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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
General Chemistry
- X Draw the major products of the elimination reaction below. If elimination would not occur at a significant rate, check the box under the drawing area instead. ది www. Cl + OH Elimination will not occur at a significant rate. Click and drag to start drawing a structure.arrow_forwardNonearrow_forward1A H 2A Li Be Use the References to access important values if needed for this question. 8A 3A 4A 5A 6A 7A He B C N O F Ne Na Mg 3B 4B 5B 6B 7B 8B-1B 2B Al Si P 1B 2B Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe * Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Rf Ha ****** Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr Analyze the following reaction by looking at the electron configurations given below each box. Put a number and a symbol in each box to show the number and kind of the corresponding atom or ion. Use the smallest integers possible. cation anion + + Shell 1: 2 Shell 2: 8 Shell 3: 1 Shell 1 : 2 Shell 2 : 6 Shell 1 : 2 Shell 2: 8 Shell 1: 2 Shell 2: 8arrow_forward
- Nonearrow_forwardIV. Show the detailed synthesis strategy for the following compounds. a. CH3CH2CH2CH2Br CH3CH2CCH2CH2CH3arrow_forwardDo the electrons on the OH participate in resonance with the ring through a p orbital? How many pi electrons are in the ring, 4 (from the two double bonds) or 6 (including the electrons on the O)?arrow_forward
- Predict and draw the product of the following organic reaction:arrow_forwardNonearrow_forwardRedraw the molecule below as a skeletal ("line") structure. Be sure to use wedge and dash bonds if necessary to accurately represent the direction of the bonds to ring substituents. Cl. Br Click and drag to start drawing a structure. : ☐ ☑ Parrow_forward
- K m Choose the best reagents to complete the following reaction. L ZI 0 Problem 4 of 11 A 1. NaOH 2. CH3CH2CH2NH2 1. HCI B OH 2. CH3CH2CH2NH2 DII F1 F2 F3 F4 F5 A F6 C CH3CH2CH2NH2 1. SOCl2 D 2. CH3CH2CH2NH2 1. CH3CH2CH2NH2 E 2. SOCl2 Done PrtScn Home End FA FQ 510 * PgUp M Submit PgDn F11arrow_forwardNonearrow_forwardPlease provide a mechanism of synthesis 1,4-diaminobenzene, start from a benzene ring.arrow_forward
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
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