Classification Of Elements And Periodic Properties - Class 11 PDF Download

THE PERIODIC TABLE OF THE ELEMENTS


PERIODIC TABLE
It is a table of elements in which the elements with similar properties are placed together.
This was the classification of elements into groups of three elements each with similar properties such that the atomic weight of the middle element was the arithmetic mean of the other two e.g. Ca, Sr, Ba; Cl, Br, I etc.
The difference between atomic weights of any two successive elements is nearly constant.

NEWLAND’S LAW OF OCTAVES
This was an arrangement of elements in order of increasing atomic weights in which it was observed that every eighth element had properties similar to those of the first just like the eight node of an octave of music.

Prouts Hypothesis: Atomic weight of an element is simple multiple of atomic weight of hydrogen.

MENDELEEV’S PERIODIC TABLE
This is based upon Mendeleev’s periodic law which states that the physical and chemical properties of the elements are a periodic function of their atomic weights.
Mendeleev named Gallium and Germanium elements as eka-aluminium and eka-silicon respectively.

STRUCTURAL FEATURES OF THE MENDELEEV’S PERIODIC TABLE
Mendeleev’s original periodic table consists eight vertical columns are called groups I-VIII & seven horizontal rows are called periods 1-7. But modified Mendeleev’s periodic table contains nine groups, i.e., I-VIII and zero (of noble gases).
All groups except VII and zero have been further divided into two sub-groups called A and B. A group of left hand side consists of normal elements while groups B of right hand side consist of transition elements.
Elements of group IA are called alkali metals while those of group IB (i.e., Cu, Ag, and Au) are called coinage metals.

MODERN PERIODIC LAW
Moseley formed the basis of the modern periodic law. He discovered that the square root of the frequency of the more prominent X-rays emitted by a metal was proportional to the atomic number and not the atomic weight of the atom of the metal. Hence atomic number should be the basis of classification of the elements.
Modern periodic law states, “That the physical and chemical properties of the elements are a periodic function of their atomic number.”
It was observed that the elements with similar properties reoccurred at regular intervals of 2, 8, 8, 18 or 32. These numbers (2, 8, 8, 18 and 32) are called magic numbers, and cause of periodicity in properties.

STRUCTURAL FEATURES OF THE LONG FORM OF THE PERIODIC TABLE
GROUPS - The 18 vertical columns, of the periodic table, are called groups
(1) Elements of groups 1, 2, 13-17 are called normal or representative elements.
(2) Elements of groups 3-12 are called transition elements
(3) The elements belonging to a particular group is known as a family and is usually named after the first element. For example, Boron family (group 13). In addition to this, some groups have typical names. For example,Elements of group 1 are called alkali metals
(4) Elements of group 2 are called alkaline earth metals
(5) Elements of group 3 are called pnicogens
(6) Elements of group 16 are called chalcogens
(7) Elements of group 17 are called halogens
(8) Elements of group 18 are called noble gases

PERIODS - The 7 horizontal columns (or rows) are called periods. The seven periods of periodic table are
(1) Shortest period - 1st period (₁H to ₂He) contains 2 elements. It is the shortest period.
(2) Short periods - 2nd period (₃Li to ₁₀Ne) and 3rd period (₁₁Na  ₁₈Ar) contains 8 elements each. These are short periods.
(3) Long periods - 4th period (₁₉K to ₃₆Kr) and 5th period (₃₇Rb  ₅₄Xe) contain 18 elements each and are called long periods.
(4) Longest period - 6th period (₅₅Cs – ₈₆Rn) contains 32 elements and is the longest period.
(5) Incomplete period - 7th period (₈₇Fr –) is, however, incomplete and contains at present only 26 elements.

BLOCKS - The periodic table is divided into four main blocks (s, p, d and f) depending upon the sub shell to which the valence electron enters into.
S-Block - Elements of group 1 and 2 constitute s-block.
P-Block - Elements of group 13 to 18 constitute p-block.
D-Block - Elements of group 3 to 12 constitute d-block. There are three complete series and one incomplete series of d-block elements. These are: 1st or 3d-transition series which contains ten elements with atomic number 21-30 (₂₁Sc–₃₀Zn).
2nd or 4d-transition series which contains ten elements with atomic number 39–48 (39Y–48Cd).
3rd or 5d-transition series which also contains ten elements with atomic n umber 57 and 72-80 (57La, ₇₂Hf–₈₀Hg).
4th or 6d-transition series which contains only ten elements.

The f-block elements comprise two horizontal rows placed at the bottom of the periodic table to avoid its un-necessary expansion and make the symmetrical nature of periodic table. The two series of f-block elements containing 14 elements each. Lanthanides - The 14 elements from 58Ce–71Lu in which 4f-subshell is being progressively filled up are called lanthanides or rare earth elements. Actinides - Similarly, the 14 elements from 90Th –103Lr in which 5f-subshell is being progressively filled up are called actinides.

Elements of s and p-blocks are called normal or representative elements, those of d-block are called transition elements while the f-block elements are called inner transition elements.

The 11 elements with Z = 93–103 (₉₃Np – ₁₀₃Lr) which occur in the periodic table after uranium and have been prepared by artificial means are called transuranics. These are all radioactive elements.

NOMENCLATURE OF ELEMENTS WITH ATOMIC NUMBER > 100
According to IUPAC the nomenclature can be derived using numerical roots for 0 and numbers 1-9 for atomic numbers of elements.
The roots are put together in order of digits which make the atomic number and ‘ium’ is added at the end. Use the following table

Example: Name the IUPAC name of the element of atomic number 108: Name will be Unniloctium and symbol - UnO
Example – Name the element with atomic number 115.
Name will be - Ununpentium and symbol UuP

DIAGONAL RELATIONSHIP

Certain elements of 2nd period i.e., Li, Be, B. show similarity with their diagonal elements in the 3rd period i.e., Mg, Al, and Si, as shown below:

 This is called diagonal relationship and is due to the reason that these pairs of elements have almost identical ionic radii and polarizing power. (i.e. charge/size ratio).

PERIODIC PROPERTIES

Properties which show a regular gradation when we move from left to right in a period or from top to bottom in a group are called periodic properties. These properties are atomic size, ionization energy electron affinity etc.

ATOMIC SIZE

(1) It refers to the distance between the centers of nucleus of atom to its outermost shell of electrons. The absolute value of atomic radius cannot be determined because
(2) It is not possible to locate the exact position of electrons in an atom as an orbital has no sharp boundaries.
(3) It is not possible to isolate an individual atom.
(4) In a group of atoms, the probability distribution of electrons is influenced by the presence of neighboring atoms.
(5) Since absolute value of atomic size cannot be determined, it is expressed in terms of the operational definitions such as covalent radius, vander Waal’s radius, ionic radius and metallic radius.

Covalent radius : It is half of the distance between the nuclei of two like atoms bonded together by a single covalent bond, hence it is also known as single bond covalent radius (SBCR). Thus, covalent radius.
 r_covalent = d/2
Where d = internuclear distance between two covalently bonded like atoms.

Van-der Waal’s radius: It is one-half of the distance between the nuclei of two adjacent atoms belonging to two neighboring molecules of an element in the solid state.
The covalent radius is always smaller than the corresponding van der Waal’s radius.

Metallic radius: It is half of the distance between two successive nuclei of two adjacent metal atoms in the metallic closed packed crystal lattice. Metallic radius of an element is always greater than its covalent radius.

Ionic radius: It is the effective distance from the nucleus of the ion up to the electrons in the outer shell to which it can influence the ionic bond.
An atom can be changed into a cations (by loss of electron) which is always much smaller than the corresponding atom, or to an anion (by gaining of electrons) which is always larger than the corresponding atom.

FACTORS INFLUENCING COVALENT RADIUS 
Multiplicity of bond: Covalent radii depend on the multiplicity of bonds. e.g., the bond length of C-C bond in methane is 0.77 Å but the length of C=C bond in ethane is 0.67 Å  
Percentage of ionic character: Covalent radius of H in HCl, HBr and HI are different.
Effective nuclear charge: Greater the effective nuclear charge, more tightly is the hold with nucleus and hence smaller the radius.

PERIODIC VARIATION OF ATOMIC RADII
On moving down the group the valence shells become far away from the nucleus and thus the atomic radius increases.
On moving along the period, the effective nuclear charge increases and thus the electron cloud are attracted more strongly towards the nucleus resulting in the contraction of atomic radius.

ISOELECTRONIC IONS OR SPECIES
These are ions of the different elements which have the same number of electrons but different magnitude of the nuclear charge. The size of isoelectronic ions decreases with the increase in the nuclear charge.

IONISATION ENERGY (I.E.)

The amount of energy (work) required to remove an electron from the last orbit of an isolated (free) atom in gaseous state is known as ionization potential or energy or better first ionization potential of the element, i.e.,  
M(g) + I.E → M⁺ (g) + e^-
The amount of energies required to remove the subsequent electrons (2nd, 3rd,) from the monovalent gaseous cations of the element one after the other are collectively called successive ionization energies. These are designated as I.E₁, I.E₂, I.E₃, I.E₄ and so on. It may be noted that. I.E₄ > I.E₃ > I.E₂> I.E₁ (for a particular element)
IE is expressed in eV/atom or kcal mol^–1 or kJ mol^–1
Note that eV atom^–1 = 23.06 kcal mol^–1 = 96.3 kJ mol^–1
In general, the first I.E. increases along the period from left to right. However there are some exceptions to the general trend -
I.E. decreases from elements of group 2 to 3.
I.E. decreases from elements of group 15  to 16.
In a group of the periodic table, the ionization energy decreases from top to bottom.

The factors which affect the ionization energy are:
(1) Atomic Size or Radius: I.E. decreases as the atomic size increases so the attractive force decreases.
(2) Number of electrons in the inner shell (screening effect):  On moving down a group, the number of inner shells increases which increase the screening effect and hence the ionization potential tends to decrease.
(3) Nuclear charge: On moving along the period, effective nuclear charge increases due to addition of electrons in same valence shell and hence ionization energy increases.
(4) Stable configuration: Half filled or completely filled sub shells possess extra stable nature and thus it is more difficult to remove electron and hence more is I.E.
(5) Penetration effect: More penetrating (i.e. more closes) are sub shells of a shell to the nucleus, more tightly the electrons are held towards the nucleus and more is I.E.
I.E.: s > p > d > f for a given shell
Penetration power is directly proportional to the surface area of a sub shell
In second period elements, the correct increasing order of ionization energies is
IE₁: Li < B < Be < C < O < N < F < Ne
IE₂: Be < C < B < N < F < O < Ne < Li

In third period elements, the correct increasing order of ionization is
IE₁: Na < Al < Mg < Si < S < P < Cl < Ar
IE₂: Mg < Si < Al < P < S < Cl < Ar < Na

ELECTRON AFFINITY (EA)
It is the amount of energy released when a gaseous atom accepts the electron to form gaseous anion.

EA values are expressed in eV/atom or kcal/mol or kJ/mol.

The energy change accompanying the addition of first, second, third etc. electrons to neutral isolated gaseous atoms are called successive electron affinities and are designated as EA₁, EA₂, EA₃ etc.
The first EA is always taken as positive. However, the addition of second electron to the negatively charged ion is opposed by columbic repulsion and hence required (absorbed) energy for the addition of second electron. Thus, second electron affinity (EA₂) of an element is taken as negative. For example
(Exothermic)
(Endothermic) 

Electron affinity increases in moving along the period from left to right due to increase in charge. But the values are unexpectedly low in elements of group 2, 15 and 18 due to stable electronic configurations of exactly half-filled and completely filled orbitals.

Within a group of the periodic table the electron affinities decreases from top to bottom.

In general, electron affinity follows the following trend:

Halogens > Oxygen family > Nitrogen family > Metals of groups 1 and 13 and Non-metals of group 14 > Metals of group 2.
The electron affinities are indirectly measured with the help of Born-Haber Cycle, i.e.,

Where, S, D, IE, EA and U are the heat of sublimation, bond dissociation energy, ionization energy, electron affinity and lattice energy respectively.

Electron affinity depends upon:-
Effective nuclear charge: Greater the effective nuclear charge more is the attraction of nucleus towards the electron and hence higher will be the value of E.A.
Atomic size: Greater the atomic radius of the atom less will be the attraction of the nucleus to the electron to be added and hence lower will be the value of E.A.
Penetrating power: Due to penetrating power, E.A. for addition of electron show the order s > p > d > f
Electronic configuration: Half filled and fully filled sub shell are extra stable and thus oppose the addition of electron which leads to lower, E.A. value e.g. EA, of C > EA, of N.

ELECTRONEGATIVITY (EN)
It is the tendency of an atom in a molecule to attract the bonded shared pair of electrons, towards itself
There are several electro negativity scales:-
Mulliken scale: On the Mulliken scale, electro negativity X is taken as average of IE and EA, i.e.,
Where IE and EA are expressed in electron volts

Orwhere IE and EA are expressed in kJ mol^–1

Orwhere IE and EA are expressed in kcal mol^–1
Pauling scale: This is the most widely used scale and is based upon bond energy data. According to Pauling, the difference in electro negativity of two atoms A and B is given by the relationship as

Where X_A and X_B are electro negativities of the atoms A and B respectively while.

Where EA–B, EA–A and EB–B represent bond dissociation energies of the bonds A-B, A-A and B-B respectively. The Pauling and the Mulliken scales are related to each other by the relation,

In a period, EN increases from left to right due to decrease in size and increase in nuclear charge of atoms.
In a group, EN decreases from top to bottom due to increase in atomic size.

Electro negativity depends on:
(1) Atomic size
(2) Nuclear charge
(3) Shielding effect
(4) Oxidation state - EN increases as the positive oxidation state increases.
(5) Hybridization - Greater is the s-character in a hybrid orbital more is electro negativity.
If electro negativity difference is greater than 1.7 the bond is ionic otherwise covalent.
In general, greater is difference of EN between two atoms smaller is the bond length.
Smaller the electro negativity, larger is the atomic size.

VALENCY
(1) Valency of an element is the number of electrons gained or lost or shared with other atoms in the formation of compounds.
(2) Valency of group 1 and 2 elements is equal to the number of electrons in the outermost shell, while that for groups 13 to 14 is group number -minus 10 and that for group 15–18 is 8 -minus the number of electrons in the outermost shell.

ATOMIC VOLUME
(1) It may be defined as the volume occupied by one mole atoms of the element at its melting point in solid state.
(2) It is obtained by dividing the gram atomic mass of the element by its density.
(3) It decreases along the period, reaches a minimum in the middle and then starts increasing, because of different packing arrangement of atoms in different elements in the solid state, i.e., P₄, S₈ etc.
(4) In moving down the group atomic volume goes on increasing gradually

ACID-BASE BEHAVIOUR OF OXIDES AND HYDROXIDES 
The oxides or hydroxide of an element may act either as base or an acid depending upon its ionization energy.
If the IE is low, it acts as a base and if the IE is high, it acts as an acid.
The IE of alkali metals is the lowest; therefore, their oxides and hydroxides are the strongest bases. The basic character of their hydroxides increases in the order:
CsOH > RbOH > KOH > NaOH > LiOH
The IE of halogens is quite high; therefore, their oxides are the strongest acids. The acidic character of their oxides and hydroxides decreases in the order:
HClO₄ > HBrO₄ > HIO₄
Within a period, the ionization energies of the elements usually increase and hence their oxides and hydroxides show a gradual variation from strongly basic through amphoteric to strongly acidic character.
The non-metallic character, oxidizing character and acidic nature of oxides of the elements increases from left to right in a period and decrease from top to bottom in a group. The stability of the metal increases and activity decreases from left to right in a period whereas the stability decreases and activity increases down the group.

SUBJECTIVE QUESTIONS (based on NCERT)

Q.1. What would be the IUPAC name and symbol for the element with atomic number 120?
Ans. The root words for 1, 2 and 0 are UN, bi and nil respectively. Hence, the symbol and the name respectively are Ubn and unbinilium.

Q.2. How would you justify the presence of 18 elements in the 5^th period of the periodic table?
Ans. When n = 5, l = 0,1,2,3. The order in which the energy of the available energy of the available orbital’s 4d, 5s and 5p increases is 5s < 4d < 5p. The total numbers of orbitals avialiable are 9. The maximum number of electrons that can be accommodated is 18 and therefore 18 elements are there in 5^th orbit.

Q.3. The elements z = 117 and 120 have not yet been discovered. In which family/group would you place these elements and also given the electronic configurations in each case.
Ans. The element with z = 117 has electronic configuration [Rn]5f146d107s27p5 so it will belongs to the halogen family (group 17) and the element with z = 120 has electronic configuration, [Uuo]8s2 so it belongs to group 2 (alkaline earth metals).

Q.4. Arrange the following species in the order of their increasing atomic radius, Mg, Mg²⁺, Al, Al³⁺
Ans. We know that cations are smaller than their parent atoms and atomic radii decreases across a period. So the order of size will be, Al³⁺ < Al < Mg²⁺ < Mg

Q.5. Predict the formulas formed by the following pairs 
(a) Silicon and bromine
(b) Aluminum and sulphur
Ans.
(a) Si has a valence 4 and Br has valence 1. So, the compound formed would be SiBr₄
(b) Al has a valence 3 and S has valence 2. So, the compound formed would be Al₂Br₃

Q.6. Among the second period elements the actual ionization enthalpies are in the order Li < B < Be <C <O < N < F < Ne. Explain why,
(a) Be has higher first ionization enthalpy than B
(b) O has lesser first ionization enthalpy than N and F
Ans. When we consider the same principle quantum number level, an s-electron is attracted to the nucleus more than a p-electron. In Be the electron removed is an s-electron whereas in B it is a p-electron. The 2p electron is more shielded from the nucleus making it easier to remove and thus B has lesser first ionization potential than Be. Also, in oxygen there are more electron-electron repulsions because two electrons reside in same orbital that in nitrogen where all electrons reside in different orbitals. So, it is easier to remove an electron from oxygen and thus it has lesser first ionization enthalpy than nitrogen.

Q.7. Which of the following pairs of elements would have a more negative electron gain enthalpy?
(a) O (or) F
(b) Cl (or) F
Ans. Electron gain enthalpy becomes more negative with increase in the atomic number across a period. So, fluorine has more negative value than oxygen. Down the period the value of enthalpy must decrease. So, fluorine must have more negative than chlorine. But it is observed that chlorine has the most negative value. The reason is that for O (or) F the added electron goes to n = 2 quantum level but for S (or) Cl the electron enters n = 3 quantum level. For n = 3 the added electron occupies a larger region of space and the electron – electron repulsion is much less.

Q.8. Consider the following species: N^3-, O^2-, F^-, Na⁺, Mg²⁺, Al³⁺
(a) What is common in them?
(b) Arrange them in order of increasing ionic radius.
Ans. All the given species are isoelectronic since they have same number of electrons (10). Since all of them have same number of electrons more is the nuclear charge smaller is the radius. So, the order of ionic radius would be N^3- > O^2- > F^- > Na⁺ > Mg²⁺ > Al³⁺

Q.9. Arrange the elements N,P,O and S in the order of
(a) increasing first ionisation enthalpy
(b) increasing non metallic character.
Ans.
(a) As discussed above, nitrogen has higher fist ionisation enthalpy than oxygen and phosphorous has lesser ionisation enthalpy than sulphur. Also, ionisation enthalpy decreases on going down a group. So the order will be N > O > S > P
(b) Order of non-metallic character : O > N > S > P

Q.10. The order of screening effect of electrons of s, p, d and f orbital’s of a given shell of an atom on its outer shell electrons is?
Ans. We know that an s-electron is more shielded than a p and d electron. So, the order of screening effect is s > p > d > f.

Q.11. The first (IE1) and the second (IE2) ionisation enthalpies (KJ mol-1) of few elements designed by Roman numerals are shown below:

Which of the above elements is likely to be?
(a) a reactive metal
(b) a reactive non metal
(c) a noble gases
(d) a metal that forms a stable binary halide of the formulae AX₂(X = halogen) 

Ans. The most reactive metal will be II since it has the least first ionisation enthalpy. The most reactive non-metal will be IV. I will be the noble gas since it has highest first ionisation enthalpy. Element III forms stable A²⁺ ion since it has considerably less first and second ionisation enthalpy.

STRAIGHT OBJECTIVE QUESTIONS 
Q.1. 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
Ans. (d)
Solution.
Non-metallic character increases from left to right across a period and decreases down the group.

Q.2. 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
Ans. (d)

Metallic character decreases from left to right across a period and increases down the group 

Q.3. Considering the elements Cl, N, F and O the correct order of their chemical reactivity in terms of oxidizing property is, 
(a) F > Cl > O > N 
(b) F > O > Cl > N
(c) Cl > F > O > N
(d) O > F > N > Cl
Ans. (b)
Solution.
More is the electronegativity of an element more will be its oxidizing power. Fluorine being the most electronegative atom has the most oxidizing character followed by oxygen, chlorine and nitrogen.

Q.4. The statement that is not correct for periodic classification of element is
(a) The properties of element are periodic function of their atomic numbers
(b) Non-metallic elements are less in number than metallic elements 
(c) For transition elements, the 3d-orbitals are filled with electrons after 3p-orbitals and before 4p-orbitals
(d) The first ionization enthalpies of elements generally increase with increase of atomic number as we go along a period
Ans. (c)
Solution.

For transition elements, the 3d-orbitals are filled with electrons after 4s-orbitals and before 4p-orbitals.

Q.5. Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy) is
(a) F > CI > Br > I
(b) F < CI < Br < I
(c) F < CI > Br > I
(d) F < CI > Br< I
Ans. (C)
Solution.
Electron gain enthalpy is expected to decrease down the group. So, fluorine is expected to have a more negative enthalpy, but chlorine has the most negative electron enthalpy (as discussed earlier)

Q.6. The period in the long form of the periodic table is equal to
(a) magnetic quantum number of any element of the period
(b) Atomic number of any element of the period
(c) Maximum principal quantum number of any element of the period
(d) Maximum azimuthal quantum number of any element of the period
Ans. (c)
Solution.

An element goes to the period whose number is equal to the highest principle quantum number (n) of the element in its electronic configuration.

Q.7. The elements in which electron are progressively filled in 4f-orbital are called 
(a) Actinoids 
(b) Transition elements 
(c) Lanthanides
(d) Halogens

Ans. (c)
Solution.
Lanthanides are the elements in which electrons are progressively filled in the 4f-orbital. In Actinides’ electrons are filled in 5-f orbital.

Q.8. Which of the following is the correct order of size of the given species 
(a) I > I- > I+  
(b) I+ > I- > I
(c) I > I+ > I-
(d) I- > I > I+
Ans. (d)
Solution.
Cations are smaller than the parent atom and the anions are larger than the parent atom. The reason is due to the nuclear charge which will be more in cations and lesser in anions.

Q.9. Consider the isoelectronic species Na⁺, Mg²⁺, F^- and O^2-. The correct order of increasing length of their radii is
(a) F^- < O^2-< Mg²⁺ < Na⁺
(b) Mg²⁺ < Na⁺ < F- < O^2-
(c) O^2-< F- < Na⁺ < Mg²⁺
(d) O^2-< F^- < Mg²⁺ < Na⁺
Ans. (b)
Solution.
The given species are isoelectronic as they all have same number of electrons (10). So, more is the nuclear charge (or more is the atomic number) smaller is the radius of the ion.

Q.12. Which of the following has the least lattice energy?
(a) Li₂CO₃
(b) Na₂CO₃
(c) K₂CO₃
(d) C_sCO₃
Ans. (c)
K₂CO₃ has the least lattice energy among all the carbonates of the group I elements.

Q.13. Order of the ionisation energy is
(a) F₂ > O₂ > N₂
(b) F₂ > N₂ > O₂
(c) F₂ > N₂ ≈ O₂
(d) F₂ > N₂ >> O₂
Ans. (b)
Solution.
Fluorine has the highest ionisation energy among the given species. Oxygen is expected to have the next highest but it is observed that nitrogen has more ionisation enthalpy than oxygen (as discussed earlier).

Q.14. Order of the ionisation enthalpy is
(a) B > Al > Ga > In > Tl
(b) B > Ga > Al > In > Tl
(c) B > Al > In > Ga > Tl
(d) B > Al > Ga > Tl > In
Ans. (d)
Solution.
Ionisation enthalpy decreases down the group.

Q.15. 3 and 6 electrons are present in the outermost orbit of A and B respectively. The chemical formula of the compound formed when these two react is
(A) A₃B₂
(B) A₂B₃
(C) A₂B
(D) AB

Ans. (b)
Solution.

A can lose 3 electrons to form a stable A³⁺ cation and B can gain 2 electrons to form a stable B^2- anion. So, the compound formed when these two react is A₂B₃.

MULTIPLE CORRECT ANSWER TYPE QUESTIONS 
Q.1. Which of the following elements can show covalency greater than 4?
(a) Be
(b) P
(c) S
(d) B
Ans. (b),(c)
Solution.
Phosphorus has a maximum covalency of 5 and sulphur has a maximum covalency of 6 because any element can have a maximum covalency equal to its group number.

Q.2. Which of the following sequences contain atomic numbers of only representative elements
(a) 3,33,53,87
(b) 2,10,22,36
(c) 7,17,25,37,48
(d) 9,35,51,88

Ans. (a),(d)
Solution.
Representative elements are elements present on the extremes of the periodic table.

Q.3. which of the following elements all gain one electron more readily in comparison to other elements of their group
(a) S(g)
(b) F(g)
(c) O(g)
(d) CI(g)
Ans. (a),(d)
Solution.
Sulphur and chlorine have higher electron gain enthalpy in their respective group although oxygen and fluorine are expected to have a higher enthalpy

Q.4. Which of the following statements are correct

(a) Helium has the highest first ionisation enthalpy in the periodic table 
(b) Chlorine has the least negative electron gain enthalpy than fluorine
(c) Mercury and bromine are liquids at room temperature
(d) In any period, atomic radius of alkali metal is the highest 
Ans. (a),(c),(d)
Solution.
Helium has the highest first ionisation enthalpy in the periodic table because of its octet configuration, high nuclear charge and least atomic size. Chlorine has more negative electron gain enthalpy than fluorine. Mercury and bromine are liquids at room temperature. In any period atomic radius of alkali metals is the highest due to the least nuclear charge in their respective period.

Q.5. Which of the following sets contain only isoelectronic ions?
(a) Zn²⁺, Ca²⁺, Ga³⁺, Al³⁺  
(b) K⁺, Ca²⁺, Sc³⁺, Cl^- 
(c) P^3-, S^2-, Cl^-, K⁺  
(d) Ti⁴⁺, Ar, Cr³⁺, V⁵⁺
Ans. (b),(c)
Solution.
Isoelectronic species are the one which have same number of electrons in them. (K⁺, Ca²⁺, Sc³⁺, Cl^-) and (P^3-, S^2-, Cl^-, K⁺) have 18 electrons each

Q.6. Ionic radii vary in 
(a) inverse proportion to the effective nuclear charge
(b) inverse proportion to the square of effective nuclear charge 
(c) direct proportion to the screening effect
(d) direct proportion to the square of screening effect 
Ans. (a),(c)
Solution.
Ionic radius is directly proportional to the screening effect and inversely proportional to the nuclear charge. More is the screening effect more is the radius and vice-versa. Similarly more is the nuclear charge less is the radius and vice-versa. Both of them come into play while determining the atomic radius of a particular element.

Q.7. In which of the following options order of arrangement does not agree with the variation of property indicated against it?
(a) Al³⁺ < Mg²⁺ < Na⁺ < F^- (increasing ionic size)
(b) B < C < N < O (increasing first ionisation enthalpy)
(c) I < Br < Cl < F (increasing electron gain enthalpy)
(d) Li < Na < K < Rb (increasing metallic radius)
Ans. (b),(c)
Solution.
B < C < O < N (increasing order of first ionisation enthalpy)
I < Br < F < Cl (increasing order of electron gain enthalpy)

Q.8. The order of reactivity is
(a) Cs > Rb > K > Na > Li
(b) F₂ > Cl₂ > Br₂ > I₂ 
(c) I₂ > Br₂ > Cl₂ > F₂
(d) O₂ > S > Se > Te 
Ans. (a),(b),(d)
Solution.
Reactivity increases down the group.

Q.9. Which is correct
(a) O^2- > F^- > Na⁺ (Ionic radii)
(b) Na > Mg > Al (Atomic radii)
(c) Cl₂ > F₂ > Br₂ (Electron enthalpy)
(d) F₂ > Cl₂ > Br₂ (Electronegativity)
Ans. (a),(b),(c),(d)
Solution.
For isoelectronic species more is the atomic number less is their radius. Across a period atomic radius decreases with increase in atomic number. Among the halides chlorine has the most negative electron affinity followed by fluorine and bromine and fluorine is the most electronegative halogen followed by chlorine and bromine.

MATIX MATCHING QUESTIONS
Q.1. Match the correct atomic radius with the element

 Ans. (A)-R,(B)-S,(C)-T,(D)-Q,(E)-P
Solution.
Across a period atomic radius decreases with increase in atomic number. So, Be should have the highest atomic radius (111 pm) and oxygen should have the least radius (66 pm).

NUMERICAL VALUE QUESTIONS

Q.1. Among the following, the number of elements showing only one non zero oxidation state is O, Cl, F, N, P, Sn, Tl, Na, Ti
Ans. 2
Solution.

Fluorine is the strongest oxidizing agent so it can show only one oxidizing state (-1). Similarly sodium can show only one oxidation state (+1) since it belongs to group I elements.

Q.2. I.E and E.A of an element are 13 eV and 3.8 eV respectively. If the electro negative of the element in Pauling’s scale is En then find [EN] (where [.] is greatest integer function)
Ans. 3
Solution.
EN = 0.187(IE + EA) + 0.17 (in Pauling’s scale)
EN = (0.187) (13.6 + 3.8) + 0.17
= 3.34
[EN] = 3

Q.3. Find out the sum of all the digits of atomic number of the element “Unq” (if your answer is greater than 9 then add all he digits
Ans. 6
Solution.

U (uni) corresponds to 1 and q (quad) corresponds to 4. So, the atomic number of element with symbol “Uuq” is 114 and 1 + 1 + 4 = 6

Q.4. From the given compounds if X number of compounds are acidic in water 
CaO, SO₂, SO₃, Fe₂O₃, Cl₂O₇, CO₂, Na₂O
Find the X
Ans. 4

Among the given oxides the acidic ones are SO₂, SO₃, Cl₂O₇ and CO₂. Fe₂O₃ is amphoteric and the rest are basic.

Q.5. Find the total number of cations for which I.P of cation is lower than corresponding atom 
K⁺, O⁺, Ne⁺, P⁺, Be⁺, Na2⁺, Fe⁺, Sn⁺
Ans. 0
Solution.
In cations the nuclear charge is always greater than in the parent atom. So, no cation can have a lesser IP value than its parent atom.

Q.6. Count the number of properties, which have higher values for ‘O’ atom than that of ‘S’
(a) Ionisation potential
(b) Electron affinity
(c) Covalent radius
(d) Electronnegativity
(e) (Proton: electron) ratio 

Ans. 2
Solution.

Oxygen has greater Ionisation potential and Electronegativity when compared to sulphur. On the other hand sulphur has greater Electron affinity and Covalent radius than oxygen. The (proton : electron) ratio is same for both atoms.

Q.7. How many of the following radius order are correct
(a) N^3- > P^3-
(b) O^2- > F^-
(c) Ca²⁺ > Sr²⁺
(d) S^- > S^2-
(e) S^2- > O^2-

Ans. 3
Solution.

N^3- < P^3- and S^- < S^2- because of more nuclear charge in the first species. Ca²⁺ < Sr²⁺ since ionic radius increases down the group.

PREVIOUS YEAR QUESTIONS

Q.1. According to the Periodic Law of elements, the variation in their properties is related to their    (AIEEE 20)
(a) Nuclear mass
(b) Atomic mass
(c) Atomic Number
(d) Neutron Proton ratio
Ans. (3)
Solution.
According to modern periodic law the variation of elements in their properties is related to their Atomic Number.

Q.2. Which one of the following is an amphoteric oxide
(a) Na₂O

(b) ZnO
(c) B2O3
(d) SO2
Ans. (b)
Solution.
The acidic nature of oxides increase from left to right across a period. The leftmost elements form basic oxides, the rightmost elements form acidic oxides and the middle group elements form amphoteric oxides. Here ZnO is amphoteric, Na₂O and B₂O₃ are basic, SO₂ is acidic.

Q.3. Which one of the following ions has the highest value of ionic radius    (AIEEE 2004)
(a) O_2-
(b) B³⁺
(c) Li⁺
(d) F^-
Ans. (a)
Solution.
Here O₂^- , F- is isoelectronic and Li⁺, B³⁺ are isoelectronic. In the first pair O_2- has larger radius and out of the second pair Li⁺ has greater radius. The final order is A > D > C > B

Q.4. Among Al₂O₃, SiO₂, P₂O₃, SO₂ the correct order of acidic strength is, 
(a) Al₂O₃ < SiO₂ < SO₂ P₂O₃ 
(b) SiO₂ < SO₂ < Al₂O₃ < P₂O₃ 
(c) SO₂ < SiO₂ < P₂O₃ < Al₂O₃
(d) Al₂O₃ < SiO₂ < P₂O₃ < SO₂ 
Ans. (d)
Solution.
Acidic nature increases on going from left to right across a period.

Q.5. Which of the following is expected to have a jump in second ionisation enthalpy (atomic number of that particular element is given inside the brackets    (AIEEE 2003)
(a) Manganese (25)
(b) Iron (26)
(c) Vanadium (23)
(d) Chromium (24)
Ans. (D)
Solution.
Chromium has an exceptional electronic configuration of [Ar]4s¹3d⁵. So, one removing an electron it will have a half filled configuration which is very much stable so its second ionisation enthalpy is very high.

Q.6. Which of the following is not correct according to the property indicated against it (AIEEE 2005) 
(a) Increasing size: Al³⁺< Mg²⁺ < Na⁺ < F^-
(b) Increasing IE₁: B < C < N < O
(c) Increasing EA₁: I < Br < F < Cl
(d) Increasing metallic radius: Li < Na < K < Rb 

Ans. (b)
Solution.
Correct order of IE₁: B < C < O < N

Q.7. In which of the following the hydrogen bond is the strongest (AIEEE 2006)
(a) O-H-N
(b) F-H-F
(c) O-H-O
(d) O-H-F
Ans. (b)
Solution.
More is the Electronegativity of the element bonded to hydrogen more will be the extent of hydrogen bonding. As fluorine is the most electronegative atom (B) will have the highest extent of hydrogen bonding.

Q.8. The ionic mobility of alkali metal ions in aqueous solution is maximum for (AIEEE 2006)
(a) Na⁺
(b) K⁺
(c) Rb⁺
(d) Li⁺
Ans. (c)
Solution.
Ionic mobility increases down the group.

Q.9. Lanthanide contraction is caused due to    (AIEEE 2006) 
(a) The imperfect shielding on outer electrons by 4-f electrons from the nuclear charge 
(b) The appreciable shielding on outer electrons by the 5d-elctrons from the nuclear charge 
(c) The appreciable shielding on outer electrons by the 4f-elctrons from the nuclear charge
(d) The same effective nuclear charge from Ce to Lu
Ans. (A)
Solution.
It is because of the poor shielding effect of 4f-electrons, the outer electrons are more drawn inwards due to the nuclear charge and thus decreasing the atomic size.

Q.10. The ionic radius (in Armstrong) of N^3-, O^2-, F^- are respectively    (JEE Main 2015) 
(a) 1.35, 1.40 and 1.71
(b) 1.36, 1.71 and 1.40
(c) 1.71, 1.40 and 1.35
(d) 1.71, 1.35 and 1.40
Ans. (c)

The given species are isoelectronic so more is the atomic number lesser is the atomic radius.

Q.11. The hydration energy of Mg²⁺ is larger than that of     (IITJEE 1984) 
(a) Al³⁺
(b) Na⁺
(c) Be²⁺
(d) Mg³⁺
Ans. (b)
Solution.
Hydration enthalpy decreases down the group and increases left to right across a period. It also increases with increase in oxidation state for a particular element. So, only Na⁺ has lesser hydration enthalpy than Mg²⁺.

Q.12. Among the following elements which is expected to be the more stable one (IITJEE 1990)
(a) [Ne]3s²3p¹ 
(b) [Ne]3s²3p²
(c) [Ne]3s²3p³
(d) [Ne]3s²3p⁴
Ans. (C)
Solution.
Half filled and full filled configurations are much stable.

Q.13. Sodium sulphate is soluble in water whereas Barium sulphate is sparingly soluble because   (IITJEE 1989) 
(a) The hydration of sodium sulphate is more than its lattice energy
(b) The lattice energy of barium sulphate is more than its hydration energy
(c) The hydration of barium sulphate is less than its lattice energy
(d) The hydration of sodium sulphate is less than its lattice energy
Ans. (a),(b)
Solution.
The criteria for solubility are less lattice energy and more hydration energy. Sodium sulphate is soluble in water because of its higher hydration energy over lattice energy but barium sulphate is sparingly soluble in water because of its higher lattice energy over hydration energy.

Q.14. On Mulliken scale the average of ionisation potential and electron affinity is known as ___________.    (IITJEE 1985)
Ans. Electronegativity

Q.15. Compounds that formally contain Pb4+ are easily reduced to Pb2+. The stability of this lower oxidation is due to _________.    (IITJEE 1997)
Ans. Inert Pair Effect
Solution.
Inert pair effect is defined as “The non-participation of the two s electrons in bonding due to the high energy needed for unpairing them”. The inert pair theory was proposed by Sedgwick. He along with Powell accounted for the shapes of several molecules and correlated the shapes with some of their physical properties.
The inert pair effect among group 4 and group 5 elements. Sn2+ and Pb2+ and Sb3+ and B3+which are the lower oxidation states of the elements are formed because of the inert pair effect. When the s electrons remain paired the oxidation state is lower than the characteristic oxidation state of the group.

Q.16. Difference between Inert Pair Effect and Shielding Effect
Ans.