CBSE Class 10 Science Periodic Classification Of Elements

Periodic classification of elements is a Class 10 Science chapter (Chapter 5) that explains how scientists arranged elements, from Dobereiner's triads and Newlands' law of octaves to Mendeleev's periodic table and the modern periodic table based on atomic number. These notes cover the merits and limitations of each classification, the groups and periods of the modern periodic table, the trends in properties, the mind map, the periodic table, a crossword and the chapter question bank with answers.

Topics Covered

1. Dobereiner's Triads
2. Newlands' Law of Octaves
3. Mendeleev's Periodic Classification
4. Modern Periodic Classification
5. Groups and Periods in the Modern Periodic Table
6. Trends in the Modern Periodic Table
7. Diagrams and Crossword
8. Question Bank

DOBEREINER'S TRIADS

Dobereiner observed that when elements were arranged into groups of three in the order of their increasing atomic masses, the atomic mass of the middle element was the arithmetic mean of the other two.

Limitation
It could be applied only to a limited number of elements.

NEWLANDS' LAW OF OCTAVES

Newlands found that when elements were arranged in increasing order of their atomic masses, every eighth element had properties similar to the first element, like the eighth note of an octave in music.

Limitations
It could be valid only up to calcium.
Newlands assumed that only 56 elements existed in nature and that no more elements would be discovered.

MENDELEEV'S PERIODIC CLASSIFICATION

Mendeleev's Periodic Law states that the properties of elements are a periodic function of their atomic masses.

Merits of Mendeleev's Periodic Table
Mendeleev left some blank spaces for undiscovered elements.
Mendeleev predicted the discovery of some elements and named them eka-boron, eka-aluminium and eka-silicon.
Noble gases discovered later could be placed without disturbing the existing order.

Limitations of Mendeleev's Periodic Table
Position of hydrogen: A correct position could not be assigned to hydrogen, as hydrogen resembles alkali metals as well as halogens.
Position of isotopes: Isotopes are placed in the same position though they have different atomic masses.
Separation of chemically similar elements while dissimilar elements are placed in the same group.

MODERN PERIODIC CLASSIFICATION

The Modern Periodic Law states that the properties of elements are a periodic function of their atomic numbers.

Groups in the Modern Periodic Table

Periods in the Modern Periodic Table

TRENDS IN THE MODERN PERIODIC TABLE

DIAGRAMS

Mind Map: Periodic Classification of Elements

[Figure: Mind map titled MIND MAP with the heading Periodic Classification of Elements. Groups (Vertical Columns) are shown on the left side with Metals and Atomic Size increases going down the group. Periods (Horizontal Columns) are shown in the middle with Non-metallic and Metallic regions, Atomic size decreases slightly across the period, Electropositive (giving electrons) and Electronegative (taking electrons). On the right side Non-metals and Noble Gases - Zero Valency are shown. A blue box reads: Properties of metals are periodic functions of their atomic numbers]

Dobereiner's Triads

Newlands' Octaves

Modern Periodic Table (Table 5.6)

[Figure: Table 5.6 Modern Periodic Table, with GROUP NUMBER 1 to 18 across the top and PERIOD NUMBER 1 to 7 down the left side. Groups 3 to 12 are the block of transition elements in the middle. Elements are shaded as Metals (yellow-green), Metalloids (pink) and Nonmetals (light yellow). A zig-zag line separates metals from nonmetals. Two rows are shown below the table: *Lanthanoids (elements 58 to 71) and **Actinoids (elements 90 to 103)]

CROSSWORD

Across
5. Vertical columns in the periodic table
6. Element with atomic number 6
7. Metalloid in Group 13
8. Variation of atomic size along the group
9. Variation of electronegativity along the group

Down
1. Elements that are electropositive in nature
2. Other name for inert gases
3. Horizontal rows in the periodic table
4. Non-metal that exists in liquid state at room temperature
10. Number of electrons in L shell

Question Bank: Periodic Classification of Elements

Very Short Answer Type Questions (1 mark)

Q1. Give an example of Dobereiner's triad. [1 Mark]

Answer: Lithium (Li), sodium (Na) and potassium (K) form a Dobereiner's triad.

Teacher's Note:
a) The atomic masses of Li, Na and K are 7, 23 and 39.

Q2. What is the basis of Mendeleev's periodic table? [1 Mark]

Answer: The basis of Mendeleev's periodic table is the atomic mass of the elements.

Teacher's Note:
a) Mendeleev arranged elements in increasing order of atomic mass and found that their properties repeat at regular intervals.

Short Answer Type Questions (2 marks)

Q1. State the modern periodic law for classification of elements. How many groups and periods are there in the modern periodic table? [2 Marks]

Answer:
1. Modern periodic law: The properties of elements are a periodic function of their atomic numbers.
2. The modern periodic table has 18 groups and 7 periods.

Teacher's Note:
a) The modern law uses atomic number, not atomic mass, as the basis of classification.
b) Groups run from 1 to 18 and periods from 1 to 7; students often mix up the two counts.

Q2. An element 'M' has atomic number 11. [2 Marks]

(i) Write its electronic configuration.

Answer: 2, 8, 1

Teacher's Note:
a) Electrons fill the shells in the order K, L, M, and the atomic number 11 gives 11 electrons in all.
b) The last shell holds only 1 electron, which tells you the valency of M.

(ii) State the group to which 'M' belongs.

Answer: Group 1 (alkali metals), in period 3.

Teacher's Note:
a) The number of valence electrons equals the group number for elements of groups 1 and 2.
b) Three shells means the element is in period 3.

(iii) Is 'M' a metal or a non-metal?

Answer: M is a metal.

Teacher's Note:
a) Metals have 1 to 3 valence electrons, which they lose easily to form positive ions.

(iv) Write the formula of its chloride.

Answer: MCl

Teacher's Note:
a) M has a valency of 1, and chlorine has a valency of 1, so one atom of each combines.
b) Element M is sodium (Na), so its chloride is sodium chloride, NaCl.

Q3. Name two elements that show chemical properties similar to bromine. Give reason. [2 Marks]

Answer:
1. Chlorine (Cl) and iodine (I) show chemical properties similar to bromine.
2. Bromine, chlorine and iodine belong to group 17 (halogens) and each has 7 valence electrons, so elements of the same group show similar chemical properties.

Teacher's Note:
a) Similar chemical properties come from the same number of valence electrons.
b) Fluorine is also a halogen, so it could be given as an answer as well.

Q4. An atom has electronic configuration 2, 8,2. [2 Marks]

(i) What is the atomic no. of this element?

Answer: 12

Teacher's Note:
a) Atomic number equals the total number of electrons, which is 2 + 8 + 2 = 12.
b) The element with atomic number 12 is magnesium (Mg).

(ii) Is it a metal or non-metal?

Answer: It is a metal.

Teacher's Note:
a) The atom has only 2 valence electrons, which it loses easily.

Short Answer Type Questions (3 marks)

Q1. The elements Li (Z = 3), Na (Z = 11) and K (Z = 19) belong to group 1. [3 Marks]

(i) Predict the periods they belong.

Answer:
1. Li has 2 shells (2, 1), so it belongs to period 2.
2. Na has 3 shells (2, 8, 1), so it belongs to period 3.
3. K has 4 shells (2, 8, 8, 1), so it belongs to period 4.

Teacher's Note:
a) The period number equals the number of shells an atom has.
b) Electron configurations are easy to build from the atomic number, and that makes the period obvious.
c) Students often count the valence electrons instead of the shells, which gives the group number, not the period.

(ii) Which one of them is least reactive?

Answer: Li (lithium) is the least reactive.

Teacher's Note:
a) Reactivity of alkali metals increases down the group.
b) Lithium is at the top of group 1, where the outer electron is held most strongly.
c) Its single valence electron sits closest to the nucleus.

(iii) Which one of them has the largest atomic radius? Give reason to justify.

Answer:
1. K (potassium) has the largest atomic radius.
2. Atomic radius increases down a group because each element has one more shell than the element above it.
3. K has 4 shells, the most among the three elements.

Teacher's Note:
a) More shells mean the outer electron lies farther from the nucleus.
b) Inner shells shield the outer electrons from the nucleus, which further increases the size.

Q2. F, Cl and Br are the elements each having seven valence electrons. [3 Marks]
Pick the element (i) with the largest atomic radius (ii) which is most reactive. Justify your answer.

(i) Pick the element with the largest atomic radius.

Answer: Br (bromine) has the largest atomic radius, because it has 4 shells while F has 2 and Cl has 3.

Teacher's Note:
a) Atomic radius increases down the halogen group as shells are added.
b) Bromine is the lowest of the three halogens in group 17.

(ii) Pick the element which is most reactive. Justify your answer.

Answer:
1. F (fluorine) is the most reactive of the three halogens.
2. Fluorine is the smallest, so its nucleus attracts an extra electron most strongly.

Teacher's Note:
a) Reactivity of halogens decreases down the group.
b) A small atom gains an electron easily, because the new electron feels a strong pull from the nucleus.
c) Fluorine is the most electronegative element, which is why it is so reactive.

Q3. Nitrogen (Z = 7) and Phosphorus (Z = 15) belong to same group-15 of the periodic table. Write the electronic configuration of these two elements. Which of these two is more electronegative? Why? [3 Marks]

Answer:
1. Electronic configuration of nitrogen (Z = 7) is 2, 5.
2. Electronic configuration of phosphorus (Z = 15) is 2, 8, 5.
3. Nitrogen is more electronegative, because its atomic size is smaller and its nucleus attracts shared electrons more strongly.

Teacher's Note:
a) Both elements have 5 valence electrons, which is why they belong to group 15.
b) Electronegativity decreases down a group, so the element higher up is more electronegative.
c) Phosphorus has an extra shell, and that extra shell shields its outer electrons from the nucleus.

Long Answer Type Questions (5 marks)

Q1. [5 Marks]

(i) How does atomic size vary along the group? Give reason.

Answer:
1. Atomic size increases down a group.
2. Each element down the group has one more shell than the one above it, so the outer electrons lie farther from the nucleus.

Teacher's Note:
a) A new shell is added at each step down the group.
b) Inner shells shield the outer electrons, so the pull of the nucleus on them is weaker.
c) Students often confuse this trend with the trend across a period, where size decreases.
d) Lithium is smaller than sodium, and sodium is smaller than potassium, which shows the trend clearly.

(ii) Why are metals electropositive in nature?

Answer: Metals have 1 to 3 valence electrons, which they lose easily to form positive ions. Hence, metals are electropositive.

Teacher's Note:
a) Electropositive means the tendency to lose electrons.
b) Metals have large atoms with few valence electrons, which are held weakly by the nucleus.

(iii) What are metalloids? Give an example.

Answer:
1. Metalloids are elements that show properties intermediate between metals and non-metals.
2. Silicon (Si) is an example of a metalloid.

Teacher's Note:
a) Metalloids such as silicon and germanium are semiconductors.
b) They are shiny like metals but brittle like non-metals.

Q2. Name- [5 Marks]

(i) Two elements that have a single electron in their outermost shells.

Answer: Hydrogen (H) and lithium (Li).

Teacher's Note:
a) Hydrogen has 1 electron in its only shell.
b) Lithium has 2, 1 as its configuration, so its outer shell has one electron.

(ii) Two elements that have two electrons in their outermost shells.

Answer: Beryllium (Be) and magnesium (Mg).

Teacher's Note:
a) Beryllium has 2, 2 and magnesium has 2, 8, 2.
b) Elements with two valence electrons belong to group 2, the alkaline earth metals.

(iii) Two elements with filled outermost shell.

Answer: Helium (He) and neon (Ne).

Teacher's Note:
a) Helium has a complete first shell with 2 electrons, which is a full outer shell.
b) Neon has 2, 8, a complete outer shell of 8 electrons.

(iv) Two elements that belong to halogen family.

Answer: Fluorine (F) and chlorine (Cl).

Teacher's Note:
a) Halogens belong to group 17 and have 7 valence electrons.
b) Bromine and iodine are also halogens.

(v) An element which is tetravalent and forms the basis of organic chemistry.

Answer: Carbon (C)

Teacher's Note:
a) Carbon has 4 valence electrons, so it forms 4 bonds.
b) Carbon's ability to form long chains is the reason it is the basis of organic chemistry.

CBSE Class 10 Science Chapter 5 Periodic classification of elements Objective Questions

Question. Two elements X and Y belong to group 1 and 2 respectively in the same period. The formulae of this oxides are
(a) XO, YO
(b) X₂O, YO
(c) X₂O, Y₂O
(d) XO, YO₂
Answer : B

Question. Which of the following has maximum non-metallic character?
(a) F
(b) Br
(c) Cl
(d) I
Answer : A

Question. Which of the following is not the characteristics of isotopes of an element?
Isotopes of an element
(a) show same atomic mass
(b) show same atomic number
(c) occupy same position in the Periodic Table
(d) show same chemical properties
Answer : A

Question. How many triads could Döbereiner identify from the existing elements then?
Answer : Döbernier could identify only three triads.

Question. On what basis did Mendeleev classified the element?
Answer : Mendeleev arranged the elements on the basis of their increasing atomic mass and similarity of chemical properties.

Question. What is the formula of oxide and hydride of Group I elements?
Answer : Oxide formula → R₂O
Hydride formula → RH.
‘R’ represents element.

Question. Write the atomic numbers of two elements ‘X’ and ‘Y’ having electronic configurations 2, 8, 2 and 2, 8, 6 respectively.
Answer. Electronic configuration of X = 2, 8, 2
 Atomic number is 2 + 8 + 2 = 12
Similarly,
Electronic configuration of Y = 2, 8, 6
 Atomic number is 2 + 8 + 6 = 16

Question. N, O, F cannot be classified as Döbereiner’s triad. Why?
Answer. Although the atomic mass of O (16 u) is approximately an average (16.5 u) of the atomic masses of N (14 u) and F
(19 u), i.e., (14 + 19)/2 = 16.5 but they cannot be regarded as a Döbereiner’s triad because their properties are altogether different.

Question. Besides gallium, which other elements have since been discovered that were left by Mendeleev in his periodic table? (any two)
Answer. Germanium, Scandium.

Question. Name two elements which show same kind of chemical reactivity as sodium.
Answer. Lithium (Li) and potassium (K) both belong to the same group as sodium (Na), So, they will show the same chemical properties as sodium.

Question. What is common among all the elements present in one period?
Answer : All the elements in same period show same number of shells e.g., all elements in period 3, show 3 electron shells each.

Question. What is atomic size?
Answer : The radius of an atom, i.e., the distance between the centre of the nucleus and the outermost shell of an atom is called atomic size.
The atomic radius is measured in picometre. (1 pm = 10^–12 m)

Question. What happens to the size of atom as we move from left to right in a period.
Answer : Ans. The atomic size in a period decreases as we move from left to right.

Question. How do you think the tendency to lose electrons will change in a group?
Answer : Down the group, the effective nuclear charge experienced by valence electrons decreases, hence they can easily lose electrons.

Question. What are noble gases/inert gases?
Answer : The element which is inactive, does not react with any other element and it has its outermost shell completely filled are called inert gases or noble gases.
e.g., He, Ne, Ar, Xe.

Question. An element ‘X’ belongs to II group and 2nd period. Write the atomic number and name of element.
Answer : K L ∴ Atomic Number = 4
2, 2 Element = Beryllium

Question. An element ‘P’ belongs to group = 2 and period = 3, state whether it is a metal or nonmetal and nature of its oxides.
Answer : Group 2 = Metals
Nature of oxide = Basic oxide

Question. Choose from the following :
₄Be, ₉F, ₁₉K, ₂₀Ca
(i) The element having one electron in the outermost shell.
(ii) Two elements of the same group.
Answer. The electronic configurations of the given elements are :
₄Be = 2, 2
₉F = 2, 7
₁₉K = 2, 8, 8, 1
₂₀Ca = 2, 8, 8, 2
(i) Potassium (K) has one electron in its outermost shell.
(ii) Be and Ca have two electrons in their outermost shells hence, they belong to same group.

Question. What is meant by periodicity of properties of elements? Why are the properties of elements placed in the same group of the periodic table similar?
Answer. When elements are arranged in increasing order of their atomic numbers, elements with similar chemical properties are repeated at definite intervals. This is known as periodicity of properties of elements.
Elements placed in the same group of the periodic table have similar properties because they have same number of outermost electrons and hence, show same valency. Thus, they all will form similar type of compounds.

Question. Rewrite the following statements after correction, if necessary
(i) Elements in the same period have equal valency.
(ii) The metallic character of elements in a period increases gradually on moving from left to right.
Answer. (i) Elements in the same group have equal valency
(ii) The metallic character of elements in a period decreases gradually on moving from left to right.

Question. In the Modern Periodic Table, the element calcium (atomic number = 20) is surrounded by elements with atomic numbers 12, 19, 21 and 38. Which of these elements have physical and chemical properties resembling those of calcium and why?
Answer. From the given data, the electronic configuration of
different elements can be written as :
Calcium (20) = 2, 8, 8, 2
Element with atomic number 12 = 2, 8, 2
Element with atomic number 19 = 2, 8, 8, 1
Element with atomic number 21 = 2, 8, 8, 3
Element with atomic number 38 = 2, 8, 18, 8, 2
It can be easily seen that elements with atomic numbers 12 and 38 have two electrons in their outermost shell thus, they belong to same group as that of calcium. So, they will show the physical and chemical properties resembling those of calcium.

Question. Carbon (atomic number = 6) and silicon (atomic number = 14) are elements in the same group of the periodic table. Give the electronic arrangements of the carbon and silicon atoms and state the group in which these elements occur.
Answer. Electronic arrangement of carbon : 2, 4
Electronic arrangement of silicon : 2, 8, 4
Both C and Si belong to group 14.

Question. From the list of the elements given below, select three elements which form a Döbereiner’s triad. F, Mg, Ca, Br, Li, Rb, Cl, Sr, I.
Answer. Cl, Br and I form a Döbereiner’s triad because atomic mass of Br is approximetly equal to the average of the atomic mass of Cl and I. Although Mg, Ca and Sr have similar properties but the atomic mass of Ca (40 u) is not an average of the atomic masses of Mg (24.31 u) and Sr (87.62 u). Therefore, these elements do not constitute a Döebereiner’s triad.

Question. What physical and chemical properties of elements were used by Mendeleev in creating his Periodic Table? List two observations which posed a challenge to Mendeleev’s Periodic Law. 
Answer : The physical property used was the atomic mass of an element.
The chemical property used was the nature of oxide and hydride formed i.e. similarity in chemical properties were used by Mendeleev. The two observations that posed challenge in Mendeleev Periodic Law were:
(i) Arranging elements according to the increasing order of atomic mass could not be maintained. Chemical properties do not depend on atomic mass.
(ii) Isotopes were not given any place in the table as they have different atomic mass but same chemical properties.

Question. Two elements A and B belong to the 3rd period of Modern Periodic Table and are in group 2 and 13 respectively. Compare their following characteristics in tabular form.
(a) Number of electrons in their atoms
(b) Size of their atoms
(c) Their tendencies to loose electrons
(d) The formula of their oxides
(e) Their metallic characters
(f) The formula of their chlorides
Answer. Electronic configuration of A = 2, 8, 2 i.e., Mg Electronic configuration of B = 2, 8, 3 i.e., Al

Question. An element X has a total of 31 nucleons, out of which 16 are neutrons.
(a) Write the electronic configuration of an atom of element X.
(b) Determine the group and period number of element X.
(c) Give the formula of the ion formed by element X.
Answer. (a) Number of protons = 31 – 16 = 15
Number of electrons = 15
Electronic configuration = 2,8,5
(b) Element X has five valence electrons.
Hence, element X is from Group V.
Element X has three occupied shells.
Hence, element X is in Period 3.
(c) Formula of the ion formed by X = X 3–

Question. (a) Consider three elements Na, Cl, Ar and answer the following.
(i) Discuss their metallic and non-metallic character.
(ii) Discuss the acid-base character of their oxides.
(b) The element with atomic number 14 is hard and forms acidic oxide and a covalent halide.To which of the categories does the element belong?
Answer. (a) The electronic configurations of these elements are
Na : 2, 8, 1 ; Cl : 2, 8, 7 and Ar : 2, 8, 8
(i) Sodium has only 1 electron in its valence shell which it can lose easily. Therefore it is a typical metal, chlorine has seven electrons in its outermost shell, it can easily accept one electron to get noble gas configuration, hence it is a typical non-metal. Ar is a noble gas.
(ii) The oxide of Na i.e. Na2O dissolves in water to form NaOH which is a strong base. The oxide of chlorine i.e. Cl₂O₇ dissolves in water to give HClO4 which is a strong acid.
Ar does not form any oxides.
Na₂O + H₂O → 2NaOH
Cl₂O₇ + H₂O → 2HClO₄
Hence, both metallic character of the elements and basic character of oxides decreases from left to right across a period.
(b) Non-metal, because it forms acidic oxide and covalent halide which are characteristics of non-metals.

Question. In the table given below, some of the elements of the periodic table with atomic numbers from 3 to 18 are given. These are represented by letters, which are not the usual symbols of the elements.

(a) Which of these
(i) is an alkaline earth metal?
(ii) are alkali metals?
(iii) is an element with valency 4?
(b) If A combines with F, what would be the formula of the resulting compound?
(c) What is the electronic arrangement of G?
Answer. (a) (i) C is an alkaline earth metal.
(ii) A and B are alkali metals.
(iii) D is an element with valency 4.
(b) If A combines with F, the formula of the resulting compound will be A+F –., i.e., AF.
(c) The electronic arrangement of G is 2, 8.

Question. The elements of the third period of the Periodic Table are given below:
Group       I      II      III    IV    V    VI   VII
                Na  Mg    Al     Si    P     S   Cl
(a) Which atom is bigger — Na or Mg? Why?
(b) Identify the most
(i) Metallic and (ii) Non-metallic element in period 3.
Answer : (a) Na atom is bigger in size this is because as we move from Na to Cl, the atomic number goes on increasing and the nuclear charge also increases. It pulls/attract the valence electrons at the centre and thus the atomic size decreases.
(b) (i) Most metallic — Na
(ii) Most non-metallic — Cl

Question. Chlorine (atomic number 17) is more electronegative than sulphur (atomic number 16).
Explain.
Answer. Chlorine (Z = 17) is placed after sulphur (Z = 16) in the same period i.e. third period. The size of chlorine is smaller than that of sulphur and its atom needs only one electron to have noble gas electronic configuration while sulphur atom needs two electrons. Therefore, chlorine has greater attraction for electrons than sulphur. It is more electronegative than sulphur.

Question. Three elements A, B and C have atomic numbers 7, 8 and 9 respectively.
(a) What would be their positions in the modern periodic table ? (Mention group and period both)
(b) Arrange A, B and C in decreasing order of their size.
(c) Which one of the three elements is most reactive and why ?
Answer. The electronic configuration of these elements are :
(a) A ( Z = 7) 2, 5;
B (Z = 8) 2, 6;
C (Z = 9) 2, 7
Position of element A = 15th group and 2nd period
Position of element B = 16th group and 2nd period ‘
Position of element C = 17th group and 2nd period.
(b) In general, atomic size decreases along a period. Therefore, decreasing order of size is A > B > C
(c) The element C (Z = 9) is fluorine. It is the most reactive element since it needs only one electron to acquire a noble gas configuration.

Question. The atomic numbers of three elements X, Y and Z are 9, 11 and 17 respectively. Which of these two elements will show similar characteristics and why ?
Answer. The elements X and Z will show similar characteristics because they differ in their atomic numbers by 8 (9, 17). Both these are halogens and belong to group 17 (halogen family). These are fluorine (Z= 9) and chlorine (Z = 17) .

Question.

The atomic number of three elements are given below :
Write the symbol of the element which belongs to
(i) group 13,
(ii) group 15, of the periodic table. State the period of the periodic table to which these elements belong. Give reason for your answer.
Answer. Electronic configuration of the elements A, B and C are as given :

Element (Symbol)         A                   B                 C
Atomic number         5 (2, 3)         7 (2, 5)       10 (2, 8)
(i) Element A belongs to group 13 (Group No. = 10 + 3 = 13). It is boron (B)
(ii) Element B belongs to group 15 (Group No. = 10 + 5 = 15). It is nitrogen (N)
Both these elements belong to second period since they have two shells.

Question. Write two reasons responsible for the late discovery of noble gases.
Answer. (i) Noble gas elements were not present in earth crust as minerals like other elements and were present in air to a very small extent.
(ii) Their atoms have stable electronic configuration of their outermost shells also called valence shells. (2 in case of He and 8 for other elements). They do not combine with atoms of other elements.
That is why noble gas elements were discovered at a later stage.

Question. The elements with atomic number 3 to 10 belong to the second period. Taking into account the trends in the general periodic properties, predict.
(a) The most electronegative element
(b) The most electropositive element
(c) The element belonging to noble gas family
(d) The element which constitutes large number of organic compounds.
Answer. (a) The most electronegative element has atomic number (Z) = 9. It is fluorine (F).
(b) The most electropositive element has atomic number (Z) = 3. It is lithium (Li)
(c) The element belonging to noble gas family has atomic number (Z) = 10. It is neon (Ne)
(d) The element which constitutes large number of organic compounds has atomic number (Z) = 6. It is carbon (C).

Question.

The atomic numbers of elements A, B, C, D and E are given below :
From the above table, answer the following questions
(a) Which two elements are chemically similar ?
(b) Which is an inert gas ?
(c) Which element belongs to 3rd period of periodic table ?
(d) Which element among these is a non-metal ?
Answer. 

The electronic configuration of the elements are as follows :
Element          A             B               C               D                 E
Atomic no. 7(2, 5)   10 (2, 8)   12 (2, 8, 2)   4 (2, 2)   19 (2, 8, 8, 1)
(a) Elements C and D are chemically similar since they have same number of valence electrons.
(b) Element B is an inert gas element since it has complete octet.
(c) Element C belongs to third period since it has three shells.
(d) Element A is a non-metal since it has five valence electrons.

Question. Account for the following :
(a) Elements C, N, O and F are placed in the second period of the periodic table.
(b) Elements of group 17 are monovalent.
Answer. (a) All these elements have two electron shells. Therefore, these are placed in the second period.
C (Z=6) 2, 4 ;
N (Z = 7) 2, 5 ;
O (Z = 8) 2, 6 ;
F (Z = 9) 2, 7
(b) All the elements included in the group 17 have 7 valence electrons in their atoms. These atoms need only one electron to acquire the electronic configuration of nearest noble gas atom. Therefore, these are monovalent

Question. “Elements in Periodic Table show periodicity of properties”. List any four properties.
Answer. Periodicity i.e., repetition of similar properties is shown by the elements present in a group and separated by definite gaps of atomic number. For example,
1. Elements in a group have same number of valence electrons and same valency.
2. Elements present in a group show similar chemical properties.
3. The atomic sizes of the elements in a group increase regularly.
4. The m.p. and b.p. of the elements in a group increase regularly.

Question. The electronic configuration of three elements X, Y and Z are given below ;
X = 2 ; Y = 2, 6 ; Z = 2, 8, 2.
1. Which element belongs to the second period ?
2. Which element belongs to the eighteenth group ?
3. Which element belongs to the second group ?
4. What is the valency of Y ?
5. Are Y and Z metals or non-metals ?
Answer. 1. The element ‘Y’ belongs to second period.
2. The element ‘X’ belongs to eighteenth group also called zero group.
3. The element ‘Z’ belongs to second group.
4. Element ‘Y’ has valency equal to 2(8 -6 = 2).
5. The element ‘Y’ is a non-metal while element ‘Z’ is a metal.

Question. Using the part of the periodic table given below, answer the questions that follow.

(i) Na has physical properties similar to which elements and why ?
(ii) Write the electronic configuration of N and P.
(iii) State one property common to fluorine and chlorine.
Answer. (i) N a has physical properties similar to Li and K. All the three elements have one electron each in the valence shell of their atoms. These are known as alkali metals. However, the element hydrogen has different physical properties.
(ii) Electronic configuration of N (Z = 7) = 2, 5
Electronic configuration of P (Z = 15) = 2, 8, 5
(iii) Both the elements have seven electrons in the valence shells as their atoms and have valency equal to one.
Fluorine (Z = 19) 2, 7 ;
Chlorine (Z = 17) 2, 8, 7.

Question. The list of the elements present in the same period but in different groups is given :
1 2 13 14 15 16 17 18
(a) Do these groups represent modern periodic table ?
(b) Which element will belong to oxygen family ?
(c) Which element will not take part in chemical combination ?
(d) The elements belonging to which groups will form ionic bonds most readily ?
Answer. (a) Yes, these groups of elements represent modern periodic table.
(b) The element present in group 16 belongs to oxygen family.
(c) The element included in group 18 (noble gas elements) will not take part in chemical combination.
(d) The elements belonging to group 1 and group 17 will form ionic bonds most readily.

Question. Atomic radii of the elements present in second period are given :

1. Arrange them in decreasing order of their atomic radii.
2. Are these elements arranged in the pattern of a period in the periodic table ?
3. Which elements have the largest and the smallest atoms ?
4. How does the atomic radius change as you move from left to right in a period ?
Answer. 1. The decreasing order of atomic radii is :
Li (152) > Be (111) > B (88) > C(77) > N(74) > O(66)
2. No, the arrangement of the elements is not systematic. The correct arrangement is as given above.
3. The element Li has the largest atoms while the element O has the smallest atoms.
4. The atomic radii of the elements decrease in moving from left to the right.

Question. “The atomic number of Cl is 17. On the basis of this information, answer the questions that follow :
(a) Write the electronic configuration of Cl.
(b) Find its valency.
(c) To which group does it belong ?
(d) Identify the type of ion it will form.
(e) Write down the formula of the compound it forms with other elements.
Answer. (a) Electronic configuration of Cl (Z = 17) = 2, 8, 7.
(b) Valency of Cl = 8 – 7 = 1
(c) Group to which Cl belongs =17 id) Type of ion formed by Cl = Anion (Cl–).
(e) Formula of compound with other elements (M) = MClx
∴ Here x is the valency of the element.

Question. An element E has following electronic configuration :
(a) To which group of the periodic table does element E belong ?
(b) To which period of the periodic table does element E belong ?
(c) State the number of valence electrons present in element E.
(d) State the valency of the element E.
Answer. The element with atomic no. (Z) = 16 is sulphur (S).
(a) It belongs to group 16 of the periodic table,
(b) It belongs to third period since it has three shells.
(c) The element has six valence electrons.
(d) The valency of the element is 2 (8 – 6 = 2).

From the part of the periodic table given, answer the following questions.
(a) Which is the most reactive metal ?
(b) Name the family of L, Q, R, T.
(c) Name one element of group 2 and 15,
(d) Name one member of group 18 other than neon.
(e) Give the name of the element S placed below carbon in group 14.
Answer. (a) The element Z is the most reactive metal.
(b) The elements are present in group 17. The family is that of halogens.
(c) One element belonging to group 2 is calcium (Ca) while one present in to group 15 is nitrogen (N).
(d) The element argon (Ar) is also present in group 18.
(e) The element is silicon (Si)

Question. Elements magnesium and oxygen respectively belong to group 2 and group 16 of the Modern Periodic Table. If the atomic numbers of magnesium and oxygen are 12 and 8 respectively,draw their electronic configuration and show the process of formation of their compound by transfer of electrons.
Answer. Atomic number of magnesium = 12
 Electronic configuration = 2, 8, 2
Valency of magnesium (Mg) = 2
Similarly, for oxygen (O) atomic number = 8
Electronic configuration = 2, 6
Valency of oxygen = 8 – 6 = 2
Formation of their compound will be as follows :