electrons). Larger nuclear charge contracts the electron cloud more, so larger Z → smaller radius. Z: Al(13) > Mg(12) > Na(11) > F(9) > O(8), giving Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻.

Question 5

What is the de Broglie wavelength of an electron (mass = 9.11 × 10⁻³¹ kg) moving at 2.00 × 10⁶ m/s? (h = 6.626 × 10⁻³⁴ J·s)

Accepted Answer

3.64 × 10⁻¹⁰ m

Answer

λ = h/mv = (6.626 × 10⁻³⁴ J·s)/((9.11 × 10⁻³¹ kg)(2.00 × 10⁶ m/s)) = 6.626 × 10⁻³⁴/(1.822 × 10⁻²⁴) = 3.64 × 10⁻¹⁰ m.

AP Practice Test

AP Practice Test

AP Atomic Structure and Periodicity 2