Exam 1-solutions

# Exam 1-solutions - Version 241 Exam 1 Sutclie(53770 This...

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

Version 241 – Exam 1 – Sutcliffe – (53770) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. THIS EXAM IS ONLY FOR STUDENTS ENROLLED IN UNIQUE NUMBER 53770 WHICH MEETS T TH 3.30-5PM 001 10.0 points How many p electrons does Se (atomic num- ber 34) possess? 1. 6 2. 16 correct 3. 34 4. 10 5. 12 6. 0 7. 4 Explanation: Se has 6 electrons in the 2 p and 3 p orbitals plus 4 more in the 4 p orbital. This gives it a total of 16 p electrons. 002 10.0 points Which quantum number is most important in determining the type, or shape of atomic orbitals? What would be its value for a 5 p orbital? 1. m , m = 1 2. ℓ, ℓ = 1 correct 3. m , m = 2 4. n, n = 5 5. n, n = 1 6. ℓ, ℓ = 5 Explanation: 003 10.0 points The length of 4 meters is the same as 1. 400 mm. 2. 40 mm. 3. 4000 mm. correct 4. 40000 mm. Explanation: ? mm = 4 m × 1000 mm 1 m = 4000 mm 004 10.0 points Calculate the frequency of a photon with a wavelength of 6240 ˚ A. 1. 4 . 81 × 10 14 Hz correct 2. 4 . 81 × 10 13 Hz 3. 3 . 19 × 10 - 39 Hz 4. 4 . 13 × 10 - 30 Hz 5. 48100 Hz 6. 3 . 19 × 10 - 19 Hz 7. 3 . 19 × 10 - 29 Hz 8. 4 . 13 × 10 - 20 Hz 9. 4 . 81 × 10 - 6 Hz Explanation: λ = 6240 ˚ A The equation that connects frequency ν to wavelength λ is ν = c λ where c is the speed of light in a vacuum (3.00 × 10 8 m/s). Remember that Hz is equal to inverse seconds (s - 1 ) and that 1 ˚ A is equal to 1 × 10 - 10 m. ν = 3 . 00 × 10 8 m / s (6240 ˚ A) 1 × 10 - 10 m 1 ˚ A

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

View Full Document
Version 241 – Exam 1 – Sutcliffe – (53770) 2 = 4 . 80769 × 10 14 Hz 005 10.0 points The quantum mechanical approach to atomic structure permits the calculation of 1. the most probable spin value that will be associated with an electron of specified energy. 2. the most probable radius of an orbit that an electron of specified energy will follow. 3. a region about the nucleus in which an electron of specified energy will probably be found. correct 4. the number of electrons in an atom. 5. the most probable distance between any two specified electrons. Explanation: According to the Heisenberg Uncertainty Principle, the momentum and the position of an electron cannot both be known simula- taneously with any accuracy. The quantum mechanical approach attempts to calculate the energy of the electron exactly by defining possible energy states for the electron. The position of the electron cannot then be known with any certainty. Instead, a region of space around the nucleus is identified in which the electron is likely to be. These regions of space are commonly known as orbitals. 006 10.0 points Which of the following sets of quantum num- bers are invalid , i.e. violate one or more boundary conditions? I) n = 3 , ℓ = 2 , m = - 2 , m s = + 1 2 II) n = 9 , ℓ = 5 , m = 6 , m s = + 1 2 III) n = 2 , ℓ = 1 , m = 0 , m s = +1 IV) n = 2 , ℓ = 0 , m = 0 , m s = + 1 2 V) n = 1 , ℓ = 0 , m = 0 , m s = - 1 2 1. II, III correct 2. I, III, IV 3. III only 4. IV only 5. I, II, IV 6. II only 7. I, IV 8. I only Explanation: Set II and III are invalid. For II, m l = 6 is disallowed because = 5. For III, m s = +1 is disallowed because m s may only be + 1 2 or - 1 2 .
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