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
CHEMISTRY: MOLECULAR...(LLF) W/CONNECT
- Part 1. Aqueous 0.010M AgNO 3 is slowly added to a 50-ml solution containing both carbonate [co32-] = 0.105 M and sulfate [soy] = 0.164 M anions. Given the ksp of Ag2CO3 and Ag₂ soy below. Answer the ff: Ag₂ CO3 = 2 Ag+ caq) + co} (aq) ksp = 8.10 × 10-12 Ag₂SO4 = 2Ag+(aq) + soy² (aq) ksp = 1.20 × 10-5 a) which salt will precipitate first? (b) What % of the first anion precipitated will remain in the solution. by the time the second anion starts to precipitate? (c) What is the effect of low pH (more acidic) condition on the separate of the carbonate and sulfate anions via silver precipitation? What is the effect of high pH (more basic)? Provide appropriate explanation per answerarrow_forwardPart 4. Butanoic acid (ka= 1.52× 10-5) has a partition coefficient of 3.0 (favors benzene) when distributed bet. water and benzene. What is the formal concentration of butanoic acid in each phase when 0.10M aqueous butanoic acid is extracted w❘ 25 mL of benzene 100 mL of a) at pit 5.00 b) at pH 9.00arrow_forwardCalculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0 Group of answer choices 0.0269 kJ/mole 2610 kJ/mole 27.6 kJ/mole 0.215 kJ/mole 20.8 kJ/molearrow_forward
- Calculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0 choices: 0.0269 kJ/mole 2610 kJ/mole 27.6 kJ/mole 0.215 kJ/mole 20.8 kJ/molearrow_forwardCalculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0arrow_forwardDon't used hand raiting and don't used Ai solutionarrow_forward
- Rank the compounds in each group below according to their reactivity toward electrophilic aromatic substitution (most reactive = 1; least reactive = 3). Place the number corresponding to the compounds' relative reactivity in the blank below the compound. a. CH₂F CH3 F b. At what position, and on what ring, is bromination of phenyl benzoate expected to occur? Explain your answer. :0: C-O phenyl benzoate 6.Consider the reaction below to answer the following questions. A B C NO₂ FeBr3 + Br₂ D a. The nucleophile in the reaction is: BODADES b. The Lewis acid catalyst in the reaction is: C. This reaction proceeds d. Draw the structure of product D. (faster or slower) than benzene.arrow_forwardPart 2. A solution of 6.00g of substance B in 100.0mL of aqueous solution is in equilibrium, at room temperature, wl a solution of B in diethyl ether (ethoxyethane) containing 25.0 g of B in 50.0 mL 9) what is the distribution coefficient of substance B b) what is the mass of B extracted by shaking 200 ml of an aqueous solution containing 10g of B with call at room temp): i) 100 mL of diethyl ether ii) 50ml of diethyl ether twice iii) 25ml of diethyl ether four timesarrow_forward- Rank the following groups of compounds from most acidic (1) to least acidic (4). Place the number corresponding to the compound's relative rank in the blank below the structure. a. NO₂ NO₂ CH2CH2CH2CH2OH CH3 CH3CH2CHOH CH3CH2CH2CH2OH NO₂ CH3CHCH2CH2OH b. OH OH CH₂OH CO₂H HC CN CN CNarrow_forward
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