Concept explainers
(a)
Interpretation:
Based on
Concept introduction:
According to valence bond theory, complex ions have coordinate covalent bonds between
Ligands (Lewis bases) and metal ions (Lewis acids).
A coordinate covalent bond is a bond formed when both electrons came from one atom.
Valence bond theory, which helps explain bonding and structure in main-group compounds (can also be used to describe bonding in complex ions.
In the formation of a complex ion, the filled ligand orbital overlaps an empty metal-ion
orbital: the ligand (Lewis base) donates an electron pair, and the metal ion (Lewis acid)
accepts it to form a covalent bond in the complex ion (Lewis adduct).
A bond in which one atom contributes both electrons is a coordinate covalent bond; once
formed, it is identical to any covalent single bond.
The VB concept of hybridization proposes mixing particular combinations of
Let’s discuss orbital combinations that lead to octahedral, square planar, and tetrahedral geometries.
(b)
Interpretation:
Based on valence bond theory, which set of orbitals is used by a period 4 metal ion in forming a tetrahedral complex has to be explained.
Concept introduction:
According to valence bond theory, complex ions have coordinate covalent bonds between
Ligands (Lewis bases) and metal ions (Lewis acids).
A coordinate covalent bond is a bond formed when both electrons came from one atom.
Valence bond theory, which helps explain bonding and structure in main-group compounds (can also be used to describe bonding in complex ions.
In the formation of a complex ion, the filled ligand orbital overlaps an empty metal-ion
orbital: the ligand (Lewis base) donates an electron pair, and the metal ion (Lewis acid)
accepts it to form a covalent bond in the complex ion (Lewis adduct).
A bond in which one atom contributes both electrons is a coordinate covalent bond; once
formed, it is identical to any covalent single bond.
The VB concept of hybridization proposes mixing particular combinations of
Let’s discuss orbital combinations that lead to octahedral, square planar, and tetrahedral geometries.
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Chapter 23 Solutions
CHEM 212:CHEMISTSRY V 2
- 1. For the four structures provided, Please answer the following questions in the table below. a. Please draw π molecular orbital diagram (use the polygon-and-circle method if appropriate) and fill electrons in each molecular orbital b. Please indicate the number of π electrons c. Please indicate if each molecule provided is anti-aromatic, aromatic, or non- aromatic TT MO diagram Number of π e- Aromaticity Evaluation (X choose one) Non-aromatic Aromatic Anti-aromatic || ||| + IVarrow_forward1.3 grams of pottasium iodide is placed in 100 mL of o.11 mol/L lead nitrate solution. At room temperature, lead iodide has a Ksp of 4.4x10^-9. How many moles of precipitate will form?arrow_forwardQ3: Circle the molecules that are optically active: ДДДДarrow_forward
- 6. How many peaks would be observed for each of the circled protons in the compounds below? 8 pts CH3 CH3 ΤΙ A. H3C-C-C-CH3 I (₁₁ +1)= 7 H CI B. H3C-C-CI H (3+1)=4 H LIH)=2 C. (CH3CH2-C-OH H D. CH3arrow_forwardNonearrow_forwardQ1: Draw the most stable and the least stable Newman projections about the C2-C3 bond for each of the following isomers (A-C). Are the barriers to rotation identical for enantiomers A and B? How about the diastereomers (A versus C or B versus C)? H Br H Br (S) CH3 (R) CH3 H3C (S) H3C H Br Br H A C enantiomers H Br H Br (R) CH3 H3C (R) (S) CH3 H3C H Br Br H B D identicalarrow_forward
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