
Concept explainers
Predict the geometries of the following species using the VSEPR method: (a) PCl3, (b) CHCl3, (c) SiH4, (d) TeCl4.
(a)

Interpretation: For the given set of molecules the molecular geometry around the central metal should be predicted using VSEPR model.
Concept Introduction:
Molecular geometry: It is defined as unique three dimensional arrangements of atoms around the central metal present in the molecule which is determined by using spectroscopic techniques and also by using Lewis structure or the valence shell electron pair repulsion theory (VSEPR).
VSEPR Theory:
As the name itself indicates that the basis for this theory is the electron pair that is bonded electron present in either single or double bonds or lone pair electrons, present in the valence shell tends to repel each other which then the tends to be in position in order to minimize the repulsions. The steps involved in the theory in describing the geometry is as follows,
- The first step is to draw the correct Lewis structure for the molecule.
- Then, the electron domain around the central atom should be counted and the geometry that matches with that type of domain in VSEPR should be determined.
- Finally, the geometry is predicted by using the orientation of atoms.
If the molecules (
The molecules of type
The molecules of type
Lewis structure for any molecule is drawn by using the following steps,
First the skeletal structure for the given molecule is drawn then the total number of valence electrons for all atoms present in the molecule is determined
The next step is to subtract the electrons present in the total number of bonds present in the skeletal structure of the molecule with the total valence electrons such that considering each bond contains two electrons with it.
Finally, the electrons which got after subtractions has to be equally distributed such that each atom contains eight electrons in its valence shell.
Electron Domain: In VSEPR theory, both the lone pair and the bonded pair are together considered as electron domain regardless of the type of bond in which the bonded pair presents.
Answer to Problem 7.9QP
(a)
Trigonal pyramidal
Explanation of Solution
To predict: The geometry for the given molecule.
Draw the Lewis structure for the molecule (a)
First the skeletal structure for the given molecule is drawn then the total number of valence electrons in the molecule is 26.
The next step is to subtract the electrons present in the total number of bonds present in the molecule with the total valence electrons such that 6 has to be subtracted with 26 as each bond contains two electrons with it and there are three bonds in the skeletal structure.
Finally, the 20 electrons got after subtractions has to be equally distributed such that each atom contains eight electrons in its valence shell.
Determine the molecular geometry for the molecule (a) using VSEPR.
The electron domain for the given molecule is obtained by viewing the Lewis structure which is of type tetrahedral since the
The molecular geometry for the given molecule is trigonal pyramidal due to the presence of one lone pair around the central atom.
(b)

Interpretation: For the given set of molecules the molecular geometry around the central metal should be predicted using VSEPR model.
Concept Introduction:
Molecular geometry: It is defined as unique three dimensional arrangements of atoms around the central metal present in the molecule which is determined by using spectroscopic techniques and also by using Lewis structure or the valence shell electron pair repulsion theory (VSEPR).
VSEPR Theory:
As the name itself indicates that the basis for this theory is the electron pair that is bonded electron present in either single or double bonds or lone pair electrons, present in the valence shell tends to repel each other which then the tends to be in position in order to minimize the repulsions. The steps involved in the theory in describing the geometry is as follows,
- The first step is to draw the correct Lewis structure for the molecule.
- Then, the electron domain around the central atom should be counted and the geometry that matches with that type of domain in VSEPR should be determined.
- Finally, the geometry is predicted by using the orientation of atoms.
If the molecules (
The molecules of type
The molecules of type
Lewis structure for any molecule is drawn by using the following steps,
First the skeletal structure for the given molecule is drawn then the total number of valence electrons for all atoms present in the molecule is determined
The next step is to subtract the electrons present in the total number of bonds present in the skeletal structure of the molecule with the total valence electrons such that considering each bond contains two electrons with it.
Finally, the electrons which got after subtractions has to be equally distributed such that each atom contains eight electrons in its valence shell.
Electron Domain: In VSEPR theory, both the lone pair and the bonded pair are together considered as electron domain regardless of the type of bond in which the bonded pair presents.
Answer to Problem 7.9QP
(b)
Tetrahedral
Explanation of Solution
To predict: The geometry for the given molecule.
Draw the Lewis structure for the molecule (b)
First the skeletal structure for the given molecule is drawn then the total number of valence electrons in the molecule is 26.
The next step is to subtract the electrons present in the total number of bonds present in the molecule with the total valence electrons such that 8 has to be subtracted with 26 as each bond contains two electrons with it and there are four bonds in the skeletal structure.
Finally, the 18 electrons got after subtractions has to be equally distributed such that each atom contains eight electrons in its valence shell.
Determine the molecular geometry for the molecule (b) using VSEPR.
The electron domain for the given molecule is obtained by viewing the Lewis structure which is of type tetrahedral as there is no lone pair of electron over the central metal atom and hence the molecular geometry for the given molecule is also Tetrahedral.
(c)

Interpretation: For the given set of molecules the molecular geometry around the central metal should be predicted using VSEPR model.
Concept Introduction:
Molecular geometry: It is defined as unique three dimensional arrangements of atoms around the central metal present in the molecule which is determined by using spectroscopic techniques and also by using Lewis structure or the valence shell electron pair repulsion theory (VSEPR).
VSEPR Theory:
As the name itself indicates that the basis for this theory is the electron pair that is bonded electron present in either single or double bonds or lone pair electrons, present in the valence shell tends to repel each other which then the tends to be in position in order to minimize the repulsions. The steps involved in the theory in describing the geometry is as follows,
- The first step is to draw the correct Lewis structure for the molecule.
- Then, the electron domain around the central atom should be counted and the geometry that matches with that type of domain in VSEPR should be determined.
- Finally, the geometry is predicted by using the orientation of atoms.
If the molecules (
The molecules of type
The molecules of type
Lewis structure for any molecule is drawn by using the following steps,
First the skeletal structure for the given molecule is drawn then the total number of valence electrons for all atoms present in the molecule is determined
The next step is to subtract the electrons present in the total number of bonds present in the skeletal structure of the molecule with the total valence electrons such that considering each bond contains two electrons with it.
Finally, the electrons which got after subtractions has to be equally distributed such that each atom contains eight electrons in its valence shell.
Electron Domain: In VSEPR theory, both the lone pair and the bonded pair are together considered as electron domain regardless of the type of bond in which the bonded pair presents.
Answer to Problem 7.9QP
Answer
(c)
Tetrahedral
Explanation of Solution
To predict: The geometry for the given molecule.
Draw the Lewis structure for the molecule (c)
First the skeletal structure for the given molecule is drawn then the total number of valence electrons in the molecule is 8.
The next step is to subtract the electrons present in the total number of bonds present in the molecule with the total valence electrons such that 8 has to be subtracted with 8 as each bond contains two electrons with it and there are four bonds in the skeletal structure.
There are no remaining electrons hence all the atoms in the molecules are fulfilled the octet rule that is each atom involves in bonding in order to fill their valence with eight electrons.
Determine the molecular geometry for the molecule (c) using VSEPR.
The electron domain for the given molecule is obtained by viewing the Lewis structure which is of type tetrahedral since central atom does not contain any lone pair of electron with it.
The molecular geometry for the molecule is also tetrahedral as there are four atoms bonded with the central metal atom and there is absence of lone pair of electrons.
(d)

Interpretation: For the given set of molecules the molecular geometry around the central metal should be predicted using VSEPR model.
Concept Introduction:
Molecular geometry: It is defined as unique three dimensional arrangements of atoms around the central metal present in the molecule which is determined by using spectroscopic techniques and also by using Lewis structure or the valence shell electron pair repulsion theory (VSEPR).
VSEPR Theory:
As the name itself indicates that the basis for this theory is the electron pair that is bonded electron present in either single or double bonds or lone pair electrons, present in the valence shell tends to repel each other which then the tends to be in position in order to minimize the repulsions. The steps involved in the theory in describing the geometry is as follows,
- The first step is to draw the correct Lewis structure for the molecule.
- Then, the electron domain around the central atom should be counted and the geometry that matches with that type of domain in VSEPR should be determined.
- Finally, the geometry is predicted by using the orientation of atoms.
If the molecules (
The molecules of type
The molecules of type
Lewis structure for any molecule is drawn by using the following steps,
First the skeletal structure for the given molecule is drawn then the total number of valence electrons for all atoms present in the molecule is determined
The next step is to subtract the electrons present in the total number of bonds present in the skeletal structure of the molecule with the total valence electrons such that considering each bond contains two electrons with it.
Finally, the electrons which got after subtractions has to be equally distributed such that each atom contains eight electrons in its valence shell.
Electron Domain: In VSEPR theory, both the lone pair and the bonded pair are together considered as electron domain regardless of the type of bond in which the bonded pair presents.
Answer to Problem 7.9QP
(d)
See-saw shaped
Explanation of Solution
To predict: The geometry for the given molecule.
Draw the Lewis structure for the molecule (d)
First the skeletal structure for the given molecule is drawn then the total number of valence electrons in the molecule is 34.
The next step is to subtract the electrons present in the total number of bonds present in the molecule with the total valence electrons such that 8 has to be subtracted with 8 as each bond contains two electrons with it and there are four bonds in the skeletal structure.
Then the 26 electrons got after the subtractions should be placed over the atoms present in the molecule such that each atom contains eight electrons in the valence shell.
Determine the molecular geometry for the molecule (d) using VSEPR.
The electron domain for the given molecule is obtained by viewing the Lewis structure shows that it contains five electron domains since it has 4 chlorine atoms and one lone pair with it.
The molecular geometry for the molecule is see-saw shape due to the present of that one lone pair of electron.
Want to see more full solutions like this?
Chapter 7 Solutions
Chemistry: Atoms First
- If I have 10 data points for variables x and y, when I represent y versus x I obtain a line with the equation y = mx + b. Is the slope m equal to dy/dx?arrow_forwardThe data for the potential difference of a battery and its temperature are given in the table. Calculate the entropy change in J mol-1 K-1 (indicate the formulas used).Data: F = 96485 C mol-1arrow_forwardIn a cell, the change in entropy (AS) can be calculated from the slope of the E° vs 1/T graph. The slope is equal to -AS/R, where R is the gas constant. Is this correct?arrow_forward
- Using the Arrhenius equation, it is possible to establish the relationship between the rate constant (k) of a chemical reaction and the temperature (T), in Kelvin (K), the universal gas constant (R), the pre-exponential factor (A) and the activation energy (Ea). This equation is widely applied in studies of chemical kinetics, and is also widely used to determine the activation energy of reactions. In this context, the following graph shows the variation of the rate constant with the inverse of the absolute temperature, for a given chemical reaction that obeys the Arrhenius equation. Based on the analysis of this graph and the concepts acquired about the kinetics of chemical reactions, analyze the following statements: I. The activation energy (Ea) varies with the temperature of the system. II. The activation energy (Ea) varies with the concentration of the reactants. III. The rate constant (K) varies proportionally with temperature. IV. The value of the…arrow_forwardIn an electrolytic cell, indicate the formula that relates E0 to the temperature T.arrow_forward-- 14:33 A Candidate Identification docs.google.com 11. Compound A can transform into compound B through an organic reaction. From the structures below, mark the correct one: HO A تھے۔ די HO B ○ A) Compounds A and B are isomers. B) Both have the same number of chiral carbons. C) Compound A underwent an addition reaction of Cl2 and H2O to form compound B. D) Compound A underwent a substitution reaction forming the intermediate chlorohydrin to obtain compound B. E) Compound A underwent an addition reaction of Cl2 forming the chloronium ion and then added methanol to obtain compound B. 60arrow_forward
- -- 14:40 A Candidate Identification docs.google.com 13. The compound 1-bromo-hex-2-ene reacts with methanol to form two products. About this reaction, mark the correct statement: OCH3 CH3OH Br OCH3 + + HBr A B A) The two products formed will have the same percentage of formation. B) Product B will be formed by SN1 substitution reaction with the formation of an allylic carbocation. C) Product A will be formed by SN1 substitution reaction with the formation of a more stable carbocation than product B. D) Product A will be formed by an SN2 substitution reaction occurring in two stages, the first with slow kinetics and the second with fast kinetics. E) The two compounds were obtained by addition reaction, with compound B having the highest percentage of formation. 57arrow_forward-- ☑ 14:30 A Candidate Identification docs.google.com 10. Amoxicillin (figure X) is one of the most widely used antibiotics in the penicillin family. The discovery and synthesis of these antibiotics in the 20th century made the treatment of infections that were previously fatal routine. About amoxicillin, mark the correct one: HO NH2 H S -N. HO Figura X. Amoxicilina A) It has the organic functions amide, ester, phenol and amine. B) It has four chiral carbons and 8 stereoisomers. C) The substitution of the aromatic ring is of the ortho-meta type. D) If amoxicillin reacts with an alcohol it can form an ester. E) The structure has two tertiary amides. 62arrow_forwardThe environmental police of a Brazilian state received a report of contamination of a river by inorganic arsenic, due to the excessive use of pesticides on a plantation on the riverbanks. Arsenic (As) is extremely toxic in its many forms and oxidation states. In nature, especially in groundwater, it is found in the form of arsenate (AsO ₄ ³ ⁻ ), which can be electrochemically reduced to As ⁰ and collected at the cathode of a coulometric cell. In this case, Potentiostatic Coulometry (at 25°C) was performed in an alkaline medium (pH = 7.5 throughout the analysis) to quantify the species. What potential (E) should have been selected/applied to perform the analysis, considering that this is an exhaustive electrolysis technique (until 99.99% of all AsO ₄ ³ ⁻ has been reduced to As ⁰ at the electrode, or n( final) = 0.01% n( initial )) and that the concentration of AsO ₄ ³ ⁻ found in the initial sample was 0.15 mmol/L ? Data: AsO ₄ 3 ⁻ (aq) + 2 H ₂ O ( l ) + 2 e ⁻ → A s O ₂ ⁻ ( a…arrow_forward
- -- 14:17 15. Water-soluble proteins are denatured when there is a change in the pH of the environment in which they are found. This occurs due to the protonation and deprotonation of functional groups present in their structure. Choose the option that indicates the chemical bonds modified by pH in the protein represented in the following figure. E CH2 C-OH CH2 H₂C H₁C CH CH3 CH3 CH CH₂-S-S-CH₂- 910 H B -CH2-CH2-CH2-CH₂-NH3* −0—C—CH₂- ○ A) A, C e D. • В) Вес ○ C) DeE ○ D) B, De E ○ E) A, B e C 68arrow_forwardSuppose sodium sulfate has been gradually added to 100 mL of a solution containing calcium ions and strontium ions, both at 0.15 mol/L. Indicate the alternative that presents the percentage of strontium ions that will have precipitated when the calcium sulfate begins to precipitate. Data: Kps of calcium sulfate: 2.4x10 ⁻ ⁵; Kps of strontium sulfate: 3.2x10 ⁻ ⁷ A) 20,2 % B) 36,6 % C) 62,9 % D) 87,5 % E) 98.7%arrow_forward14:43 A Candidate Identification docs.google.com 14. The following diagrams represent hypothetical membrane structures with their components numbered from 1 to 6. Based on the figures and your knowledge of biological membranes, select the correct alternative. | 3 5 || 人 2 500000 6 A) Structures 1, 3, 5, 2 and 4 are present in a constantly fluid arrangement that allows the selectivity of the movement ○ of molecules. Structure 4, present integrally or peripherally, is responsible for this selection, while the quantity of 6 regulates the fluidity. B) The membranes isolate the cell from the environment, but allow the passage of water-soluble molecules thanks to the presence of 2 and 3. The membrane in scheme is more fluid than that in 55arrow_forward
- Chemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning

