{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW4 - padilla(tp5647 Homework 4 Sutclie(53770 This...

This preview shows pages 1–3. Sign up to view the full content.

padilla (tp5647) – Homework 4 – Sutcliffe – (53770) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. READ THIS: Note that SOME of these questions require you to understand expanded or incomplete octets and/or VSEPR theory in order to do them. You may find it advisable to wait until this is covered in class before submitting them, but you should look at them before class. 001 10.0 points How many unshared (non-bonded) pairs of electrons are in NF 3 ? 1. thirteen 2. no pairs 3. twenty 4. ten correct 5. one 6. three 7. four 8. two Explanation: The total number of valence electrons is N 1 × 5 e = 5 e P 3 × 7 e = 21 e 26 e and the Lewis structure is N F F F 002 10.0 points How many valence electrons must be shown in the dot formula for the C 3 H 8 N 2 O molecule? 1. 64 2. 28 3. 36 correct 4. 44 5. 30 Explanation: To find the total number of valence elec- trons available for the dot formula we sum the number of valence electrons in each atom of the molecule. Carbon has 4 valence electrons and there are 3 C atoms. Hydrogen has 1 valence elec- trons and there are 8 H atoms. Nitrogen has 5 valence electrons and there are 2 N atoms. Oxygen has 6 valence electrons and there is 1 O atom. Total valence e = 4 × 3 (C atom) + 1 × 8 (H atom) + 5 × 2 (N atom) + 6 × 1 (O atom) = 36 003 10.0 points NOTE: This question assumes you follow the octet rule; do not use an expanded octet. The Lewis dot structure of BrO + 3 has how many total electrons and how many shared electrons? 1. 24; 6 2. 25; 7 3. 24; 8 correct 4. 25; 8 5. 26; 8 6. 26; 6 7. 25; 6 8. 26; 7 Explanation:

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
padilla (tp5647) – Homework 4 – Sutcliffe – (53770) 2 The total number of valence electrons is 1 × 7 e (for Br) + 3 × 6 e (for O) - 1 e (from + 1 charge) = 24 e Br is the central atom, with the O atoms ar- ranged around it. Br will have an incomplete octet if we show one shared pair of electrons (a single bond) with each O, so one double bond is necessary: Br O O O + 004 10.0 points From what column must atom A come in the Lewis dot structure for CH 3 A? The central atom is carbon. It is bonded to three hy- drogen atoms and single-bonded to A. Both carbon and A follow the octet rule. 1. II 2. IV 3. VII correct 4. III 5. VI 6. VIII 7. V Explanation: Since A follows the octet rule, it must have three lone pairs of electrons around it: C H H H A · · · · ·· This structure has 14 valence electrons. H 3 × 1 e = 3 e C 1 × 4 e = 4 e A 1 × xe = xe (7 + x ) e 14 = 7 + x , so A has 7 electrons, and A comes from group VII. 005 10.0 points In most Lewis dot structures of neutral com- pounds, carbon as the central atom will be bound to a maximum of how many atoms, ni- trogen will be bound to a maximum of how many atoms, and oxygen will be bound to a maximum of how many atoms?
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern