Chemistry: Principles and Practice
3rd Edition
ISBN: 9780534420123
Author: Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher: Cengage Learning
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Chapter 10, Problem 10.3QE
Interpretation Introduction
Interpretation:
Based on VSPER theory, a molecule with a carbon centre that forms four bonds and has a trigonal planar structure and a molecule with a central carbon atom that forms four bonds and has a linear strucure has to be given
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What effect does the presence of lone-pair electrons have on the bond angles in a molecule?
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It is recommended that you draw the Lewis Structure of the molecules and then complete this table.
electronic
Molecular
Molecule
Geometry
Geometry
CH3F
tetrahedral
tetrahedral
NH3
tetrahedral
trigonal pyramidal
C2F4
trigonal planar
trigonal planar
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intermolecular
moment
force
CH3F
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yes
NH3
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yes
C2F4
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no
CO2
( Select |
no
>
>
Describe the bond angles to be found in each of the following molecular structures: (a) trigonal planar, (b) tetrahedral, (c) octahedral, (d) linear.
Chapter 10 Solutions
Chemistry: Principles and Practice
Ch. 10 - Prob. 10.1QECh. 10 - Prob. 10.2QECh. 10 - Prob. 10.3QECh. 10 - Prob. 10.4QECh. 10 - Prob. 10.5QECh. 10 - Prob. 10.6QECh. 10 - Prob. 10.7QECh. 10 - Prob. 10.8QECh. 10 - Prob. 10.9QECh. 10 - Prob. 10.10QE
Ch. 10 - Which atomic orbitals overlap to form the bonds in...Ch. 10 - Prob. 10.12QECh. 10 - Identify the hybrid orbitals used by boron in BCl3...Ch. 10 - Identify the hybrid orbitals used by antimony in...Ch. 10 - Prob. 10.15QECh. 10 - Prob. 10.16QECh. 10 - Prob. 10.17QECh. 10 - Prob. 10.18QECh. 10 - Prob. 10.19QECh. 10 - Prob. 10.20QECh. 10 - Compare and contrast the molecular orbital and...Ch. 10 - Describe the bonding in molecular orbital terms...Ch. 10 - Prob. 10.23QECh. 10 - Prob. 10.24QECh. 10 - Prob. 10.25QECh. 10 - Prob. 10.26QECh. 10 - Prob. 10.27QECh. 10 - Prob. 10.28QECh. 10 - Prob. 10.29QECh. 10 - Prob. 10.30QECh. 10 - Prob. 10.31QECh. 10 - Prob. 10.32QECh. 10 - Prob. 10.33QECh. 10 - Prob. 10.34QECh. 10 - Prob. 10.35QECh. 10 - Prob. 10.36QECh. 10 - Prob. 10.37QECh. 10 - Prob. 10.38QECh. 10 - Prob. 10.39QECh. 10 - Use the VSEPR model to predict the bond angles...Ch. 10 - Prob. 10.41QECh. 10 - Prob. 10.42QECh. 10 - For each of the following molecules, complete the...Ch. 10 - Prob. 10.44QECh. 10 - Prob. 10.45QECh. 10 - Prob. 10.46QECh. 10 - Indicate which molecules are polar and which are...Ch. 10 - Prob. 10.48QECh. 10 - Indicate which of the following molecules are...Ch. 10 - Prob. 10.50QECh. 10 - Prob. 10.51QECh. 10 - Prob. 10.52QECh. 10 - Prob. 10.53QECh. 10 - Prob. 10.54QECh. 10 - Prob. 10.55QECh. 10 - Prob. 10.56QECh. 10 - Prob. 10.57QECh. 10 - Prob. 10.58QECh. 10 - Prob. 10.59QECh. 10 - Prob. 10.60QECh. 10 - Prob. 10.61QECh. 10 - Prob. 10.62QECh. 10 - Prob. 10.63QECh. 10 - Prob. 10.64QECh. 10 - Prob. 10.65QECh. 10 - Prob. 10.66QECh. 10 - Prob. 10.67QECh. 10 - Prob. 10.68QECh. 10 - Prob. 10.69QECh. 10 - Prob. 10.70QECh. 10 - Prob. 10.71QECh. 10 - Prob. 10.72QECh. 10 - Identify the orbitals on each of the atoms that...Ch. 10 - Prob. 10.74QECh. 10 - Prob. 10.75QECh. 10 - How many sigma bonds and how many pi bonds are...Ch. 10 - Give the hybridization of each central atom in the...Ch. 10 - Prob. 10.78QECh. 10 - Prob. 10.79QECh. 10 - Prob. 10.80QECh. 10 - Prob. 10.81QECh. 10 - Predict the hybridization at each central atom in...Ch. 10 - Prob. 10.83QECh. 10 - Tetrafluoroethylene, C2F4, is used to produce...Ch. 10 - Prob. 10.85QECh. 10 - Prob. 10.86QECh. 10 - Prob. 10.87QECh. 10 - Prob. 10.88QECh. 10 - Prob. 10.89QECh. 10 - Prob. 10.90QECh. 10 - Prob. 10.91QECh. 10 - Prob. 10.92QECh. 10 - Prob. 10.93QECh. 10 - Prob. 10.94QECh. 10 - Prob. 10.95QECh. 10 - Prob. 10.96QECh. 10 - Prob. 10.97QECh. 10 - Prob. 10.98QECh. 10 - The molecular orbital diagram of NO shown in...Ch. 10 - The molecular orbital diagram of NO shown in...Ch. 10 - The molecular orbital diagram of NO shown in...Ch. 10 - Prob. 10.102QECh. 10 - Prob. 10.103QECh. 10 - Prob. 10.104QECh. 10 - Prob. 10.105QECh. 10 - Following are the structures of three isomers of...Ch. 10 - The ions ClF2 and ClF2+ have both been observed....Ch. 10 - Aspirin, or acetylsalicylic acid, has the formula...Ch. 10 - Aspartame is a compound that is 200 times sweeter...Ch. 10 - Prob. 10.110QECh. 10 - Prob. 10.111QECh. 10 - Calcium cyanamide, CaNCN, is used both to kill...Ch. 10 - Histidine is an essential amino acid that the body...Ch. 10 - Formamide, HC(O)NH2, is prepared at high pressures...Ch. 10 - Prob. 10.115QECh. 10 - Prob. 10.116QECh. 10 - Prob. 10.117QECh. 10 - Prob. 10.118QECh. 10 - Prob. 10.119QECh. 10 - Prob. 10.120QECh. 10 - Prob. 10.121QECh. 10 - Prob. 10.122QECh. 10 - Prob. 10.123QECh. 10 - Prob. 10.124QECh. 10 - Two compounds have the formula S2F2. Disulfur...Ch. 10 - Prob. 10.126QECh. 10 - Prob. 10.127QE
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- • identify sigma and pi bonds in a molecule and explain the difference between them.arrow_forward(b) Arrange the following in decreasing order of bond length and bond angle with explanation C2H6, C2H4, C¿H2arrow_forwardExplain what is wrong with each molecular geometry and provide the correct molecular geometry, given the numbers of lone pairs and bonding groups on the central atom. Match the words in the left column to the appropriate blanks in the sentences on the right. a bent a trigonal bipyramidal an octahedral a trigonal planar a square planar a trigonal pyramidal a tetrahedral a seesaw a linear In structure (a), four pairs of electrons give cause lone pair-bonded pair repulsions and would have Reset electron geometry. The lone pair would molecular geometry. Help In structure (b), five pairs of electrons give electron geometry. The lone pair occupies an equatorial position to minimize lone pair-bonded pair repulsions, and the molecule would have molecular geometry. In structure (c), six pairs of electrons give electron geometry. The two lone pairs would occupy opposite positions to minimize lone pair-lone pair repulsions, and the molecule would have molecular geometry.arrow_forward
- Use VSEPR to predict the geometry (including bond angles) about each interior atom of methyl azide (CH3N3) and draw the molecule. Would you expect the bond angle between the two interior nitrogen atoms to be the same or different? Would you expect the two nitrogen–nitrogen bond lengths to be the same or different?arrow_forwardHow many lone pairs should be present on the central Xe atom?arrow_forwardPredict the geometry and bond angles around atom A for molecules with the general formulas AB4, AB3E, and AB2E2, where B represents another atom and E represents an electron pair.arrow_forward
- This is not a real molecule, A is not an element, but if it was an molecule what would the molecular geometry of atom A be? Linear Bent Trigonal pyramidal Trigonal planar Tetrahedralarrow_forwardGive the number of lone pairs around the central atom and the molecular geometry of CBr4. help me to grasp the concept behind it.arrow_forwardThe concept of valence-shell electron-pair repulsion (VSEPR) is presented in introductory organic chemistry as a way to predict molecular geometries. The idea be- hind VSEPR is that areas of electron density repel each other so that the geometry of bonds and/or lone pairs of electrons around any one atom places these areas as far apart as possible. For four areas of electron density, a tetrahedral geometry is predicted. For three areas of electron density, a trigonal planar geometry is predicted. Two areas of electron density lead to a linear geometry. VSEPR is simply a predictive tool, but in some cases, it gives an incorrect prediction. In these instances, additional insights into bonding are necessary, such as resonance. Interestingly, several important situations are critical to biochemistry where VSEPR breaks down. Two examples are shown here. :O: CH3 H,N H3N CH3 H :0: :0: Histidine The circled nitrogen atoms are predicted by VSEPR to be tetrahedral in geometry because each…arrow_forward
- List the correct electron and molecular geometries that correspond to set of electron groups around the central atom of a molecule. six electron groups overall; four bonding groups and two lone pairsarrow_forward0=c=0; Lewis electron-dot diagrams for CO2 and SO2 are given above. The molecular geometry and polarity of the two substances are the same because the molecular formulas are similar the same because C and S have similar electronegativity values different because the lone pair of electrons on the S atom make it the negative end of a dipole D different because S has a greater number of electron domains (regions of electron density) surrounding it than C hasarrow_forwardUse VSEPR theory to predict the geometry (including bond angles) about each interior atom of methyl azide (CH3N3), and make a sketch of the molecule. Would you expect the bond angle between the two interior nitrogen atoms to be the same or different? Would you expect the two nitrogen–nitrogen bond lengths to be the same or different?arrow_forward
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