ch 3 classification of elements and periodicity in properties

Chapter 3
Classification of elements and
periodicity in properties

Why to classify elements:
• Elements are the basic units of matter
• At present 118 elements are identified
• First 94 occur naturally on earth & recently discovered (24 elements) are man-made
• Efforts to synthesize new elements are still continue
• It is difficult to study individually the chemistry of all these elements and their
compounds
• To solve this problem, all these elements are organized in a systematic way by classifying
them into groups and tables in Periodic Table

History of classification of elements
Döbereiner’s Triads (1817-1829) :
• Arrange the elements with similar properties into groups having three
elements each. He called these groups ‘triads’
• The three elements in a triad are written in the order of increasing atomic masses.
• The atomic mass of the middle element was roughly the average of the atomic masses of the other two
elements.
• For example, triad consisting of lithium (Li), sodium (Na) and potassium (K) having atomic masses 7, 23.0
and 39.0
• Döbereiner could identify only three triads from the 53 elements known at that time
• Hence, this system of classification into triads was not useful

Newlands’ Law of Octaves (By John Newlands In 1866)
• The elements were arrangedin the order of increasing atomic masses.
• He found that every eighth element had properties similar to that of the first.
• He compared this to the octaves found in music. Therefore, he called it the ‘Law of Octaves’. Also known as
‘Newlands’ Law of Octaves’.
• In Newlands’ Octaves, the properties of lithium and sodium were found to be the same. Sodium is the eighth
element after lithium.
• Similarly, beryllium and magnesium resemble each other.
• A part of the original form of Newlands’ Octaves is given as

• The Law of Octaves was found to be applicable only upto calcium, as after calcium every eighth element did not
possess properties similar to that of the first.
• It was assumed by Newlands that only 56 elements existed in nature and no more elements would be
discovered in the future.
• But, later on, several new elements were discovered, whose properties did not fit into the Law of Octaves.
• In order to fit elements into his Table, Newlands adjusted two elements in the same slot, but also put some
unlike elements under the same note.
• Example, cobalt and nickel are in the same slot and these are placed in the same column as fluorine, chlorine
and bromine which have very different properties than these elements.
• Iron, which resembles cobalt and nickel in properties, has been placed far away from these elements.
• Thus, Newlands’ Law of Octaves worked only with lighter elements only

MENDELÉEV’S CLASSIFICATION (1869)
• Mendeléev gave the most important contributor to the early development of a Periodic Table of
elements
• He arranged the 63 elements on the basis of their atomic mass, and also on the similarity of
chemical properties.
• He examined the relationship between the atomic masses of the elements and their physical and
chemical properties.
• Chemical properties were concentrated on the compounds formed by elements with oxygen and
hydrogen.
• He selected Hydrogen and oxygen because they are very reactive and formed compounds with
most elements
• The formulae of the hydrides and oxides formed by an element were treated as one of the basic
properties of an element for its classification.
• On arranging elements in the order of their increasing atomic masses, a periodic repetition
of elements was observed with similar physical and chemical properties.
• On this basis, Mendeléev formulated a Periodic Law, which states that ‘the properties of
elements are the periodic function of their atomic masses’.
• Mendeléev’s Periodic Table contains vertical columns called ‘groups’ and horizontal rows
called ‘periods’

Strengths in Mandeleev Periodic table:
• Mendeléev also predicted the existence of some elements that had not been discoveredat that time.
• Mendeléev named them by prefixing a Sanskrit numeral, Eka (one) to the name of preceding element in the
same group.
• For instance, scandium, gallium and germanium, discovered later, have properties similar to Eka–boron,
Eka–aluminium and Eka–silicon,respectively. The properties of Eka–Aluminium predicted by Mendeléev
and those of the element, gallium which was discovered later and replaced Ekaaluminium, are listed as:
• This indicated the correctness and usefulness of Mendeléev’s Periodic Table.
• Thus, Mandeleev was recongised as the originatorof the concept for classification of elements.
• Noble gases like helium (He), neon (Ne) and argon (Ar) gases were discovered very late because they are
very inert and present in extremely low concentrationsin our atmosphere.
• One of the strengths of Mendeléev’s Periodic Table was that, when these gases were discovered, they could
be placed in a new group without disturbing the existing order.

Limitations of Mendeléev’sClassification
• While grouping elements with similar properties, Mendeléev had to place an element with a slightly greater
atomic mass before an element with a slightly lower atomic mass.
• For example, cobalt (atomic mass 58.9) appeared before nickel (atomic mass 58.7)
• Thus, the sequence of arranging elements in order of increasing atomic masses gets inverted.
• Electronic configurationof hydrogen resembles that of alkali metals. Like alkali metals, hydrogen combines with
halogens, oxygen and sulphur to form compounds having similar formulae
• Also, just like halogens, hydrogen also exists as diatomic molecules and it combines with metals and non-metals
to form covalent compounds. Therefore, no fixed position can be given to hydrogen in the Periodic Table.
• Isotopes were discovered long after Mendeléev had proposed his periodic classification of elements.
• We know,isotopes of an element have similar chemical properties, but different atomic masses
• Thus, isotopes of all elements posed a challenge to Mendeleev’s Periodic Law.
• Another problem was that the atomic masses do not increase in a regular manner in going from one element to
the next. So, it was not possible to predict how many elements could be discovered between two elements —
especially when we consider the heavier elements.

Question: Use Mendeleev’sPeriodic Table to predict the
formulae for the oxides of the following elements: K, C, Al,
Si, Ba.
Solution: Potassium is in group 1. The oxide will be K2O.
Carbon is in group 4. The oxide will be CO2. Aluminum is in
group 3. The oxide will be Al2O3. Silicon is in group 4. The
oxide will be SiO2. Barium is in group 2. The oxide will be
BaO.
Question: Besides gallium, which other elements have since
been discovered that were left by Mendeleev in his Periodic
Table?
Answer: Germanium and Scandium
Question: What were the criteria used by Mendeleev in
creating his Periodic Table?
Solution: Mendeleev’s periodic table was based on the
atomic masses of the elements. The properties of the
elements are the periodic function of their atomic masses.
Elements are arranged in the increasing atomic mass and
their properties get reoccur after regular intervals
Question: Why do you think the noble gases are placed in a
separate group?
Solution: Noble gases are inert gases. Noble gases does not
form any compounds due to inert nature. Their periodic
properties are different from all other elements.

Modern Periodic Law (by Henry Moseley in 1913
• Atomic number is more fundamental property of an
element
• The physical and chemical properties of the
elements are periodic functions of their atomic
numbers
• Periodic Law is essentially the consequence of the
periodic variation in electronic configurations
• Helps in determining the physical and chemical
properties of elements and their compounds
• In present periodic table, the horizontal rows are
called periods and the vertical columns are groups.
• Elements having similar outer electronic
configurations in their atoms are arranged in vertical
columns (Groups)
• There are total seven periods, the first period
contains 2 elements. The subsequent periods
consists of 8, 8, 18, 18, 32 and 32 elements,
respectively.
• 14 elements of both sixth and seventh periods are
placed in separate panels at the bottom (known as
lanthanoids and actinoids)
Modern periodic Law & present form of periodic table

Electronic configurationof elements
• Electrons are filled in different subshells (also known as orbitals)
• The distribution of electrons into orbitals of an atom is electronic configuration
(a) Period wise electronic configuration
• period indicates the value of n for the outermost or valence shell
• The first period (n = 1) starts with the filling of the lowest level (1s) which has two elements — hydrogen(1s1 )
and helium (1s2 )
• The second period (n = 2) starts with lithium and the third electron enters the 2s orbital. (1s2 2s1)
• The next element, beryllium has four electrons and has the electronic configuration 1s2 2s2 .
• In next element boron, the electron starts enter in 2p orbitals and at neon, all the 2p orbitals are filled (2s2 2p6 )
• The third period (n = 3) begins at sodium, and the added electron enters a 3s orbital.
• The fourth period (n = 4) starts at potassium, and the added electrons fill up the 4s orbital.
• filling up of 3d orbitals becomes energetically favorable before the 4p orbital which gives the 3d transition
series of elements (from scandium (Z = 21) having electronic configuration 3d1 4s2 to zinc (Z=30) with electronic
configuration 3d10 4s2 . The fourth period ends at krypton with the filling up of the 4p orbitals.
• The fifth period (n = 5) beginning with rubidium which contains the 4d transition series starting at yttrium (Z =
39) and ends at xenon with the filling up of the 5p orbitals.
1s< 2s< 2p< 3s< 3p< 4s< 3d< 4p< 5s< 4d<5p<6s< 4f<5d< 6p<7s...

• The sixth period (n = 6) contains 32 elements and successive electrons enter 6s, 4f, 5d and 6p orbitals
• Filling of the 4f orbitals begins with cerium (Z = 58) and ends at lutetium (Z = 71) to give the 4f-inner transition series
(also called lanthanoid series)
• The seventh period (n = 7) begins with the successive filling up of the 7s, 5f, 6d and 7p orbitals and includes most of the
man-made radioactive elements. This period will end at the element with atomic number 118 which would belong to
the noble gas family.
• Filling up of the 5f orbitals after actinium (Z = 89) gives the 5f-inner transition series known as the actinoid series
• 4f and 5f-inner transition series of elements are placed separately in the Periodic Table
(b) Groupwise Electronic Configurations
• Elements in the same vertical column or group have similar valence shell electronic configurations due to the
same number of electrons in the outer orbitals, and similar properties.
For example, the Group 1 elements, all elements have ns1 electronic configuration

Classification of Elements into s, p, d and f blocks
• Depends on the type of atomic orbitals (s, p, d, f) that are being filled with electrons
• Two exceptions:
1) Hydrogen: placed separately at the top of the Periodic Table.
It has only one s-electron and can be placed in group 1 (alkali metals). It can also gain an
electron to achieve a noble gas arrangement and hence it can behave similar to group 17
(Halogen family)
2) Helium: It belongs to the s-block but placed in the p-block along with other group 18
elements. because it has a completely filled valence shell (1s2 ) and thus, exhibits properties
characteristic of other noble gases. Groups
Periods
s-Block Elements:
• These are Group 1 (alkali metals) and Group 2 (alkaline earth metals) which have ns1 and ns2
outermost electronic configuration
• They are all reactive metals with low ionization enthalpies.
• They lose the outermost electron(s) readily to form 1+ ion (in the case of alkali metals) or 2+ ion
(in the case of alkaline earth metals).
• The metallic character and the reactivity increase as we go down the group. Because of high
reactivity they are never found pure in nature.
• The compounds of the s-block elements, with the exception of those of lithium and beryllium
are predominantly ionic.
s-block elements

p-Block Elements (belongs to Group 13 to 18)
• P- block elements together with the s-Block Elements are called the
Representative Elements or Main Group Elements.
• The outermost electronic configuration varies from ns2np1 to ns2np6 in each
period.
• At the end of each period is a noble gas element having valence shell ns2np6
configuration.
• All the orbitals in the valence shell of the noble gases are completely filled by
electrons and it is very difficult to alter this stable arrangement by the addition
or removal of electrons.
• The noble gases thus exhibit very low chemical reactivity.
• Group 17 elements are known as halogens and the Group 16 as chalcogens
• Groups 16 and 17 elements have highly negative electron gain enthalpies
• They readily add one or two electrons respectively to attain the stable noble gas
configuration.
• The non-metallic character increases as we move from left to right across a
period
• metallic character increases as we go down the group.

d-Block Elements (Transition Elements):
• These are the elements of Group 3 to 12.
• These are characterized by the filling of inner d orbitals by electrons and are therefore referred to as d-Block
Elements.
• These elements have the general outer electronic configuration (n-1)d1-10ns0-2 .
• They are all metals.
• They mostly form coloured ions, exhibit variable valence (oxidation states), paramagnetism and oftenly used as
catalysts.
• Zn, Cd and Hg which have the electronic configuration, (n-1) d10ns2 do not show most of the properties of
transition elements.
• Transition metals thus form a bridge between the chemically active s-block elements and the less active
elements of Groups 13 and 14. Thus, named as “Transition Elements”

f-Block Elements (Inner-Transition Elements):
• Contains two rows of elements at the bottom of the Periodic Table
• called the Lanthanoids, Ce(Z = 58) – Lu(Z = 71) and Actinoids, Th(Z = 90) – Lr (Z = 103)
• outer electronic configuration (n-2)f1-14 (n-1)d 0–1ns2 . The last electron added to each element is filled
in f- orbital.
• The Lanthanoid and Actinoid series are called the Inner Transition Elements
• They are all metals.
• Within each series, the properties of the elements are quite similar.
• The chemistry of the early actinoids is more complicated than the corresponding lanthanoids, due to
the large number of oxidation states possible for these actinoid elements.
• Actinoid elements are radioactive. Many of the actinoid elements have been made only in nanogram
quantities
• The elements after uranium are called Trans uranium Elements.

Classification of elements into Metals, Non-metals and Metalloids
Metals
• appear on the left side of the Periodic Table (comprise more than 78% of all known elements)
• Metals are usually solids at room temperature
mercury is an exception (liquid at room temperature; gallium and cesium also have very low melting points (303K and 302K)
•
•
•
Metals usually have high melting and boiling points.
They are good conductors of heat and electricity.
They are malleable (can be flattened into thin sheets by hammering)
& ductile (can be drawn into wires).
• The elements become more metallic as we go down a group;
non-metals
• located at the top right hand side of the Periodic Table.
• Property of elements change from metallic to non-metallic as we moves from
• left on the right.
• Non-metals are usually solids or gases at room temperature
• Have low melting and boiling points (boron and carbon are exceptions).
• They are poor conductors of heat and electricity.
• Most nonmetallic solids are brittle and are neither malleable nor ductile.
• Nonmetallic character increases as one goes from left to right across the Periodic Table.
• The elements (e.g., silicon, germanium, arsenic, antimony and tellurium) bordering
the zig zag line and running diagonally across the Periodic Table show properties that are characteristic of both metals and
nonmetals.
• These elements are called Semi-metals or Metalloids.

Problem: Considering the atomic number and position in the periodic table, arrange the following elements in the
increasing order of metallic character : Si, Be, Mg, Na, P.
Solution. Si (Atomic no 14)
Be (Atomic no 4)
Mg (Atomic no. 12)
Na(11)
P(15)
Their position in periodic table is
Be
Na Mg Al Si P
We know, metallic character increases down the group
And decreasing on moving from left to right on periodic table
Hence the order of increasing metallic character is: P < Si < Be < Mg < Na

• There are various trends in physical & chemical properties of elements as we descend in a group or move across a period
• Within a period, chemical reactivity tends to be high in Group 1 metals which becomes lower towards the middle of the
table, and increases to a maximum in the Group 17 (non-metals)
• Similarly, within a group of alkali metals, reactivity increases on moving down the group
• whereas within a group of non-metals (say halogens), reactivity decreases down the group.
Why do the properties of elements follow these trends?
• Trends in physical properties (atomic and ionic radii, ionization enthalpy, electron gain enthalpy and electronegativity):
Trends in properties of elements across the period and moving down the group
Across the period
Down the group
Atomic radius: Total distance from the nucleus of an atom to
the outermost shell of electron
For non-metal, it is called covalent radius For metals, it is called ionic radius
Atomic radius decreases across the period
It is because within the period the electrons are in the same valence shell and the effective nuclear charge increases which
results in the increased attraction of electrons to the nucleus Atomic radius increases down the group
as we move down the groups, the principal quantum number
(n) increases and the valence electrons are farther from the nucleus. This happens because the inner energy levels are filled
with electrons, which shield the outer electrons from the nucleus.

Variation of atomic radius across period (n=2) Variation of atomic radius down the group
(Group 1 & group 17)

Ionic radius:
• ionic radius is the distances between cations and anions in ionic crystals
• Ionic radius of elements follows the same trends as atomic radius of element
• Cation results from the removal of an electron from an atom
• Anion results from the gain of an electron
• A cation is smaller than its parent atom because it has fewer electrons while its nuclear charge remains the same.
• The size of an anion will be larger than that of the parent atom because the addition of one or more electrons would
result in increased repulsion among the electrons and a decrease the effectivenuclear charge.
• For example, the ionic radius of fluoride ion (F– ) is 136 pm whereas the atomic radius of fluorine is only 64 pm.
• The atomic radius of sodium is 186 pm compared to the ionic radius of 95 pm for Na+
Trends among isoelectronic series (O2–, F– , Na+ and Mg2+ (number of electrons: 10)
The cation with the greater positive charge will have a smaller radius
Because positive charge means removal of electron, when electron removes from an atom, there will be greater attraction
between electrons and nucleus which reduce the size
Anion with the greater negative charge will have the larger radius
Because negative charge means addition of electrons, when electron added to an atom, the net repulsion among the
electrons increase and effective nuclear charge decreases. Thus, size will increase.
Problem: Which of the following species will have the largest and the smallest size? Mg, Mg2+, Al, Al3+ .
Solution Atomic radius decrease across a period. Cations are smaller than their parent atoms. Among isoelectronicspecies,
the one with the larger positive nuclear charge will have a smaller radius. Hence the largest species is Mg; the smallest one is
Al3+

Trend in Ionization Enthalpy:
• It measures the tendency of an element to lose electron.
• It represents the energy required to remove an electron from an atom in its ground state
• For an element X, the first ionization enthalpy is the enthalpy change (∆iH) for the reaction given below:
X(g) → X+ (g) + e–
• ionization enthalpy is expressed in units of kJ mol–1
• Second ionization enthalpy is the energy required to remove the second most loosely bound electron
• This energy is represented by equation below:
X+ (g) → X2+(g) + e–
• Ionization enthalpies are always positive because energy is always required to remove electrons from an
atom
• The second ionization enthalpy will be higher than the first ionization enthalpy because it is more difficult to
remove an electron from a positively charged ion than from a neutral atom.
• Similarly, third ionization enthalpy will be higher than the second.
The first ionization enthalpies of elements having atomic numbers up to 60
• maxima at the noble gases which have closed electron shells and very stable
electron configurations.
• minima occur at the alkali metals which have only one electron in their valence
shell and have high tendency to loose that electron to attain noble gas
configuration. Due to this, they are highly reactive and low ionization enthalpy

Trend across the period and down the group:
• As we move across the period, the first ionization enthalpy increases
• As we move down the group, the ionization energy decreases.
• This trend in ionization enthalpy is closely related to atomic radius
Trends in ionization energy across period (n=2)
Reason:
When we move from lithium to fluorine across the second period
• the atomic radius decreases
• Thus the atom gets smaller and the outer shell electrons becomes more closer to
nucleus and thus strongly attracted to nucleus at the center.
• Therefore it becomes more difficult to remove outermost electron
When we move down the group
• the outer shell electron being increasingly farther from the nucleus
• Inner shell electrons reduces the force of attraction between outer shell electron
and nucleus
• Therefore, it becomes easier to remove outer shell electrons i.e. it requires less
ionization energy
Trends in ionization energy across the group (group 1)

• Ionization enthalpy of boron (Z = 5) is slightly less than that of beryllium (Z = 4)
• In Be, the electron removed during the ionization from 2s shell whereas the
electron removed in B is from 2p shell. For the same principal quantum level, an s-
electron is more closely attracted to the nucleus than a p-electron.
• the 2p electron of boron is more shielded from the nucleus by the inner core of
electrons than the 2s electrons of beryllium.
• Thus. 2p electron from B is easier to remove than 2s electron from Be
Ionization enthalpy of oxygen is smaller as compared to nitrogen.
• Nitrogen has electronic configuration 1s2, 2s2, 2p3
• In nitrogen atom, three 2p-electrons reside in different atomic orbitals (Hund’s rule)
• whereas in the oxygen atom has electronic configuration 1s2, 2s2, 2p4
• two of the four electrons in 2p shell must occupy the same 2p-orbital
• It results in an increased electron-electron repulsion.
• Thus, it is easier to remove the fourth 2p-electron from oxygen than to remove one
of the three 2p-electrons from nitrogen

Trend in Electron Gain Enthalpy
• When an electron is added to a neutral gaseous atom (X) to convert it into a negative ion, the enthalpy change
accompanying the process is defined as the Electron Gain Enthalpy (∆egH).
• Electron gain enthalpy provides a measure of the ease with which an atom adds an electron to form anion
X(g) + e– → X– (g)
• The process of adding an electron to the atom can be either endothermic or exothermic which depends on the element
• For most of the elements, energy is released when an electron is added to the atom and the ∆egH is negative
• For group 17 elements (the halogens) have very high negative electrongain enthalpies because they can attain stable
noble gas electronic configurations by picking up an electron.
• On the other hand, noble gases have large positive electron gain enthalpies because the electron has to enter the next
higher principal quantum level leading to a very unstable electronic configuration.
• It may be noted that electron gain enthalpies have large negative values toward the upper right of the periodic table
preceding the noble gases.

• As we move across a period, electron gain enthalpy becomes more negative
• As we go down the group, electron gain enthalpy become less negative
Reason:
• As we move from left to right across a period, the atomic size decrease and the effective nuclear charge increases.
These factors increase the force of attraction between nucleus and incoming electron. Thus, it will be easier to add an
electron to a smaller atom
• As we go down the group, the size of the atom increases and the added electron would become farther from the
nucleus.
• Electron gain enthalpy of O or F is less negative than that of the succeeding element.
This is because when an electron is added to O or F, the added electron goes to the smaller n = 2 quantum level and
suffers significant repulsion from the other electrons present in this level.
For the n = 3 quantum level (in case of S or Cl), the added electron occupies a larger region of space and the electron-
electron repulsion is much less
Trend across the period and down the group:

Problem (based on Ionization energy):
The first ionization enthalpy (∆i H ) values of third period elements, Na, Mg and Si are respectively 496, 737 and 786 kJ mol–1.
Predict whether the first ∆iH value for Al will be more close to 575 or 760 kJ mol–1 ?
Solution Across the period from Na, Mg, Al, Si
The value for Al should be lower than that of Mg because electronic configuration of Mg is 1s2, 2s2, 2p6, 3s2 and electronic
configuration of Al is 1s2, 2s2, 2p6, 3s2, 3p1. In case of Mg, electron has to be removed from 3s orbital, which has paired
electrons and has stable electronic configuration. In case of Al, electron has to be removed from highest energy orbital 3p.
Therefore, it is easier to remove electron from 3p orbital as compared to 3s orbital.
Therefore, ∆i H value for Al will be more close to 575 kJ mol–1
Problem (based on electron gain enthalpy):
Which of the following will have the most negative electron gain enthalpy and which the least negative?
P, S, Cl, F.
Solution Arrange these elements in form of periodic table
F
P S Cl
Electron gain enthalpy generally becomes more negative across a period as we move from left to right. Within a group,
electron gain enthalpy becomes less negative down a group.
Since, adding an electron to the 2p-orbital leads to greater repulsion than adding an electron to the larger 3p-orbital.
Hence the element with most negative electron gain enthalpy is Cl;
the element with the least negative electron gain enthalpy is P.

Trends in Electronegativity
• Electronegativity measure of the ability of an atom in a chemical compound to attract shared pair of electrons to itself.
• Unlike ionization enthalpy and electron gain enthalpy, it is not a measurable quantity
• There are different types of numerical scales of electronegativity (Pauling scale, Mulliken-Jaffe scale, Allred-Rochow scale)
Pauling scale is most widely used (by Linus Pauling)
• According to this scale, fluorine is considered to have the greatest ability to attract electrons and assigned it value 4
• Electronegativity is not measurable quantity but It predicts the nature of force that holds a pair of atoms together.
• Electronegativity values depends on the element to which it is bound
• Trend in electronegativity is related to atomic radius
• Electronegativity generally increases across a period from left to right
• It decreases down the group
• The trend is similar to that of ionization enthalpy
• As we go across the period, the atomic radius decreases and the outer shell becomes more closer to nucleus and it can
attractshared pair of electron more towards itself. Thus the tendency to attract the electron increase which leads to
increase in electronegativity.
• Similarly, electronegativity values decrease down the group, as the atomic radius increases

• Electronegativity is directly related to that non-metallic properties of elements.
• Because non-metallic elements have strong tendency to gain electrons.
• Thus, electronegativity is inversely related to the metallic properties of elements.
• Thus, the increase in electronegativities across a period is due to increase in non-metallic
properties (or decrease in metallic properties) of elements.
• Similarly, the decrease in electronegativity down a group is due to decrease in non-metallic
properties (or increase in metallic properties) of elements.

Periodic trends in chemical properties:
Periodicity of Valence/ valency/ Oxidation States
• This property can be studied in terms of electronic configurations.
• In representative elements, the valence or oxidation state is equal to the number of electrons in the outermost orbitals
and / or equal to eight minus the number of outermost electrons as shown below.
• Consider the two oxygen containing compounds: OF2 and Na2O.
• In OF2 ,electronic configuration of O is 2s22p4 and F is 2s22p5
• Since there are two F atoms in this molecule, each of the fluorine atom in OF2 shares one electron
with oxygen. Similarly, oxygen shares two electrons with fluorine atoms
• The order of electronegativity of the three elements involved in these compounds is F > O > Na.
• Fluorine is more electronegative than Oxygen. So, Fluorine can easily take electron from O
• So, Fluorine can be represented as F–1 which means that oxidation state of F= –1
• Since, oxygen give one electrons to each F atom (total 2 electrons), so oxidation state of O= +2.
• In Na2O, electronic configuration of O is 2s22p4 and sodium is 3s1
• oxygen being more electronegative than sodium. So, Oxygen accepts two electrons, one from each
of the two sodium atoms to complete its shell. Thus, oxygen shows oxidation state –2. On the
other hand sodium can easily loses one electron to oxygen and is given oxidation state +1.
• Thus, the oxidation state of an element in a particular compound can be defined as the charge
acquired by its atom on the basis of electronegative consideration from other atoms in the
molecule
x
F
x
x
F x
x x
x x x
x
x
Na
Nax

Problem: Using the Periodic Table, predict the formulas of compounds which might be formed by the following pairs of
elements; (a) silicon and bromine
Solution (a) Silicon is group 14 element,
electronic configuration of Si is 1s2, 2s2, 2p6, 3s2, 3p2 or [Ne] 3s2, 3p2
So, valence of Si =4;
bromine belongs to the halogen family ns2, np5
Valence of Br=1.
Since, Br is more electronegative than Si, So, we can write Si as Si4+ and Br as Br-
After balancing the charge, the formula of the compound can be written as SiBr4.
(b) aluminium and sulphur
aluminium belongs to group 13,
valence of Al =3
sulphur belongs to group 16 elements ns2, np4
Valence of S =2.
So, we can write as Al3+ and S2-
To balance the charges, 2x Al3+ and 3x S2-
Hence, the formula of the compound formed would be Al2S3

Periodic trends for formationof hydrides and oxides

Exception properties or abnormal Properties of elements of second period Li, Be, B, C, N, O, F,
• These elements differs in many respects from the other members of their respective group.
• For example, lithium unlike other alkali metals, and beryllium unlike other alkaline earth metals,
form covalent compounds; the other members of these groups predominantly form ionic
compounds.
• In fact the behaviour of lithium and beryllium is more similar with the second element of the
following group i.e., magnesium and aluminium, respectively. This sort of similarity is commonly
referred to as diagonal relationship
What are the reasons for the different chemical behaviour of the first member of a group of
s- and p-block elements compared to that of the subsequent members in the same group?
• The different behaviour is attributed to their small size, large charge/ radius ratio and high
electronegativity of the elements.
• The first member of group has only four valence orbitals (2s and 2p) available for bonding,
• They have maximum 4 orbitals to form bonds so they can form only four covalent bonds. So
their maximum valency is 4.
• whereas the second member of the groups have nine valence orbitals (3s, 3p, 3d).
• These members of the group have d orbitals in their valence shell so they can accommodate
more than four pairs of electrons.
• The first member of p-block elements displays greater ability to form pπ – pπ multiple bonds to
itself (e.g., C = C, C ≡ C, N = N, N ≡ Ν) and to other second period elements (e.g., C = O, C = N, C ≡
N, N = O) compared to subsequent members of the same group
Ques: Boron can only form BF4
− and aluminium forms AlF6
3 −?

• There is high chemical reactivity at the two extremes and the lowest in the centre
• Maximum chemical reactivity at the extreme left (among alkali metals) is due to the loss of an electron leading to the
formation of a cation
• The maximum reactivity at the extreme right (among halogens) is due to the gain of an electron forming an anion.
• This property is related to the metallic and non-metallic character of elements.
• the metallic character of an element decreases from left to right
• the non-metallic character increases while moving from left to right across the period.
• The chemical reactivity of an element can be best shown by its reactions with oxygen
• Elements on two extremes of a period easily combine with oxygen to form oxides. The oxide formed by the element
on extreme left is the most basic (e.g., Na2O), whereas that formed by the element on extreme right is the most
acidic (e.g., Cl2O7 ).
• Oxides of elements in the centre are amphoteric (e.g., Al2O3 , As2O3) or neutral (e.g., CO, NO, N2O).
• Amphoteric oxides behave as acidic with bases and as basic with acids, whereas neutral oxides have no acidic or basic
properties
• The basic nature of Na2O and acidic nature of Cl2O7 can also show by their chemical reaction with water
• Solution Na2O with water forms a strong base NaOH (Na2O + H2O → 2NaOH)
• whereas Cl2O7 forms strong acid HClO4 ( Cl2O7 + H2O → 2HClO4 )
• This basic or acidic nature can be tested with litmus paper (Litmus turns blue to red in acidic solution, while its
turns red to blue in basic solution )

• Among transition metals (3d series), the change in atomic radius is much smaller as compared to those of
representative elements across the period.
• The change in atomic radius is still smaller among inner-transition metals (4f series).
• The ionization enthalpies are intermediate between those of s- and p-blocks. Thus, they are less electropositive
than group 1 and 2 metals.
• In the case of main group elements , on moving down the group, the atomic and ionic radius increase with increase
in atomic number which results in gradual decrease in ionization enthalpies and electron gain enthalpies. Thus, the
metallic character increases down the group and non-metallic character decreases.
• In the case of transition elements, however, different trend is observed. This can be explained in terms of atomic
size and ionization enthalpy.

Problems:
1. What is the basic theme of organization in the periodic table?
Solution: In the periodic table, elements are classified into groups and periods on the basis of their properties. Elements with
similar properties are placed in the same group. This classification helps us in systematic study of elements and their
compounds
2. Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?
Solution: Mendeleev arranged the elements in his periodic table in increasing order of atomic weight. He also arranged the
elements in periods and groups and placed the elements with similar properties in the same group.
He didn’t stick to this arrangement because some elements didn’t fit this scheme of classification
3. What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?
Solution: According to Mendeleev, physical & chemical properties of elements are periodic functions of their atomic weights.
Accordingto modern periodic law, physical & chemical properties of elements are periodic functions of their atomic numbers
4. On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements
Solution: We know that in periodic table, number of period= the value of principal quantum number (n) for the outermost
shells. Each period begins with filling of n shell.
for value of n=6, possible values of l=0, 1, 2, 3, 4, 5
This corresponds to 6s, 6p, 6d, 6f, 6g, 6h subshells
In 6th period, filling of electrons starts from 6s shell, but we know
energy of 6d subshell is higher than 7s.
So, electron are filled in 6s, 4f, 5d, and 6p subshells
6s has 1 orbital (2e-), 4f has 7 orbital (14e-), 5d has 5 orbital (10e-), 6p has 3 orbitals (6e-).

5. Why do elements in the same group have similar physical and chemical properties?
Solution Physical and chemical properties of the elements depends on the number of valence electrons.Elements present on
same group have same number of valence electrons. Therefore, elements in same group have similar properties.
6. What do you understandby isoelectronic species? Name a species that will be isoelectronic with each of the following
atoms or ions. (i) F– (ii) Ar (iii) Mg2+ (iv) Rb+
Try yourself
7. Consider the following species : N3– , O2–, F– , Na+ , Mg2+ and Al3+ (a) What is common in them? (b) Arrange them in the
order of increasing ionic radius
(a) Each species has same number of electrons. Hence, these species are isoelectronic.
(b) We know ionic radius increases with decrease in nuclear charge.
Solution: Nuclear charge=no of protons in the nucleus= atomic number
In above species number of protons are
N3– (7), O2–(8), F– (9), Na+ (11), Mg2+ (12) and Al3+ (13)
So, ionic radius follows the order N3– > O2–> F– > Na+ > Mg2+ > Al3+
8. How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its
second ionization enthalpy is higher than that of magnesium?
Solution: The first ionization enthalpy of sodium is lower than that of magnesium because atomic size of sodium is greater
than that of Mg. Thus, effective nuclear charge is higher is Mg. therefore, the energy required to remove electron from Mg is
more than that required in sodium.
However, second ionization enthalpy of Na is higher than that of Mg because after loosing one electron, Na acquires noble gas
configuration. It becomes very difficult to remove second electron from Na. therefore, ots second ionization enthalpy is more
than Mg.

9. The first ionization enthalpy values (in kJ mol–1) of group 13 elements are :
B Al Ga In Tl
801 577 579 558 589
How would you explain this deviation from the general trend ?
Solution: On increasing down the group, the ionization enthalpy generally decreases due to increase in atomic size and
shielding of electrons.
On moving down the group, the ionization enthalpy decreases from B to Al. but Ga has higher ionization enthalpy than Al,
because Ga follows after d-block elements and the shielding of d-electrons is not very effective. These electrons do not shield
the valence electrons very effectively, so, valence electron of Ga experiences greater effective nuclear charge than Al. thus,
there is strong interaction between valence electron of Ga and nucleus, which decreases size of Ga than Al and increases the
Ionization enthalpy of Ga.
The ionization enthalpy decreases from Ga to In due to increase in atomic size and shielding of electrons.
But on moving from In to Tl, ionization enthalpy again increases because Tl follows after 4f and 5d electrons and the shielding
by electrons in this orbitals is not very effective. Therefore valence electrons in Tl is strongly held with nucleus and increase its
ionization enthalpy.
10. Which of the following pairs of elements would have a more negative electron gain enthalpy? (i) O or F (ii) F or Cl
Try yourself (Explained in class)
11. Would you expect the second electron gain enthalpy of O as positive, more negativeor less negative than the first? Justify
your answer.
Solution: When an electron is added to O atom to form O-, energy is released, thus, first electron gain enthalpy is negative.
Whereas, when an electron is added to O- to form O2-, energy has to be given out in order to overcome the strong electronic
repulsion. Thus, the second electron gain enthalpy is positive.

12. Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different?
Justify your answer.
Solution: The ionization enthalpy depends on nuclear charge which in turn depends on the number of electrons and proton.
In isotopes, number of electrons and protons are equal. So first ionization energy for isotopes of same element is same.
13. Use the periodic table to answer the following questions.
(a) Identify an element with five electrons in the outer subshell. (ns2np5 (Group 17 elements or halogens eg. F, Cl)
(b) Identify an element that would tend to lose two electrons. (ns2) alkaline earth metals or group 2 metals ( have low ∆iH)
eg. Mg, Ca)
(c) Identify an element that would tend to gain two electrons. (ns2np4 Group 16 elements eg. O, S,)
(d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature
Group 17 has metals, non-metals, liquid as well as gas at room temperature. Cl, Br are non-metals, Iodine is a metal,
fluorides are liquid in state while Cl, Br and I are gases.
14. The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that among group 17
elements is F>CI > Br > I. Explain
Solution: Reactivity of group 1 elements lies in terms of their tendency to loose electron (i.e. Ionization energy, more easily
the electron remove higher the reactivity) whereas reactivity of group 17 elements lies in terms of their tendency to gain
electron (i.e. electron gain enthalpy, more easily the electron gain, higher the reactivity)

15. Write the general outer electronic configurationof s-, p-, d- and f- block elements
Solution:
16. Assign the position of the element having outer electronic configuration (i) ns2np4 for n=3 (ii) (n-1)d2ns2 for n=4, and
(iii) (n-2)f 7(n-1)d1ns2 for n=6, in the periodic table.
Solution: n=3 indicates element belongs to third period
ns2np4 indicates a p-block element which is in 16th group (sulphur)
(ii) n=4 indicates 4th period,
(n-1)d2ns2 indicates a d-block element which has 2 electrons in d-orbital (titanium)
(iii) n=6 indicates 6th period
(n-2)f 7(n-1)d1ns2 indicates a f-block element having 7 electrons in f orbital.
Its electronic configuration is [Xe]4f7 5d1 6s2 (atomic number=54+7+2+1=64)

17 The first (∆iH1 ) and second (∆iH2 ) ionization enthalpies (in kJ mol–1) and the (∆egH) electron gain enthalpy (in kJ mol–1) of
a few elements are given below:
Elements ∆H1 ∆H2 ∆egH
I 520 7300 –60
II 419 3051 –48
III 1681 3374 –328
IV 1008 1846 –295
V 2372 5251 +48
VI 738 1451 –40
Which of the above elements is likely to be :
(a) the least reactive element. (b) the most reactive metal. (c) the most reactive non-metal. (d) the least reactive non-
metal. (e) the metal which can form a stable binary halide of the formula MX2 (X=halogen).(f) the metal which can form a
predominantly stable covalent halide of the formula MX (X=halogen)?
Solution: (a) Element V. A least reactive metal must have high ionization enthalpy and having least negative or positive
electron gain enthalpy
(b) Element II. Most reactive metal have lowest first ionization enthalpy and low negative electron gain enthalpy
(c) Element III. Most reactive non-metals have high first ionization enthalpy and highest negative electron gain enthalpy
(d) Element V is least reactive non metal
(e) Element VI has low negative electron gain enthalpy and less positive second ionization energy
(f) Element I has lowest first ionization enthalpy and high second ionization enthalpy.
Hint: More chemical reactive element when it easily loose electron (ionization enthalpy is least +ve quantity
Or, More chemical reactive element when it easily gain electron (electron gain enthalpy is highest –ve quanity)

18. Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs
(a) Lithium and oxygen
(b) Magnesium and nitrogen
(c) Aluminium and iodine
(d) Silicon and oxygen
(e) Phosphorus and fluorine
(Li2O)
(Mg3N2)
(AlI3)
(SiO2)
(PF3, PF5)
19. Considering the elements B, Al, Mg, and K, the correct order of their metallic character is :
(a) B > Al > Mg > K (b) Al > Mg > B > K (c) Mg > Al > K > B (d) K > Mg > Al > B
20. Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is :
(a) B > C > Si > N > F (b) Si > C > B > N > F (c) F > N > C > B > Si (d) F > N > C > Si > B
B
Na Mg Al
K

Tricks to memorize main group elements (s and p-block elements)
Groups/
(block)
Tricks Elements
1 (s) HaLiNa K Rb Csey Fryad H, Li, Na, K, Rb, Cs, Fr
2 (s) Beta Mange Car Scoter Bap Razi Be, Mg, Ca, Sr, Ba, Ra
13(p) Began, Aaloo, Gajar In Thela B, Al, Ga, In, Tl
14(p) Cute Sisters Get Small (TINy)
Problems
C, Si, Ge, Sn, Pb
15(p) Nahi Pasand Aise Sab Bhai N, P, As, Sb, Bi
16(p) Oh, Seema Sent The Post O, S, Se, Te, Po
17(p) First Class Biriyani In Austria F, Cl, Br, I, At
18(p) He Never Argue; Kal Xero Run
pe out hua
He, Ne, Ar, Kr, Xe, Rn

ch 3 classification of elements and periodicity in properties

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