Alert if there are more than four atoms attached to

Info icon This preview shows pages 9–19. Sign up to view the full content.

View Full Document Right Arrow Icon
Alert: If there are more than four atoms attached to the central atom, such as SF 6 , attach all six F atoms to the S. There will be 12 electrons around S and it will violate the octet rule by having 12 around the central atom. This happens occasionally. Step 3 Put eight dots around ALL atoms except H, which only needs 2. Again using water as an example, put two dots above the O and two dots below the O. With the two (-), that makes 8 electrons around the O and two around both H's. If Possible, your goal is to keep 8 electrons around each atom. 9
Image of page 9

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

View Full Document Right Arrow Icon
Step 4 Count the number of electrons used in the drawing and compare it to the number of available valence electrons. A. The dots in the drawing equal the number of of valence electrons. If this happens, the drawing is correct. Using H 2 O as an example, the drawing has 8 dots and the valence electrons equal 8 so the drawing is correct. B. There are fewer dots in the drawing than the number of valence electrons. Add pairs of dots to the central atom (attach them to the chemical symbol which is in the middle of your drawing) until the dots on the drawing equal the number of valence electrons. C. The drawing has more dots than the number of valence electrons. Erase four dots( two dots from the central atom and two from one of the other atoms). Now put a second line between the central atom and the atom from which you erased two dots. The two lines between these atoms represent a double bond (four dots) and are shared by both atoms. If you still need to reduce the number of dots, repeat this procedure. 10
Image of page 10
Draw Lewis Dot Drawing Name the molecular geometry ( example: Tetrahedral) _______________________________________________________________________ CH 4 ____valence electrons _______________________________________________________________________ _ ClF 3 ____valence electrons _______________________________________________________________________ _ CO 2 _____valence electrons _______________________________________________________________________ _ NH 3 _____valence electrons _______________________________________________________________________ _ PCl 5 _____valence electrons _______________________________________________________________________ _ H 2 O_____valence electrons
Image of page 11

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

View Full Document Right Arrow Icon
_______________________________________________________________________ _ SF 4 _____valence electrons 11 _______________________________________________________________________ _ SiO 2 _____valence electrons _______________________________________________________________________ _ XeF 4 _____valence electrons _______________________________________________________________________ _ BrF 5 _____valence electrons _______________________________________________________________________ _ N 2 _____valence electrons _______________________________________________________________________ _ NO 3 - ( when determining valence electrons, add on extra for the (-) charge) ______valence electrons _______________________________________________________________________ _
Image of page 12
SO 4 2- ( add two extra valence electrons for the 2- ) _____valence electrons _______________________________________________________________________ _ SF 6 _____valence electrons 12
Image of page 13

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

View Full Document Right Arrow Icon
Image of page 14
Answers CH 4 Tetrahedral ClF 3 T-shaped CO 2 Linear NH 3 Pyramidal PCl 5 Trigonal Bipyramidal H 2 O Bent SF 4 See-Saw SiO 2 Linear XeF 4 Square Planar BrF 5 Square Pyramidal N 2 Linear NO 3 - Trigonal Planar SO 4 2- Tetrahedral SF 6 Octahedral
Image of page 15

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

View Full Document Right Arrow Icon
VSEPR Geometries “A” is the central atom, “X” is a bonded atom, and “E” is a nonbonded pair of electrons NOTE: “E” can be a nonbonded pair of electrons on the central atom or an extra pair of electrons which was added to the central atom. 13
Image of page 16
Practice Test One Chapter 7: Lewis Dot and Molecular Geometry Molecule Lewis Dot Molecular Geometry ___________________________________________________________ 1. SiH 4 ____________________________________________________________ 2. NH 3 _____________________________________________________________ 3. KrF 4 ______________________________________________________________ 4. BiBr 5 _______________________________________________________________ 5. NO 3 - ________________________________________________________________ 6. ICl 3 _________________________________________________________________ 7. SiO 2 __________________________________________________________________ 8. SF 4 ___________________________________________________________________ 14
Image of page 17

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

View Full Document Right Arrow Icon
Answers: ( Test one starting with SiH 4 ) 1.
Image of page 18
Image of page 19
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern