
a.
Interpretation: The geometry of
Concept Introduction:
Hybridization: It relates to the mixing of atomic orbitals into new hybrid orbitals that have varied energies and shapes. It is appropriate for the pairing of the electrons for forming
The main concept behind this theory is that the electron pairs are always present in the outermost shell i.e. valence shell of an atom of a molecule and they repel each other due to which they try to attain the best possible position so that the value of their repulsion is the least. Hence, the electrons occupy such positions around the atom that reduces their repulsion and provides a molecule to their shape.
Here the electrons that take part in the bonding of a molecule are known as the bonding pair and the electrons that do not take part in the bonding are known as the lone pairs. The bond pairs are in the influence of the two bonding atoms whereas the lone pairs are in the influence of only of the atom.
Due to the presence of lone pairs, there is more space occupied between the atoms of the molecules. Now they suffer the repulsion between the lone pair-lone pair and bond pair-lone pair. Their repulsion can be represented as:-
lp-lp>lp-bp>bp-bp
a.

Answer to Problem 7.65PAE
Solution: The geometry of
Explanation of Solution
The electronic configuration of I is
Structure of
The geometry of
b.
Interpretation:
The geometry of
Concept Introduction
Hybridization: It relates to the mixing of atomic orbitals into new hybrid orbitals that have varied energies and shapes. It is appropriate for the pairing of the electrons for forming chemical bonds in the Valence Bond Theory. We can predict the shape of a particular molecule by the knowledge of their atomic numbers and VSEPR theory according to which the atoms take such a position that there is a minimum possible repulsion between the bonded atoms and the lone pair of electrons if any.
The main concept behind this theory is that the electron pairs are always present in the outermost shell i.e. valence shell of an atom of a molecule and they repel each other due to which they try to attain the best possible position so that the value of their repulsion is the least. Hence, the electrons occupy such positions around the atom that reduces their repulsion and provides a molecule to their shape.
Here the electrons that take part in the bonding of a molecule are known as the bonding pair and the electrons that do not take part in the bonding are known as the lone pairs. The bond pairs are in the influence of the two bonding atoms whereas the lone pairs are in the influence of only of the atom.
Due to the presence of lone pairs, there is more space occupied between the atoms of the molecules. Now they suffer the repulsion between the lone pair-lone pair and bond pair-lone pair. Their repulsion can be represented as:-
lp-lp>lp-bp>bp-bp.
b.

Answer to Problem 7.65PAE
Solution:
The geometry of
Explanation of Solution
The electronic configuration of Cl is
Structure of
The geometry of
c.
Interpretation:
The geometry of
Concept Introduction
Hybridization: It relates to the mixing of atomic orbitals into new hybrid orbitals that have varied energies and shapes. It is appropriate for the pairing of the electrons for forming chemical bonds in the Valence Bond Theory. We can predict the shape of a particular molecule by the knowledge of their atomic numbers and VSEPR theory according to which the atoms take such a position that there is a minimum possible repulsion between the bonded atoms and the lone pair of electrons if any.
The main concept behind this theory is that the electron pairs are always present in the outermost shell i.e. valence shell of an atom of a molecule and they repel each other due to which they try to attain the best possible position so that the value of their repulsion is the least. Hence, the electrons occupy such positions around the atom that reduces their repulsion and provides a molecule to their shape.
Here the electrons that take part in the bonding of a molecule are known as the bonding pair and the electrons that do not take part in the bonding are known as the lone pairs. The bond pairs are in the influence of the two bonding atoms whereas the lone pairs are in the influence of only of the atom.
Due to the presence of lone pairs, there is more space occupied between the atoms of the molecules. Now they suffer the repulsion between the lone pair-lone pair and bond pair-lone pair. Their repulsion can be represented as:-
lp-lp>lp-bp>bp-bp
c.

Answer to Problem 7.65PAE
Solution:
The geometry of
is trigonal pyramidal as the hybridization of I is sp3d
Explanation of Solution
The electronic configuration of Tellurium is
Structure of
The geometry of
d.
Interpretation:
The geometry of
Concept Introduction:
Hybridization: It relates to the mixing of atomic orbitals into new hybrid orbitals that have varied energies and shapes. It is appropriate for the pairing of the electrons for forming chemical bonds in the Valence Bond Theory. We can predict the shape of a particular molecule by the knowledge of their atomic numbers and VSEPR theory according to which the atoms take such a position that there is a minimum possible repulsion between the bonded atoms and the lone pair of electrons if any.
The main concept behind this theory is that the electron pairs are always present in the outermost shell i.e. valence shell of an atom of a molecule and they repel each other due to which they try to attain the best possible position so that the value of their repulsion is the least. Hence, the electrons occupy such positions around the atom that reduces their repulsion and provides a molecule to their shape.
Here the electrons that take part in the bonding of a molecule are known as the bonding pair and the electrons that do not take part in the bonding are known as the lone pairs. The bond pairs are in the influence of the two bonding atoms whereas the lone pairs are in the influence of only of the atom.
Due to the presence of lone pairs, there is more space occupied between the atoms of the molecules. Now they suffer the repulsion between the lone pair-lone pair and bond pair-lone pair. Their repulsion can be represented as:-
lp-lp>lp-bp>bp-bp
d.

Answer to Problem 7.65PAE
Solution:
The geometry of
Explanation of Solution
The electronic configuration of Tellurium is
Structure of
The geometry of
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Chapter 7 Solutions
EBK CHEMISTRY FOR ENGINEERING STUDENTS,
- Experiment 27 hates & Mechanisms of Reations Method I visual Clock Reaction A. Concentration effects on reaction Rates Iodine Run [I] mol/L [S₂082] | Time mo/L (SCC) 0.04 54.7 Log 1/ Time Temp Log [ ] 13,20] (time) / [I] 199 20.06 23.0 30.04 0.04 0.04 80.0 22.8 45 40.02 0.04 79.0 21.6 50.08 0.03 51.0 22.4 60-080-02 95.0 23.4 7 0.08 0-01 1970 23.4 8 0.08 0.04 16.1 22.6arrow_forward(15 pts) Consider the molecule B2H6. Generate a molecular orbital diagram but this time using a different approach that draws on your knowledge and ability to put concepts together. First use VSEPR or some other method to make sure you know the ground state structure of the molecule. Next, generate an MO diagram for BH2. Sketch the highest occupied and lowest unoccupied MOs of the BH2 fragment. These are called frontier orbitals. Now use these frontier orbitals as your basis set for producing LGO's for B2H6. Since the BH2 frontier orbitals become the LGOS, you will have to think about what is in the middle of the molecule and treat its basis as well. Do you arrive at the same qualitative MO diagram as is discussed in the book? Sketch the new highest occupied and lowest unoccupied MOs for the molecule (B2H6).arrow_forwardQ8: Propose an efficient synthesis of cyclopentene from cyclopentane.arrow_forward
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- Predict major product(s) for the following reactions. Note the mechanism(s) of the reactions (SN1, E1, SN2 or E2).arrow_forwardPredict major product(s) for the following reactions. Note the mechanism(s) of the reactions (SN1, E1, SN2 or E2).arrow_forwardQ3: Rank the following compounds in increasing reactivity of E1 and E2 eliminations, respectively. Br ca. go do A CI CI B C CI Darrow_forward
- Q5: Predict major product(s) for the following reactions. Note the mechanism(s) of the reactions (SN1, E1, SN2 or E2). H₂O דיי "Br KN3 CH3CH2OH NaNH2 NH3 Page 3 of 6 Chem 0310 Organic Chemistry 1 HW Problem Sets CI Br excess NaOCH 3 CH3OH Br KOC(CH3)3 DuckDuckGarrow_forwardQ4: Circle the substrate that gives a single alkene product in a E2 elimination. CI CI Br Brarrow_forwardPlease calculate the chemical shift of each protonsarrow_forward
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