Chemistry ch 8

Radii Part A:Place the following elements in order of decreasing atomic size: arsenic, sulfur, neon, cesium, calcium, and phosphorus. Ne-S-Cs-P-Ca-As Rank from largest to smallest.

Cs>Ca>As>P>S>Ne

Part 6: Radii

Each element in the periodic table has a distinctive atomic radius.

Part 13: Electron Affinity

Electron affinity is the measure of the attraction of an electron toward an isolated gaseous atom. When an electron is added to an isolated gaseous atom or ion energy is either released or absorbed; this energy change is known as electron affinity. Electron affinity is positive when energy is absorbed, and it is negative when energy is released.

Part 9: Analysis of Metallic Oxides

Metals often react with oxygen to produce ionic oxides such as K2O, CaO, TiO2, and V2O5.

Electron Configurations of Atoms and Ions Part C:Which element has the following configuration: [Xe]6s24f4? Enter the chemical symbol for the element.

Nd-Neodymium

Electron Configurations of Atoms and Ions Part B:What is the ground-state electron configuration of the sulfide ion S2−? Express your answer in condensed form, in order of increasing orbital energy. For example, [He]2s22p2 should be entered as [He]2s^22p^2 .

[Ar] This is correct, but you should still write out the configuration.

Ionic Radii Part C:The following ions contain the same number of electrons. Rank them in order of decreasing ionic radii. S2- P3- Ca2+ Sc3+ K+ Cl- Rank from largest to smallest radius.

P3->S2->Cl->K+>Ca2+>Sc3+ In a group of ions with the same number of electrons, the most negative ion is the largest and the most positive ion is the smallest.

Electron Configurations of Atoms and Ions Part A: What is the ground-state electron configuration of a neutral atom of titanium? Express your answer in condensed form, in order of increasing orbital energy. For example, [He]2s22p2 would be entered as [He]2s^22p^2 .

[Ar]4s^23d^2

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

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.

Electron Affinity and Electron Configurations Part B:Arrange the following elements greatest to least tendency to accept an electron. Mg-Si-Na-S-Cl Rank from greatest to least tendency to accept an electron

The tendency to gain an electron is quantitatively measured by the electron affinity, the amount of energy involved in the addition of an electron to a neutral gaseous atom. Ordering these elements by the electron affinity provides an identical order: Cl>S>Si>Na>Mg

Analysis of Metallic Oxides Part A:Give the oxidation state for the metal ion in each of the following compounds, respectively: K2O,CaO,TiO2,V2O5 Enter the oxidation states of the metals in the order given, separated by commas (e.g., +1,+3,-2,+2).

Transition metals such as Ti and V tend to have multiple possible ion charges. Thus, TiO2 and V2O5 are named titanium(IV) oxide and vanadium(V) oxide, respectively, where the Roman number specifies the oxidation state of the metal. In contrast, metals from groups 1 and 2 of the periodic table, such as K and Ca, have only one possible ion charge and so the names of the compounds K2O and CaO are potassium oxide and calcium oxide, respectively, without Roman numerals. ox.state of O in oxide = -2, so that, K2O , (x)*2+(-2) = 0 x = +1 ox.state of K in K2O = +2/2 = +1 ox.state of Ca in CaO = +2 ox.state of Ti in TiO2 = +4 ox.state of V in V2O5 = +5

Analysis of Metallic Oxides Part B:Match each ion to its electron configuration. K^+, Ca^2+, Ti^4+, V^5+ 1)[Ar] 2)[Ar]4s1 3)[Ar]4s2 4)[Ar]4d2 5)[Ar]4d3 6)[Ar]4s23d2 7)[Ar]4s23d3

All elements go to the number 1, as each element loses their own electron number, which makes configuration as [Ar]

Magnetic Properties Part B:Classify the following substances as diamagnetic or paramagnetic based on its magnetic properties. 1)The substance that is weakly repelled by the external magnetic field 2)A substance that has no net magnetic moment 3)A substance that has only paired electrons and no unpaired electrons. 4)A substance that is attracted to an external magnetic field. 5)A substance that has unpaired electrons

Diamagnetic: The substance that is weakly repelled by the external magnetic field A substance that has no net magnetic moment A substance that has only paired electrons and no unpaired electrons. Paramagnetic: A substance that is attracted to an external magnetic field. A substance that has unpaired electrons

Ionic Radii Part B:Rank the following items in order of decreasing radius: Na, Na+, and Na−. Rank from largest to smallest radius.

Na->Na>Na+

Electron Affinity Part A:Watch the video on electron affinity and identify which of the following statements matches each reaction. ResetHelp Cl(g)+e−→Cl−(g) O(g)+e−→O−(g) O−(g)+e−→O2−(g) All three reactions None of the reactions

Electron affinity is the energy change that occurs when an electron is added to a gaseous atom or ion. The electron affinity of an O atom is −142 kJ, meaning that when an electron is added to an O atom, energy is released and the O− ion is stable. However, when a second electron is added to an O− ion, energy is required. Since the O− ion is already negatively charged, adding another negatively charged electron is difficult. Thus, the electron affinity of an O− ion is +710 kJ.

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

Part 5: Periodic Trends in Relative Electron Affinity

Electron affinity, EA, is the energy required to add an electron to a neutral gaseous atom and is related to an element's position on the periodic table. Electron affinities can be positive, negative, or zero, as shown in the table. Element Electron affinity (kJ/mol) N (nitrogen) >0 Br (bromine) −325 Sr (strontium) −5

Pat 11: First Ionization Energy

Ionization energy is the energy required to remove an electron from an atom or ion. First ionization energy refers to the energy required to remove an electron from an electrically neutral atom in the gas phase. Subsequent ionization energies reflect the energies needed to strip successive electrons off an increasingly positively charged ion.

Periodic Trends in Relative Electron Affinity Part B: Rank the following elements by electron affinity, from most positive to most negative EA value. K-Ne-I-Sb-O Rank from most positive to most negative

Ne<K<Sb<O<I The reason why group 8A elements have a positive EA value is because they are incredibly stable in their neutral form with an octet of electrons in the outermost energy level. They have little tendency to gain an electron. The reason that group 5A elements have an EA value that is less negative than expected is because a half-filled p subshell is particularly stable.

Electron Affinity Part C:Consider the addition of an electron to the following atoms from the fifth period. Rank the atoms in order from the most negative to the least negative electron affinity values based on their electron configurations. Atom or ion Electron configuration I [Kr]4d105s25p5 Sn [Kr]4d105s25p2 Xe [Kr]4d105s25p6 Rank the electron affinity from most negative to least negative.

The electron affinity values for I and Sn are −295 kJ/mol and −200 kJ/mol, respectively. The noble gas (group 8A) has a positive electron affinity. This suggests that I has the most negative electron affinity value, whereas Xe has the least negative electron affinity value.

Periodic Trends in Relative Electron Affinity Part A: For the elements with the electron affinities given in the table in the introduction, which element is most likely to accept an electron? a)N b)Br c)Sr

b) Br Because bromine is only one electron away from noble-gas configuration, it becomes incredibly stable upon the addition of that one electron. Even though nitrogen also needs to gain electrons to achieve a noble-gas configuration, N− is nowhere near as stable as Br−. Strontium must lose electrons to achieve noble-gas configuration. Therefore, adding an electron to a neutral strontium atom makes it less stable.

Part 12: Magnetic Properties

Learning Goal: To use electron configuration to explain magnetic behavior. An important property that results from the electron configuration of an atom or an ion is behavior in the presence of an external magnetic field. Due to the random orientations of electron spins most materials have no permanent magnetic effect without an external magnetic field. Electron configuration The magnetic behavior of an atom or ion depends on its electron configuration.

Ionic Radii Part A:Rank the following ions in order of decreasing radius: O2−,S2−,Se2−,Te2−, and Po2−. Use the periodic table as necessary. Rank from largest to smallest radius

O2−<S2−<Se2−<Te2−<Po2−

Magnetic Properties Part C:Sort the following atom or ions as paramagnetic or diamagnetic according to the electron configurations determined in Part A. O^2- Ag^+ Cs Ni C

Paramagnetic: Cs Ni C Diamagnetic: O^2- Ag^+ if the atom or ion having unpaired electrons in its valence shell, having paramagnetic nature. if the atom or ion having paired electrons in its valence shell, having diamagnetic nature. In this example, the ions contain only paired electrons, and the atoms contain only unpaired electrons. This is not always the case. For example, all noble gases have paired electrons and are diamagnetic. Some ions, such as Cl+ and Mn2+, have unpaired electrons and are paramagnetic.

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.

Example 8.5: Atom Size Part B: Arrange these elements in order of decreasing radius: S, Ca, F, Rb, and Si. Rank from largest to smallest atomic radius

Rb>Ca>Si>S>F There are significant enough differences in the positions of these elements in the periodic table for you to reliably compare their atomic radii using periodic trends. The atomic radius decreases moving left to right and moving from the bottom to the top, so this ranking corresponds to the order of elements along these trends.

Reduxion of Metallic Oxides

Reduction of metallic oxides The reduction of a metallic oxide is represented by this generic equation MxOy(s)+yH2(g)→xM(s)+yH2O(g) where M represents the symbol for the metal and x and y are the subscripts in the formula.

Ionization Energy Part A:Based on position in the periodic table and electron configuration, arrange these elements in order of decreasing Ei1: Rank the elements from highest to lowest ionization energy.

Removal of successive electrons always requires more energy (e.g., Ei3>Ei2>Ei1) because a negative electron is being removed for a successively higher positive charge. This pattern of increasing values for successive ionization energies shows a drastic increase in magnitude when an electron is removed from a completely filled energy level or one that is identical to that of a noble gas.

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.

Electron affinity and electron configuration of the atoms

The electron affinity of elements is a periodic property, so you can predict whether the electron affinity of an element is positive or negative based on its electron configuration. In general, you can use the following rules to correlate electron affinity and electron configuration: The addition of an electron to a neutral atom that can complete any subshell, such as an s, p, d, or f subshell, will be favorable, and the electron affinity will be negative. For example, the I atom has five electrons in the 5p subshell. The addition of an extra electron will result in a completely filled 5p subshell. The addition of an electron will be favorable, and, therefore, the electron affinity is negative. I(g) + e− → I−(g), ΔE<0[Kr]4d105s25p5 [Kr]4d105s25p6 The addition of an electron to any negatively charged ion will need to overcome the repulsive force of the negative ion. This is generally less favorable than adding an electron to a neutral atom, and the electron affinity will be positive. Even though the additional electron will complete the octet, as in the case of an O− ion where the addition of electron will complete the 2p subshell, the electron affinity is a positive value. O−(g) + e− → O2−(g), ΔE>01s22s22p5 1s22s22p6 The addition of an electron to a neutral atom that has a completely filled subshell will result in a loss of stability associated with completely filled subshells. This process is often less favorable, and the electron affinity will be positive. For example, the Xe atom has a completely filled 5p subshell. The addition of an extra electron will be unfavorable, and, therefore, the electron affinity is positive. Xe(g) + e− → Xe−(g), ΔE>0[Kr]4d105s25p6 [Kr]4d105s25p66s1

Relating electron configuration and magnetism

The electron configuration of an atom can be used to predict whether it will display paramagnetic or diamagnetic behavior in the presence of an applied magnetic field. A spinning electron is an electric charge in motion. It induces a magnetic field. In a diamagnetic atom or ion, all electrons are paired, and the individual magnetic effects cancel out. A diamagnetic species is weakly repelled by an external magnetic field. A paramagnetic atom or ion has unpaired electrons, and the individual magnetic effects do not cancel out. Each unpaired electron possesses a magnetic moment that causes the atom or ion to be attracted to an external magnetic field.

Part 10: Electron Configurations of Atoms and Ions

The electron configuration of an atom tells us how many electrons are in each orbital. For example, helium has two electrons in the 1s orbital. Therefore the electron configuration of He is 1s2.

Magnetism

The overall magnetic behavior of a material can be predicted from its electron configuration, which dictates the magnetic properties observed when a material is placed in a magnetic field. The origin of magnetism lies in the orbital and spin motions of electrons and how the electrons interact with one another. Each electron in an atom moving about the nucleus while spinning about its own axis gives rise to an electric field and a consequent magnetic field. A magnetic moment is assigned to each electron to describe the magnetic field produced. The total magnetic moment of the substance is the vector sum of individual magnetic moments of individual electrons. If two electrons are oriented in the opposite direction, their vectors are in equal in magnitude but opposite direction; therefore their vector sum will be zero. All matter exhibits magnetic properties when placed in an external magnetic field, and matter can be classified according to this interaction. Application of an external magnetic field, to a substance with unpaired electron spins causes the random orientation of unpaired electron spins to become aligned with the external field. This alignment causes a change in the magnetic moment that results in a slight attraction to the external field. These substances are termed paramagnetic. On the other hand, application of an external magnetic field to a substance whose electron spins are all paired causes a change in the magnetic moment that results in a magnetic field that is in opposition to the external field and a slight repulsion to the external field. These substances are termed diamagnetic.

Part 14: Ionic Radii

The size of ions as measured by ionic radii varies in a systematic manner. The size of the ion can be explained in part by effective nuclear charge, Zeff, which is the net nuclear charge felt by an electron. The effective nuclear charge takes into account the actual nuclear charge and the shielding of this charge by inner electrons. When an atom loses electrons, the resulting cation is smaller both because the remaining electrons experience a larger Zeff and because these electrons are usually in orbitals closer to the nucleus than the electrons that were lost. The more electrons that are lost, the smaller the ion becomes. Similarly, when an atom gains electrons, the resulting anion is larger owing to both increased electron-electron repulsions and a reduction in Zeff. The more electrons that are gained, the larger the ion becomes

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.

Periodic Trends and Effective Nuclear Charge Part A: Which of the following trends is indirectly proportional to effective nuclear charge, Zeff? a)Atomic size b)Ionization energy c)Electron affinity d)All of the answers are correct.

a)Atomic size When the effective nuclear charge goes up, the atomic size decreases (such as across a period on the periodic table).

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.

Electron Affinity and Electron Configurations Part A: Consider the following neutral electron configurations in which n has a constant value. Which configuration would belong to the element with the most negative electron affinity, E(ea)? a)5s^2 b)5s^2,5p^2 c)5s^2,5p^5 d)5s^2,5p^6

c)5s^2,5p^5

First Ionization Energy Part B:For the series of elements X, Y, and Z all in the same period (row), arrange the elements in order of decreasing first ionization energy. Element Radius (pm) X 125 Y 189 Z 271 Rank from highest to lowest first ionization energy

ionisation energy is the energy required to remove an outer most electron from an atom.It decreases down the group and increases across the period because the nuclear charge increases.In these elements radius is given.The smallest be the radius higher will be the ionisation energy because more will be nuclear charge hence the electron removal will become difficult.so; decreasing order of ionisation energy ; X>Y>Z The inward "pull" on the electrons from the nucleus is called the effective nuclear charge. When looking at a series of atoms within the same period, the greater the pull on the electrons, the smaller the atom and the harder it is to remove an electron.

Part 7:Example 8.5: Atom Size Part A:On the basis of periodic trends, choose the larger atom from each pair (if possible): Match the elements in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer. Of Sn or I, the larger atom is Of S n or I , the larger atom is_____ Of Cr or W, the larger atom is Of C r or W , the larger atom is_____ Of F or Se, the larger atom is Of F or S e , the larger atom is_____ Of Ge or Po, the larger atom is Of G e or P o , the larger atom is_____

According to periodic trends, atomic radius decreases as you move right across a period (row), and atomic radius increases as you move down a group (column). Therefore, predictable comparisons can be made for elements belonging to the same period (excepting transition metals), elements belonging to the same group (including transition metals), or elements residing along the direction moving up as well as to the right in the periodic table, or conversely, the direction moving down as well as to the left in the periodic table. However, for elements that reside along the direction moving down and to the right in the periodic table, the trends contradict each other, and predictions cannot be reliably made. Of Sn or I, the larger atom is SnOf S n or I , the larger atom is S n .. Of Cr or W, the larger atom is WOf C r or W , the larger atom is W .. Of F or Se, the larger atom is SeOf F or S e , the larger atom is S e .. Of Ge or Po, the larger atom is not predictable Of G e or P o , the larger atom is not predictable..

Magnetic Properties Part A:Select the correct electron configurations from the list below. You can refer to the periodic table for atomic numbers. Check all that apply. 1)The electron configuration of C is [He]2s22p2. 2)The electron configuration of Ni is [Ar]4s23d8. 3)The electron configuration of Ag+ is [Kr]5s14d9. 4)The electron configuration of O2− is [He]2s22p6. 5)The electron configuration of Cs is [Xe]6s05d1.

Answer: 1)The electron configuration of C is [He]2s22p2. 2)The electron configuration of Ni is [Ar]4s23d8. 4)The electron configuration of O2− [He]2s22p6. The electron configuration of an atom is a designation of how electrons are distributed among various orbitals. The electron configurations for the atoms and the ions are as shown in the following table: Atom or ion Electron configuration Ni [Ar]4s23d8 Ag+ [Kr]4d10 Cs [Xe]6s1 C [He]2s22p2 O2− [He]2s22p6 Many of the physical and chemical properties (such as magnetic properties) of atoms or ions can be correlated with electron configurations.

First Ionization Energy Part A:Arrange the elements in decreasing order of first ionization energy. Rank from highest to lowest first ionization energy. Cs - Ge - In - Se

Cs = 375.7 kJ /mol Ge = 762 KJ /mol In = 558.3 KJ / mol Se = 941.0 KJ / mol Se>Ge>In>Cs

Part 4: Electron Affinity and Electron Configurations

Electron affinity, E(ea), is the change in energy that occurs when an electron is added to a neutral isolated atom. This can be represented by the following equation: X(g)+e−→X−(g) Most electron affinity values are negative because energy is usually released when a neutral atom gains an electron. Eea values become more negative with increasing tendency of the atom to accept an electron and increasing stability of the resulting anion. Eea shows a periodic trend that is related to electron configuration. Elements with less than an octet and with high effective nuclear charge (Zeff) tend to have large negative Eea values. Elements with filled valence shells or subshells and low Zeff tend to have Eea values near zero.

Analysis of Metallic Oxides Part D:Repeat the procedure from Part C for the other three oxides, Oxide ΔH∘f (kJ/mol) K2O(s) −363.2 CaO(s) −635.1 TiO2(s) −938.7 then rank all four oxides according to their enthalpy of reduction. V2O5 - TiO2 - CaO - K2O Rank from greatest (most positive) to least enthalpy of reduction.

Even though ΔH∘f for the oxides steadily increases with the position of the metal in period 4, the enthalpies of reduction do not follow a regular pattern because of the stoichiometry of the reactions. TO2>CaO>V2O5>K2O

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.