rank these transition metal ions in order of decreasing number of unpaired electrons. if two ions have the same number of unpaired electrons, overlap them such that the two appear in a single column.
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Chemistry ch 8 Flashcards
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Chemistry ch 8
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Part 1:Atomic Radii and Effective Nuclear Charge
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The atomic radius of an element can be predicted based on its periodic properties. Atomic radii increase going down a group in the periodic table, because successively larger valence-shell orbitals are occupied by electrons. Atomic radii generally decrease moving from left to right across a period because the effective nuclear charge increases.
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Atomic Radii and Effective Nuclear Charge
Part A: Rank the following elements in order of decreasing atomic radius: Be - Mg - Sr - Ca
Rank from largest to smallest radius.
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Sr>Ca>Mg>Be
Atomic radii increase going down a group, because successively larger valence-shell orbitals are occupied by electrons. For example, strontium has electrons in the fifth shell, which contains much larger orbitals than the fourth, third, second, or first shells.
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Part 1:Atomic Radii and Effective Nuclear Charge
The atomic radius of an element can be predicted based on its periodic properties. Atomic radii increase going down a group in the periodic table, because successively larger valence-shell orbitals are occupied by electrons. Atomic radii generally decrease moving from left to right across a period because the effective nuclear charge increases.
Atomic Radii and Effective Nuclear Charge
Part A: Rank the following elements in order of decreasing atomic radius: Be - Mg - Sr - Ca
Rank from largest to smallest radius.
Sr>Ca>Mg>Be
Atomic radii increase going down a group, because successively larger valence-shell orbitals are occupied by electrons. For example, strontium has electrons in the fifth shell, which contains much larger orbitals than the fourth, third, second, or first shells.
Atomic Radii and Effective Nuclear Charge
Part B:Rank the following elements in order of decreasing atomic radius. Na - Si - Mg - Al
Rank from largest to smallest radius.
Na>Mg>Al>Si
Atomic radii decreases as you go right, but increases as you go down or left
Atomic Radii and Effective Nuclear Charge
Part C:The shielding of electrons gives rise to an effective nuclear charge, Z(eff), which explains why boron is larger than oxygen.
Estimate the approximate Z(eff) felt by a valence electron of boron and oxygen, respectively?
a)+5 and +8 b)+3 and +6 c)+5 and +6 d)+3 and +8 e)+1 and +4 b)+3 and +6
The valence electrons in an oxygen atom (box) are attracted to the nucleus by a positive charge nearly double, 1s^2-1box,2s^2-1box,2p^3-3boxes, that of boron. Therefore, the electrons in oxygen are held closer to the nucleus, giving it a smaller radius.
Part 2:Electron Configurations of Ions
When an atom forms an ion, it will gain or lose electrons to attain a more stable electron configuration, frequently that of a noble gas. Nonmetals tend to form anions by gaining electrons, which enter the lowest energy unoccupied orbital.
Metals tend to form cations by losing electrons. Main group metals lose electrons in the reverse order of filling.
Transition metals, however, lose s electrons first.
Electron Configurations of Ions
Part A:In the ground-state electron configuration of Fe^3+, how many unpaired electrons are present?
Express your answer numerically as an integer.
5 unpaired electrons
Fe^3+ lost electron, as charge is positive, so the configuration is counted as element Mn, which is before Fe. The configuration is:
Fe^3+ - [Ar] 4s^2,3d^5 - 5 unpaired electrons
Electron Configurations of Ions
Part B:Build the orbital diagram for the ion most likely formed by phosphorus.
Phosphorous forms an ion with a charge of -3, which means it has the same electron configuration as Argon. That electron configuration is: 1s^2 2s^2 2p^6 3s^2 3p^6,
a total of 18 electrons.
Part 3: Ionization Energy
Ionization energy (Ei) is the amount of energy required to remove an electron from a neutral gaseous atom or gaseous ion. Electrons are attracted to the positively charged nucleus; therefore removing an electron requires energy. The process is endothermic, and so ionization energies have a positive value. The first ionization energy (Ei1) is the energy associated with the removal of an electron from the neutral gaseous atom. The reaction is represented for the generalized atom X as
X→X^+ + e−
The amount of energy required to remove an electron is related to the effective nuclear charge and the stability of the electron configuration of the atom. It therefore shows periodic variation generally increasing from left to right in a period and from bottom to top of a group. In general, metals have lower Ei1 values than nonmetals. Exceptions to this general trend from left to right occur when a completely filled s subshell or half-filled p subshell is encountered. These stable configurations have larger than expected Eil values.
8.4 Molecular Orbital Theory – Chemistry
8.4 MOLECULAR ORBITAL THEORY
Learning Objectives
By the end of this section, you will be able to:
Outline the basic quantum-mechanical approach to deriving molecular orbitals from atomic orbitals
Describe traits of bonding and antibonding molecular orbitals
Calculate bond orders based on molecular electron configurations
Write molecular electron configurations for first- and second-row diatomic molecules
Relate these electron configurations to the molecules’ stabilities and magnetic properties
For almost every covalent molecule that exists, we can now draw the Lewis structure, predict the electron-pair geometry, predict the molecular geometry, and come close to predicting bond angles. However, one of the most important molecules we know, the oxygen molecule O2, presents a problem with respect to its Lewis structure. We would write the following Lewis structure for O2:
This electronic structure adheres to all the rules governing Lewis theory. There is an O=O double bond, and each oxygen atom has eight electrons around it. However, this picture is at odds with the magnetic behavior of oxygen. By itself, O2 is not magnetic, but it is attracted to magnetic fields. Thus, when we pour liquid oxygen past a strong magnet, it collects between the poles of the magnet and defies gravity, as in Figure 1 in Chapter 8 Introduction. Such attraction to a magnetic field is called paramagnetism, and it arises in molecules that have unpaired electrons. And yet, the Lewis structure of O2 indicates that all electrons are paired. How do we account for this discrepancy?
Magnetic susceptibility measures the force experienced by a substance in a magnetic field. When we compare the weight of a sample to the weight measured in a magnetic field (Figure 1), paramagnetic samples that are attracted to the magnet will appear heavier because of the force exerted by the magnetic field. We can calculate the number of unpaired electrons based on the increase in weight.
Experiments show that each O2 molecule has two unpaired electrons. The Lewis-structure model does not predict the presence of these two unpaired electrons. Unlike oxygen, the apparent weight of most molecules decreases slightly in the presence of an inhomogeneous magnetic field. Materials in which all of the electrons are paired are diamagnetic and weakly repel a magnetic field. Paramagnetic and diamagnetic materials do not act as permanent magnets. Only in the presence of an applied magnetic field do they demonstrate attraction or repulsion.
Water, like most molecules, contains all paired electrons. Living things contain a large percentage of water, so they demonstrate diamagnetic behavior. If you place a frog near a sufficiently large magnet, it will levitate. You can see videos of diamagnetic floating frogs, strawberries, and more.
Molecular orbital theory (MO theory) provides an explanation of chemical bonding that accounts for the paramagnetism of the oxygen molecule. It also explains the bonding in a number of other molecules, such as violations of the octet rule and more molecules with more complicated bonding (beyond the scope of this text) that are difficult to describe with Lewis structures. Additionally, it provides a model for describing the energies of electrons in a molecule and the probable location of these electrons. Unlike valence bond theory, which uses hybrid orbitals that are assigned to one specific atom, MO theory uses the combination of atomic orbitals to yield molecular orbitals that are delocalized over the entire molecule rather than being localized on its constituent atoms. MO theory also helps us understand why some substances are electrical conductors, others are semiconductors, and still others are insulators. Table 2 summarizes the main points of the two complementary bonding theories. Both theories provide different, useful ways of describing molecular structure.
Valence Bond Theory Molecular Orbital Theory
considers bonds as localized between one pair of atoms considers electrons delocalized throughout the entire molecule
creates bonds from overlap of atomic orbitals (s, p, d…) and hybrid orbitals (sp, sp2, sp3…) combines atomic orbitals to form molecular orbitals (σ, σ*, π, π*)
forms σ or π bonds creates bonding and antibonding interactions based on which orbitals are filled
predicts molecular shape based on the number of regions of electron density predicts the arrangement of electrons in molecules
needs multiple structures to describe resonance
Table 2. Comparison of Bonding TheoriesMolecular orbital theory describes the distribution of electrons in molecules in much the same way that the distribution of electrons in atoms is described using atomic orbitals. Using quantum mechanics, the behavior of an electron in a molecule is still described by a wave function, Ψ, analogous to the behavior in an atom. Just like electrons around isolated atoms, electrons around atoms in molecules are limited to discrete (quantized) energies. The region of space in which a valence electron in a molecule is likely to be found is called a molecular orbital (Ψ2). Like an atomic orbital, a molecular orbital is full when it contains two electrons with opposite spin.The correct order of a number of unpaired electrons is :
Click here👆to get an answer to your question ✍️ The correct order of a number of unpaired electrons is :
Question
The correct order of a number of unpaired electrons is :
ACu2+ >Ni 2+ >Cr 3+ >Fe 3+
BNi2+ >Cu 2+ >Fe 3+ >Cr 3+
CFe3+ >Cr 3+ >Ni 2+ >Cu 2+
DCr3+ >Fe 3+ >Ni 2+ >Cu 2+ Medium Open in App Solution Verified by Toppr
Correct option is C)
The no. of unpaired e
−
in the given ions is:
Cr 3+ - [Ar] 3d 3 4s 0 Fe 3+ - [Ar] 3d 5 4s 0 Ni 2+ - [Ar] 3d 8 4s 0 Cu 2+ - [Ar] 3d 9 4s 0 Fe 3+ −3d 5
No. of unpaired electrons = 5
Cr 3+ −3d 3
No. of unpaired electrons = 3
Ni 2+ −3d 8
No. of unpaired electrons = 2
Cu 2+ −3d 9
No. of unpaired electrons = 1
The order is: Fe 3+ > Cr 3+ > Ni 2+ >Cu 2+
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