Concept explainers
Videos
(a)
Interpretation:
The molecular shape of
Concept introduction:
The following steps are used to determine the molecular shape and electron-group arrangement of the given molecule.
Step 1: Write the Lewis structure of the given formula.
Step 2: Count all electron groups including bonding and non-bonding pairs around the central atom and use it to assign one of the five electron-group arrangement.
Step 3: Write the ideal bond angle related to the electron-group arrangement and the ideal angles can be predicted by using VSEPR.
Step 4: Count the bonding and non-bonding electron groups and draw the molecular shape accordingly.
(b)
Interpretation:
The molecular shape of
Concept introduction:
The following steps are used to determine the molecular shape and electron-group arrangement of the given molecule.
Step 1: Write the Lewis structure of the given formula and place the atoms.
Step 2: Count all electron groups including bonding and non-bonding pairs around the central atom and use it to assign one of the five electron-group arrangement.
Step 3: Write the ideal bond angle related to the electron-group arrangement and the ideal angles can be predicted by using VSEPR.
Step 4: Count the bonding and non-bonding electron groups and draw the molecular shape accordingly.
(c)
Interpretation:
The molecular shape of
Concept introduction:
The following steps are used to determine the molecular shape and electron-group arrangement of the given molecule.
Step 1: Write the Lewis structure of the given formula and place the atoms.
Step 2: Count all electron groups including bonding and non-bonding pairs around the central atom and use it to assign one of the five electron-group arrangement.
Step 3: Write the ideal bond angle related to the electron-group arrangement and the effect of any deviation can be predicted by using VSEPR.
Step 4: Count the bonding and non-bonding electron groups and draw the molecular shape accordingly.
(d)
Interpretation:
The molecular shape of
Concept introduction:
The following steps are used to determine the molecular shape and electron-group arrangement of the given molecule.
Step 1: Write the Lewis structure of the given formula and place the atoms.
Step 2: Count all electron groups including bonding and non-bonding pairs around the central atom and use it to assign one of the five electron-group arrangement.
Step 3: Write the ideal bond angle related to the electron-group arrangement and the ideal angles can be predicted by using VSEPR.
Step 4: Count the bonding and non-bonding electron groups and draw the molecular shape accordingly.
Want to see the full answer?
Check out a sample textbook solution
Chapter 10 Solutions
MOLECULAR NATURE OF MATTER ALEKS ACCESS
- 2. Write a complete mechanism for the reaction shown below. NaOCH LOCH₁ O₂N NO2 CH₂OH, 20 °C O₂N NO2arrow_forward4. Propose a synthesis of the target molecules from the respective starting materials. a) b) LUCH C Br OHarrow_forwardThe following mechanism for the gas phase reaction of H2 and ICI that is consistent with the observed rate law is: step 1 step 2 slow: H2(g) +ICI(g) → HCl(g) + HI(g) fast: ICI(g) + HI(g) → HCl(g) + |2(g) (1) What is the equation for the overall reaction? Use the smallest integer coefficients possible. If a box is not needed, leave it blank. + → + (2) Which species acts as a catalyst? Enter formula. If none, leave box blank: (3) Which species acts as a reaction intermediate? Enter formula. If none, leave box blank: (4) Complete the rate law for the overall reaction that is consistent with this mechanism. (Use the form k[A][B]"..., where '1' is understood (so don't write it) for m, n etc.) Rate =arrow_forward
- Please correct answer and don't use hand rating and don't use Ai solutionarrow_forward1. For each of the following statements, indicate whether they are true of false. ⚫ the terms primary, secondary and tertiary have different meanings when applied to amines than they do when applied to alcohols. • a tertiary amine is one that is bonded to a tertiary carbon atom (one with three C atoms bonded to it). • simple five-membered heteroaromatic compounds (e.g. pyrrole) are typically more electron rich than benzene. ⚫ simple six-membered heteroaromatic compounds (e.g. pyridine) are typically more electron rich than benzene. • pyrrole is very weakly basic because protonation anywhere on the ring disrupts the aromaticity. • thiophene is more reactive than benzene toward electrophilic aromatic substitution. • pyridine is more reactive than nitrobenzene toward electrophilic aromatic substitution. • the lone pair on the nitrogen atom of pyridine is part of the pi system.arrow_forwardThe following reactions are NOT ordered in the way in which they occur. Reaction 1 PhO-OPh Reaction 2 Ph-O -CH₂ heat 2 *OPh Pho -CH2 Reaction 3 Ph-O ⚫OPh + -CH₂ Reaction 4 Pho Pho + H₂C OPh + CHOPh H₂C -CH₂ Reactions 1 and 3 Reaction 2 O Reaction 3 ○ Reactions 3 and 4 ○ Reactions 1 and 2 Reaction 4 ○ Reaction 1arrow_forward
- Select all possible products from the following reaction: NaOH H₂O a) b) ОН HO O HO HO e) ОН f) O HO g) h) + OHarrow_forward3. Draw diagrams to represent the conjugation in these molecules. Draw two types of diagram: a. Show curly arrows linking at least two different ways of representing the molecule b. Indicate with dotted lines and partial charges (where necessary) the partial double bond (and charge) distribution H₂N* H₂N -NH2arrow_forwardQuestion 2 of 25 point Question Attempt 3 of Ulimited Draw the structure for 3-chloro-4-ethylheptane. Part 2 of 3 Click and drag to start drawing a structure. Draw the structure for 1-chloro-4-ethyl-3-lodooctane. Click and drag to start drawing a structure. X G X B c Part 3 of 30 Draw the structure for (R)-2-chlorobutane. Include the stereochemistry at all stereogenic centers. Check Click and drag to start drawing a structure. G X A 。 MacBook Pro G P Save For Later Submit Assignment Privacyarrow_forward
- Please correct answer and don't used hand raitingarrow_forwardIn a silicon and aluminum alloy, with 12.6% silicon, what are the approximate percentages of the phases present in the constituent that is formed at the end of solidification? Temperature (°C) 1500 1000 L B+L 1415- α+L 577' 500 1.65 12.6 99.83 α+B B 0 Al 20 40 60 Weight percent silicon 80 Siarrow_forwardPlease correct answer and don't used hand raitingarrow_forward
ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
ChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill Education
Principles of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill Education
Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
Elementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY