
Concept explainers
Interpretation:
The molecular geometry that is not possible about the central atom that contains four electron groups has to be chosen from the given options.
Concept Introduction:
Information about the number of bonds and types of bonds can be obtained from Lewis structure but the molecular geometry cannot be obtained. Three dimensional arrangement of atoms in a molecule can be given by molecular geometry. Physical and chemical properties are determined by the molecular geometry of the molecule.
Using VSEPR theory and Lewis structure, the molecular geometry of the molecule that contain less number of atoms can be predicted. VSEPR theory uses the information from Lewis structure of the molecule to predict the molecular geometry of the molecule. Main concept of VSEPR theory is that electron pairs that are present in the valence shell adopt arrangement in a way that minimize the repulsion between like charges.
If the central atom contains two electron pairs, then it has to be far apart means, it has to be on opposite side of the nucleus. This means the angle has to be
If the central atom contains three electron pairs, then it has to be far apart means, it has to be on corner of a triangle. This means the angle has to be
If the central atom contains four electron pairs, then it has to be far apart means, it has to be in a tetrahedral arrangement. This means the angle has to be
The collection of valence electron that is present in localized region about central atom in a molecule is known as VSEPR electron group. This may contain two electrons, four electrons, or six electrons. The electron group that contain four and six electrons repel each other.
Tetrahedral VSEPR electron group:
The four electron pairs can be of three VSEPR electron groups. They are 4 bonding electron groups, 3 bonding and 1 nonbonding electron groups, and 2 bonding and 2 nonbonding electron groups. The molecular geometry that is associated with 4 bonding electron groups is tetrahedral. The molecular geometry that is associated with 3 bonding and 1 nonbonding electron groups is trigonal pyramidal. The molecular geometry that is associated with 2 bonding and 2 nonbonding electron groups is angular.
Trigonal planar VSEPR electron group:
The three electron pairs can be of two VSEPR electron groups. They are 3 bonding electron groups, and 2 bonding and 1 nonbonding electron groups. The molecular geometry that is associated with 3 bonding electron groups is trigonal planar. The molecular geometry that is associated with 2 bonding and 1 nonbonding electron groups is angular.
Linear VSEPR electron group:
The two electron pairs can be of only one VSEPR electron groups. It is only 2 bonding electron groups and the geometry associated with it is linear geometry.

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Chapter 5 Solutions
General, Organic, and Biological Chemistry
- A small artisanal cheesemaker is testing the acidity of their milk before it coagulates. During fermentation, bacteria produce lactic acid (K₁ = 1.4 x 104), a weak acid that helps to curdle the milk and develop flavor. The cheesemaker has measured that the developing mixture contains lactic acid at an initial concentration of 0.025 M. Your task is to calculate the pH of this mixture and determine whether it meets the required acidity for proper cheese development. To achieve the best flavor, texture and reduce/control microbial growth, the pH range needs to be between pH 4.6 and 5.0. Assumptions: Lactic acid is a monoprotic acid H H :0:0: H-C-C H :0: O-H Figure 1: Lewis Structure for Lactic Acid For simplicity, you can use the generic formula HA to represent the acid You can assume lactic acid dissociation is in water as milk is mostly water. Temperature is 25°C 1. Write the K, expression for the dissociation of lactic acid in the space provided. Do not forget to include state symbols.…arrow_forwardCurved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. :0: :0 H. 0:0 :0: :6: S: :0: Select to Edit Arrows ::0 Select to Edit Arrows H :0: H :CI: Rotation Select to Edit Arrows H. < :0: :0: :0: S:arrow_forward3:48 PM Fri Apr 4 K Problem 4 of 10 Submit Curved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. Mg. :0: Select to Add Arrows :0: :Br: Mg :0: :0: Select to Add Arrows Mg. Br: :0: 0:0- Br -190 H 0:0 Select to Add Arrows Select to Add Arrows neutralizing workup H CH3arrow_forward
- Iarrow_forwardDraw the Markovnikov product of the hydrobromination of this alkene. Note for advanced students: draw only one product, and don't worry about showing any stereochemistry. Drawing dash and wedge bonds has been disabled for this problem. + Explanation Check 1 X E 4 1 1 1 1 1 HBr Click and drag to start drawing a structure. 80 LE #3 @ 2 $4 0 I அ2 % 85 F * K M ? BH 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use Privacy Center & 6 27 FG F10 8 9 R T Y U D F G H P J K L Z X C V B N M Q W A S H option command H command optiarrow_forwardBe sure to use wedge and dash bonds to show the stereochemistry of the products when it's important, for example to distinguish between two different major products. Predict the major products of the following reaction. Explanation Q F1 A Check F2 @ 2 # 3 + X 80 F3 W E S D $ 4 I O H. H₂ 2 R Pt % 05 LL ee F6 F5 T <6 G Click and drag to start drawing a structure. 27 & A 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use Privacy Center Acce Y U H DII 8 9 F10 4 J K L Z X C V B N M T H option command F11 P H commandarrow_forward
- Curved arrows are used to illustrate the flow of electrons. Use the reaction conditions provided and follow the arrows to draw the intermediate and product in this reaction or mechanistic step(s). Include all lone pairs and charges as appropriate. Ignore stereochemistry. Ignore inorganic byproducts. H :0: CH3 O: OH Q CH3OH2+ Draw Intermediate protonation CH3OH CH3OH nucleophilic addition H Draw Intermediate deprotonation :0: H3C CH3OH2* protonation H 0: H CH3 H.arrow_forwardPredicting the reactants or products of hemiacetal and acetal formation uentify the missing organic reactants in the following reaction: H+ X+Y OH H+ за Note: This chemical equation only focuses on the important organic molecules in the reaction. Additional inorganic or small-molecule reactants or products (like H2O) are not shown. In the drawing area below, draw the skeletal ("line") structures of the missing organic reactants X and Y. You may draw the structures in any arrangement that you like, so long as they aren't touching. Explanation Check Click and drag to start drawing a structure. ? olo 18 Ar © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibilityarrow_forwardcan someone please answer thisarrow_forward
- Please, please help me figure out the the moles, molarity and Ksp column. Step by step details because I've came up with about three different number and have no idea what I'm doing wrong.arrow_forwardwhat reagents are used to get this product from this reactant? Br OCH3arrow_forwardcan someone answer this pleasearrow_forward
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