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Worksheet for Class 10 Science Chapter 5 Periodic Classification of Elements
Class 10 Science students should refer to the following printable worksheet in Pdf for Chapter 5 Periodic Classification of Elements in Class 10. This test paper with questions and answers for Class 10 will be very useful for exams and help you to score good marks
Class 10 Science Worksheet for Chapter 5 Periodic Classification of Elements
MCQ Questions for NCERT Class 10 Science Periodic Classification Of Elements
Question. At the time of Mendeleev, the number of elements known was
(a) 63
(b) 65
(c) 62
(d) 64
Answer : A
Question. Which of the following set of elements is written in order of their increasing metallic character?
(a) Na Li K
(b) C Q N
(c) Mg Al Si
(d) Be Mg Ca
Answer : D
Question. Which of the following is the outermosta shell for elements of period 2 ?
(a) K shell
(b) L shell
(c) M shell
(d) N shell
Answer : B
Question. Which of these belong to the same period?
Element A B C Atomic number 2 10 5
(a) A, B
(b) B, C
(c) C, A
(d) A, B and C
Answer : B
Question. A metal ‘M’ is in the first group of the Periodic Table. What will be the formula of its oxide?
(a) MO
(b) M2O
(C) M2O3
(d) MO2
Answer : B
Question. Which of the following elements will form an acidic oxide ?
(a) An element with atomic number 7
(b) An element with atomic number 3
(c) An element with atomic number 12
(d) An element with atomic number 19
Answer : A
Question. Carbon belongs to the second period and Group 14. Silicon belongs to the third period and Group 14. If atomic number of carbon is 6, the atomic number of silicon is
(a) 7
(b) 14
(c) 24
(d) 16
Answer : B
Question. Which of the given elements A, B, C, D and E with atomic number 2, 3, 7, 10 and 30 respectively belong to the same period?
(a) A, B, C
(b) B, C, D
(c) A, D, E
(d) B, D, E
Answer : B
Question. An element has 12 protons. The group and period to which this element belongs to is
(a) 2nd group, 3rd period
(b) 2nd group, 2nd period
(c) 3rd group, 2nd period
(d) 3rd group, 3rd period
Answer : A
Question. An element X from group 2 of the Periodic Table reacts with Y from group 17 to form a compound. Give the formula of the compound.
(a) XY2
(b) XY
(c) X2Y
(d) (XY)2
Answer : A
Question. Consider the following elements 20Ca, 8Or 18Ar, 16S, 4Be, 2He Which of the above elements would you expect to be in group 16 of the Periodic Table?
(a) 20Ca and 16S
(b) 20Ca and 8O
(c) 18Ar and 16S
(d) 8O and 16S
Answer : D
Question. WTiich of the following elements does not lose an electron easily ?
(a) Na
(b) F
(c) Mg
(d) Al
Answer : B
Question. Which of the following is correct order of atomic size ?
(a) Li < Na < K < Rb < Cs
(b) Li > Na > K > Rb > Cs
(c) Na < K < Li < Rb < Cs
(d) K < Na < Li < Rb < Cs
Answer : A
Question. An element ‘A’ belongs to the third period and group 16 of the Periodic Table. Find out the valency of A.
(a) Valency = 6
(b) Valency = 2
(c) Valency = 1
(d) Valency = 3
Answer : B
Question. Silicon(14) belongs to class of
(a) Metal
(b) Non-metal
(c) Metalloids
(d) Ionic compound
Answer : C
Question. An element X has mass number 40 and contains 21 neutrons in its atom. To which group of the Periodic Table does it belong?
(a) Group 1
(b) Group 4
(c) Group 2
(d) Group 3
Answer : A
Question. Chlorine (17) belongs to which group and period
(a) 7, 3
(b) 17, 3
(c) 1, 3
(d) 16, 3
Answer : B
Question. Which of the following is correct order of size ?
(a) I+ > I > I–
(b) I– > I > I+
(c) I > I+ > I–
(d) I > I– > I+
Answer : B
Question. The atom of an element has electronic con-figuration 2, 8, 7. To which of the following elements would it be chemically similar?
(a) N(7)
(b) P(15)
(c) Na(11)
(d) F (9)
Answer : D
Question. The properties of eka-aluminium predicted by Mendeleev are the same as the properties of later discovered element:
(a) Scandium
(b) Germanium
(c) Gallium
(d) Aluminium
Answer : C
Very Short Answers :
Question. How did Dobereiner classify the elements?
Answer: He classified elements in group of three, such that atomic weight of middle element is average of 1st and 3rd element.
Question. Out of Li and K, which have stronger metallic character and why?
Answer: K because it has larger size and more tendency to lose electron.
Question. Why is atomic number more important?
Answer: Chemical properties depend upon number of valence electrons which depend upon atomic number.
Question. On what basis we can classify elements?
Answer: Similarity in properties.
Question. Could Modem Periodic Table solve all the problems?
Answer: Yes, except position of hydrogen.
Question. List any two properties of the elements belonging to the first group of the modem periodic table.
Answer: First group elements are also known as alkali metals.
(a) These elements exhibit +1 valency.
(b) These are very reactive and not found freely in nature.
Question. Write the formula used to determine the maximum number of electrons which a shell in an atom can accommodate.
Answer: Each shell has a maximum number of ‘2n2’ electrons, where n is the shell number.
Question. How many elements are known the date?
Answer: 118 elements are known till date
Question. Why did Mendeleev’s left some gaps?
Answer: These gaps were left for undiscovered elements.
Question. Which chemical property of element was used by Mendeleev for classification?
Answer: Formula of hydrides and oxides.
Question. Compare the radii of two species X and Y. Give reasons for your answer. • X has 12 protons and 12 electrons • Y has 12 protons and 10 electrons
Answer : Here, X has the same number of electrons and protons, therefore, it is a neutral atom, whereas Y contains 12 protons and 10 electrons, so there are 2 protons extra, giving Y a charge of +2. The electronic configurations of the two species can be expressed as:
Answer : Second period has 8 elements. There are only two metals in Second period. (C,N,OF,Ne) Lithium and Beryllium
Question. State Modern Periodic Law of classification of elements.
Answer : The Modern Periodic Law states that ‘Properties of elements are a periodic function of their atomic number.’
Short Answers :
Question. Based on the group valency of element write the molecular formula of the following compounds giving justification for each:
(A) Oxide of first group element.
(B) Halide of the element of group thirteen, and
(C) Compound formed when an element, , A of group 2 combines with element, B of group seventeen.
Answer : While writing the formula of compounds, we need to take the valencies of the combining elements into account.
(A) First group elements have a valency one, as they have one valence electron. Oxygen belongs to group 16 and has a valency of 2 as it has 6 valence electrons.
The molecular formula of compound formed by oxides of first group element will be X2O, where X is the first group element. Example of such a compound is Na2O.
(B) Elements of group 13 have a valency 3, as they have 3 valence electrons. Halogens belong to group 17 and have a valency of 1, as they have 7 valence electrons.
The molecular formula of halide of elements of group 13 will be YX3, where Y is the element of group 13 and X is a halogen.
Example of such a compound is AlCl3.
(C) All elements of group 2 have a valency 2, as they have 2 valence electrons. Group 17 elements are halogens, having a valency of 1. Therefore, valency of A is 2 and that of B is 1. The molecular formula of compound formed when A combines with B will be AB2. Example of such a compound is MgCl2.
Question. Two elements A and B have atomic numbers 11 and 19 respectively.
(A) State their position in the moden periodic table.
(B) Which element has a bigger atomic radius?
(C) What is the nature of their oxides?
Answer : (A) Electronic configuration of A: 2, 8, 1 Electronic configuration of B: 2, 8, 8, 1 From its electronic configuration, it is clear that element A belongs to Group 1 (since valence electron is 1) and Period 3 (since it has 3 electron shells).
Similarly, we can say that the element B belongs to Group 1 and Period 4.
(B) The atomic radius of element B is bigger as compared to element A. This is because the atomic radius increases (due to the presence of more shells) as we move down in a group.
(C) They form basic oxides.
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. Chlorine is an element in period 3 of the Periodic Table. Bromine is found in period 4 of the Periodic Table. These two elements may be from different periods of the periodic table, but they have many similar properties. 
(a) Complete the given table.
(b) Explain why the properties of chlorine and bromine closely resemble one another.
(c) Lithium is an element from Group I of the Periodic Table. Write the formula of the compound formed between lithium and
(i) chlorine
(ii) bromine.
(iii) What type of bonding is found in these compounds? Give reason.
Answer: (a) 
(b) Chlorine and bromine are in the same group i.e., Group VII of the Periodic Table.
Chlorine and bromine have seven valence electrons each.
Since both have the same number of valence electrons, they have similar chemical properties.
Both readily gain or share one electron to achieve a stable octet configuration.
(c) (i) LiCl (ii) LiBr
(iii) Ionic bonds
Lithium has one valence electron.
Lithium readily loses this valence electron to achieve a noble gas configuration similar to helium.
Li → Li+ + e–
Chlorine and bromine have seven valence electrons each.
Chlorine readily gains one electron to achieve a noble gas configuration similar to argon.
Cl + e– → Cl–
Bromine readily gains one electron to achieve a noble gas configuration similar to krypton.
Br + e– → Br–
Oppositely charged ions are attracted together by strong electrostatic forces of attraction to form ionic bonds.
Question. (a) Which is more basic (i) K2O or Na2O
(ii) K2O or CaO?
(b) Name a species that will be isoelectronic with each of the following atoms or ions :
(i) Ne (ii) Cl–
(iii) Ca2+ (iv) Rb
Answer: (a) (i) K2O
Potassium (K) and sodium (Na) belong to same group and the basic nature of oxides increases down the group. Therefore, K2O is more basic than Na2O.
(ii) K2O
Potassium (K) and calcium (Ca) belong to the same period and basic nature of oxides decreases from left to right in a given period. Therefore, K2O is more basic than CaO.
(b) (i) Na+ (ii) S2–
(iii) K+ (iv) Sr2+
Question. How many elements can be accommodated in each period of the periodic table? What are these periods called on the basis of number of elements?
Answer: Based on the maximum capacity of a shell according to the formula 2n2 the number of elements in each period can be given as follows :
n = 1 (Maximum no. of elements 2). Thus, first period has 2 elements. It is called very short period.
n = 2 (Maximum no. of elements 8). Thus, 2nd period has 8 elements. It is called short period.
n = 3 (Maximum no. of elements 8). Thus, 3rd period has 8 elements. It is called short period.
n = 4 (Maximum no. of elements 18). Thus, 4th period has 18 elements. It is called long period.
n = 5 (Maximum no. of elements 18). Thus, 5th period has 18 elements. It is called long period.
n = 6 (Maximum no. of elements 32). Thus, 6th period has 32 elements. It is called very long period.
n = 7 (Maximum no. of elements 32). Thus, 7th period has 32 elements. It is also called very long period.
Question. Na, Mg and Al are the elements of the 3rd periods of the Modern Periodic Table having group number 1, 2 and 13 respectively. Which one of these elements has the
(a) highest valency,
(b) largest atomic radius, and
(c) maximum
chemical reactivity? Justify your answer stating the reason for each.
Answer: Period number of Na, Mg and Al is 3 Group number of Na, Mg and Al are 1, 2 and 13 respectively.
(a) Aluminium (Al) will show highest valency of 3 as it belongs to group number 13 (valency = 13 – 10 = 3).
Moreover, along the period from left to right valency first increases to maximum and then decreases.
(b) Sodium (Na) will have the largest atomic radius because as we move along the period from left to right, the atomic radius decreases.
(c) Sodium (Na) will have maximum chemical reactivity because as we move along the period from left to right, chemical reactivity decreases.
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. 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. The second period of the long form of periodic table contains the following elements :
Li B e B C N O F N e
(a) Write down their electronic configurations.
(b) Do they contain the same number of valence electrons?
(c) Do they contain the same number of shells?
Answer: (a) Electronic configurations of the elements :
Li Be B C N O F Ne
2, 1 2, 2 2, 3 2, 4 2, 5 2, 6 2, 7 2, 8
(b) These elements do not contain same number of valence electrons.
(c) They contain same number of shells (K and L).
Question. The atomic number of an element ‘X’ is 20.
(i) Determine the position of the element ‘X’ in the periodic table.
(ii) Write the formula of the compound formed when ‘X’ reacts/combines with another elements ‘Y’ (atomic number 8).
(iii) What would be the nature (acidic or basic) of the compound formed? Justify your answer.
Answer: Atomic number of element X is 20 so, it is calcium (Ca).
Electronic configuration of Ca = 2, 8, 8, 2
(i) As calcium has two valence electrons in its outermost shell, so it belongs to group 2.
Moreover, it has four shells which indicates that it belongs to period number 4.
(ii) Calcium forms a basic oxide having the formula : 
(iii) When calcium oxide is treated with water then calcium hydroxide (Basic oxide) is formed.
CaO+H2O → Ca(OH)2
Calcium hydroxide
Question. In the following table six elements A, B, C, D, E and F (here letters are not the usual symbols of the elements) of the Modern Periodic Table with atomic number 3 to 18 are given : 
(a) Which of these
(i) a noble gas,
(ii) a halogen?
(b) If B combines with F, what would be the formula of the compound formed?
(c) Write the electronic configurations of C and E.
Answer: (a) (i) Noble gas = G
(ii) Halogen = F
(b) B(11) = 2, 8, 1
F(17) = 2, 8, 7
Valency of B = 1
Valency of F = 8 – 7 = 1
Formula of the compound formed :
(c) Electronic configuration of C(12) = 2, 8, 2
Electronic configuration of E(8) = 2, 6
Question. The positions of three elements A, B and C in the periodic table are indicated below :
Group 16 Group 17
– – (First period)
– A (Second period)
– – (Third period)
B C (Fourth period
(a) State whether element C would be a metal or a non-metal? Why?
(b) Which is the more active element A or C? Why?
(c) Which type of ion (cation or anion) will be formed by the element C? Why?
Answer: (a) C belongs to group 17 and hence, it will have 7 valence electrons in the outermost shell and has a tendency to gain electrons thus, it is a non-metal.
(b) Among A and C, A will be more reactive as the reactivity decreases down the group. So, A has more tendency to gain electrons.
(c) C will form negatively charged ion which is known as anion because group 17 elements have seven electrons in their outermost shell so, they have strong tendency to gain
an electron to attain the noble gas configuration.
Question. An element ‘X’ is placed in the 3rd group and 3rd period of the Modern Periodic Table.
Answer the following questions stating reason for your answer in each case :
(a) Write the electronic configuration of the element ‘X’.
(b) Write the formula of the compound formed when the element ‘X’ reacts with another element ‘Y’ of atomic number 17.
(c) Will the oxide of this element be acidic or basic ?
Answer: X is placed in 3rd group (IIIA) and 3rd period of the Modern periodic table then it must be aluminium (Al).
As it belongs to 3rd group so it will have 3 electrons in its outermost shell.
Also it belongs to 3rd period, so it will have 3 shells.
(a) Electronic configuration of X = 2, 8, 3
(b) Atomic number of Y is 17
Electronic configuration is 2, 8, 7
Valency of Y = 8 – 7 = 1
∴ Formula of compound formed when X reacts with Y is
(c) Al2O3 is amphoteric in nature i.e., acidic as well as basic oxide.
Question. Element Y is given the chemical symbol 40 20Y.
(a) What is the electronic configuration of element Y?
(b) Determine the position of element Y in the periodic table.
(c) Explain how an atom of element Y can form an ion.
Answer: (a) Proton number = 20 = number of electrons Electronic configuration = 2, 8, 8, 2
(b) Element Y is in group II and period 4 because it has 2 electrons in outermost shell and four occupied shells.
(c) It is easier for an atom of element Y to lose the two valence electrons to achieve an electronic configuration similar to argon (2, 8, 8). Hence, an atom of element Y will
form a positive ion with charge equal to its group number, i.e. 2. The formula of the ion is Y2+.
Question. The atomic number of an element is 16.
Predict
(i) the number of valence electrons in its atom
(ii) its valency
(iii) its group number
(iv) whether it is a metal or a non-metal
(v) the nature of oxide formed by it
(vi) the formula of its chloride.
Answer: Atomic number of element (E) is 16
∴ Electronic configuration = 2, 8, 6
(i) Number of valence electrons in its atom = 6
(ii) Valency = 8 – 6 = 2
(iii) As there are 6 valence electrons thus, its group number is 10 + 6 = 16
(iv) This element is a non-metal.
(v) The nature of oxide formed by this element is acidic.
(vi) The formula of the chloride of non-metal ‘E’ will be
Question. Choose from the following:
20Ca, 3Li, 11Na, 10Ne
a. An element having two shells completely filled with electrons.
b. Two elements belonging to same group of the periodic table.
Answer : a. 10Ne (2, 8)
b. 3Li(2, 1) and 11Na(2, 9, 1) belong to the same group.
Question. Write the electronic configuration of two elements X’ and ‘Y’ whose atomic numbers are 20 and 17 respectively. Write the molecular formula of the compound formed when element X’ reacts with element ‘Y’. Draw electron-dot structure of the product and also state the nature of the bond formed between both the elements.
Answer :
Question. The atomic number of an element is 12.
a. Write its electronic configuration and determine its valency.
b. Is it more reactive or less reactive than Ca(20)?
c. Is it a metal or a non-metal?
d. Write the formula of its oxide.
Answer : a. Mg(12): 2, 8, 2. Its valency is equal to 2.
b. It is less reactive than Ca.
c. It is a metal.
d. MgO
Question. a. How many periods are there in the Modem Periodic Table of elements?
b. How do atomic radius, valency and metallic character vary down a group?
c. How do atomic size and metallic character of elements vary as we move from left to right in a period?
Answer : a. There are 7 periods.
b. Atomic radius and Metallic character increases down the group. Valency remains the same in a group.
c. Atomic size and metallic character decrease along a period from left to right in a period.
Question. Give reasons:
a. Elements in a group have similar chemical properties.
b. Elements of Group 1 form ions with a charge of +1.
Answer : a. It is due to the same number of valence electrons.
b. Group-1 elements can lose one electron to formpositive ions with charge equal to +1.
Question. a. State the main characteristic of elements on which modern periodic table is based.
b. No fixed position is assigned to hydrogen in the periodic table, why?
Answer : a. Modem periodic table is based on the trend of increasing order of atomic number. Elements of same group have same number of valence electrons.
b. Hydrogen resembles with both group 1 and group 17 elements, therefore it does not have a fixed position.
Question. An element ‘E’ has the following electronic configuration:
K L M
2 8 6
a. To which group of the periodic table does element E belong to?
b. To which period of the periodic table does element E belong to?
c. State the number of valence electrons present in element E.
d. State the valency of the element E.
Answer : a. E (2, 8, 6) belongs to group 16,
b. It belongs to 3rd period.
c. It has 6 valence electrons.
d. The valency of E is 2.
Question. Give reasons for the following:
a. Lithium atom is smaller than sodium atom.
b. Chlorine (Atomic number 17) is more electronegative than Sulphur (Atomic number 16)
Answer : a. Li(3): 2, 1; Na(11): (2, 8, 1).
Li has two shells, therefore it is smaller than Na which has 3 shells.
b. Chlorine (17) is smaller than sulphur (16), therefore it is more electronegative than sulphur.
Question. An element ‘P’ (atomic number 20) reacts with an element ‘Q’ (atomic number 17) to form a compound.
Answer the following questions giving reason:
Write the position of ‘P’ and ‘Q’ in the Modem Periodic Table and the molecular formula of the compound formed when ‘P* reacts with ‘Q’.
Answer : P(20): 2, 9, 9, 2; Q(17): 2, 8, 7
‘P’ belongs to group 2 and 4th period.
‘Q’ belongs to group 17 and 3rd period.
PQ2 is the molecular formula of the compound formed.
Question. Would you place the two isotopes Cl-35 and Cl-37 in different slots because of their different atomic masses or in the same slot because their chemical properties are same? Justify your answer.
Answer : They will be placed in the same slot because they have similar chemical properties due to same number of valence electrons.
Question. Write the names given to the vertical columns and horizontal rows in the Modern Periodic Table. How does the metallic character of elements vary on moving down a vertical column? How does the size of atomic radius vary on moving from left to right in a horizontal row? Give reason in support of your answer in the above two cases.
Answer : Vertical columns are called groups. Horizontal rows are called periods. Metallic character of elements increases down the group because tendency to lose electrons increases down the group due to increase in atomic size. Atomic size goes on decreasing along the period due to increase in effective nuclear charge due to increase in number of protons and electrons.
Question. What is meant by periodicity of properties of elements? Why are the properties of elements placed on the same group of the periodic table similar?
Answer : The repetition of similar properties after a definite interval is called periodicity of properties. It is due to the same number of valence electrons.
Question. The atomic number of these elements are given below:
Element B O N C
Atomic radius (in pm) 86 66 74 97
Arrange this elements in increasing order of their atomic numbers. Give reason for your answer.
Answer : B(5), C(6), N(7), 0(8) is the increasing order of their atomic numbers.
It is because when we move along a period, atomic radii decreases due to increase in effective nuclear charge due to increase in number of protons and electrons continuously.
Question. Justify the following with suitable reasons:
a. Cations are smaller than the corresponding atoms.
b. Size of atom increases as we move down the group.
c. Atomic size decreases as we move across a period.
Answer : a. Cations are formed by loss of electrons, therefore effective nuclear charge increases, size of atom decreases.
b. It is because number of shells goes on increasing down the group.
c. It is because effective nuclear charge increases along a period.
Question. Write the number of periods in the modern periodic table. State the changes in valency and metallic character of elements as we move from left to right in a period. Also state the changes, if any, in the valency and atomic size of elements as we move down the group.
Answer : There are 7 periods in Modern Periodic Table. Valency first increases and then decreases. Metallic character decreases along a period from left to right. There is no change in valency down the group. Atomic size increases down the group.
Question. F, Cl and Br are the elements each having seven valence electrons. Which of these (a) has the largest atomic radius, (b) is most reactive? Justify your answer stating reason for each.
Answer : a. Br has the largest atomic size because it has 4 shells.
b. F is the most reactive v it can give electrons most easily due to its smallest size.
Question. Elements Mg and O respectively belong to group 2 and group 16 of the modem periodic table. If the atomic number of Mg and O are 12 and 8 respectively, draw their electronic structures and show the process of formation of the compound by transfer of electrons between them.
Answer : Mg(12): 2,8,2
O(8): 2,6
Question. How does tendency to lose electrons change in the Modern Periodic Table in (a) a group, (b) a period and why?
Answer : a. In a group, tendency to lose electrons increases down the group because atomic size increases, forces of attraction between the valence electron and nucleus decreases.
b. In a period, tendency to lose electrons decreases due to decrease in atomic size due to more effective nuclear charge.
Question. Given below are four elements with their atomic numbers:
A(16), B(11), C(3), D(14)
a. Identify the elements which belong to the same group of the Modem periodic table.
b. Arrange the elements in decreasing order of the atomic size.
c. Write the formula of oxide of B.
d. Which of these elements is a metalloid?
Answer : a. C(3) and B( 11) belongs to the same group.
b. A
d. D is a metalloid.
Question. (a) Identify the element that have two completely filled shells and the number of valence electrons in each case if atomic numbers are: (i) 1, (ii) 2, (iii) 7, (iv) 8
(b) Analyse which amongst them is inert.
Answer : a. (i) 2, 8, 1: Sodium
(ii) 2, 8, 2: Magnesium
(iii) 2, 8, 7: Chlorine
(iv) 2, 8, 8: Argon
(b) Argon is inert.
Question. Table given below shows a part of the modern periodic table.
Using this table, explain why
a. Li and Na are considered as active metals.
b. Atomic size of Mg is less than that of Na.
c. Fluorine is more reactive than chlorine.
Answer : a. Li and Na have largest atomic size in respective period, therefore they can lose an electron easily, hence they are active metals.
b. Mg has 12 protons and 12 electrons which has more forces of attraction, therefore, it is smaller in size than Na which is having 11 protons and 11 electrons.
c. F is smaller in size, it can gain electrons easily, therefore, it is more reactive than Cl.
Question. The elements 4Be, 12Mg and 20Ca having two valence electrons in their valence shells are in periods 2, 3 and 4 respectively of the modern periodic table. Answer the following questions associated with these elements, giving reason in each case:
a. In which group should they be placed?
b. Which one of them is least reactive?
c. Which one of them has the largest atomic size?
Answer : a. They belong to group 2.
b. Be is least reactive.
c. Ca has largest atomic size.
Question. State Mendeleev’s periodic law. Write two achievements of Mendeleev’s periodic table.
Answer : Properties of elements are a periodic function of their atomic mass.
Achievements:
a. It could arrange all the elements discovered at that time.
b. It left some gaps for the elements to be discovered and helped in their discovery by predicting their properties.
Question. Arrange the following elements in increasing order of their atomic radii:
a. Li, Be, F, N
b. Cl, At, Br, I
Answer : a. F < N < Be < Li
b. Cl < Br < I < At
Question. In the following table, the position of six elements A, B, C, D, E and F are given as they are in the Modern Periodic Table:
On the basis of the above table, answer the following questions:
a. Name the element which forms only covalent compounds.
b. Name the element which is a metal with valency of three.
c. Name the element which is a non-metal with valency of three.
d. Out of B and C, whose atomic radius is bigger and why?
e. Write the common name of the family to which the elements D and F belongs to?
Answer : a. E is an element which form covalent compounds.
b. B is a metal with a valency of 3.
c. C is a non-metal with a valency of 3.
d. B has bigger size.
e. D and F belong to Noble gases.
Question. Mendeleev predicted the existence of certain elements not known at that time and named two of them as eka-silicon and eka-aluminium.
(A) Name the elements which have taken the place of these elements.
(B) Mention the group and the period of these elements in the modern periodic table.
(C) Classify these elements as metals, non- metals or metalloids.
(D) How many valence electrons are present in each one of them?
Answer : (A) The two elements that have taken the place of eka-silicon and eka-aluminium are germanium (Ge) and gallium (Ga), respectively. (B) Germanium is placed in group 14 and period 4 in the modern periodic table. Gallium is placed in group 13 and period 4 in the modern periodic table. (C) Germanium (Ge) is a metalloid. gallium (Ga) is a metal. (D) Valence electrons in germanium (Ge) are 4. Valence electrons in gallium (Ga) are 3.
Answer : The electronic configuration of element X having atomic number 13 is: 2, 8, 3 It is a metal having valency = 3 (number of valence electrons). It is electropositive in nature and can form X3+ ions. It is placed in 3rd period and 13th group of the Modern Periodic table. Formula of its oxide will be X2O3. Therefore, atomic number of an element is a more fundamental property and prediction of properties of elements could be made more accurately when elements were arranged on the basis of increasing atomic number.
Long Answers
Question. Explain giving justification the trends in the following properties of elements, on moving from left to right in a period, in the Modern periodic Table.
(A) Variation of valency.
(B) Change of atomic radius.
(C) Metallic to non-metallic character.
(D) Electronegative character.
(E) Nature of oxides.
Answer : The trends in the properties of elements, on moving from left to right in a period, in the Modern periodic Table are given below:
(A) Variation of valency: On moving from left to right, the valency increases from 1 to 4 and then decreases to 0 as the valency is determined by the number of valence electrons.
(B) Change of atomic radius: On moving from left to right in a period, the atomic radius decreases because the number of electrons and protons increases. Due to the large positive charge on the nucleus, electrons are pulled more strongly towards the nucleus.
(C) Metallic to non-metallic character: The metallic character decreases and non- metallic character increases as we move from left to right in a period because the tendency to lose electrons decreases as the effective nuclear charge on the valence shell electrons increases.
(D) Electronegative character: The electronegative character, or the tendency to gain electrons, increases on moving from left to right along a period because of the increase in effective nuclear charge on the valence shell.
(E) Nature of oxides: The basic nature of oxides decreases and the acidic nature increases as we move from left to right since metals form basic oxides whereas non-metals form acidic oxides.
Question. a. The modern periodic table has been evolved through the early attempts of Dobereiner, Newlands and Mendeleev. List one advantage and one limitation of all the three attempts.
b. Name the scientists who first of all stated that atomic number of an element is a more fundamental than its atomic mass.
c. State modern periodic law.
Answer : (a) Dobereiner Periodic Table
Advantage: It could predict the atomic mass of middle elements quite correctly. Limitations: He could identify only three triads of elements.
Newlands Periodic Table
Advantage: Every eight element had properties similar to the first if elements are arranged in increasing order of atomic mass.
Limitations: It was applicable only upto calcium only. No future elements could fit into it.
Mendeleev’s Periodic Table
Advantage: He could classify all the elements discovered at that time into groups and periods.He also predicted the existence of new elements which were not discovered at time.
Limitations: No fixed position of hydrogen. Position of isotopes could not be sorted out.
(b) Moseley:
Properties of elements are a periodic function of their atomic numbers.
Question. Name the element which has
a. the electronic configuration 2, 8, 1.
b. a total of two shells, with 4 electrons in the valence shell.
c. a total of three shells, with 3 electrons in the valence shell.
d. one shell which is completely filled with electrons.
e. twice as many electrons in the second shell as in the first shell.
Answer : a. Sodium (2, 8, -1)
b. Carbon (2, 4)
c. Aluminium (2, 8, 3) i
d. Helium (2)
e. Carbon (2, 4)
Question. a. Why do we classify elements?
b. What were the two criteria used by Mendeleev in creating his periodic table.
c. Why did Mendeleev left some gaps in his periodic table?
d. In Mendeleev’s periodic table, why was there no mention of nobles gases like He, Ne and Ar?
e. Would you place two isotopes of Cl-35 and Cl-37 in different slots because of their different atomic mass or in the same slot because their chemical properties are same? Justify your answer.
Answer : a. It makes their study easier.
b. (i) Increasing order of atomic mass, (ii) Formula of oxides and hydrides.
c. The gaps were left for the elements to be discovered.
d. Noble gases were not invented at that time.
e. They will be placed at the same slot as they have the same atomic number and same chemical properties.
Question. Atoms of eight elements A, B, C, D, E, F, G and H have the same number of electronic shells but different
number of electrons in their outermost shell. It was found that elements A and G combine to form an ionic compound. This compound is added in a small amount to almost all vegetable dishes during cooking. Oxides of elements A and B are basic in nature while those of E and F are acidic. The oxide of D is almost neutral. Based on the above information answer the following questions:
a. To which group or period of the Periodic Table do the listed elements belong to?
b. What would be the nature of the compoundformed by the combination of elements ‘B’ and ‘F’?
c. Which two of these elements could definitely be metals?
d. Which one of the eight elements is most likely to be found in gaseous state at room temperature?
e. If the number of electrons in the outermost shell of elements ‘C’ and ‘G’ be 3 and 7 respectively,write the formula of the compound formed by the combination of‘C’and ‘G’.
Answer : a. A and B belongs to group-1 and group-2 respectively because they form basic oxides. ‘C’ belongs to group-13, ‘D’ belongs to group-14 which forms almost neutral oxide (actually amphoteric oxide), E and F belong to group-15, 16 forming acidic oxides. ‘G’ belongs to group-17 because NaCl is used in cooking. ‘H’ belongs to group 18. they belong to third period of the periodic table.
b. B and F will form ionic compound because ‘B’ is a metal and ‘F’ is a non-metal.
c. A and B are definitely metals.
d. H is most likely to be found in gaseous state at room temperature.
Question. Atoms of eight elements A, B, C, D, E, F, G and H have the same number of electronic shells but different number of electrons in their outermost shell. It was , found that elements A and G combine to form an ionic compound which can also be extracted from sea water. Oxides of the elements A and B are basic in nature while those of E and F are acidic. The oxide of element D is almost neutral. Answer the following questions based on the information given herein:
a. To which group or period of the periodic table do the listed elements belong?
b. Which one of the eight elements is likely to be a noble gas?
c. Which of the eight elements would have the largest atomic radius?
d. Which two elements amongst these are likely to be non-metals?
e. Which one of these eight elements is likely to be a semi-metal or a metalloid?
Answer : a. A and B belongs to group-1 and. group-2 respectively because they form basic oxides. ‘C’ belongs to group-13, ‘D’ belongs to group-14 which forms almost neutral oxide (actually amphoteric oxide), E and F belong to group-15, 16 forming acidic oxides. ‘G’ belongs to group-17 because NaCl is used in cooking. ‘H’ belongs to group 18. they belong to third period of the periodic table.
b. H belongs to noble gas elements.
c. A will have largest atomic radius.
d. E and F are likely to be non-metals,
e. D is likely to be a metalloid or semi metal
Question. Which elements are known as Chalcogens ?
Question. How can you locate the element in the periodic table in terms of group and period with atomic no 16 ?
Question. Give the names and electronic configuration of the following elements
a) the third alkali metal
b) the second noble gas
c) the first chalcogen
d) the second halogen
Question. Predict the atomic number of noble gas next to Radon if discovered.
5. Which one in each of the following pairs has larger size ?
a) K or K+ b) Br or Br-
c) O2- or F- d) Br or Br-
Question. Account for the difference in size of Na+ and Mg2+ both of which have same noble gas configuration.
Question. Select the species which have same noble gas configuration.
Question. Why has Chlorine higher electron affinity than Flourine ?
Question. Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.
a) Lithium and Oxygen
b) Magnesium and nitrogen
c) Aluminium and lodine
d) Phosphorous and fluorine
Question. Which element has the highest melting point among nonmetals.
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