CBSE Class 10 Science Periodic Classification Of Elements Worksheet

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) 2^nd group, 3^rd period
(b) 2^nd group, 2^nd period
(c) 3^rd group, 2^nd period
(d) 3^rd group, 3^rd 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

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:

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. 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) O^2- 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.