Concept explainers
Interpretation:
The shape of the xenon tetrafluoride molecule has to be predicted.
Concept Introduction:
Valence Shell Electron Pair Repulsion model predicts shape by inclusion of bond angles and most distant arrangement of atoms that leads to minimum repulsion. For the molecules that have no lone pairs around the central atom the bonded-atom unshared -pair arrangement is decided by the table as follows:
In order to determine the shape the steps to be followed are indicated as follows:
- 1. Lewis structure of the molecule should be written.
- 2. The type of electron arrangement around the central atom should be identified around the central atom. This essentially refers to the determination of bond pairs and unshared or lone pairs around central atoms.
- 3. Then bonded-atom unshared -pair arrangement that can maximize the distance of electron pairs about central atom determines the shape.
For molecules that have lone pairs around central atom, lone pairs influence shape, because there are no atoms at the positions occupied by these lone pairs. The key rule that governs the molecular shape, in this case, is the extent of lone –lone pair repulsions are far greater than lone bond pair or bond pair-bond pair repulsions. The table that summarized the molecular shapes possible for various combinations of bonded and lone pairs are given as follows:

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Chapter 2 Solutions
CHEMICAL PRINCIPLES PKG W/SAPLING
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- V Highlight all of the carbon atoms that have at least one beta (B) hydrogen, using red for one ẞ hydrogen, blue for two ẞ hydrogens, and green for three ẞ hydrogens. If none of the carbon atoms have ẞ hydrogens, check the box underneath the molecule. ED X None of the carbon atoms have ẞ hydrogens. Explanation esc 2 Check * F1 F2 1 2 80 # 3 Q W tab A caps lock shift fn control F3 N S option O 694 $ F4 F5 F6 005 % E R D F LL 6 olo 18 Ar B © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility A DII F7 F8 87 & * 8 T Y U G H 4 F9 F10 ( 9 0 E F11 F12 உ J K L + || X C V B N M H H command option commandarrow_forwardConsider the reaction below and answer the following questions. Part 1 of 4 Br NaOCH2CH3 Identify the mechanisms involved. Check all that apply. SN 1 SN 2 E1 E2 None of the above Part 2 of 4 Skip Part Check esc F1 F2 lock 1 2 Q W A S #3 80 F3 F4 F5 F6 Save For © 2025 McGraw Hill LLC. All Rights Reserved. Terms ˇˇ % & 4 5 6 89 7 IK A 分 བ F7 F8 F9 F * E R T Y U 8 9 D F G H K V B N M 0 Oarrow_forwardWhat kind of holes are not generated when solid-state particles adopt a close packing pattern? Group of answer choices tetrahedral cubic octahedral None of the other choices are correctarrow_forward
- For the reaction below: 1. Draw all reasonable elimination products to the right of the arrow. 2. In the box below the reaction, redraw any product you expect to be a major product. 田 Major Product: Check ☐ + I Na OH esc F1 F2 2 1 @ 2 Q W tab A caps lock S #3 80 F3 69 4 σ F4 % 95 S Click and drag to sta drawing a structure mm Save For Later 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use GO DII F5 F6 F7 F8 F9 F10 6 CO 89 & 7 LU E R T Y U 8* 9 0 D F G H J K L Z X C V B N M 36arrow_forwardProblem 7 of 10 Draw the major product of this reaction. Ignore inorganic byproducts. S' S 1. BuLi 2. ethylene oxide (C2H4O) Select to Draw a Submitarrow_forwardFeedback (4/10) 30% Retry Curved arrows are used to illustrate the flow of electrons. Use the reaction conditions provided and follow the arrows to draw the reactant and missing intermediates involved in this reaction. Include all lone pairs and charges as appropriate. Ignore inorganic byproducts. Incorrect, 6 attempts remaining :0: Draw the Reactant H H3CO H- HIO: Ö-CH3 CH3OH2* protonation H. a H (+) H Ο CH3OH2 O: H3C protonation CH3OH deprotonation > CH3OH nucleophilic addition H. HO 0:0 Draw Intermediate a Xarrow_forward
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
