[기출문제] 울산과학기술원(unist) 일반화학 2020-1 기말 기출문제 (정답 포함)

울산과학기술원(unist) 일반화학 2020-1 기말 기출문제 (정답 포함)

 

 

 

1. 시험 정보

 

학교/과목	울산과학기술원(unist) 일반화학
시험명	2020-1 기말
교수명	-
문항수/형식	풀이형 11문제
정답/해설	✅ 있음
파일형식	-

 

 

 

2. 출제 범위 & 키워드

 

일반화학의 전자기파, 원자 구조, 주기율, 화학결합, 분자구조(VSEPR), 분자궤도 이론, 기체 법칙, 상평형 및 분자 간 힘을 종합적으로 다루는 문제

 

 

📚 키워드

 

 

 

3. 기출 미리보기

 

 

파일로 첨부

 

 

 

 

4. 자료 보기

 

[기출 문제] 

 

 

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[정답] 

 

Ch6 1. Electromagnetic waves A and B (The total horizontal length shown is 1.6 × 10⁻⁷ m) (a) Wavelength From the diagram: Wave A completes 5 full wavelengths over 1.6 × 10⁻⁷ m → Wavelength of A = 3.2 × 10⁻⁸ m Wave B completes 2 full wavelengths over 1.6 × 10⁻⁷ m → Wavelength of B = 8.0 × 10⁻⁸ m (b) Frequency Using the speed of light, c = 3.00 × 10⁸ m/s Wave A: Frequency = 3.00 × 10⁸ m/s ÷ 3.2 × 10⁻⁸ m = 9.4 × 10¹⁵ s⁻¹ Wave B: Frequency = 3.00 × 10⁸ m/s ÷ 8.0 × 10⁻⁸ m = 3.75 × 10¹⁵ s⁻¹ (c) Region of the electromagnetic spectrum Wave A: Extreme ultraviolet (EUV) Wave B: Ultraviolet (UV) 2. Hydrogen emission line at 434 nm The line is in the Balmer series, so the final energy level is n = 2. Using known Balmer lines: 434 nm corresponds to the transition n = 5 → n = 2 Answer: Beginning level: n = 5 Ending level: n = 2 3. Validity of quantum number sets Rules: n = 1, 2, 3, ... l = 0 to (n − 1) mₗ = −l to +l mₛ = +1/2 or −1/2 (a) n = 3, l = 3, mₗ = 2, mₛ = +1/2 Invalid — l cannot equal n Invalid quantum number: l (b) n = 4, l = 3, mₗ = −3, mₛ = +1/2 Valid (c) n = 3, l = 1, mₗ = 2, mₛ = +1/2 Invalid — mₗ must be −1, 0, or +1 Invalid quantum number: mₗ (d) n = 5, l = 0, mₗ = 0, mₛ = 0 Invalid — mₛ must be ±1/2 Invalid quantum number: mₛ (e) n = 2, l = 1, mₗ = 1, mₛ = −1/2 Valid Ch7 1. Atomic radius (decreasing order) 기본 경향:아래로 갈수록 증가, 오른쪽으로 갈수록 감소 (a) bromine, chlorine, iodine I > Br > Cl (b) gallium, selenium, arsenic 같은 주기(4주기), 왼쪽이 큼 Ga > As > Se (c) calcium, potassium, zinc 같은 주기(4주기), 왼쪽이 큼 K > Ca > Zn (d) barium, calcium, strontium 같은 족(2족), 아래가 큼 Ba > Sr > Ca 2. Ionization energy (decreasing order) 기본 경향:위로 갈수록 증가, 오른쪽으로 갈수록 증가 (a) selenium, oxygen, tellurium 같은 족(16족) O > Se > Te (b) gold, tantalum, osmium 같은 주기(6주기), 오른쪽이 큼 Os > Au > Ta (c) lead, barium, cesium 같은 주기(6주기) Pb > Ba > Cs 3. Electron configuration and unpaired electrons (a) Sb³⁺ Sb: [Kr] 4d¹⁰ 5s² 5p³ Sb³⁺: [Kr] 4d¹⁰ 5s² Unpaired electrons: 0 (b) Sn⁴⁺ Sn: [Kr] 4d¹⁰ 5s² 5p² Sn⁴⁺: [Kr] 4d¹⁰ Unpaired electrons: 0 (c) W²⁺ W: [Xe] 4f¹⁴ 5d⁴ 6s² W²⁺: [Xe] 4f¹⁴ 5d⁴ Unpaired electrons: 4 (d) Br⁻ Br: [Ar] 3d¹⁰ 4s² 4p⁵ Br⁻: [Ar] 3d¹⁰ 4s² 4p⁶ Unpaired electrons: 0 (e) Ni²⁺ Ni: [Ar] 3d⁸ 4s² Ni²⁺: [Ar] 3d⁸ Unpaired electrons: 2 4. Metals forming M²⁺ with given configurations (a) [Ar] 3d⁷ → Co²⁺ (from Co) (b) [Ar] 3d⁶ → Fe²⁺ (from Fe) (c) [Kr] 4d⁴ → Mo²⁺ (from Mo) (d) [Kr] 4d³ → Nb²⁺ (from Nb) Ch8 1. Born–Haber cycle for CaH₂(s) (a) Lattice energy of CaH₂(s) Reaction of formation: Ca(s) + H₂(g) → CaH₂(s)  ΔH = −186.2 kJ/mol Given data: Sublimation of Ca(s): +178.2 kJ/mol Bond dissociation of H₂(g): +435.9 kJ/mol Electron affinity of H(g): −72.8 kJ/mol (per H, ×2 = −145.6 kJ/mol) First ionization energy of Ca(g): +589.8 kJ/mol Second ionization energy of Ca(g): +1145 kJ/mol Born–Haber energy balance: ΔHf = ΔHsubl + D(H₂) + IE₁ + IE₂ + 2(EA) + lattice energy −186.2 = 178.2 + 435.9 + 589.8 + 1145 − 145.6 + lattice energy Sum of known terms = 2203.3 kJ/mol Lattice energy = −186.2 − 2203.3 = −2389.5 kJ/mol Answer: Lattice energy of CaH₂(s) = −2.39 × 10³ kJ/mol (b) Born–Haber cycle (description) Steps in the cycle: Ca(s) → Ca(g)  (sublimation) H₂(g) → 2 H(g)  (bond dissociation) Ca(g) → Ca⁺(g) + e⁻  (first ionization) Ca⁺(g) → Ca²⁺(g) + e⁻  (second ionization) 2 H(g) + 2 e⁻ → 2 H⁻(g)  (electron affinity) Ca²⁺(g) + 2 H⁻(g) → CaH₂(s)  (lattice formation) The sum of all steps equals the enthalpy of formation of CaH₂(s). 2. Polarizing power and polarizability (a) Cations: Rb⁺, Be²⁺, Sr²⁺ Polarizing power increases with higher charge and smaller ionic radius. Order (increasing polarizing power): Rb⁺ < Sr²⁺ < Be²⁺ (b) Cations: K⁺, Mg²⁺, Al³⁺, Cs⁺ Charge density increases strongly with higher charge and smaller size. Order (increasing polarizing power): Cs⁺ < K⁺ < Mg²⁺ < Al³⁺ (c) Anions: Cl⁻, Br⁻, N³⁻, O²⁻ Polarizability increases with larger size and higher negative charge. Order (increasing polarizability): Cl⁻ < Br⁻ < O²⁻ < N³⁻ (d) Anions: N³⁻, P³⁻, I⁻, At⁻ Down a group, size and polarizability increase significantly. Order (increasing polarizability): N³⁻ < P³⁻ < I⁻ < At⁻ Ch9 1. Lewis structure, geometry, polarity (a) ICl₄⁻ i) Lewis structure Central I atom with four I–Cl single bonds and two lone pairs on I ii) Electron domains 6 (4 bonding + 2 lone pairs) iii) Geometry Electron domain geometry: octahedral Molecular geometry: square planar iv) Formal charges I: −1 each Cl: 0 v) Oxidation number of I +III vi) Polarity Nonpolar (symmetrical square planar) (b) TeCl₄ i) Lewis structure Central Te with four Te–Cl bonds and one lone pair ii) Electron domains 5 (4 bonding + 1 lone pair) iii) Geometry Electron domain geometry: trigonal bipyramidal Molecular geometry: seesaw iv) Formal charges All atoms: 0 v) Oxidation number of Te +IV vi) Polarity Polar (asymmetric shape) (c) AsF₃ i) Lewis structure Central As with three As–F bonds and one lone pair ii) Electron domains 4 (3 bonding + 1 lone pair) iii) Geometry Electron domain geometry: tetrahedral Molecular geometry: trigonal pyramidal iv) Formal charges All atoms: 0 v) Oxidation number of As +III vi) Polarity Polar (d) XeOF₄ i) Lewis structure Central Xe with four Xe–F single bonds, one Xe=O double bond, and one lone pair ii) Electron domains 6 (5 bonding + 1 lone pair) iii) Geometry Electron domain geometry: octahedral Molecular geometry: square pyramidal iv) Formal charges All atoms: 0 v) Oxidation number of Xe +VI vi) Polarity Polar 2. Germide ion Ge₄ⁿ⁻ (a) Lewis structure Four Ge atoms form a tetrahedron Each Ge bonded to the other three Ge atoms Each Ge has one lone pair (b) Value of n Each Ge: 3 bonds + 1 lone pair → formal charge = −1 Total charge = 4 × (−1) = −4 Answer: Ge₄⁴⁻ (c) Hybridization of Ge Each Ge has 4 electron domains → sp³ hybridization (d) Polarity Tetrahedral arrangement with identical atoms → Nonpolar 3. Molecular orbital theory (i) Paramagnetism Rules: paramagnetic if unpaired electrons present (a) N₂⁻ One extra electron enters π* orbital Paramagnetic, 1 unpaired electron (b) F₂⁺ One electron removed from π* orbital Paramagnetic, 1 unpaired electron (c) O₂²⁺ Two electrons removed from π* orbitals Diamagnetic, 0 unpaired electrons (ii) Lowest ionization energy Ionization energy decreases as it becomes easier to remove an electron. C₂⁺: already electron-poor C₂: neutral C₂⁻: extra electron in antibonding orbital Lowest ionization energy: C₂⁻ Ch10 1. Exhaled air Given composition (mole percent): N₂ 74.8 %, O₂ 15.3 %, CO₂ 3.7 %, H₂O 6.2 % Total pressure = 99.8 kPa (a) Partial pressures Partial pressure = mole fraction × total pressure N₂: 0.748 × 99.8 = 74.7 kPa O₂: 0.153 × 99.8 = 15.3 kPa CO₂: 0.037 × 99.8 = 3.69 kPa H₂O: 0.062 × 99.8 = 6.19 kPa (b) Moles of CO₂ exhaled Given: Volume = 455 mL = 0.455 L Temperature = 37 °C = 310 K Partial pressure of CO₂ = 3.69 kPa Gas constant R = 8.314 L·kPa·mol⁻¹·K⁻¹ Using PV = nRT: n = (3.69 × 0.455) ÷ (8.314 × 310) n = 6.52 × 10⁻⁴ mol CO₂ (c) Grams of glucose metabolized Combustion reaction: C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O Mole ratio: 1 mol glucose → 6 mol CO₂ Moles of glucose: 6.52 × 10⁻⁴ ÷ 6 = 1.09 × 10⁻⁴ mol Molar mass of glucose = 180.16 g/mol Mass of glucose: 1.09 × 10⁻⁴ × 180.16 = 1.96 × 10⁻² g Answer: 0.0196 g glucose 2. Properties of gases at STP Gases: Ne, SF₆, N₂, CH₄ (a) Most likely to deviate from ideal behavior SF₆ (Largest size, strongest intermolecular attractions) (b) Closest to an ideal gas Ne (Monatomic, very weak intermolecular forces) (c) Highest root-mean-square speed CH₄ (Lowest molar mass) (d) Highest total molecular volume relative to container SF₆ (Largest molecules) (e) Highest average kinetic energy All are the same (Depends only on temperature) (f) Effuses more rapidly than N₂ Effusion is faster for gases with lower molar mass than N₂ (28 g/mol): Ne CH₄ (g) Largest van der Waals b parameter SF₆ (b represents molecular size) Ch11 1.(i) Substance with the higher normal boiling point (a) H₂S vs H₂O Answer: H₂O Reason: H₂O molecules form hydrogen bonds. H₂S lacks significant hydrogen bonding and mainly exhibits weak dispersion forces. Therefore, H₂O has a much higher boiling point. (b) NH₃ vs PH₃ Answer: NH₃ Reason: NH₃ can form hydrogen bonds through N–H bonds. PH₃ has low electronegativity on P and cannot form significant hydrogen bonds. Therefore, NH₃ has stronger intermolecular forces. (c) KBr vs CH₃Br Answer: KBr Reason: KBr is an ionic compound with strong electrostatic interactions in the solid state. CH₃Br is molecular and exhibits only weak intermolecular forces. Ionic bonding is stronger than molecular forces, so KBr has a much higher boiling point. (d) CH₄ vs SiH₄ Answer: SiH₄ Reason: Both molecules are nonpolar and exhibit only dispersion forces. SiH₄ has a larger molar mass and molecular size, leading to stronger dispersion forces. Therefore, SiH₄ has a higher boiling point. 1.(ii) Substance with stronger intermolecular forces (a) Ne vs Ar Answer: Ar Reason: Ar has more electrons, larger atomic size, and stronger dispersion forces than Ne. (b) NF₃ vs BF₃ Answer: NF₃ Reason: NF₃ is polar, so it exhibits dipole-dipole interactions. BF₃ is nonpolar due to its trigonal planar geometry. Therefore, NF₃ has stronger intermolecular forces. (c) SiH₄ vs GeH₄ Answer: GeH₄ Reason: GeH₄ is heavier and larger, resulting in stronger dispersion forces. (d) NaF vs HF Answer: NaF Reason: NaF exhibits ionic interactions, while HF has strong hydrogen bonds but these are molecular forces. Ionic bonding is stronger than molecular forces. 1.(iii) VSEPR structure and boiling point comparison (a) BF₃ vs ClF₃ BF₃: trigonal planar, nonpolar ClF₃: T-shaped, polar Answer: ClF₃ Reason: ClF₃ is polar and has a higher molar mass, resulting in stronger intermolecular forces and a higher boiling point. (b) SF₄ vs CF₄ SF₄: seesaw, polar CF₄: tetrahedral, nonpolar Answer: SF₄ Reason: SF₄ is polar and larger in size, leading to stronger intermolecular forces and a higher boiling point. (c) cis-CHCl=CHCl vs trans-CHCl=CHCl cis isomer: polar (dipoles do not cancel) trans isomer: nearly nonpolar (dipoles cancel) Answer: cis isomer Reason: The cis isomer exhibits dipole-dipole interactions, which are stronger than the interactions in the trans isomer, resulting in a higher boiling point. 2. Phases of Argon (a) Liquid – At 100 K, argon is above its melting point but the pressure is high. (b) Gas – At 150 K, near the critical temperature, but 8 atm is much lower than the critical pressure. (c) Solid – Low temperature and moderate pressure. (d) Gas – High temperature and very low pressure. 3. Phase Changes of Water (a) Liquid – At 298 K (25°C) and 1 atm, water is in its normal liquid state. (b) Gas (Water Vapor) – At 400 K (127°C) and 1 atm, water is above its boiling point (100°C), so it exists as a gas. (c) No abrupt phase change – The process passes around the critical point. Compressing water to 500 atm (above critical pressure) and heating it produces a supercritical fluid. Upon decompression and cooling, it transitions continuously into a gas without crossing distinct phase boundaries, so no abrupt phase change occurs.