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Concept explainers
Interpretation:
Distribution of d-electrons in the complex ion
Concept Introduction:
The term Crystal field splitting refers to segregation of d-orbitals in to higher energy orbitals and lower energy orbitals when ligands approaches metal ion to co-ordinate. Crystal field splitting is said to be larger when the energy gap between higher energy d-orbitals and lower energy d-orbitals is larger. Crystal field splitting is said to be smaller when the energy gap between higher energy d-orbitals and lower energy d-orbitals is smaller.
Spectrochemical series refers to the arrangement of ligands with respect to crystal field splitting they cause in the d-orbitals of a metal ion. The ligand that causes larger splitting pattern is referred to as stronger ligand.
The ligands are arranged in increasing order of crystal field splitting they produce.
The ligand that causes larger crystal field splitting is strong ligand and the ligand that causes smaller splitting is termed as weak ligand.
Pairing energy refers to the energy required to pair the electrons - that is energy required to make two electrons to be paired in same orbital with opposite spins.
If the pairing energy of a complex is high it means the electrons are difficult to be paired and so the complex will be high spin complex with unpaired electrons. If the pairing energy of a complex is low it means the electrons are readily paired and so the complex will be low spin complex with paired electrons.
A complex with high splitting energy will have low pairing energy because high splitting energy indicates larger energy gap between lower and higher energy levels that the electrons need lesser energy to be paired and they tend to remain paired in lower energy orbitals. A complex with low splitting energy will have high pairing energy because low splitting energy indicates smaller energy gap between lower and higher energy levels that the electrons disperse among orbitals more randomly and they need higher energy to be paired.

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Chapter 22 Solutions
OWLv2 with Student Solutions Manual eBook for Ebbing/Gammon's General Chemistry, 11th Edition, [Instant Access], 4 terms (24 months)
- Identify the unknown compound from its IR and proton NMR spectra. C4H6O: 'H NMR: 82.43 (1H, t, J = 2 Hz); 8 3.41 (3H, s); 8 4.10 (2H, d, J = 2 Hz) IR: 2125, 3300 cm¹ The C4H6O compound liberates a gas when treated with C2H5 MgBr. Draw the unknown compound. Select Draw с H Templates Morearrow_forwardPlease help with number 6 I got a negative number could that be right?arrow_forward1,4-Dimethyl-1,3-cyclohexadiene can undergo 1,2- or 1,4-addition with hydrogen halides. (a) 1,2-Addition i. Draw the carbocation intermediate(s) formed during the 1,2-addition of hydrobromic acid to 1,4-dimethyl-1,3-cyclohexadiene. ii. What is the major 1,2-addition product formed during the reaction in (i)? (b) 1,4-Addition i. Draw the carbocation intermediate(s) formed during the 1,4-addition of hydrobromic acid to 1,4-dimethyl-1,3-cyclohexadiene. ii. What is the major 1,4-addition product formed from the reaction in (i)? (c) What is the kinetic product from the reaction of one mole of hydrobromic acid with 1,4-dimethyl-1,3-cyclohexadiene? Explain your reasoning. (d) What is the thermodynamic product from the reaction of one mole of hydrobro-mic acid with 1,4-dimethyl-1,3-cyclohexadiene? Explain your reasoning. (e) What major product will result when 1,4-dimethyl-1,3-cyclohexadiene is treated with one mole of hydrobromic acid at - 78 deg * C ? Explain your reasoning.arrow_forward
- Give the product of the bimolecular elimination from each of the isomeric halogenated compounds. Reaction A Reaction B. КОВ CH₂ HotBu +B+ ко HOIBU +Br+ Templates More QQQ Select Cv Templates More Cras QQQ One of these compounds undergoes elimination 50x faster than the other. Which one and why? Reaction A because the conformation needed for elimination places the phenyl groups and to each other Reaction A because the conformation needed for elimination places the phenyl groups gauche to each other. ◇ Reaction B because the conformation needed for elimination places the phenyl groups gach to each other. Reaction B because the conformation needed for elimination places the phenyl groups anti to each other.arrow_forwardFive isomeric alkenes. A through each undergo catalytic hydrogenation to give 2-methylpentane The IR spectra of these five alkenes have the key absorptions (in cm Compound Compound A –912. (§), 994 (5), 1643 (%), 3077 (1) Compound B 833 (3), 1667 (W), 3050 (weak shoulder on C-Habsorption) Compound C Compound D) –714 (5), 1665 (w), 3010 (m) 885 (3), 1650 (m), 3086 (m) 967 (5), no aharption 1600 to 1700, 3040 (m) Compound K Match each compound to the data presented. Compound A Compound B Compound C Compound D Compoundarrow_forward7. The three sets of replicate results below were accumulated for the analysis of the same sample. Pool these data to obtain the most efficient estimate of the mean analyte content and the standard deviation. Lead content/ppm: Set 1 Set 2 Set 3 1. 9.76 9.87 9.85 2. 9.42 9.64 9.91 3. 9.53 9.71 9.42 9.81 9.49arrow_forward
- Draw the Zaitsev product famed when 2,3-dimethylpentan-3-of undergoes an El dehydration. CH₂ E1 OH H₁PO₁ Select Draw Templates More QQQ +H₂Oarrow_forwardComplete the clean-pushing mechanism for the given ether synthesia from propanol in concentrated sulfurica140°C by adding any mining aloms, bands, charges, nonbonding electron pairs, and curved arrows. Draw hydrogen bonded to cayan, when applicable. ore 11,0 HPC Step 1: Draw curved arrows Step 2: Complete the intend carved Q2Q 56 QQQ Step 3: Complete the intermediate and add curved Step 4: Modify the structures to draw the QQQ QQQarrow_forward6. In an experiment the following replicate set of volume measurements (cm3) was recorded: (25.35, 25.80, 25.28, 25.50, 25.45, 25.43) A. Calculate the mean of the raw data. B. Using the rejection quotient (Q-test) reject any questionable results. C. Recalculate the mean and compare it with the value obtained in 2(a).arrow_forward
- A student proposes the transformation below in one step of an organic synthesis. There may be one or more reactants missing from the left-hand side, but there are no products missing from the right-hand side. There may also be catalysts, small inorganic reagents, and other important reaction conditions missing from the arrow. • Is the student's transformation possible? If not, check the box under the drawing area. • If the student's transformation is possible, then complete the reaction by adding any missing reactants to the left-hand side, and adding required catalysts, inorganic reagents, or other important reaction conditions above and below the arrow. • You do not need to balance the reaction, but be sure every important organic reactant or product is shown. + T G OH де OH This transformation can't be done in one step.arrow_forwardMacmillan Leaming Draw the major organic product of the reaction. 1. CH3CH2MgBr 2. H+ - G Select Draw Templates More H о QQarrow_forwardDraw the condensed structure of 3-hydroxy-2-butanone. Click anywhere to draw the first atom of your structure.arrow_forward
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
