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a)
Interpretation: The molecular electronic configuration of
Concept introduction: Two or more than two atomic orbitals overlap to form bond, these orbitals are called molecular orbitals. Count of molecular orbital obtained is same as count of atomic orbitals mixed.
Two forms of molecular orbital are obtained. These are bonding and antibonding orbital.
Bonding orbitals are those in that electrons are in between nucleus of two atom. Antibonding orbitals are those in which electrons are away from nucleus of two-atom. Also, electrons in antibonding orbital have energy high as compared to bonding orbital.
In sigma
In pi
In pi
b)
Interpretation: The bond order of
Concept introduction: Two or more than two atomic orbitals overlap to form bond, these orbitals are called molecular orbitals. Count of molecular orbital obtained is same as count of atomic orbitals mixed.
Two forms of molecular orbital are obtained. These are bonding and antibonding orbital.
Bonding orbitals are those in that electrons are in between nucleus of two atom. Antibonding orbitals are those in which electrons are away from nucleus of two-atom. Also, electrons in antibonding orbital have energy high as compared to bonding orbital.
In sigma
In pi
In pi
c)
Interpretation: The species that is paramagnetic should be determined.
Concept introduction: Two or more than two atomic orbitals overlap to form bond, these orbitals are called molecular orbitals. Count of molecular orbital obtained is same as count of atomic orbitals mixed.
Two forms of molecular orbital are obtained. These are bonding and antibonding orbital.
Bonding orbitals are those in that electrons are in between nucleus of two atom. Antibonding orbitals are those in which electrons are away from nucleus of two-atom. Also, electrons in antibonding orbital have energy high as compared to bonding orbital.
In sigma
In pi
In pi
d)
Interpretation: The ion that has high
Concept introduction: Two or more than two atomic orbitals overlap to form bond, these orbitals are called molecular orbitals. Count of molecular orbital obtained is same as count of atomic orbitals mixed.
Two forms of molecular orbital are obtained. These are bonding and antibonding orbital.
Bonding orbitals are those in that electrons are in between nucleus of two atom. Antibonding orbitals are those in which electrons are away from nucleus of two-atom. Also, electrons in antibonding orbital have energy high as compared to bonding orbital.
In sigma
In pi
In pi
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Chapter 6 Solutions
PRINCIPLES OF MODERN CHEMISTRY-OWLV2
- 3. Propose a synthesis for the following transformation. Do not draw an arrow-pushing mechanism below, but make sure to draw the product of each proposed step (3 points). CN + En CNarrow_forward3) Propagation of uncertainty. Every measurement has uncertainty. In this problem, we'll evaluate the uncertainty in every step of a titration of potassium hydrogen phthalate (a common acid used in titrations, abbreviated KHP, formula CsH5KO4) with NaOH of an unknown concentration. The calculation that ultimately needs to be carried out is: concentration NaOH 1000 x mass KHP × purity KHP molar mass KHP x volume NaOH Measurements: a) You use a balance to weigh 0.3992 g of KHP. The uncertainty is ±0.15 mg (0.00015 g). b) You use a buret to slowly add NaOH to the KHP until it reaches the endpoint. It takes 18.73 mL of NaOH. The uncertainty of the burst is 0.03 mL.. c) The manufacturer states the purity of KHP is 100%±0.05%. d) Even though we don't think much about them, molar masses have uncertainty as well. The uncertainty comes from the distribution of isotopes, rather than random measurement error. The uncertainty in the elements composing KHP are: a. Carbon: b. Hydrogen: ±0.0008…arrow_forwardDon't used hand raiting and don't used Ai solutionarrow_forward
- How would you use infrared spectroscopy to distinguish between the following pairs of constitutional isomers? (a) CH3C=CCH3 || and CH3CH2C=CH (b) CH3CCH=CHCH3 and CH3CCH2CH=CH2 Problem 12-41 The mass spectrum (a) and the infrared spectrum (b) of an unknown hydrocarbon are shown. Propose as many structures as you can. (a) 100 Relative abundance (%) 80 60 60 40 200 20 (b) 100 Transmittance (%) 10 20 20 80- 60- 40- 20 40 60 80 100 120 140 m/z 500 4000 3500 3000 2500 2000 1500 Wavenumber (cm-1) 1000arrow_forwardPropagation of uncertainty. You have a stock solution certified by the manufacturer to contain 150.0±0.03 µg SO42-/mL. You would like to dilute it by a factor of 100 to obtain 1.500 µg/mL. Calculate the uncertainty in the two methods of dilution below. Use the following uncertainty values for glassware: Glassware Uncertainty (assume glassware has been calibrated and treat the values below as random error) 1.00 mL volumetric pipet 0.01 mL 10.00 mL volumetric pipet 0.02 mL 100.00 mL volumetric flask 0.08 mL Transfer 10.00 mL with a volumetric pipet and dilute it to 100 mL with a volumetric flask. Then take 10.00 mL of the resulting solution and dilute it a second time with a 100 mL flask. 2. Transfer 1.00 mL with a volumetric pipet and dilute it to 100 mL with a volumetric flask.arrow_forwardDraw all resonance structures for the following ion: CH₂ Draw all resonance structures on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars, including charges where needed. The single bond is active by default. 2D ד CONT HD EXP CON ? 1 [1] Α 12 Marvin JS by Chemaxon A DOO H C N Br I UZ OSPFarrow_forward
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