Unit 03 Classification Of Elements And Periodicity In Properties

With the discovery of a large number of elements, it became difficult to study the elements individually, so classification of elements was done to make the study easier.

Dobereiner’s Triads

He arranged similar elements in group of 3 and showed that atomic weight (arranged in increasing order) of middle element was approximately the arithmetic mean of the other two.

Drawback of Dobereiner’s Triads

The concept was applicable to limited elements. He could find only 3 such triads.

New lands law of octaves. He arranged the elements in increasing order of their atomic weights and showed that the properties of every eight elements were similar to those of the first one.

Drawbacks

1. Do not account the noble gases.

2. Applicable to only lighter elements

Mendeleev’s Periodic Law -

Physical and chemical properties of elements are a periodic function of their atomic masses.

Mendeleev’s periodic table correlated atomic weights with the periodicity of physical and chemical properties of the elements. Since the atomic number and not the atomic weight characterizes an element, Mendeleev’s table had several drawbacks, which were removed in the modern periodic table. The modern periodic table is based on the modern periodic law which states that the properties of elements are a periodic function of their atomic numbers. (Moselay gave the concept of atomic number)

Drawbacks of Mendeelev’s periodic table

1. Anomalous position of hydrogen-It resembles both halogens as well as alkali metals. It is placed above alkali metals. So, its position is not justified.

2. Anomalous pairs of elements - Some elements with higher atomic weight precede elements with lower atomic weight eg. Ar (39.9) precedes potassium (39.1), Co (58.9) precedes Ni (58.7).

3. No place for isotopes.

4. Dissimilar elements arranged together in agroup.

5. No proper place for Lanthanones and Actinones.

Modern Periodic Law -

Physical and chemical properties of elements are a periodic function of their atomic number and classified the elements according to their atomic number.

Long form of Periodic Table (Bohr’s table)

1. It is divided into two categories

a. Vertical columns-groups

b. Horizontal rows - periods

2. There are 18 groups

3. There are 7 periods

The groups are numbered 1 to 18 . The first period consists of two elements. The second and the third contain eight elements each. $4^{\text {th }}$ and $5^{\text {th }}$ periods contain 18 elements each. The sixth period contains 32 elements. The seventh period is incomplete and contains only 27 elements, of which the last five are highly unstable and have been synthetically produced and detected. 14 elements of each of the $6^{\text {th }}$ and $7^{\text {th }}$ periods are shown separately below the main periodic table.

Division of Elements into s, p, d, f blocks

Depending on the particular subshell which is filled up, the 113 elements are divided into $s, p, d$ and $f$ block elements. The $s$ and $p$ block elements, corresponding to filling up of the outermost shell (the valence shell), show the maximum variation in properties and are placed in the main groups of the periodic table (groups 1,2 and 13 to 18) and are called representative elements.

s-block

Elements in which last electron enters the s orbital of their outermost shell. Since there is only one orbital in s block which can accommodate two electrons so there are two groups in s block.

Group 1 (alkali metals), group 2 (alkaline earth metals)

General electronic configuration: $n s^{1-2}, n=2-7$

p-block

Elements in which last electron enters any one of the three p orbitals. Since p subshell has three orbitals which can accommodate six electrons so there are 6 groups in periodic table.

Group 13-18

General electronic configuration : $n s^{2} n p^{1-6}, n=2-7$

group 16 : chalcogens, group 17 : halogens

group 18 : noble gases

Elements of $s$ and $p$ block collectively are called representative elements.

d-block

Elements in which last electron enters any one of the five $d$ orbitals of their respective penultimate shells. Since d subshell has five d orbitals, which can accommodate ten electrons so there are 10 groups in the periodic table.

Group 3-12

General electronic configuration : $(n-1) d^{1-10} n s^{2}, n=4-7$. Since properties of these elements are midway between $s$ and $p$ block. They are called transition elements. An incomplete $d$ subshell in the atom or its stable ion is a pre -Condition for transition elements.

The $\mathrm{f}$ block elements correspond to filling up of the prepenultimate shell, show very little variation in properties and are placed separately as lanthanoids and actinoids outside the periodic table. Lanthanum is not a lanthanoid and actinium is not an actinoid.

General electronic configuration : $(n-2) f^{1-14}(n-1) d^{0-1} n s^{2}, n=6-7$

There are two series in f block

Lanthanoids $\left(\mathrm{Ce} _{58}-\mathrm{Lu} _{71}\right)=4 \mathrm{f}$ orbitals are filled

Actinoids $\left(\mathrm{Th} _{90}-\mathrm{Lr} _{103}\right)=5 \mathrm{f}$ orbitals are filled. They are known as inner transition elements since they form transition series within $\mathrm{d}$ block.

IUPAC nomenclature of elements

Digit root abbreviation
0 nil $\mathrm{n}$
1 un $\mathrm{u}$
2 bi $\mathrm{b}$
3 tri $\mathrm{t}$
4 quad $\mathrm{q}$
5 pent $\mathrm{p}$
6 hex $\mathrm{h}$
7 sept $\mathrm{s}$
8 oct 0
9 enn $\mathrm{e}$

addition of ium at the end

example - 109-un nil ennium

116 - ununhexium

The number of electrons in the valence shell and the inner shells ( $\mathrm{d}$ and $\mathrm{f}$ subshells) help to assign the group number. The principal quantum number of the valence shell specifies the period.

Prediction of Period, Group and Block of a given element

1. Period $=$ Principal quantum number of valence shell

2. Block = Orbital in which last electron enters

3. Group

Block group no.

$\mathrm{S}$

number of valence $e^{-1} s$

p $\quad 12+$ number of valence $e^{-} s$

d number of $e^{-1} s$ in $(n-1) d+$ electron in ns orbitals

Atomic Radius

The size of atoms may be calculated by determining the internuclear distances between two atoms and this depends on their state of bonding.

Types of atomic radii

1. Covalent radius - One half of the distance between the nuclei of two covalently bonded atoms of same element in a molecule (used for non metals).

2. Van der waals radius-One half of the distance between nuclei of two identical atoms.

3. Metallic radius - One half of the internuclear distance between the two adjacent metal ions in the metallic lattice.

Van der waals radius $>$ metallic radius $>$ covalent radius

The size of the cations or anions having the same charge increases with atomic number. The $3+$ cations of lanthanoides ( $Z=58$ to $Z=71)$ however, show a steady decrease with atomic number, due to inefficient shielding by the f subshells (lanthanoid contraction).

Variation of atomic radius

Along a period

size decreases from left to right due to increase in effective nuclear charge.

Size of inert gases are larger than halogens.

In inert gases all orbitals are completely filled, inter electronic repulsions are maximum. Measured in terms of Vanderwaals radius

Down the group

size increases due to addition of a new energy shell at each succeeding element.

Ionic radius

effective distance from the centre of the nucleus of the ion upto which it exerts its influence on its electronic cloud.

Radius of cation is always smaller than parent atom

size of $\mathrm{Mg}^{2+}$ is smaller than $\mathrm{Mg}$ due to increase in effective nuclear charge.

Radius of anion is always larger than parent atom

size of $\mathrm{Cl}^{-}$is larger than $\mathrm{Cl}$ due to decrease in effective nuclear charge.

Variation of ionic radii among isoelectronic species

$$ \mathrm{Al}^{3+}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{F}^{-}<\mathrm{O}^{2-}<\mathrm{N}^{3-} $$

As the effective nuclear charge increases, size decreases

Ionization Enthalpy

Minimum amount of energy required to remove the most loosely bound valence electron from an isolated gaseous atom so as to convert it into gaseous cation.

Successive ionization enthalpies

$$ \mathrm{IE} _{3}>\mathrm{IE} _{2}>\mathrm{IE} _{1} $$

After the removal of one electron from neutral gaseous atom, the electrostatic attraction between the positively charged ion formed and the electrons increases.

Factor affecting lonization enthalpy

1. Nuclear charge-IE increases as the nuclear charge increases.

2. Atomic size-IE decreases as the size increases

3. Electronic configuration - If an atom contains exactly half filled or completely filled orbitals then it has extra stability

$\mathrm{N}: 1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{3}$

$0: 1 s^{2} 2 s^{2} 2 p^{4}$

$\mathrm{N}$ has half filled p orbitals, such an arrangement imparts extra stability to it so removal of electron is difficult.

Noble gases have completely filled orbitals so large amount of energy is required to remove electron from them.

Due to stability of the $n s^{2}$ and $n s^{2} n p^{3}$ configurations, the $1 s$ t ionization enthalpies of group 2 and 15 elements have higher values than expected from the general trend causing an anomaly.

4. Penetration effect of electrons - IE increases as the penetration effect of electrons increases.

$s>p>d>f$

If the penetration effect of the electron is more, it will be closer to the nucleus, hence it is more firmly held by the nucleus.

5. Shielding / Screening effect of electrons - As the shielding effect of inner electrons increases, ionization enthalpy decreases.

In multi electron atoms, electrons in the valence shell experience an attractive force from the nucleus and repulsive force from electrons in the inner shells. As a result attractive force exerted by the nucleus on the valence shell electrons gets reduced by the repulsive force exerted by the inner shell electrons.

$ \begin{array}{clllll} & \text{B} & \text{Al} & \text{Ga} & \text{In} & \text{TI} \\ \text{IE}_{1} & 801 & 577 & 579 & 558 & 589 \\ \end{array} $

(KJ/mol)

Ga and In follow after $d$ block elements, Ti after $d$ and $f$ block elements. Since $d$ and $f$ electrons have poor shielding effect as compared to $s$ and p electrons so $\mathrm{IE} _{1}$ of $\mathrm{Ga}$ is higher than Al.

Solved Examples

Question 1. The correct order of second ionization potential of carbon, nitrogen, oxygen and fluorine is

a) $\mathrm{C}>\mathrm{N}>\mathrm{O}>\mathrm{F}$

b) $\mathrm{O}>\mathrm{N}>\mathrm{F}>\mathrm{C}$

c) $\mathrm{O}>\mathrm{F}>\mathrm{N}>\mathrm{C}$

d) $\mathrm{F}>\mathrm{O}>\mathrm{N}>\mathrm{C}$

Show Answer

Answer. After removal of first electron the configuration of these elements is

$ \begin{array}{cccc} \mathrm{C}^{+} & 1 \mathrm{~s}^{2} & 2 \mathrm{~s}^{2} & 2 \mathrm{p}^{1} \\ \mathrm{O}^{+} & 1 \mathrm{~s}^{2} & 2 \mathrm{~s}^{2} & 2 \mathrm{p}^{3} \\ \mathrm{N}^{+} & 1 \mathrm{~s}^{2} & 2 \mathrm{~s}^{2} & 2 \mathrm{p}^{2} \\ \mathrm{F}^{+} & 1 \mathrm{~s}^{2} & 2 \mathrm{~s}^{2} & 2 \mathrm{p}^{4} \\ \end{array} $

Out of these, 0 has half filled $p$ subshell so to remove e will be most difficult.

0 will have maximum ionization potential then $C$ will leave their readily so as to have empty $p$ subshell thus the potential will be least. Out of $\mathrm{N}$ and $\mathrm{F}$, $\mathrm{F}$ is more electronegative than $\mathrm{N}$ and $\mathrm{N}$ is more electronegative than $\mathrm{C}$ thus their ionization potential will be in the same order. Thus the order will be $\mathrm{C}) \mathrm{O}>\mathrm{F}>\mathrm{N}>\mathrm{C}$

Question 2. The first ionization polential of $\mathrm{Na}, \mathrm{Mg}, \mathrm{Al}$ and $\mathrm{Si}$ are in the order

a) $\mathrm{Na}<\mathrm{Mg}>\mathrm{Al}<\mathrm{Si}$

b) $\mathrm{Na}>\mathrm{Mg}>\mathrm{Al}>\mathrm{Si}$

c) $\mathrm{Na}<\mathrm{Mg}>\mathrm{Al}<\mathrm{Si}$

d) $\mathrm{Na}>\mathrm{Mg}>\mathrm{Al}>\mathrm{Si}$

Show Answer Answer. Across the period the ionization potential increases as the electron enters the same shell so the effective nuclear charge increases thus the order for $\mathrm{Na}, \mathrm{Al}, \mathrm{Si}$ is $\mathrm{Na}<\mathrm{Al}<\mathrm{Si}$ but $\mathrm{Mg}$ has the maximum ionization potential because its configuration is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}$ i.e. completely filled $3 s$ subshell thus more stable have the order is $\mathrm{Na}<\mathrm{Mg}>\mathrm{Al}>\mathrm{Si}$

Question 3. Which electronic configuration of an element has abnormally high difference between second and third ionization energy

a) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$

b) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}$

c) $1 s^{2} 2 s^{2} 2 p^{6}$

d) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}$

Show Answer Answer. $1 s^{2} 2 s^{2} 2 p^{6}$, the both electrons are removed from $2 p$ so not much difference in the 2 potentials as the electrons are removed from the same subshell, same for $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}, 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}$ but for $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}$ the second and third election are removed from different subshells that is $3 s$ and $2 p$ thus the high difference will be there.

Practice Questions

1. The set representing the correct order of ionic radius is

(a) $\mathrm{Li}^{+}>\mathrm{Be}^{2+}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}$

(b) $\mathrm{Na}^{+}>\mathrm{Li}^{+}>\mathrm{Mg}^{2+}>\mathrm{Be}^{2+}$

(c) $\mathrm{Li}^{+}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Be}^{2+}$

(d) $\mathrm{Mg}^{2+}>\mathrm{Be}^{2+}>\mathrm{Li}^{+}>\mathrm{Na}^{+}$

Show Answer Answer: (b)

2. Atomic radii of fluorine and neon in Angstrom units are respectively given by

(a) $0.72,1.60$

(b) $1.60,1.60$

(c) $0.72,0.72$

(d) None of these

Show Answer Answer: (a)

3. Which one of the following is the smallest in size?

(a) $\mathrm{N}^{3}$

(b) $0^{2}$

(c) $\mathrm{F}$

(d) $\mathrm{Na}^{+}$

Show Answer Answer: (d)

4. The correct order of atomic radii is

(a) $\mathrm{N}<\mathrm{Be}<\mathrm{B}$

(b) $\mathrm{F}^{-}<\mathrm{O}^{2-}<\mathrm{N}^{3-}$

(c) $\mathrm{Na}<\mathrm{Li}<\mathrm{K}$

(d) $\mathrm{Fe}^{3+}<\mathrm{Fe}^{2+}<\mathrm{Fe}^{4+}$

Show Answer Answer: (b)

5. The correct sequence which shows decreasing order of the ionic radii of the elements is:

(a) $\mathrm{Na}^{+}>\mathrm{F}^{-}>\mathrm{Mg}^{2+}>\mathrm{O}^{2-}>\mathrm{Al}^{3+}$

(b) $\mathrm{O}^{2-}>\mathrm{F}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}$

(c) $\mathrm{Al}^{3+}>\mathrm{Mg}^{2+}>\mathrm{Na}^{+}>\mathrm{F}>\mathrm{O}^{2-}$

(d) $\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}>\mathrm{O}^{2-}>\mathrm{F}$

Show Answer Answer: (b)

6. Among the elements $\mathrm{Ca}, \mathrm{Mg}, \mathrm{P}$ and $\mathrm{Cl}$, the order of increasing atomic radii is:

(a) $\mathrm{Ca}<\mathrm{Mg}<\mathrm{P}<\mathrm{Cl}$

(b) $\mathrm{Mg}<\mathrm{Ca}<\mathrm{Cl}<\mathrm{P}$

(c) $\mathrm{Cl}<\mathrm{P}<\mathrm{Mg}<\mathrm{Ca}$

(d) $\mathrm{P}<\mathrm{Cl}<\mathrm{Ca}<\mathrm{Mg}$

Show Answer Answer: (c)

7. The correct order of the decreasing ionic radii among the following isoelectronic species are

(a) $\mathrm{Ca}^{2+}>\mathrm{K}^{+}>\mathrm{S}^{2-}>\mathrm{Cl}$

(b) $\mathrm{Cl}^{-}>\mathrm{S}^{2-}>\mathrm{Ca}^{2+}>\mathrm{K}^{+}$

(c) $\mathrm{S}^{2-}>\mathrm{Cl}^{-}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}$

(d) $\mathrm{K}^{+}>\mathrm{Ca}^{2+}>\mathrm{Cl}^{-}>\mathrm{S}^{2-}$

Show Answer Answer: (c)

8. The element with the highest firstionization potential is

(a) boron

(b) carbon

(c) nitrogen

(d) oxygen

Show Answer Answer: (c)

9. The first ionization potential in electron volts of nitrogen and oxygen atoms are respectively given by

(a) $14.6,13.6$

(b) $13.6,14.6$

(c) $13.6,13.6$

(d) $14.6,14.6$

Show Answer Answer: (a)

10. Which of the following represents the correct order of increasing first ionization enthalpy for $\mathrm{Ca}$, $\mathrm{Ba}, \mathrm{S}, \mathrm{Se}$ and $\mathrm{Ar}$ ?

(a) $\mathrm{Ca}<\mathrm{S}<\mathrm{Ba}<\mathrm{Se}<\mathrm{Ar}$

(b) $\mathrm{S}<\mathrm{Se}<\mathrm{Ca}<\mathrm{Ba}<\mathrm{Ar}$

(c) $\mathrm{Ba}<\mathrm{Ca}<\mathrm{Se}<\mathrm{S}<\mathrm{Ar}$

(d) $\mathrm{Ca}<\mathrm{Ba}<\mathrm{S}<\mathrm{Se}<\mathrm{Ar}$

Show Answer Answer: (c)

11. Amongst the following elements (whose electronic configurations are given below), the one having the highest ionization energy is

(a) $[\mathrm{Ne}] 3 \mathrm{~s}^{2}, 3 p^{1}$

(b) $[\mathrm{Ne}] 3 \mathrm{~s}^{2}, 3 \mathrm{p}^{3}$

(c) $[\mathrm{Ne}] 3 \mathrm{~s}^{2} 3 \mathrm{p}^{2}$

(d) $[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^{2} 4 \mathrm{p}^{3}$

Show Answer Answer: (b)

12. Which of the following has the maximum number of unpaired electrons?

(a) $\mathrm{Mg}^{2+}$

(b) $\mathrm{Ti}^{3+}$

(c) $\mathrm{V}^{3+}$

(d) $\mathrm{Fe}^{2+}$

Show Answer Answer: (d)

13. The incorrect statement among the following is

(a) The first ionization potential of Al is less than the first ionization potential of $\mathrm{Mg}$

(b) The second ionization potential of $\mathrm{Mg}$ is greater than the second ionization potential of $\mathrm{Na}$

(c) The first ionization potential of $\mathrm{Na}$ is less than the first ionization potential of $\mathrm{Mg}$

(d) The third ionization potential of $\mathrm{Mg}$ is greater than third ionization potential of $\mathrm{Na}$

Show Answer Answer: (b)

14. The set representing the correct order of first ionization potential is

(a) $\mathrm{K}>\mathrm{Na}>\mathrm{Li}$

(b) $\mathrm{Be}>\mathrm{Mg}>\mathrm{Ca}$

(c) $\mathrm{B}>\mathrm{C}>\mathrm{N}$

(d) $\mathrm{Ge}>\mathrm{Si}>\mathrm{C}$

Show Answer Answer: (b)

15. Which of the following represents the correct order of increasing first ionization enthalpy for $\mathrm{Ca}$, $\mathrm{Ba}, \mathrm{S}, \mathrm{Se}$ and Ar?

(a) $\mathrm{Ca}<\mathrm{S}<\mathrm{Ba}<\mathrm{Se}<\mathrm{Ar}$

(b) $\mathrm{S}<\mathrm{Se}<\mathrm{Ca}<\mathrm{Ba}<\mathrm{Ar}$

(c) $\mathrm{Ba}<\mathrm{Ca}<\mathrm{Se}<\mathrm{S}<\mathrm{Ar}$

(d) $\mathrm{Ca}<\mathrm{Ba}<\mathrm{S}<\mathrm{Se}<\mathrm{Ar}$

Show Answer Answer: (c)

16. The five successive ionization energies of an element are $800,2427,3658,25024$ and $32824 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$ respectively. The number of valence electrons is

(a) 3

(b) 5

(c) 4

(d) 2

Show Answer Answer: (a)

17. The second ionization energies of $\mathrm{Li}, \mathrm{Be}, \mathrm{B}$ and $\mathrm{C}$ are in the order

(a) $\mathrm{Li}>\mathrm{C}>\mathrm{B}>\mathrm{Be}$

(b) $\mathrm{Li}>\mathrm{B}>\mathrm{C}>\mathrm{Be}$

(c) $\mathrm{B}>\mathrm{C}>\mathrm{Be}>\mathrm{Li}$

(d) $\mathrm{Be}>\mathrm{C}>\mathrm{B}>\mathrm{Li}$

Show Answer Answer: (b)

18. Ionisation energies of the following elements increases in the order

(a) $\mathrm{N}<\mathrm{O}<\mathrm{P}<\mathrm{S}$

(b) $0<\mathrm{N}<\mathrm{S}<\mathrm{P}$

(c) $\mathrm{P}<\mathrm{S}<\mathrm{N}<0$

(d) $\mathrm{S}<\mathrm{P}<0<\mathrm{N}$

Show Answer Answer: (d)

PERIODICITY AND CLASSIFICATION OF ELEMENTS - II

Electron Gain Enthalpy

Energy released when a neutral isolated gaseous atom accepts an extra electron to form a gaseous negative ion. $X _{(g)}+\mathrm{e} \longrightarrow \mathrm{X} _{(9)}^{-}$

Greater the amount of energy released, higher is the electron gain enthalpy of the element.

Formation of 0 is exothermic but $0^{2-}$ is endothermic.

$\mathrm{O}(\mathrm{g})+\mathrm{e} \longrightarrow 0^{-}(\mathrm{g}) ; \Delta _{\mathrm{eg}} \mathrm{H}=-141 \mathrm{KJ} \mathrm{mol}^{-1}$

$0(\mathrm{~g})+\mathrm{e} \longrightarrow \mathrm{O}^{2-}(\mathrm{g}) ; \Delta _{\text {eg }} \mathrm{H}=+780 \mathrm{KJ} \mathrm{mol}^{-1}$

After the addition of one electron the atom becomes negatively charged and second electron is to be added to a - vely charged ion. Addition of second electron is opposed by electrostatic repulsion, hence energy has to be supplied for addition of second electron to overcome the strong electrostatic repulsion between the negatively charged $0^{-}$ion and second electron being added.

Down the group

As the size increases, tendency to add the electron decreases hence electron gain enthalpy becomes less-ve.

Electron gain enthalpy of $\mathrm{F}$ is less negative than $\mathrm{CI}$

Due to the small size, electron - electron repulsions in the compact $2 p$ subshell are large, hence, incoming electron is not accepted with the same ease as in $\mathrm{Cl}$.

Halogens have most-ve electron gain enthalpy

Valence shell configuration of halogens is $n s^{2} n p^{5}$, so they require only one more electron to acquire stable noble gas configuration. So, they have a strong tendency to accept an additional electron.

Noble gases have positive electron gain enthalpy

They have completely filled orbitals. Additional electron will be placed in next higher shell. As a result energy has to be supplied to add an additional electron.

Electronegativity

It is the tendency of an atom to attract the shared pair of electrons towards itself. It decreases down the group and increases along a period.

Electron gain enthalpy Electronegativity
1 Tendency of an isolated gaseous atom to
attract an electron.
1 Tendency of an atom to attract shared
pair of electrons in a bond.
2 Can be measured experimentally. 2 Cannot be measured experimentally.
3 Property of an isolated atom. 3 Property of bonded atom.
4 It is constant. 4 It is not constant. It depends on the
hybridization of atom and nature of
other bonded atom.

Bond Formation

Bonds formed between electropositive atoms are metallic, whereas those formed between electronegative atoms are covalent. If there is a difference in electronegativity values of two bonded atoms a polar covalent bond is formed where percent ionic character increases with the increase in the difference in electronegativity. When $\Delta \mathrm{EN}>$ 1.7 , the bond is said to have an ionic character greater than or equal to $50 %$ and is referred to as an ionic bond (a bond between an electropositive and an electronegative atom).

Polarisation of the anion by the cation in an ionic compound leads to the development of a covalent character in ionic bonds. Fajan’s rules for estimating covalent character states that small cations, large anions, large charge on either ion and cations with non-noble gas configuration give rise to the covalent character of ionic bonds.

Diagonal relationship

Some elements of second period show similarities with elements of third period placed diagonally to each other due to same charge / radius ratio.

Anomalous behavior of first element of every group is due to

1. Small size and high electronegativity

$N$ can form $p \pi-p \pi$ multiple bonds whereas $P$ cannot.

2. High IE

They form only covalent compounds and not ionic compounds.

3. Absence of vacant d orbitals

$\mathrm{N}$ cannot form $\mathrm{NCl} _{5}$ or $\mathrm{R} _{3} \mathrm{~N}=0$ since it cannot expand its covalence beyond 4 whereas $\mathrm{P}$ can form $\mathrm{PCl} _{5}$ and $\mathrm{R} _{3} \mathrm{P}=0$

Solved Examples

Question 1. Which of the following processes involves absorption of energy?

a) $\mathrm{Cl}+\mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-}$

b) $0^{-}+\mathrm{e}^{-} \rightarrow 0^{2-}$

c) $0+\mathrm{e}^{-} \rightarrow 0^{-}$

d) $\mathrm{S}+\mathrm{e}^{-} \rightarrow \mathrm{S}^{-}$

Show Answer Answer. b) $\quad 0^{-}+\mathrm{e}^{-} \rightarrow 0^{2-}$

As electron is being added to anion, whereas in all rest of three, electron is being added to a neutral atom which will not have as much repulsion as in case of $0^{-}$

Question 2. Electronic Configuration of most electro negative element is

a) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$

b) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}$

c) $1 s^{2} 2 s^{2} 2 p^{5}$

d) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$

Show Answer

Answer. c) $1 s^{2} 2 s^{2} 2 p^{5}$

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$ has completely filled subshell so will not attract $e^{-}$

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$ will rather leave e- to achieve stable electronic configuration in $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}$ and $1 s^{2} 2 s^{2} 2 p^{5}$ will attract election to achieve completely filled $p$ subshell and $2 p$ will do it more strongly as it is more closer to the nucleus.

Practice Questions

Question 1. The correct order of electron gain enthalpy with negative sign of $\mathrm{F}, \mathrm{Cl}, \mathrm{Br}$ and $\mathrm{I}$, having atomic number $9,17,35$ and 53 respectively is

(a) $\mathrm{I}>\mathrm{Br}>\mathrm{Cl}>\mathrm{F}$

(b) $\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$

(c) $\mathrm{Cl}>\mathrm{F}>\mathrm{Br}>\mathrm{I}$

(d) $\mathrm{Br}>\mathrm{Cl}>\mathrm{I}>\mathrm{F}$

Show Answer Answer: (c)

Question 2. The element with positive electron gain enthalpy is

(a) hydrogen

(b) sodium

(c) oxygen

(d) none

Show Answer Answer: (d)

Question 3. Which of the following species has the highest electron affinity ?

(a) $\mathrm{F}^{-}$

(b) 0

(c) $0^{-}$

(d) $\mathrm{Na}^{+}$

Show Answer Answer: (b)

Question 4. The highest electron affinity is shown by

(a) $0^{-}$

(b) $\mathrm{F}^{-}$

(c) $\mathrm{Cl} _{2}$

(d) $\mathrm{F} _{2}$

Show Answer Answer: (a)

Question 5. Which of the following represents the correct order of increasing electron gain enthalpy with negative sign for the elements $\mathrm{O}, \mathrm{S}, \mathrm{F}$ and $\mathrm{Cl}$ ?

(a) $\mathrm{S}<0<\mathrm{Cl}<\mathrm{F}$

(b) $\mathrm{F}<\mathrm{O}<\mathrm{P}<\mathrm{N}$

(c) $\mathrm{F}>\mathrm{O}>\mathrm{N}>\mathrm{P}$

(d) $\mathrm{N}>0<\mathrm{F}<\mathrm{P}$

Show Answer Answer: (c)

Question 6. The order of decreasing negative electron gain enthalpy of $0, S, S e$ is

(a) $0>\mathrm{S}>\mathrm{Se}$

(b) $\mathrm{S}>0>\mathrm{Se}$

(c) $\mathrm{Se}>0>\mathrm{S}$

(d) $\mathrm{S}>\mathrm{Se}>0$

Show Answer Answer: (d)

Question 7. The correct order of electron gain enthalpy with negative sign of $\mathrm{F}, \mathrm{Cl}, \mathrm{Br}$ and $\mathrm{I}$, having atomic number $9,17,35$ and 53 respectively is

(a) $\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$

(b) $\mathrm{Br}>\mathrm{Cl}>$ I $>\mathrm{F}$

(c) $\mathrm{Cl}>\mathrm{F}>\mathrm{Br}>\mathrm{I}$

(d) I $>\mathrm{Br}>\mathrm{Cl}>\mathrm{F}$

Show Answer Answer: (c)

Question 8. Elements of which of the following groups will form anions most readily?

(a) Oxygen family

(b) Nitrogen family

(c) Halogens

(d) Alkalimetals

Show Answer Answer: (c)

Question 9. Which of the following represents the correct order of increasing electron gain enthalpy with negative sign for the elements $\mathrm{O}, \mathrm{S}, \mathrm{F}$ and $\mathrm{Cl}$ ?

(a) $\mathrm{Cl}<\mathrm{F}<0<\mathrm{S}$

(b) $0<\mathrm{S}<\mathrm{F}<\mathrm{Cl}$

(c) $\mathrm{F}<\mathrm{S}<0<\mathrm{Cl}$

(d) $\mathrm{S}<0<\mathrm{Cl}<\mathrm{F}$

Show Answer Answer: (b)

Question 10. Which one of the following ionic species has the greatest proton affinity to form stable compound?

(a) $\mathrm{NH} _{2}^{-}$

(b) $\mathrm{F}^{-}$

(c) $\mathrm{I}^{-}$

(d) $\mathrm{HS}^{-}$

Show Answer Answer: (a)

Question 11. Electronegativity of the following elements increases in the order

(a) $\mathrm{C}, \mathrm{N}, \mathrm{Si}, \mathrm{P}$

(b) $\mathrm{N}, \mathrm{Si}, \mathrm{C}, \mathrm{P}$

(c) $\mathrm{Si}, \mathrm{P}, \mathrm{C}, \mathrm{N}$

(d) P, Si, N, C

Show Answer Answer: (c)

Question 12. The correct order of electronegativities of $\mathrm{N}, 0, \mathrm{~F}$ and $\mathrm{P}$ is

(a) $\mathrm{F}>\mathrm{N}>\mathrm{P}>0$

(b) $\mathrm{F}>\mathrm{O}>\mathrm{P}>\mathrm{N}$

(c) $\mathrm{F}>\mathrm{O}>\mathrm{N}>\mathrm{P}$

(d) $\mathrm{N}>\mathrm{O}>\mathrm{F}>\mathrm{P}$

Show Answer Answer: (c)

Question 13. Among the elements $\mathrm{B}, \mathrm{Mg}, \mathrm{Al}$ and $\mathrm{K}$, the correct order of increasing metallic character is

(a) $\mathrm{B}<\mathrm{Al}<\mathrm{Mg}<\mathrm{K}$

(b) $\mathrm{B}<\mathrm{Mg}<\mathrm{Al}<\mathrm{K}$

(c) $\mathrm{Mg}<\mathrm{B}<\mathrm{Al}<\mathrm{K}$

(d) $\mathrm{Mg}<\mathrm{Al}<\mathrm{B}<\mathrm{K}$

Show Answer Answer: (a)

Question 14. In which of the following options order of arrangement does not agree with the variation of property indicated against it?

(a) $\mathrm{Al}^{3+}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{F}^{-}$(increasing ionic size)

(b) $\mathrm{B}<\mathrm{C}<\mathrm{N}<\mathrm{O}$ (increasing first ionization enthalpy)

(c) $\mathrm{I}<\mathrm{Br}<\mathrm{F}<\mathrm{Cl}$ (increasing electron gain enthalpy)

(d) $\mathrm{Li}<\mathrm{Na}<\mathrm{K}<\mathrm{Rb}$ (increasing metallic radius)

Show Answer Answer: (b)

Question 15. Which one of the following orders presents the correct sequence of the increasing basic nature of the given oxides?

(a) $\mathrm{Al} _{2} \mathrm{O} _{3}<\mathrm{MgO}<\mathrm{Na} _{2} \mathrm{O}<\mathrm{K} _{2} \mathrm{O}$

(b) $\mathrm{MgO}<\mathrm{K} _{2} \mathrm{O}<\mathrm{Al} _{2} \mathrm{O} _{3}<\mathrm{Na} _{2} \mathrm{O}$

(c) $\mathrm{Na} _{2} \mathrm{O}<\mathrm{K} _{2} \mathrm{O}<\mathrm{MgO} _{2}<\mathrm{Al} _{2} \mathrm{O} _{3}$

(d) $\mathrm{K} _{2} \mathrm{O}<\mathrm{Na} _{2} \mathrm{O}<\mathrm{Al} _{2} \mathrm{O} _{3}<\mathrm{MgO}$

Show Answer Answer: (a)

Question 16. Which of the following sets has strongest tendency to forms anions?

(a) Ga, In, TI

(b) $\mathrm{Na}, \mathrm{Mg}, \mathrm{Al}$

(c) $\mathrm{N}, 0, \mathrm{~F}$

(d) $\mathrm{V}, \mathrm{Cr}, \mathrm{Mn}$

Show Answer Answer: (c)

Question 17. On the basis of the modern theory regarding the structure of the nucleus, it has been predicted that an element with $Z=114$ and $A=298$ will be quite stable. Which of the following statements regarding this element is likely to be true?

(a) It should belong to the $8^{\text {th }}$ period

(b) It should be a transition metal

(c) It should be a metal in group 14

(d) It should be an inert gas

Show Answer Answer: (c)

Question 18. The order of decreasing polarity in the compounds $\mathrm{CaO}, \mathrm{CsF}, \mathrm{KCl}$ and $\mathrm{MgO}$ is

(a) $\mathrm{MgO}>\mathrm{CaO}>\mathrm{KCl}>\mathrm{CsF}$

(b) $\mathrm{MgO}>\mathrm{KCl}>\mathrm{CaO}>\mathrm{CsF}$

(c) $\mathrm{KCl}>\mathrm{CaO}>\mathrm{CsF}>\mathrm{MgO}$

(d) $\mathrm{CsF}>\mathrm{KCl}>\mathrm{CaO}>\mathrm{MgO}$

Show Answer Answer: (d)

Question 19. To which block of the periodic table does the element with atomic number 56 belong?

(a) sblock

(b) pblock

(c) dblock

(d) fblock

Show Answer Answer: (a)

Question 20. The electronic configurations of four elements are given below. Which elements does not belong to the same family as others?

(a) $[\mathrm{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{~d}^{10} 6 \mathrm{~s}^{2}$

(b) $[\mathrm{Kr}] 4 \mathrm{~d}^{10} 5 \mathrm{~s}^{2}$

(c) $[\mathrm{Ne}] 3 \mathrm{~s}^{2} \mathrm{p}^{5}$

(d) $[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^{2}$

Show Answer Answer: (c)

Question 21. The similarity between $\mathrm{Li}$ and $\mathrm{Mg}$ is due to

(a) similar electronic configuration

(b) same charge

(c) same principal quantum number

(d) similar electronegativity

Show Answer Answer: (d)


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