Inorganic Chemistry
By
Dr. Khalil K. Abid
Lecture 6
The Periodic Table
1
IONIZATION ENERGY
The ionization energy (also known as the ionization potential) is the energy required to remove
an electron from a gaseous atom or ion:
A n+(g)
A (n+1) + (g) + e –
ionization energy = ΔU
where n = 0 (first ionization energy), 1, 2, . . . (second, third, . . . )
As would be expected from the effects of shielding, the ionization energy varies with different
nuclei and different numbers of electrons. Trends for the first ionization energies of the early
elements in the periodic table are shown in figure. The general trend across a period is an increase
in ionization energy as the nuclear charge increases. A plot of Z"/r, the potential energy for attraction
between an electron and the shielded nucleus, is nearly a straight line, with approximately the same
slope as the shorter segments (boron through nitrogen, for example) shown in figure . However, the
experimental values show a break in the trend at boron and again at oxygen. Because the new
electron in B is in a new p orbital that has most of its electron density farther away from the nucleus
than the other electrons, its ionization energy is smaller than that of the 2s2 electrons of Be. At the
fourth p electron, at oxygen, a similar drop in ionization energy occurs. Here, the new electron
shares an orbital with one of the previous 2p electrons, and the fourth p electron has a higher energy
than the trend would indicate because it must be paired with another in the same p orbital.
The pairing energy, or repulsion between two electrons in the same region of space, reduces
the ionization energy. Similar patterns appear in lower periods. The transition elements have smaller
differences in ionization energies, usually with a lower value for heavier atoms in the same family
because of increased shielding by inner electrons and increased distance between the nucleus and
the outer electrons.
2
So now let’s consider what happens for the case of sodium (Na), for which Z = 11. If we consider
the effective potential of electron 1 by averaging over the motion of the remaining 10 electrons (this
can be done relatively easily using a computer), we obtain a value of Zeff ≈ 1. It’s not exactly 1, but
close enough to 1 that we will call it 1. For electron 2, we now remove electron 1, and compute an
effective potential by averaging over the motion of the remaining 9 electrons, and we obtain a value of
Zeff ≈ 5. It’s not exactly 5 but pretty close. Repeating this procedure for electrons 3, 4, 5,... up to 9, we
find that Zeff does not change much from 5. It’s a little different for each electron, but the value for
electrons 2 – 9 is fairly close to 5. However, if we now consider electrons 10, the value of Zeff changes
suddenly to a value quite close to 11. Obviously, for electron 11, the value of Zeff = 11 exactly because
the only charge electron 11 sees is the nuclear charge (which is exactly 11). Thus, we see very
roughly, that three effective charge values Zeff = 1, 5, 11 emerge from the shell model, and these
values correspond to the shells observed in the energy plot above.
Electrons in the innermost shells with the highest values of Zeff are called core electrons, and they
tend to be unimportant for chemical reactivity. Electrons in the outermost shells, with low values of
Zeff are called valence electrons, and these are the most important in chemical reactions. The
importance of electrons in shells with intermediate values of Zeff in chemical reactions depends on
the chemical process under consideration. Their role is likely to be important but indirect in that they
can affect the distributions of valence electrons and influence their role in chemical bonding without,
themselves, participating in chemical bonds.
3
The First Ionization Energy
The energy needed to remove one or more electrons from a neutral atom to form a positively
charged ion is a physical property that influences the chemical behavior of the atom. By definition,
the first ionization energy of an element is the energy needed to remove the outermost, or highest
energy, electron from a neutral atom in the gas phase. The process by which the first ionization
energy of hydrogen is measured would be represented by the following equation.
H(g)
H+(g) + e-
ΔHo = -1312.0 kJ/mol
The first ionization energy for helium is slightly less than twice the ionization energy for hydrogen
because each electron in helium feels the attractive force of two protons, instead of one.
He(g)
He+(g) + e-
ΔHo = 2372.3 kJ/mol
It takes far less energy, however, to remove an electron from a lithium atom, which has three
protons in its nucleus.
Li(g)
Li+(g) + e-
ΔHo = 572.3 kJ/mol
This can be explained by noting that the outermost, or highest energy, electron on a lithium atom is
in the 2s orbital. Because the electron in a 2s orbital is already at a higher energy than the electrons in
a 1s orbital, it takes less energy to remove this electron from the atom. The first ionization energies for
the main group elements are given in the two figures below.
4
Two trends are apparent from these data.
1 – In general, the first ionization energy increases as we go from left to right across a row of the
periodic table.
2 – The first ionization energy decreases as we go down a column of the periodic table.
The first trend isn't surprising. We might expect the first ionization energy to become larger as
we go across a row of the periodic table because the force of attraction between the nucleus and an
electron becomes larger as the number of protons in the nucleus of the atom becomes larger.
The second trend results from the fact that the principal quantum number of the orbital holding
the outermost electron becomes larger as we go down a column of the periodic table. Although the
number of protons in the nucleus also becomes larger, the electrons in smaller shells and subshells
tend to screen the outermost electron from some of the force of attraction of the nucleus.
Furthermore, the electron being removed when the first ionization energy is measured spends less
of its time near the nucleus of the atom, and it therefore takes less energy to remove this electron
from the atom.
Although there is a general trend toward an increase in the first ionization energy as we go from
left to right across the row, there are two minor inversions in this pattern. As example the first
ionization energy of boron is smaller than beryllium, and the first ionization energy of oxygen is
smaller than nitrogen. These observations can be explained by looking at the electron configurations
of these elements. The electron removed when a beryllium atom is ionized comes from the 2s orbital,
but a 2p electron is removed when boron is ionized.
Be: [He] 2s2
B: [He] 2s2 2p1
5
But there is an important difference in the way electrons are distributed in these atoms. Hund's
rules predict that the three electrons in the 2p orbitals of a nitrogen atom all have the same spin, but
electrons are paired in one of the 2p orbitals on an oxygen atom.
N
O
Hund's rules can be understood by assuming that electrons try to stay as far apart as possible to
minimize the force of repulsion between these particles. The three electrons in the 2p orbitals on
nitrogen therefore enter different orbitals with their spins aligned in the same direction. In oxygen,
two electrons must occupy one of the 2p orbitals. The force of repulsion between these electrons is
minimized to some extent by pairing the electrons. There is still some residual repulsion between
these electrons, however, which makes it slightly easier to remove an electron from a neutral oxygen
atom than we would expect from the number of protons in the nucleus of the atom.
Much larger decreases in ionization energy occur at the start of each new period, because the
change to the next major quantum number requires that the new s electron have a much higher
energy. The maxima at the noble gases decrease with increasing Z because the outer electrons are
farther from the nucleus in the heavier elements. Overall, the trends are toward higher ionization
energy from left to right in the periodic table (the major change) and lower ionization energy from top
to bottom (a minor change).
6
Second, Third, Fourth and Higher Ionization Energies
By now you know that sodium forms Na+ ions, magnesium forms Mg2+ ions, and aluminum
forms Al3+ ions. But have you ever wondered why sodium doesn't form Na2+ ions, or even Na3+ ions?
The answer can be obtained from data for the second, third, and higher ionization energies of the
element.
ELECTRON AFFINITY
Ionization energies measure the tendency of a neutral atom to resist the loss of electrons. It takes
a considerable amount of energy, for example, to remove an electron from a neutral fluorine atom to
form a positively charged ion.
F(g)
F+(g) + e-
ΔHo = 1681.0 kJ/mol
The electron affinity of an element is the energy given off when a neutral atom in the gas phase
gains an extra electron to form a negatively charged ion. A fluorine atom in the gas phase, for
example, gives off energy when it gains an electron to form a fluoride ion.
F(g) + e-
F-(g)
ΔHo = -328.0 kJ/mol
The symbol is EA, and the unit is kJ/mol. The electron affinity, EA is defined to be the negative of
∆E for this reaction:
EA = −∆E = E(reactants) − E(products)
EA can be either positive or negative. If EA > 0, attachment is stable. If EA < 0, attachment is
unstable, and the electron must be forced to stick to the atom. Energy potential can be calculate by
the relation:
7
Because of the similarity of this reaction to the ionization for an atom, electron affinity is
sometimes described as the zeroth ionization energy. This reaction is endothermic (+ ΔU )except for
the noble gases and the alkaline earth elements. The pattern of electron affinities with changing Z
shown in previous figure is similar to that of the ionization energies, but for one larger Z value (one
more electron for each species) and with much smaller absolute numbers. For either of the
reactions, removal of the first electron past a noble gas configuration is easy, so the noble gases
have the lowest electron affinities.
Several patterns can be found when we look at the Periodic Table.
1 – Electron affinities generally become smaller as we go down a column of the periodic table for two
reasons. First, the electron being added to the atom is placed in larger orbitals, where it spends less
time near the nucleus of the atom. Second, the number of electrons on an atom increases as we go
down a column, so the force of repulsion between the electron being added and the electrons
already present on a neutral atom becomes larger.
2 – Electron affinity data are complicated by the fact that the repulsion between the electron being
added to the atom and the electrons already present on the atom depends on the volume of the
atom. Among the nonmetals in Groups VIA and VIIA, this force of repulsion is largest for the very
smallest atoms in these columns: oxygen and fluorine. As a result, these elements have a smaller
electron affinity than the elements below them in these columns as shown in the figure below. From
that point on, however, the electron affinities decrease as we continue down these columns.
8
Ionization Energies and Electron Affinities.
9
The electron affinities are all much smaller than the corresponding ionization energies because
electron removal from a negative ion is easier than removal from a neutral atom. The electronegativity
measures the ability of an atom to draw electrons to itself in a chemical bond. While it is reported as
an atomic property, it is concerned with bonding within a molecule. The larger the electronegativity,
the greater the tendency for an atom to draw electrons to itself in a chemical bond. In general, if EA
is large, giving up an electron is unlikely, but gaining an electron is likely, and the atom tends to act
as an electron acceptor, or is “electronegative.” if EA is small or negative, giving up an electron is
likely, but gaining an electron is unlikely, and the atom tends to act as an electron donor, or is
“electropositive.” Elements toward the left of the periodic table have low electron affinities, so they
tend to act as electron donors, while atoms to the right of the periodic table have high electro
negativities, and they tend to act as electron acceptors.
When data observed were listed along with the electron configurations of these elements,
however, they make sense. These data can be explained by noting that electron affinities are much
smaller than ionization energies. As a result, elements such as helium, beryllium, nitrogen, and neon,
which have unusually stable electron configurations, have such small affinities for extra electrons
that no energy is given off when a neutral atom of these elements picks up an electron. These
configurations are so stable that it actually takes energy to force one of these elements to pick up an
extra electron to form a negative ion.
10
H
He
2.1
-
Li
Be
F
Ne
1.0
1.5
4.0
-
Na
Mg
Cl
Ar
0.9
1.2
3.0
-
K
Ca
Ti
Cr
Fe
Ni
Zn
As
Br
Kr
0.8
1.0
1.5
1.6
1.8
1.8
1.8
2.0
2.8
-
Rb
Sr
I
Xe
0.8
1.0
2.5
-
Cs
Ba
At
Rn
0.7
0.9
2.2
-
Fr
Ra
0.7
0.9
Smaller electronegativity
Larger electronegativity
11
Electron Affinities and Electron Configurations for the First 10 Elements in the Periodic Table
Element E. A. (kJ/mol)
Electron Configuration Element E. A. (kJ/mol)
Electron Configuration
H
72.8
1s1
C
122.3
[He] 2s2 2p2
He
<0
1s2
N
<0
[He] 2s2 2p3
Li
59.8
[He] 2s1
O
141.1
[He] 2s2 2p4
Be
<0
[He] 2s2
F
328.0
[He] 2s2 2p5
B
27
[He] 2s2 2p1
Ne
<0
[He] 2s2 2p6
Consequences of the Relative Size of Ionization Energies and Electron Affinities
Sodium reacts with chlorine to form Na+ and Cl- ions because chlorine atoms "like" electrons
more than sodium atoms do. There is no doubt that sodium reacts vigorously with chlorine to
form NaCl. Furthermore, the ease with which solutions of NaCl in water conduct electricity is
evidence for the fact that the product of this reaction is a salt, which contains Na+ and Cl- ions.
2Na(s) + Cl2(g)
NaCl(s) + H2O(aq)
2 NaCl(s)
Na+(aq) + Cl-(aq)
The only question is whether it is legitimate to assume that this reaction occurs because
chlorine atoms "like" electrons more than sodium atoms. The first ionization energy for sodium is
one and one-half times larger than the electron affinity for chlorine.
Na: 1st IE = 495.8 kJ/mol
Cl: EA = 328.8 kJ/mol
Thus, it takes more energy to remove an electron from a neutral sodium atom than is given off
when the electron is picked up by a neutral chlorine atom.
12
Electronegativity vs Electron Affinity
The transfer of one electron from one atom to another is a very common occurrence that we do not
notice. In order to achieve a transfer, the electron affinity should be met. Electron affinity is a measure
of how much energy is released when an electron is gained by a certain atom, conversely, it is also the
amount of energy required in order to detach the electron from the other atom.
Sometimes the electron doesn’t have to be detached from one molecule in order to form a bond
with another. In these cases the electron is shared and a covalent bond is formed. Electronegativity is
the quantification of a molecules ability to attract an electron and form a covalent bond. Therefore, the
higher the electronegativity value of a specific molecule, the stronger it pulls electrons towards it.
Electron affinity is a truly quantifiable value that can be measured by specific scientific means
after an electron has attached itself to the atom. Electronegativity on the other hand cannot be
measured by scientific means. It is a number derived from computations involving an atom’s
characteristics in the molecule. It should then be understood that electronegativity values can vary
depending on the molecule that it is bonded to. There are also a number of ways to calculate the
electronegativity value of a certain atom due to the fact that it cannot be verified. Some scientists
believe that they have come up with a formula that best describes electronegativity, thus the number
of differing equations. Electronegativity is not really that different from Electron Affinity. It’s just that
Electron affinity is a fixed value of an individual atom, while electronegativity is much more often
associated with molecules rather than each atom and the values could vary to some degree, although
a single value is commonly used for most calculations in inorganic chemistry.
13
Summary:
1. Electronegativity is a numerical value associated with an atoms ability to form a covalent bond
2. Electron affinity is the amount of energy that is released when an electron attaches to the atom.
3. Electron affinity is a fixed and measurable value
4. Electronegativity cannot be measured and needs to be computed from other atomic properties by
a few differing equations depending on which scientist you adhere to
5. Electronegativity values vary with different chemical environments
6. Electron Affinity is a value associated with individual atoms while Electronegativity is the
derived value of a molecule
Atomic Size
Another important periodic property is atomic radius. This parameter can be measured a number
of ways. Interatomic spacing in crystals (metals, rare gases) determined by spectroscopy or by X-ray
crystallography can give atomic separations. From experimental data like these, a consistent set of
atomic data can be determined.
General trends:
• Across a row, the principal quantum number is fixed. The atomic size decreases because of the
increase in nuclear charge.
• Down a column, the outermost electrons are, on average, further from the nucleus that the species
immediately above it in the Periodic Table. Thus, the atomic size increases down a column.
14
Electronegativities of selected elements relative to the position of the elements in the periodic table
15
Atom Size and Ionization Energy
Some important, qualitative statements regarding the dependencies of the size of the atoms and
of the ionization energies of the atoms on the atomic number can be made on the basis of the
systematization introduced by the Periodic Table. First it should be remarked that the notion “size”
of the atom or size of the “ion” (ion = atom with one or more “outer” electrons taken away by a
process of adding energy enough to deboned one or more electrons from the atom, which is called
“ionization”) is unclear. As there is a finite probability to find an electron, thus also an “outer”
electron of an atom, at any location in space (apart from at infinity where the probability is zero), the
“size” of any atom appears to be of “infinite” nature.
However, for example, on the basis of considering interatomic distances (that is the distance
between the centric positions of atomic mass) for specific types of bonding between the atoms, one
can define “atomic size”. A further complication arises because it has been tacitly assumed in the
above lines that the atom is a sphere. In a specific kind of chemical bonding the bonding is not of
isotropic nature, which holds for so-called covalent bonding. Then “size” refers to a certain direction
in space as well.
(1). It is obvious that for the same number of electrons and increasing Z the size of the atom/ion will
decrease. This is a straightforward consequence of the Coulomb interaction of the positive nucleus
and the negative electrons. An “outer” electron does not experience the full nuclear charge: the
nucleus charge is “screened” by the inner electrons.
16
Because electrons are not confined to specific parts of space outside the nucleus, this
screening is not 100% (i.e. for a nucleus of charge +Ze the screening by m “inner” electrons leads
to an effective (i.e. “felt” by the “outer” electron) nuclear charge larger than +(Z − m)e). And also,
the “outer” electron can “penetrate” the “cloud” of “inner” electrons. It appears that the lower the
second quantum number l the larger the probability to find the electron close to the nucleus and at
the same time the larger the probability to find the electron at relatively large distances from the
nucleus . Thus s electrons have the largest penetrative power and the smallest screening effect:
– For the same principal quantum number n, s electrons experience more Coulomb attraction by
the nucleus than p, d, f, etc. electrons do.
– Similarly, for the same principal quantum number n, p, d, f, etc. electrons screen more effectively
from the nuclear charge than s electrons do. (
2). Going from the left to the right in a period of the Periodic Table the number of “outer” electrons
in the same shell increases. Because the screening constant of these electrons is smaller than
100% and at the same time the nuclear charge increases with one for each electron added upon
increasing the atomic number, it follows that, the atomic size decreases going from the left to the
right in a period of the Periodic Table.
Because s electrons screen less good than p electrons, the size changes between, for
example, considering the second period, Li, Be and B are larger than those between B, C, N, O and
F. Further, considering the transition series, upon increasing the atomic number “inner” electrons
of relatively high second quantum number l and thus relatively good “screening” power are added.
Hence, the atomic size is practically constant in a transition series.
17
(3). Going from top to bottom in a column of the Periodic Table the number of “outer” electrons is
constant for increasing principal quantum number. The highest probability for finding this “outer”
electron occurs at a distance from the nucleus increasing with principal quantum number n . Thus
the atomic size increases going from top to bottom in a column of the Periodic Table. The “inner”
electrons “screen” the nuclear charge. But, this “screening” is not 100%. Hence, the effective nuclear
charge experienced by the “outer” electrons in a column increases for increasing principal quantum
number. This effect causes the increase of atomic size from top to bottom in a column of the Periodic
Table to be moderate. The ionization energies for removal of a first electron (i.e. from the neutral
atom) as function of position in the Periodic Table can be discussed in a similar way.
(4). Going from the left to the right in a period of the Periodic Table the number of “outer” electrons in
the same shell increases. Because the screening constant of these electrons is smaller than 100%
and at the same time the nuclear charge increases with one for each electron added upon increasing
the atomic number, it follows that the ionization energy will overall increase going from the left to the
right in a period of the Periodic Table.
Further, because the d and f electrons added in the transition series have a relatively high
“screening” power, it follows that the ionization energies of a transition series are rather equal.
Because the p electrons are less “penetrating” than s electrons, their Coulomb interaction with the
nucleus is less, their energy level is higher and consequently they are relatively easily ionizable. This
explains the drop in ionization energy experienced going, for example, in the second period, from Be
to B.
18
(5). Going from top to bottom in a column of the Periodic Table the number of “outer” electrons is
constant for increasing principal quantum number n. The energy of the “outer” electrons increases
with n . The “inner” electrons “screen” the nuclear charge. But, this “screening” is not 100%. Hence,
the effective nuclear charge experienced by the “outer” electrons in a column increases for increasing
principal quantum number. The net effect (relative change of effective Z2 is smaller than the relative
change of n2; is a decrease of the ionization energy going from top to bottom in a column of the
Periodic Table.
Exercises
1 – Of the choices below, which gives the order for first ionization energies?
Cl, S, Al, Ar, Si
2 – Which of the following sets contains species that are isoelectronic?
a)Cl, Ar, K
b) F-, Ne, Na+
c) F, Ne, Na
d) Al3+ , S2-, Ar
e) P3-, S2-, Ar-.
3 – Which of the following elements has the most negative electron affinity? Be, N, F, Li, Na.
4 – Of the elements below, which is the most metallic? sodium, barium, calcium, cesium, magnesium
5 – Which one of the following has the smallest radius? a) P,
b) Na,
c) Br,
d) Cl,
e) Fe
6 – What factors influence the ease of removing an electron from an element?
• the effective charge of a nucleus on the specific electron
• the electron configuration
• the size of the element
19
COVALENT AND IONIC RADII
The sizes of atoms and ions are also related to the ionization energies and electron affinities.
As the nuclear charge increases, the electrons are pulled in toward the center of the atom, and the
size of any particular orbital decreases. On the other hand, as the nuclear charge increases, more
electrons are added to the atom and their mutual repulsion keeps the outer orbitals large. The
interaction of these two effects (increasing nuclear charge and increasing number of electrons)
results in a gradual decrease in atomic size across each period.
Electronegative element is an element whose atoms want to participate in chemical interactions
by accepting electrons and are therefore also highly reactive. Electronegativity: relative measure of
ability of an atom to attract electrons to form an anion, in HCl, Cl is more electronegative than H and,
therefore, attracts the electron more.
Compared to their neutral atoms, metal cations have fewer electrons being pulled in by the
same number of protons. Less shielding…smaller radius.
Compared to their neutral atoms, anions have more electrons being pulled in by the same
number of protons. More shielding…bigger radius.
Nonpolar covalent radii, calculated for ideal molecules with no polarity shown in table. There are
other measures of atomic size, such as the van der Waals radius, in which collisions with other atoms
are used to define the size. It is difficult to obtain consistent data for any such measure, because the
polarity, chemical structure, and physical state of molecules change drastically from one compound
to another. The numbers shown here are sufficient for a general comparison of one element with
another.
20
There are similar problems in determining the size of ions. Because the stable ions of the
different elements have different charges and different numbers of electrons, as well as different
crystal structures for their compounds, it is difficult to find a suitable set of numbers for comparison.
Earlier data were based on Pauling's approach, in which the ratio of the radii of isoelectronic ions
was assumed to be equal to the ratio of their effective nuclear charges. More recent calculations are
based on a number of considerations, including electron density maps from X-ray data that show
larger cations and smaller anions than those previously found.
Factors that influence ionic size include the coordination number of the ion, the ; covalent
character of the bonding, distortions of regular crystal geometries, and delocalization of electrons .
The radius of the anion is also influenced by the size and charge of the cation (the anion
exerts a smaller influence on the radius of the cation).
The values in Tables below showed that anions are generally larger than calions with similar
numbers of electrons (F – and Na + differ only in nuclear charge, but the radius of fluoride is 37%
larger). The radius decreases as nuclear charge increases for ions with the same electronic structure,
such as O –2, F – , Na+ and Mg +2, with a much larger change with nuclear charge for the cations.
Within a family, the ionic radius increases as Z increases because of the larger number of electrons in
the ions and, for the same element, the radius decreases with increasing charge on the cation.
21
Ion Protons Electrons Radius
Ion Protons Eectrons Radius Ion Protons Electrons radius
O –2
8
10
126
O –2
8
10
126
Ti+2
22
20
100
F–
9
10
119
S –2
16
18
170
Ti+3
22
19
81
Na+
11
10
116
Se – 2
34
36
184
Ti+4
22
18
75
Mg+2 12
10
86
Te – 2
54
207
Crystal radius and nuclear charge
52
Crystal radius and no. of e
Crystal radius and ionic charge
Isoelectronic Species
A very useful comparison with atoms (and molecules too) is the look at properties of systems
with the same electron configuration. Such species are isoelectronic. Let’s look at Ne, with a 2p6
electron configuration:
Other species with that same configuration are: Na+, Mg2+ . We get these species by removing
the 3s electron(s) from the first two atoms in the next row of the periodic table. Also, F – , and O2have the same rare gas configuration. So, for the following isoelectronic series O2-, F-, Ne, Na+, Mg2+
the nuclear charge increases left to right. So, we expect the size to decrease left to right. Why?
How about the energy to remove the last 2p electron?
Should be highest for Mg2+ and lowest for O2-, right?
22
Summary of Periodic Trends and Atomic Sizes
1.Atomic radius increases as you go top to bottom through a group of elements
2.Atomic radius decreases from left to right as you cross a period of main group elements.
3.The atomic radii of transition metals in the same period is approximately equal.
4. Metal cations are smaller than the atoms from which they are formed.
5. Non-metals form anions that are larger than the atoms from which they are formed.
6. The isoelectronic species with the larger number of protons has the smallest radius.
The Size of Atoms
The covalent radius is one-half the distance
between the two nuclei of two identical atoms
joined by a covalent bond.
The ionic radius is based on the distance between
the nuclei of ions joined by an ionic bond. Not equal
to ½ the distance between the cation nucleus and
the anion nucleus !
The metallic radius is one-half the distance between
the nuclei of two atoms in contact in a crystalline,
solid, metal.
23
Question and solutions
Q1: Calculate the ionization energy of the hydrogen atom on the basis of the Bohr theory.
Solution: 1 – Given: hydrogen atom.
2 – Required: ionization energy
The energy required to remove the electron from the lowest energy level in hydrogen n1=1 to infinity
n2=∞is the ionization energy. We use Eq. ῡ=1/λ = RH ( 1/n12 – 1/n22 ) , E = ɦc / λ = ɦcῡ so we can say
that E = ɦcRH ( 1/n12 – 1/n22 ) to solve. E = 2.179 x 10 – 18 J. To convert to ev divided on 1.6 x 10 – 19
Q2: Calculate, in joules the potential energy of an electron in the n = 2 orbit of the hydrogen atom.
Solution: 1 – Given: n = 2, hydrogen atom
Required: Ep
We use Eq. ,
with Z = 1 for a hydrogen atom to solve for Ep, in erg.
Q3: 1) Electron Affinities of two elements 'A' and 'B' are given below: A = 3. 79 eV,
B = 3. 56 eV
Which of them will ionize more easily and why?
2) Why inert gases have zero electron affinity.
A:
1) Element 'A' will ionize more easily than element 'B'. The electron affinity of 'A' being higher
means it will accept electrons more easily than 'B'. Thus element 'A' will form negatively charged ions
more easily than element 'B‘.
2) Inert gases have completely filled outmost shells (valence shells). They have a stable
configuration and need not accept any electrons. Thus they have zero electron affinity.
24
Q4: What factors influence the ease of removing an electron from an element?
• the effective charge of a nucleus on the specific electron
• the electron configuration
• the size of the element
Q5: Why does the element with the lowest atomic number have the greatest atomic radius in each
period?
A: Lower the atomic number of the element, lower is the nuclear charge. The attraction by the
nucleus on the outer shell electrons is therefore less. As a result the atomic radius is the greatest of
the element with lowest nuclear charge in each period.
Q6: Explain why the electron affinity of fluorine is less than that of chlorine though fluorine precedes
chlorine in group VIIA.
A: Fluorine is a smaller atom compared to chlorine. The 7 valence electrons in fluorine are
accommodated in a smaller volume compared to chlorine. On accepting an electron, there are 8
electrons in the valence shell. This leads to increased electron repulsion in fluorine compared to the
bigger atom chlorine. Hence tendency to accept electrons i. e., electron affinity is lesser in fluorine
than in chlorine.
Q7: Explain why metallic character increases down a group?
A: Metallic character depends on atomic size and ionization potential. The atomic size increases
while the ionization potential decreases down a group. The more easily an element loses electrons,
the more metallic is its nature, and hence the metallic character increases down a group.
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