Understanding Structure and Properties in SCH4U Chemistry
School
St. Thomas More Catholic School**We aren't endorsed by this school
Course
CHE SCH4U1
Subject
Chemistry
Date
Dec 9, 2024
Pages
3
Uploaded by DrWombat2331
St. Thomas More Catholic Secondary School Department of Science SCH4U - Unit One Test - Structure and PropertiesΞ±#:_____ Name: ____________________________ October 7, 2024 πΎ??????π?: πΌ?????πππ?π??: πΆ?????π?π?π??: TOTAL: % KNOWLEDGE: [25] Multiple Choice- Answer on your Ziprade sheet.Short Answer[6] 1. Identify the following:a) The shorthand electron configuration for molybdenum, ππ¨ππππ: ____________________________________ b) Draw the Lewis dot diagramof the neutral atom that has the electron configuration [Ar]4s23d104p4_________ c) The symbol of the atom represented by the following excited state: 1s22s22p63s23p64s23d24p2_________ d) The element represented by βXβ is found in period 4 of the Periodic Table. If the compound which has the formula XF5and has a square pyramid shape, identify βXβby giving its symbol: _________ e) Give the symbol(s) of the element(s) that have 1 unpaired electron in the 4p energy level. _________ [3] 2. Consider the diagram showing four Energy Level Diagrams. Which diagram represents the element with the highest electronegativity value? EXPLAIN your choice fully. [4] 3. State onesignificantsimilarity and state onesignificantdifference between the following pairs. Be clear and concise. Your answer must reflect material taught in SCH4U1 Chemistry. SIMILARITY DIFFERENCE Bohr-Rutherford model and quantum model of the atom AX3E and AX2E225 +13 βΉ 38 16 6 +10 βΉ 16 70 13 [Kr] 5s14d5An ANAMOLY, NOT [Kr] 5s24d42+2+6+2+6+2+2+2 = 24 Se I 24Cr Ga, Br Diagram III -Represents a halogen (fluorine). In its period, it has the greatest number of protons and will attracts electrons in a bond most strongly. Both describe the atom with a dense core named NUCLEUS . In B-R model, electrons move in distinct paths called orbits. In quantum model, electrons move in areas of high probability, called orbitals. Both have an ELECTRONICgeometry of tetrahedron. AX3E has triangular planar molecular geometry. AX2E2has a Vβshape molecular geometry. Answers
INVESTIGATION:[10total] 1. Complete the table below for ONEion only (If both are answered, only the first one will be marked.): PARTICLE π―??π(H is not acidic)πΊππ?πβ[3] Lewis Structural Diagram (Include Formal Charges, Partial Charges and Dipoles as applicable) [1] AXnEmNotation AX3AX3[2] Diagram of Actual Geometry Name of Actual Geometry triangular planartriangular planar[1] Polar or Nonpolar? polarpolar[1] Do Resonance Structures Exist? If Yes, what is the Bond Order? Yes: 32= 1.5Yes: 32= 1.5[1] Strongest Intermolecular Force of Attraction present dipole-dipole H-bonding[1] Hybridization of orbitals of the central atom sp2sp22. For a model building activity, a grade 12 chemistry student is to build the molecule shown on the right.[3] (a) When building this molecule, which sphere should the student use (i.e.how many holes) to represent the central atom βXβ? Justify your answer.[1] (b) If βXβ is in the third period of the periodic table, identify the element. ____________________ [2] 3. A molecule has the molecular formula QR3. Although the Q-R bond is polar, the molecule is nonpolar. Identify the molecular shape(s) this molecule can andcannot be. (Note:marks will be deducted for listing too many) Can be: _________________________________________________________________ Cannot be: _______________________________________________________________ 16 6 holes β4 for the bonds 2 for lone pair (paddle) chlorine Group 17 Triangular planar Triangular pyramid, T-shape
COMMUNICATION:βFor full marks, you must answer in complete sentences with proper spelling and grammar.Also, your answer must include relevant information that was covered in your SCH 4U1 lessons.[3] 1. Consider the three molecules below. Rank them by increasing boiling points (LOWEST to HIGHEST). Justify your choice.βΆβ·βΈ2. Below is a diagram showing the hybridization of the orbitals in a molecule . [3] (a) Translate the diagram into a Lewis Structural diagram. [1] (b) Label the indicated overlap. [3]4. Below are the structures of normal pentane (n-pentane) and neopentane (2,2-dimethylpropane), both with the formula C5H12. Which substance would you expect to have the higher melting point? Explain your choice.10 ππΆβπβΈ< βΆ< β·βΈis a polar molecule and therefore has London dispersion forces and dipole-dipole forces. βΆis a polar molecule with one βO-Hβ and therefor in addition to London dispersion forces and dipole-dipole forces, it has one sitefor H-bonding. β·is a polar molecule with TWOβO-Hβ and therefor in addition to London dispersion forces and dipole-dipole forces, it has two sitesfor H-bonding, yielding the strongest IMFβs. Neopentane would have a higher melting point than n-pentane. The molecules of neopentane fit into the lattice points better since they are more compact, causing the crystal lattice to be more stable and more rigid. As a result, more energy will be needed to separate the molecules, and the melting point will be higher.