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1Which set of quantum numbers (n, l, m_l, m_s) is permissible for an electron occupying a 3d orbital in a transition metal atom?
A.n = 3, l = 2, m_l = -2, m_s = +1/2
B.n = 3, l = 3, m_l = +1, m_s = -1/2
C.n = 3, l = 1, m_l = 0, m_s = +1/2
D.n = 2, l = 2, m_l = +2, m_s = -1/2
Explanation: For a 3d orbital, the principal quantum number is n = 3 and the azimuthal (angular momentum) quantum number is l = 2. The magnetic quantum number m_l can range from -l to +l (-2, -1, 0, +1, +2), and the spin quantum number m_s can be either +1/2 or -1/2.
2What is the ground-state electron configuration of a neutral copper atom (Cu, Z = 29)?
A.[Ar] 3d^10 4s^1
B.[Ar] 3d^9 4s^2
C.[Ar] 3d^8 4s^2 4p^1
D.[Kr] 3d^10 4s^1
Explanation: Copper exhibits an anomalous electron configuration because a fully filled 3d subshell (3d^10) confers extra thermodynamic stability. One electron from the 4s orbital is promoted to complete the 3d subshell, yielding [Ar] 3d^10 4s^1 instead of the expected [Ar] 3d^9 4s^2.
3Which of the following correctly orders the period 3 elements Na, Mg, Al, and Si by increasing first ionization energy (IE_1)?
A.Na < Al < Mg < Si
B.Na < Mg < Al < Si
C.Al < Na < Mg < Si
D.Si < Mg < Al < Na
Explanation: First ionization energy generally increases across a period due to increasing effective nuclear charge. However, aluminum (Al, [Ne] 3s^2 3p^1) has a slightly lower IE_1 than magnesium (Mg, [Ne] 3s^2) because removing the single 3p electron from Al requires less energy than removing a paired 3s electron from the full 3s subshell of Mg. Thus, the order is Na < Al < Mg < Si.
4According to VSEPR theory, what is the molecular geometry and approximate bond angle of the ammonia molecule (NH_3)?
A.Trigonal pyramidal, ~107°
B.Tetrahedral, ~109.5°
C.Trigonal planar, 120°
D.T-shaped, ~90°
Explanation: Nitrogen in NH_3 has four electron domains (three N-H single bonds and one lone pair), giving a tetrahedral electron-pair geometry. Because lone pair-bonding pair repulsion is stronger than bonding pair-bonding pair repulsion, the bond angle is compressed from 109.5° to approximately 107°, and the molecular geometry is trigonal pyramidal.
5Why does water (H_2O, boiling point 100 °C) have a significantly higher boiling point than hydrogen sulfide (H_2S, boiling point -60 °C)?
A.H_2O molecules form strong intermolecular hydrogen bonds, whereas H_2S experiences weaker dipole-dipole interactions.
B.The H-O covalent bond inside the H_2O molecule is much stronger than the H-S bond inside H_2S.
C.H_2O has a higher molar mass than H_2S, increasing its London dispersion forces.
D.H_2S forms ionic lattice networks in the liquid state that decompose upon boiling.
Explanation: Oxygen is highly electronegative, allowing H_2O molecules to form extensive intermolecular hydrogen bonding networks. Sulfur is less electronegative, so H_2S cannot form strong hydrogen bonds and relies primarily on dipole-dipole interactions and London dispersion forces, resulting in a much lower boiling point.
6What is the hybridization state of the central carbon atom in carbon dioxide (CO_2) and the number of sigma (σ) and pi (π) bonds in the molecule?
A.sp hybridization; 2 σ bonds and 2 π bonds
B.sp^2 hybridization; 3 σ bonds and 1 π bond
C.sp^3 hybridization; 4 σ bonds and 0 π bonds
D.sp hybridization; 4 σ bonds and 0 π bonds
Explanation: In CO_2 (O=C=O), the central carbon atom forms two double bonds and has zero lone pairs, giving it two electron domains and sp hybridization. Each double bond consists of one σ bond and one π bond, yielding a total of 2 σ bonds and 2 π bonds.
7Calculate the energy of a photon of violet light with a wavelength λ = 400 nm. (h = 6.626 × 10^-34 J·s, c = 3.00 × 10^8 m/s)
A.4.97 × 10^-19 J
B.1.99 × 10^-25 J
C.2.65 × 10^-19 J
D.7.95 × 10^-19 J
Explanation: Using Planck's equation E = hc/λ: converting 400 nm to meters gives λ = 4.00 × 10^-7 m. Substituting the constants yields E = (6.626 × 10^-34 J·s × 3.00 × 10^8 m/s) / (4.00 × 10^-7 m) = 1.9878 × 10^-25 / 4.00 × 10^-7 = 4.97 × 10^-19 J.
8What is the de Broglie wavelength of an electron (m_e = 9.11 × 10^-31 kg) moving at a velocity of 2.00 × 10^6 m/s? (h = 6.626 × 10^-34 J·s)
A.0.364 nm
B.3.64 nm
C.0.0364 nm
D.3.64 × 10^-7 m
Explanation: De Broglie wavelength is given by λ = h / (m·v). Substituting the values: λ = (6.626 × 10^-34 J·s) / (9.11 × 10^-31 kg × 2.00 × 10^6 m/s) = 6.626 × 10^-34 / 1.822 × 10^-24 = 3.6366 × 10^-10 m = 0.364 nm.
9Which of the following represents the correct order of decreasing ionic radius for the isoelectronic species O^2-, F^-, Na^+, and Mg^2+?
A.O^2- > F^- > Na^+ > Mg^2+
B.Mg^2+ > Na^+ > F^- > O^2-
C.F^- > O^2- > Mg^2+ > Na^+
D.Na^+ > Mg^2+ > O^2- > F^-
Explanation: All four species are isoelectronic with 10 electrons (configuration 1s^2 2s^2 2p^6). As the nuclear charge Z increases (O: Z=8, F: Z=9, Na: Z=11, Mg: Z=12), the nucleus exerts a stronger electrostatic pull on the electron cloud, shrinking the ionic radius. Thus, ionic radius decreases as O^2- > F^- > Na^+ > Mg^2+.
10In the standard resonance structures of the nitrate anion (NO_3^-), what is the formal charge on the central nitrogen atom?
A.+1
B.0
C.-1
D.+2
Explanation: Nitrogen has 5 valence electrons. In the Lewis structure of NO_3^-, nitrogen forms one double bond and two single bonds to oxygen atoms (sharing 4 bonding electron pairs = 8 electrons, 0 lone pairs). Formal charge = V - N_nonbonding - 1/2(N_bonding) = 5 - 0 - 4 = +1.
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