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(a)
Interpretation: The electron configuration for the boron atom needs to be written and the number of unpaired electrons present in it needs to be determined.
Concept Introduction: The electron configuration explains the electron distribution in atomic orbitals. There is a standard notation to write an electron configuration. The atomic shell is written in a sequence with the number of electrons in superscript.
(a)
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Answer to Problem 9SP
Explanation of Solution
The given atom is boron. It belongs to group 13 with
To determine the number of unpaired electrons, the atomic orbitals can be drawn as follows:
Now according to Hund’s rule, the filling of electrons in orbitals takes place in such a way that every orbital is singly occupied before any orbital is doubly occupied by electrons. Also, according to the Pauli-exclusion principle, two electrons in the same orbital must have opposite spins.
Now, electrons are filled as follows:
There is 1 unpaired electron in the 2p orbital; thus, the number of unpaired electrons will be 1.
(b)
Interpretation: The electron configuration for the silicon atom needs to be written and the number of unpaired electrons present in it needs to be determined.
Concept Introduction: The electron configuration explains the electron distribution in atomic orbitals. There is a standard notation to write an electron configuration. The atomic shell is written in a sequence with the number of electrons in superscript.
(b)
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Answer to Problem 9SP
Explanation of Solution
The given atom is silicon. It belongs to group 14 with atomic number 14. Since the maximum electrons hold by s and p orbitals are 2 and 6 respectively. The electronic configuration is represented as follows:
To determine the number of unpaired electrons, the atomic orbitals can be drawn as follows:
Now according to Hund’s rule, the filling of electrons in orbitals takes place in such a way that every orbital is singly occupied before any orbital is doubly occupied by electrons. Also, according to the Pauli-exclusion principle, two electrons in the same orbital must have opposite spins.
Now, electrons are filled as follows:
There are 2 unpaired electrons in the 3p orbital; thus, the number of unpaired electrons will be 2.
(c)
Interpretation: The electron configuration for the sulfur atom needs to be written and the number of unpaired electrons present in it needs to be determined.
Concept Introduction: The electron configuration explains the electron distribution in atomic orbitals. There is a standard notation to write an electron configuration. The atomic shell is written in a sequence with the number of electrons in superscript.
(c)
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Answer to Problem 9SP
Explanation of Solution
The given atom is sulfur. It belongs to group 16 with the atomic number 16. Since the maximum electrons hold by s and p orbitals are 2 and 6 respectively. The electronic configuration is represented as follows:
To determine the number of unpaired electrons, the atomic orbitals can be drawn as follows:
Now according to Hund’s rule, the filling of electrons in orbitals takes place in such a way that every orbital is singly occupied before any orbital is doubly occupied by electrons. Also, according to the Pauli-exclusion principle, two electrons in the same orbital must have opposite spins.
Now, electrons are filled as follows:
There are 2 unpaired electrons in the 3p orbital; thus, the number of unpaired electrons will be 2.
Chapter 5 Solutions
Chemistry 2012 Student Edition (hard Cover) Grade 11
- option choice: Isoleucine Histidine Threonine Alanine Lysine Aspartate Tryptophan Tyrosine Leucine Arginine Cysteine Asparagine Valine Glutamine Glycine Methionine Serine Proline Phenylalanine Glutamatearrow_forwardsketch the nature of the metal-alkylidene bonding interactions.arrow_forwardPart C The perspective formula of isoleucine, an amino acid, is provided below. HOOC H₂NIC H 川 CH3 CH,CH3 Draw the Newman projection in staggered conformation for isoleucine by viewing the molecule along the C-2-C-3 bond. 1. Edit the Newman projection on the canvas. 2. Replace the appropriate hydrogens with the appropriate -CH3 or other groups. 3. If you need to start over, Undo or choose a Newman projection from the Templates toolbar (bottom). Important: Never delete the hydrogen atoms or bonds directly attached to the template, and do not move them by dragging or dropping them. That will break the projections structures. Only replace them! ▸ View Available Hint(s) 0 2 H± 3D EXP. L ד י CONT. 2 H 0 N оarrow_forward
- Use the literature Ka value of the acetic acid, and the data below to answer these questions. Note: You will not use the experimental titration graphs to answer the questions that follow. Group #1: Buffer pH = 4.35 Group #2: Buffer pH = 4.70 Group #3: Buffer pH = 5.00 Group #4: Buffer pH = 5.30 Use the Henderson-Hasselbalch equation, the buffer pH provided and the literature pKa value of acetic acid to perform the following: a) calculate the ratios of [acetate]/[acetic acid] for each of the 4 groups buffer solutions above. b) using the calculated ratios, which group solution will provide the best optimal buffer (Hint: what [acetate]/[acetic acid] ratio value is expected for an optimal buffer?) c) explain your choicearrow_forwardHow would you prepare 1 liter of a 50 mM Phosphate buffer at pH 7.5 beginning with K3PO4 and 1 M HCl or 1 M NaOH? Please help and show calculations. Thank youarrow_forwardDraw the four most importantcontributing structures of the cation intermediate thatforms in the electrophilic chlorination of phenol,(C6H5OH) to form p-chlorophenol. Put a circle aroundthe best one. Can you please each step and also how you would approach a similar problem. Thank you!arrow_forward
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