Concept explainers
(a)
Interpretation:
The hybrid orbitals used by the central atom bromine and the types of bonds that are formed in
Concept introduction:
The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.
Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.
Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.
Sigma bond is formed by the end to end overlapping of atomic or hybrid orbitals. All single bonds are sigma bonds. One double bond is made up of one sigma bond and one pi bond. One triple bond is made up of one sigma bond and two pi bonds.
(b)
Interpretation:
The hybrid orbitals used by the central atom carbon and the types of bonds that are formed in propyne is to be determined.
Concept introduction:
The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.
Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.
Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.
Sigma bond is formed by the end to end overlapping of atomic or hybrid orbitals. All single bonds are sigma bonds. One double bond is made up of one sigma bond and one pi bond. One triple bond is made up of one sigma bond and two pi bonds.
(c)
Interpretation:
The hybrid orbitals used by the central atom sulfur and the types of bonds that are formed in
Concept introduction:
The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.
Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.
Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.
Sigma bond is formed by the end to end overlapping of atomic or hybrid orbitals. All single bonds are sigma bonds. One double bond is made up of one sigma bond and one pi bond. One triple bond is made up of one sigma bond and two pi bonds.
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Chapter 11 Solutions
LL CHEM: MOL NAT CHNG W/CNCT AC
- Transmitance 3. Which one of the following compounds corresponds to this IR spectrum? Point out the absorption band(s) that helped you decide. OH H3C OH H₂C CH3 H3C CH3 H3C INFRARED SPECTRUM 0.8- 0.6 0.4- 0.2 3000 2000 1000 Wavenumber (cm-1) 4. Consider this compound: H3C On the structure above, label the different types of H's as A, B, C, etc. In table form, list the labeled signals, and for each one state the number of hydrogens, their shifts, and the splitting you would observe for these hydrogens in the ¹H NMR spectrum. Label # of hydrogens splitting Shift (2)arrow_forwardNonearrow_forwardDraw the Lewis structure of C2H4Oarrow_forward
- a) 5. Circle all acidic (and anticoplanar to the Leaving group) protons in the following molecules, Solve these elimination reactions, and identify the major and minor products where appropriate: 20 points + NaOCH3 Br (2 productarrow_forwardNonearrow_forwardDr. Mendel asked his BIOL 260 class what their height was and what their parent's heights were. He plotted that data in the graph below to determine if height was a heritable trait. A. Is height a heritable trait? If yes, what is the heritability value? (2 pts) B. If the phenotypic variation is 30, what is the variation due to additive alleles? (2 pts) Offspring Height (Inches) 75 67.5 60 52.5 y = 0.9264x + 4.8519 55 60 65 MidParent Height (Inches) 70 75 12pt v V Paragraph B IUA > AT2 v Varrow_forward
- Experiment: Each team will be provided with 5g of a mixture of acetanilide and salicylic acid. You will divide it into three 1.5 g portions in separate 125 mL Erlenmeyer flasks savıng some for melting point analysis. Dissolve the mixture in each flask in ~60mL of DI water by heating to boiling on a hotplate. Take the flasks off the hotplate once you have a clear solution and let them stand on the bench top for 5 mins and then allow them to cool as described below. Sample A-Let the first sample cool slowly to room temperature by letting it stand on your lab bench, with occasional stirring to promote crystallization. Sample B-Cool the second sample 1n a tap-water bath to 10-15 °C Sample C-Cool the third sample in an ice-bath to 0-2 °C Results: weight after recrystalization and melting point temp. A=0.624g,102-115° B=0.765g, 80-105° C=1.135g, 77-108 What is the percent yield of A,B, and C.arrow_forwardRel. Intensity Q 1. Which one of the following is true of the compound whose mass spectrum is shown here? Explain how you decided. 100 a) It contains chlorine. b) It contains bromine. c) It contains neither chlorine nor bromine. 80- 60- 40- 20- 0.0 0.0 TT 40 80 120 160 m/z 2. Using the Table of IR Absorptions how could you distinguish between these two compounds in the IR? What absorbance would one compound have that the other compound does not? HO CIarrow_forwardIllustrate reaction mechanisms of alkenes with water in the presence of H2SO4, detailing each step of the process. Please show steps of processing. Please do both, I will thumb up for sure #1 #3arrow_forward
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