Periodic Classification of Elements Class 10 Important Questions with Answers Science Chapter 5

Solved the very best collection of Periodic Classification of Elements Class 10 Science Important Questions and Answers Chapter 5 Pdf from the latest NCERT edition books, It will help you in scoring more marks in CBSE Exams.

Class 10 Science Chapter 5 Important Questions with Answers Periodic Classification of Elements

Class 10 Chemistry Chapter 5 Important Questions with Answers Periodic Classification of Elements

Periodic Classification of Elements Class 10 Important Questions Very Short Answer Type

Question 1.
Write the name and formula of a molecule made up of three atoms of oxygen. (2012 OD)
Answer:
O₃ → Ozone

Question 2.
How many vertical columns are there in the modern periodic table and what are they called?  (2013 D, 2014 D)
Answer:
There are 18 vertical columns in the modern periodic table and they are called groups.

Question 3.
How many horizontal rows are there in the modern periodic table and what are they called?  (2013 D, 2014 D)
Answer:
There are seven horizontal rows in the modern periodic table. The horizontal rows in a periodic table are called periods.

Question 4.
Write two reasons responsible for late discovery of noble gases?  (2013 D)
Answer:
The noble gases were discovered very late because they are very unreactive and present in extremely low concentration in the atmosphere.

Question 5.
State modem periodic law of classification of elements.  (2012 D, 2013 OD)
Answer:
Modem periodic law states “that the properties of elements are a periodic function of their atomic numbers”.

Question 6.
The atomic numbers of three elements X, Y and Z are 3, 11 and 17 respectively. State giving reason which two elements will show similar chemical properties.  (2013 OD)
Answer:

Elements	Atomic Number	Electronic configuration
X	3	2, 1
Y	11	2, 8, 1
Z	17	2, 8, 7

X & Y will show similar chemical properties as both elements have same number of valence electrons, i.e., 1.

Question 7.
The atomic numbers of three elements A, B and C are 11,17 and 19 respectively. State giving reason which two elements will show similar chemical properties. (2013 OD)
Answer:

Elements	Atomic Number	Electronic configuration
A	11	2, 8, 1
B	17	2, 8, 7
C	19	2, 8, 8, 1

A and C will show similar chemical properties as both elements have the same number of valence electrons, i.e., 1.

Question 8.
Write any one difference in the electronic configurations of group 1 and group 2 elements? (2014 D)
Answer:

1. Elements of group 1 have one electron in their outermost shell.
2. Elements of group 2 have two electrons in their outermost shell.

Question 9.
List any two properties of the elements belonging to the first group of the modern periodic table.  (2014 OD)
Answer:
Elements of first group have one valence electron each in their atoms.
All the elements of group 1 have the same valency of 1.

Question 10.
Write the atomic numbers of two elements ‘X’ and ‘Y’ having electronic configuration 2,8,2 and 2,8,6 respectively. (2014 OD)
Answer:

Elements	Electronic configuration	Atomic numbers
X	2, 8, 2	12
Y	2, 8, 6	16

Question 11.
The atomic numbers of three elements A, B and C are 12, 18 and 20 respectively. State, giving reason, which two elements will show similar properties. (2014 OD)
Answer:

Elements	Atomic Number	Electronic configuration
A	12	2, 8, 2
B	18	2, 8, 8
C	20	2, 8, 8, 2

Elements A and C will show similar properties as both have same number of valence electrons.

Periodic Classification of Elements Class 10 Important Questions Short Answer Type I

Question 1.
(i) How do you calculate the possible valency of an element from the electronic configuration of its atoms?
(ii) Calculate the valency of an element X whose atomic number is 9. (2011 D)
Answer:
(i) The valency of an element is determined by the number of valence electrons present in the atom of the element. Electronic configuration gives the number of valence electrons in its atom. The number of valence electrons decides the number of electrons lost or gained by one atom of an element to achieve the nearest inert gas electron configuration and gives the valency of element.

(ii) Atomic number of X = 9
Electronic configuration =
K, L
2, 7
X gains 1 electron to achieve the nearest inert gas configuration , i.e., 2, 8. Therefore the valency of X is ‘1’.

Question 2.
How does the electronic configuration of an atom of an element relate to its position in the modem periodic table? Explain with one example.  (2011 D)
Answer:
The electronic configuration of an atom of an element gives its position in the modern periodic table.
(i) The ‘period number’ of an element is equal to the number of electron shells in its atom.
(ii)

• The group number of an element having upto two valence electrons is equal to the number of valence electrons.
  The group number of an element having more than 2 valence electrons is equal to the number of valence electrons plus 10.

Example: If the electronic configuration of an element is 2, 8, 7
Then its period number is 3 as it has three electron shells.
Its group number is 17 as it has 7 valence electrons. (∵ Group no. = 7 + 10 = 17)

Question 3.
The atomic numbers of three elements, X, Y and Z are 9,11 and 17 respectively. Which two of these elements will show similar chemical properties? Why?  (2011 D)
Answer:

Element	Atomic number	Electronic configuration	Valence electrons	Valency
X	9	2, 7	7	1
Y	11	2, 8, 1	1	1
Z	17	2, 8, 7	7	1

Element X and Z show similar chemical properties because both have same number of valence electrons, i.e., 7 as chemical properties of the element depends on the valence electrons.

Question 4.
How does the valency of elements vary (i) in going down a group, and (ii) in going from left to right in a period of the periodic table? (2011 OD)
Answer:
(i) The valency of elements remains the same in going down a group.
(ii) On moving from left to right in a period, the valency of elements increases from 1 to 4 and then decreases to 0 (zero).

Question 5.
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 has physical and chemical properties resembling those of Calcium and why?  (2011 OD)
Answer:
Ca (atomic number = 20) (electronic configuration = 2, 8, 8, 2)
Element with
atomic number = 12  (electronic configuration = 2, 8, 2)
atomic number = 19  (electronic configuration = 2, 8, 8, 1)
atomic number = 21  (electronic configuration = 2, 8, 8, 3)
atomic number = 38  (electronic configuration = 2, 8, 18, 8, 2)

Elements with atomic number 12 and 38 have physical and chemical properties resembling with Ca because valence electrons in these elements are same as Ca, i.e., 2. As properties of an element depend on its valence electrons.

Question 6.
How can the valency of an element be determined if its electronic configuration is known? What will be the valency of an element of atomic number 9 (nine)?  (2012 D)
Answer:
Valency of an element can be determined by knowing its valence electrons which can be determined by knowing the electronic configuration.
The elements with valence electrons 1 to 4; Their valencies are also equivalent to their respective valence electrons.

The elements with valence electrons 5 to 8: Their valencies correspond to (8 – V.E.).
Atomic number = 9
Valence electrons = 7
Electronic configuration = 2, 7
∴ Valancy = 8 – 7 = 1

Question 7.
An element ‘X’ has atomic number 13:  (2012 D)
(a) Write its electron configuration.
(b) State the group to which ‘X’ belongs?
(c) Is ‘X’ a metal or a non-metal?
(d) Write the formula of its bromide.
Answer:
Element ‘X’ that has atomic number = 13
(a) Electronic configuration =

(b) ‘X’ belongs to 13^th group.
(c) ‘X’ is a metal as its valence electrons are 3.

Question 8.
An element ‘M’ has atomic number 11:  (2012 D)
(a) Write its electron configuration.
(b) State the group to which ‘M’ belongs?
(c) Is ‘M’ a metal or a non-metal?
(d) Write the formula of its chloride
Answer:
Element ‘M’, atomic number = 11
(a) Electronic configuration =

(b) ‘M’ belongs to 1^st group as it has one valence electron.
(c) ‘M’ is a metal as atoms with valence electron 1, 2, 3 are metals.

Question 9.
Choose from the following: ₆C, ₈O, ₁₀Ne, ₁₁Na, ₁₄Si   (2012 OD)
(i) Elements that should be in the same period.
(ii) Elements that should be in the same group.
State reason for your selection in each case.
Answer:
Electronic configuration.

(i) Elements of same period ₆C, ₈O, ₁₀Ne (i.e. 2nd period)
Because all have same number of energy shells (two).
(ii) Elements of same group ₁₁Na, ₁₄Si (i.e., 14th)
Because both have same valance electrons (4)

Question 10.
An element ‘X’ belongs to 3^rd period and group 17 of the periodic table. State its (i) electronic configuration, (ii) valency. Justify your answer with reasoning.  (2012 OD)
Answer:
Element → ‘X’
Period → 3^rd
Group → 17^th
(i) Electronic configuration. ‘X’ has three energy shells – K, L, M (as it belongs to the 3^rd period and the period number of an element is equal to the number of shells in its atom).

‘X’ has 7 valence electrons since the group number of an element having more than 2 valence electrons is equal to the valence electrons plus 10.
10 + V.E. = group no. ⇒ 10 + V.E. = 17
∴ V.E. = 17 – 10 = 7
∴ Electronic configuration:

(ii) Valency = 8 – V.E. (If V.E. > 4) = 8 – 7 = 1

Question 11.
The formula of magnesium oxide is MgO. State the formula of barium nitrate and barium sulphate, if barium belongs to the same group as magnesium.  (2012 OD)
Answer:
Magnesium oxide is MgO

But valancy of oxygen is 2

∴ Valency of Mg = 2 and
Valancy of Ba is also 2

Question 12.
Choose from the following:  (2012 OD)
₂₀Ca, ₃Li, ₁₁Na, ₁₀Ne
(i) An element having two shells completely filled with electrons.
(ii) Two elements belonging to the same group of the periodic table.
Answer:

(i) Ne has two completely filled shells.
(ii) Li and Na belong to the same group as both have same number of valence electrons (1).

Question 13.
Why do all the elements of the (a) same group have similar properties, (b) same period have different properties? (2012 OD)
Answer:
(a) All the members of the same group have similar properties because the elements of a group have the same valence electrons.
(b) All the members of the same period have different properties because the elements of a period have different valence electrons.

Question 14.
An element E has following electronic configuration:  (2012 OD)

(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:
Element ‘E’, electronic configuration

(a) E belongs to 16^th group.
(b) E belongs to 3^rd period as it has 3 energy shells.
(c) Valence electrons present in E = 6.
(d) Valency of E = 8 –  6 = 2(8 – V.E.), if V.E. > 4 = 2

Periodic Classification of Elements Class 10 Important Questions Short Answer Type II

Question 1.
Two elements X and Y belong to group 1 and 2 respectively in the same period of periodic table. Compare them with respect to: (2011 D)
(i) the number of valence electrons in their atoms;
(ii) their valencies;
(iii) metallic character;
(iv) the sizes of their atoms;
(v) the formulae of their oxides;
(vi) the formulae of their chlorides.
Answer:

• X and Y belong to same period.
• X belongs to group ‘1’.
• Y belongs to group ‘2’.

(i) Valence electron in X is 1 whereas valence electrons in Y are 2.
(ii) The valency of X is 1 whereas valency of Y is 2.
(iii) X is more metallic than Y because metallic character decreases on moving from left to right in a period.
(iv) The size of X is more than Y because size of the atom decreases on moving from left to right in a period.
(v) Oxide of X = X₂O,  Oxide of Y = YO
(vi) Chloride of X = XCl,  Chloride of Y = YCl₂

Question 2.
On the basis of electronic configuration, how will you identify the first and the last element of a period?  (2011 D)
Answer:
The number of valence electrons increases from 1 to 8 on going from left to right in a period. Therefore the first element in every period has 1 valence electron and the last element in every period has 8 valence electrons (except in the first period where last element helium has only 2 valence electrons).

Question 3.
The atomic number of an element is 16. Predict   (2011 OD)
(i) the number of valence electrons in its atom;
(ii) its valencies;
(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:
The atomic number, Z = 16
Electronic configuration:

(i) Number of valence electrons in its atom = 6
(ii) Valency = 2
(iii) Group number = 16
(iv) It is a non-metal.
(v) The nature of oxide formed by it would be acidic.
(vi) Let this element be represented by the symbol ‘X’. Formula of its chloride is XCl₂.

Question 4.
(a) How does the metallic character of elements change along a period of the periodic table from the left to the right and why?
(b) How does the size of the atom change along a period of the periodic table from left to right and why? (2011 OD)
Answer:
(a) On moving from left to right in a period the metallic character of elements decreases. This is because on moving from left to right in a period of the periodic table, the nuclear charge increases due to gradual increase in the number of protons. Due to the increase in nuclear charge, the valence electrons are pulled in more strongly by the nuclear and it becomes more and more difficult for the atoms to lose electrons therefore on moving from left to right in a period thus metallic character decreases from left to right in a period.

(b) On moving from left to right in a period the size of the atom decreases because number of energy shells remains same in a period but nuclear charge increases from left to right and thus the force of attraction between the nucleus and the electrons increases which reduces the size of the atoms on moving from left to right in a period.

Question 5.
(a) How does the size of atoms of elements vary down a group in the periodic table? Why is it so?
(b) What does the tendency to lose electrons vary down a group in the periodic table? Give reason.  (2011 OD)
Answer:
(a) On going down in a group of the periodic table, the size of atoms increases because on going down in a group, a new shell of electrons is added to the atoms at every step due to which the size of atoms also increases.

(b) Tendency to lose electrons goes on increasing down the group because one energy shell is added at every step on going down in a group therefore the distance between the valence electrons and the nucleus and the force of attraction between the nucleus and the valence electrons decreases, therefore, tendency to lose electrons increases down the group.

Question 6.
Explain the variation of the following properties of the element in the periodic table.
(i) Atomic radius in a period.
(ii) Metallic character in a period.
(iii) Valency in a group.  (2012 D)
Answer:
(i) On moving from left to right in a period of the periodic table, the atomic radius of elements decreases.
(ii) On moving from left to right in a period, the metallic character of elements decreases and non- metallic character increases.
(iii) As the number of valance electrons in a group is the same, all the elements in a group have the same valency.

Question 7.
F, Cl and Br are the elements each having seven valence electrons. Which of these (i) has the largest atomic radius, (ii) is most reactive? Justify your answer stating reason for each.  (2012 D)
Answer:
F, Cl and Br belong to the same group (17th) as each has same number of valence electrons ‘7’.
Out of F, Cl and Br, Bromine has the largest atomic radius because on going down in a group of the periodic table, the size of atoms increases as number of electron shell of electrons is added to the atoms at every step.

Out of F, Cl and Br, Fluorine is the most reactive because the chemical reactivity of non-metals decreases on going down in a group of the periodic table as moving down in a group of non-metals, the size of the atoms goes on increasing. Thus the nucleus of atoms goes more and more deep inside it and hence attraction for the incoming electrons decreases. Therefore the tendency of atoms to gain electrons decreases, due to which their reactivity decreases.

Question 8.
An element ‘M’ has atomic number 12: (2012 D)
(a) Write its electronic configuration.
(b) State the group to which’ belongs.
(c) State the period to which’ belongs.
(d) State the valency of ‘M’.
(e) Is ‘M’ a metal or a non-metal?
(f) Write the formula of its oxide.
Answer:
Given: Element ‘M’, atomic number = 12
(a) Electronic configuration =

(b) ‘M’ belongs to the 2^nd group.
(c) ‘M’ belongs to the 3^rd period.
(d) Valency of ‘M’ is 2.
(e) ‘M’ is a metal as atoms with valence electron 1, 2, 3 are metals.

Question 9.
Na, Mg and Al are the elements having one, two and three valence electrons respectively. Which of these elements (i) has the largest atomic radius, (ii) is least reactive? Justify your answer stating reason for each. (2012 OD, 2015 D, 2017 D)
Answer:
Na, Mg and Al are the elements of same period where Na is the first, Mg is the second and Al is the third member.

(i) Na has the largest atomic radius because on moving from left to right in a period, the atomic radius of elements decreases. As we move from left to right in a period, the atomic number of elements increases and the extra electrons are added to the same shell. Due to large positive charge on the nucleus, the electrons are pulled in more close to the nucleus and the size of the atom decreases.

(ii) Sodium is very reactive, Mg is less reactive whereas Al is still less reactive because on moving from left to right in a period, the chemical reactivity of metals decreases.

Question 10.
(a) Which two criteria did Mendeleev use to classify the elements in his periodic table?
(b) Which element of period ‘3’ of the modern periodic table
(i) is the most reactive non-metal?
(ii) is the most reactive metal?
(iii) forms ion with -2 charge?
(iv) forms ion with +2 charge?   (2012 OD)
Answer:
(a) Mendeleev was guided by two factors:

1. Increasing atomic masses.
2. Grouping together of elements having similar properties, of making oxides and hydrides.

(b) (i) Cl is the most reactive non-metal.
(ii) Na is the most reactive metal.
(iii) S forms -2 charged ion.
(iv) Mg forms +2 charged ion.

Question 11.
(a) How are the following related?  (2012 OD)
(i) Number of valence electrons of different elements in the same group.
(ii) Number of shells of elements in the same period.
(b) How do the following change?
(i) Number of shells of elements as we go down a group.
(ii) Number of valence electrons of elements on moving from left to right in a period.
(iii) Atomic radius in moving from left to right along a period.
(iv) Atomic size down a group.
Answer:
(a) (i)

• All the elements of a group of the periodic table have the same number of valence electrons.
• Group number of an element upto 2 valance electrons (VE) is equal to the number of valence electrons and group number of an element having valance electrons (VE) more than 2 is equal to (10 + V.E.).

(ii)

• All the elements of a period of the periodic table have same number of shells.
• The period number of an element is equal to the number of electron shells in its atom.

(b) (i) As we go down in a group, a new shell of electrons is added to the atoms at every step.
(ii) On moving from left to right in a period, the number of valence electrons increases from 1 to 8.
(iii) On moving from left to right in a period the atomic radius of elements decreases due to the large positive charge on the nucleus, the electrons are pulled in more close to the nucleus.
(iv) On going down in a group, the size of the atoms increases as a new shell of electrons is added to the atoms at every step.

Question 12.
Given below are some elements of the modern periodic table:  (2013 D)
₄Be, ₉Fe, ₄₄Si, ₁₉K, ₂₀Ga
(i) Select the element that has one electron in the outermost shell and write its electronic configuration.
(ii) Select two elements that belong to the same group. Give reason for your answer.
(i) 19K has one electron in its outermost shell.Electronic configuration: K L M N
(iii) Select two elements that belong to the same period. Which one of the two has bigger atomic size?
Answer:

(i) ₁₉K has one electron in its outermost shell.
Electronic configuration:

(ii) ₄Be and ₂₀Ca belong to the same group as both have same number of valence electrons (2).
(iii) ₁₉K and ₂₀Ca belong to the same period as both have same number of energy shells (K, L, M, N).
₁₉K has bigger atomic size as size decreases on moving from left to right in a period.

Question 13.
Write the number of periods the modem periodic table has. How do the valency and metallic character of elements vary on moving from left to right in a period? How do the valency and atomic size of elements vary down a group? (2013 D, 2015 D)
Answer:

• There are seven periods in the modem periodic table.
• On moving from left to right in a period, the valency of elements increases from 1 to 4 and then decreases to 0.
• On moving from left to right in a period, the metallic characters of elements decreases as the electropositive character of elements decreases on moving from left to right in a period.
• All the elements in a group have same valency because the number of valence electrons in a group is same.
• On going down in a group of the periodic table, the size of atoms increases because a new shell of electrons is added to the atoms at every step on moving down in a group.

Question 14.
An element X (atomic number 17) reacts with an element Y (atomic number 20) to form a compound.
(a) Write the position of these elements in the modern periodic table.
(b) Write the formula of the compound formed.
Justify your answer in each case.  (2013 D)
Answer:

(a)

• Element ‘X’ has seven electrons in its outermost shell therefore it belongs to 17^th group.
• Element ‘X’ has 3 shells (energy) thus it belongs to 3^rd period.
• Element ‘Y’ has 2 electrons in its outermost shell therefore it belongs to 2^nd group.
• Element ‘Y’ has 4 energy shells thus it belongs to 4^th period.

(b) Valency of element ‘X’ is 1 and it is a non-metal whereas the valency of element ‘Y’ is 2 and it is a metal. Thus the bond between Y and X is ionic and formula is given by

Question 15.
The elements Li, Na and K, each having one valence electron, are in period 2, 3 and 4 respectively of modem periodic table. (2013 D)
(a) In which group of the periodic table should they be?
(b) Which one of them is least reactive?
(c) Which one of them has the largest atomic radius? Give reason to justify your answer in each case.
Answer:

	Valence electron	Period number
Li	1	2
Na	1	3
K	1	4

(a) All these elements belong to 1^st group because all elements have one electron in their respective outermost shell.
(b) Li is least reactive as reactivity of metals increases on going down in a group.
(c) K has the largest atomic radius because Li belongs to 2^nd period thus has 2 energy shells, Na belongs to the 3^rd period thus has 3 energy shells and K belongs to 4^th period thus has 4 energy shells. So as the number of shells increases so does the size of the atom.

Question 16.
The electronic configuration of an element ‘X’ is 2, 8, 8, 2. To which (a) period and (b) group of the modern periodic table does ‘X’ belong? State its valency. Justify your answer in each case. (2013 OD)
Answer:
Electronic configuration of ‘X’ is 2, 8, 8, 2.
(a) It belongs to the 4^th period because it has four energy shells and period number of an element is equal to the number of energy shells.
(b) It belongs to 2^nd group because it has 2 valence electrons and the group number of an element having upto two valence electrons is equal to the number of valence electrons.

Its valency is ‘2’. To acquire inert gas configuration ‘X’ either loses 2 electrons or gains 6 electrons. Since the shorter route is to lose 2 electrons for obtaining inert gas configuration, hence its valency is 2.

Question 17.
Four elements P, Q, R and S have atomic numbers 12,13,14 and 15 respectively.   (2013 OD)
Answer the following questions giving reasons:
(i) What is the valency of Q?
(ii) Classify these elements as metals and non-metals.
(iii) Which of these elements will form the most basic oxide?
Answer:

Element	Atomic number	Electronic configuration	Valence electrons
P	12	2, 8, 2	2
Q	13	2, 8, 3	3
R	14	2, 8, 4	4
S	15	2, 8, 5	5

(i) Valency of Q is 3 as Q loses 3 electrons to obtain inert gas configuration.
(ii) P and Q are metals as these elements lose 2 and 3 electrons respectively to obtain inert gas configuration and those elements which lose electrons are metals.
R and S are non-metals as these elements either share or gain electrons for attaining inert gas configuration and those elements which accept electrons either by sharing or gaining are non-metals.
(iii) Element P will form the most basic oxide because P is the most metallic element among the 4 given elements and thus forms the most basic oxide.

Question 18.
The atomic number of an element is 17. Predict (a) its valency, (b) whether it is a metal or non-metal, (c) its relative size with respect to other members of its group. Justify your answer in each case. (2013 OD)
Answer:
Atomic number of given element = 17
∴ Electronic configuration of given element = 2, 8, 7

(a) Valency = 1
Since this element requires one electron to complete its octet (outermost shell).
(b) This element is a non-metal as it gains one electron to complete its outermost shell and elements which gain electrons are non-metals.
(c) This element is the 2^nd member of this group. Its size is greater than the 1^st member (2, 7) and smaller than the rest of the other members of that group because size of the atom increases on going down in a group as one energy shell is added at every step.

Question 19.
The electronic configuration of an element ‘X’ is 2, 8, 8, 2. To which (a) period and (b) group of the modem periodic table does ‘X’ belong? State its valency. Justify your answer in each case.  (2013 OD)
Answer:
Element: ‘X’,
Electronic configuration: 2, 8, 8, 2
(a) ‘X’ belongs to 4^th period because ‘X’ has four energy shells and energy shell number corresponds to the period number.
(b) ‘X’ belongs to 2^nd group because ‘X’ has 2 electrons in its outermost shell and valence electrons (1, 2) corresponds to the group number of the element.
Valency: 2

Question 20.
The atomic number of an element ‘X’ is 20. Write
(a) its valency,
(b) whether it is a metal or non-metal,
(c) the formula of compound formed when the element ‘X’ reacts with an element ‘Y’ of atomic number 8.
Justify your answer in each case.  (2013 OD)
Answer:

Element	Atomic number	Electronic configuration	Valance electrons
X	20	2, 8, 8, 2	2

(a) Valency of X is 2 as it loses two electrons to acquire nearest inert gas configuration.
(b) X is a metal as it loses electrons to acquire inert gas configuration and an element which loses
electrons for acquiring inert gas configuration is a metal.
(c)

Element	Atomic number	Electronic configuration	Valance electrons
Y	8	2, 6	2

Formula:

Question 21.
Based on the group valency of elements state the formula for the following giving justification for each:
(i) Oxides of 1^st group elements,
(ii) Halides of the elements of group 13, and
(iii) Compounds formed when an element of group 2 combines with an element of group 16.   (2014 D)
Answer:
(i) Oxides of 1^st group element (Let it be A):

∴ Chemical Formula:

(ii) Halides of the elements of group 13 (Let it be M):

∴ Chemical Formula:

(iii) Compounds of element of group 2 combines with an element of group 16 (Y):

∴ Chemical Formula:

Question 22.
(a) Define the following terms: (2014 D)
(i) Valency
(ii) Atomic size
(b) How do the valency and the atomic size of the elements vary while going from left to right along a period in the modem periodic table?
Answer:
(a) (i) Valency. The combining capacity of an element is called its valency.

• The valency of an element is determined by the number of valence electrons present in the atom of the element.
• The number of electrons lost or gained or shared by one atom of an element to achieve the nearest inert gas configuration gives the valency of the element.

(ii) Atomic size. The atomic size of an atom is the distance between the centre of the nucleus and outermost electron shell of an isolated atom.

(b)

• On moving from left to right along a period in the modern periodic table, the valency of
  elements increases from 1 to 4 and then decreases to zero.
• On moving from left to right in a period of the periodic table, the size of the atoms decreases.

Question 23.
Consider two elements ‘A’ (Atomic number 17) and ‘B’ (Atomic number 19):
(i) Write the positions of these elements in the modem periodic table giving justification.
(ii) Write the formula of the compound formed when ‘A’ combines with ‘B’.
(iii) Draw the electron dot structure of the compound and state the nature of the bond formed between the two elements.  (2014 D)
Answer:
(i) Element ‘A’:
Atomic number = 17
Electronic configuration = 2, 8, 7
‘A’ belongs to 17^ht group as it has ‘7’ electrons in its outermost shell.
‘A’ belongs to 3^rd period as it has three shells in its atom.

Element ‘B’:
Atomic number = 19
Electronic configuration = 2, 8, 8, 1
‘B’ belongs to 1^st group as it has one electron in its outermost shell.
‘B’ belongs to 4^th period as it has four shells in its atom.

(ii) Valency of A = -1
Valency of B = +1

(iii)

Ionic bond is formed between B and A.

Question 24.
The electrons in the atoms of four elements A, B, C and D are distributed in three shells having 1, 3, 5 and 7 electrons in the outermost shell respectively. State the period in which these elements can be placed in the modern periodic table. Write the electronic configuration of the atoms of A and D and the molecular formula of the compound formed when A and D combine.   (2014 OD)
Answer:

Elements	Shells K, L, M	Atomic number
A	2, 8,1	11
B	2,8,3	13
C	2, 8, 5	15
D	2, 8, 7	17

All these elements belong to the 3^rd period as all these elements have three shells.

Element A:
Electronic Configuration 2, 8,1
Valence electrons: 1
Valency: 1

Element D:
Electronic Configuration 2, 8, 7
Valence electrons: 7
Valency: 1

Molecular formula:

Nature of the compound is ionic.

Question 25.
Study the following table in which positions of six elements A, B, C, D, E and F are shown as they are in the modern periodic table:

Group → Period ↓	1	2	3 – 12	13	14	15	16	17	18
2	A					B			C
3				D	E				F

On the basis of the above table, answer the following questions:
(i) Name the element which forms only covalent compounds.
(ii) Name the element which is a metal with valency three.
(iii) Name the element which is a non-metal with valency three.
(iv) Out of D and E, which is bigger in size and why?
(v) Write the common name for the family to which the elements C and F belong.  (2014 OD)
Answer:
(i) E
(ii) D
(iii) B
(iv) D is bigger in size because as we move from left to right in a period, the size of the atom decreases as positive charge increases but the number of shells remains the same. Thus the electrons are pulled in more close to the nucleus.
(v) Noble gases

Question 26.
What is meant by ‘group’ in the modem periodic table? How do the following change on moving from top to bottom in a group?
(i) Number of valence electrons
(ii) Number of occupied shells
(iii) Size of atoms
(iv) Metallic character of elements
(v) Effective nuclear change experienced by valence electrons   (2014 OD)
Answer:
The vertical columns in a periodic table are called groups.
(i) All the elements of a group have the same number of valence electrons.
(it) On moving down in a group the number of occupied (filled) shells increases gradually.
(iii) On going down in a group the size of atoms increases because a new shell of electrons is added to the atoms at every step.
(iv) On going down in a group the metallic character of elements increases.
(v) On moving down in a group, one more electron shell is added at every stage and size of the atom increases. Thus valence electrons move more and more away from the nucleus and hold of the nucleus or nuclear charge on valence electrons decreases.

Question 27.
The elements Be, Mg and Ca each having two electrons in their outermost shells are in periods 2, 3, and 4 respectively of the modem periodic table. Answer the following questions, giving justification in each case:
(i) Write the group to which these elements belong.
(ii) Name the least reactive element.
(iii) Name the element having largest atomic radius.  (2014 OD)
Answer:

Element	Valance electrons	Period	Electronic configuration
Be	2	2	2, 2
Mg	2	3	2, 8, 2
Ca	2	4	2, 8, 8, 2

(i) All these elements belong to the 2^nd group as all have two electrons in their outermost shell.

(ii) Be is the least reactive metal because reactivity of metals increases in a group as the tendency to lose electrons in a group increases.
Therefore Be being the smallest in all the given elements of a group has its valence electrons nearest to the nucleus. So the removal of electrons from its valence shell will be difficult.

(iii) Ca has the largest atomic radius because it has maximum number of shells, i.e., 4.

Question 28.
How many groups and periods are there in the modern periodic table? How do the atomic size and metallic character of elements vary as we move:
(i) down a group and
(ii) from left to right in a period?  (2015 D)
Answer:
There are 18 groups and 7 periods in the modern periodic table.
(i) Atomic size and metallic character of the elements increases down a group.
(ii)

• Atomic size and metallic character of elements decreases from left to right in a period.
• Metallic character of the element is decreased.

Question 29.
From the following elements: (2015 D)
₄Be; ₉F ; ₁₉K; ₂₀Ca
(i) Select the element having one electron in the outermost shell.
(ii) two elements of the same group.
Write the formula of and mention the nature of the compound formed by the union of 19K and element X (2, 8, 7).
Answer:
(i) ₁₉K
(ii) ₄Be, ₂₀Ca belong to the same group as both have 2 electrons in their outermost shells.
Electronic configuration of K (19) = 2, 8, 8,1
Valency of K = 1
Electronic configuration of X (2, 8, 7)
Valency of X = 1

The formula of the compound formed is KX. The compound KX is of ionic nature.
The bond is formed by transference of electrons.

Question 30.
Two elements ‘P’ and ‘Q’ belong to the same period of the modem periodic table and are in Group-1 and Group-2 respectively. Compare their following characteristics in tabular form:
(a) The number of electrons in their atoms
(b) The sizes of their atoms
(c) Their metallic characters
(d) Their tendencies to lose electrons
(e) The formula of their oxides
(f) The formula of their chlorides   (2015 OD)
Answer:

Characteristics	P	Q
(a) No. of electrons in their atoms	Less than Q 3 11 19	More than P 4 12 20 (any one pair)
(b) Size of the atom	Bigger	Smaller
(c) Metallic character	More metallic	Less metallic
(d) Tendency to lose electrons	More	Less
(e) Formula of their oxides	p2O	QO
(f) Formula of their chlorides	PCI	QCl2

Question 31.
Taking the example of an element of atomic number 16, explain how the electronic configuration of the atom of an element relates to its position in the modern periodic table and how valency of an element is calculated on the basis of its atomic number.  (2015 OD)
Answer:
Atomic number of the element = 16
Electronic configuration =

Since this element has 3 shells, the period number will be 3 as period number is equal to the number of shells that start filling up in it.
No. of valence electrons = 6
∴ The group number will be 10 + 6 = 16

The valency of an element is determined by the number of valence electrons present in the outermost shell.
Valency of the element will be = 8 – valence electrons = 8 – 6 = 2

Question 32.
The element ₄Be, ₁₂Mg and ₂₀Ca, each having two valance electrons in their valance shells, are in periods2, 3 and 4 respectively of the modern periodic table. Answer the following questions associted with these elements, giving reason in each case:
(a) In which group should they be?
(b) Which one of them is least reactive?
(c) Which one of them has the largest atomic size? (2015 OD)
Answer:

(a) They all belong to 2^nd group of the periodic table as all the elements have two valance electrons in their valance shell.
(b) Be is least reactive as it contains the least number of shells (i.e., 2). the tendaency of an atom to lose electrons on moving down in a group increases as valance electrons move more and more away from the nucleus and neuclear charge on electrons decreases.
(c) Ca has the largest atomic size as it has the maximum number of shells.

Question 33.
Given below are some elements of the modern periodic table. Atomic number of the element is given in the parentheses: A(4), B(9), C(14), D(19), E(20)
(a) Select the element that has one electron in the outermost shell. Also write the electronic configuration of this element.
(b) Which two elements amongst these belong to the same group? Give reason for your answer.
(c) Which two elements amongst these belong to the same period? Which one of the two has bigger atomic radius? (2015 OD)
Answer:

Elements	Atomic number	Electronic configuration(K, L, M, N)
A	4	2, 2
B	9	2, 7
C	14	2, 8, 4
D	19	2, 8, 8, 1
E	20	2, 8, 8, 2

(a) D has 1 electron in the outermost shell. Atomic no. = 19, Elec. Configuration = 2, 8, 8,1
(b) A and E belong to the same group, i.e., 2^nd group of the periodic table as both have same number of valence electrons (i.e., 2) in their outermost shell.

(c) A & B and D & E belong to the same period as they have the same no. of valence shells.
A and B have 2 shells thus belong to 2^nd period of the periodic table.
D and E have 4 shells thus belong to 4^th period of the periodic table.

A has the bigger atomic radius among A and B and D has the bigger atomic radius among D and E as atomic size decreases on moving from left to right in a period (or atomic size decreases as the atomic number increases in a period).

Question 34.
Calcium is an element with atomic number 20. Stating reason answer each of the following questions:
(i) Is calcium a metal or non-metal?
(ii) Will its atomic radius be larger or smaller than that of potassium with atomic number 19?
(iii) Write the formula of its oxide.  (2016 D)
Answer:
Calcium (Ca)
Atomic number = 20
Electronic configuration = 2, 8, 8, 2
(i) It is a metal as it has two electrons in its outermost shell and it loses these two electrons to acquire inert gas configuration and form +2 ion.

(ii) Potassium (atomic number 19) is placed before Calcium (atomic number 20) in the same period (IV period). Therefore, size of Ca is smaller than K since on moving from left to right in a period of the periodic table, the size of the atoms decreases. As on moving from left to right in a period, the atomic number of the elements increases which means that the number of protons and electrons in the atoms increases (the extra electrons being added to the same shell). Due to large positive charge on the nucleus, the outermost shell is pulled with more force by the nucleus, it moves closer to the nucleus and the size of the atom decreases.

(iii) The oxide of calcium is calcium oxide (CaO).

Question 35.
An element ‘M’ with electronic configuration (2, 8, 2) combines separately with (NO₃)^–, (SO₄)^2- and (PO₄)^3- radicals. Write the formula of the three compounds so formed. To which group and period of the Modern Periodic Table does the element ‘M’ belong? Will ‘M’ form covalent or ionic compounds? Give reason to justify your answer. (2016 D)
Answer:

• The electronic configuration (2, 8, 2) of the element ‘M’ suggests that it belongs to group 2 and period 3 of the Modern Periodic Table and its valency is 2.
• The chemical formulae of the compounds formed are:
  (i) M(NO₃)₂ (ii) MSO₄ (iii) M₃(PO₄)₂
• ‘M’ will form ionic compounds by losing two valence electrons to achieve a noble gas configuration, that is, a stable octet in the valence shell.

Question 36.
Name any two elements of group one and write their electronic configurations. What similarity do you observe in their electronic configurations? Write the formula of oxide of any of the aforesaid element. (2016 D)
Answer:
Two elements of group 1
Na = Sodium
K = Potassium

Electronic configuration of:
Na = 2,8,1
K = 2,8,8,1
Similarity between Na and K: Both have one electron in their outermost shell. Thus both have one valence electron and valency one.
Oxide of Sodium ⇒ Na₂O Oxide of Potassium ⇒ K₂O

Question 37.
Two elements ‘A’ and ‘B’ belong to the 3rd period of Modem 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 lose electrons
(d) The formula of their oxides
(e) Their metallic character
(f) The formula of their chlorides (2016 D)
Answer:
Electronic configuration of A = 2, 8, 2
Electronic configuration of B = 2, 8, 3

Characteristics	A	B
(a) No. of electrons in their atom	12	13
(b) Size of their atoms	Bigger	Smaller
(c) Their tendencies to lose electrons	More	Less
(d) The formula of their oxides	AO or A2 O2	B2O3
(e) Their metallic character	More metallic	Less metallic
(f) The formula of their chlorides	ACl2	BCl3

Question 38.
An element ‘X’ belongs to 3^rd period and group 16 of the Modern Periodic Table.
(a) Determine the number of valence electrons and the valency of ‘X’.
(b) Molecular formula of the compound when ‘X’ reacts with hydrogen and write its electron dot structure.
(c) Name the element ‘X’ and state whether it is metallic or non-metallic.  (2016 OD)
Answer:
(a) An element ‘X’ belongs to 3^rd period, thus it has three energy shells. It also belongs to 16^th group, thus it has 6 valence electrons.
Electronic configuration of X =

∴ Valence electrons = 6
∴ Valency = 8 – 6 = 2

(b) Molecular Formula:

Electron dot structure:

(c) The element ‘X’ is Sulphur. It is a non-metal.

Question 39.
An element ‘X’ has mass number 35 and number of neutrons 18. Write atomic number and electronic configuration of ‘XAlso write group number, period number and valency of ‘X’.  (2016 OD)
Answer:
‘X’
Mass number = 35
No. of neutrons = 18
∴ Atomic number of X = Mass number – Number of neutrons = 35 – 18 = 17
Thus, electronic configuration of X =

Group number = 17^th
Period number = 3^rd
Valency = 8 – 7 = 1

Question 40.
Three elements ‘X’, ‘Y’ and ‘Z’ have atomic numbers 7, 8 and 9 respectively.
(a) State their positions (Group number and period number both) in the Modern Periodic Table.
(b) Arrange these elements in the decreasing order of their atomic radii.
(c) Write the formula of the compound formed when ‘X’ combines with ‘Z’.   (2016 OD)
Answer:

	Atomic number	Electronic configuration	Valance electrons	Valancy
X	7	2, 5	5	3
Y	8	2, 6	6	2
Z	9	2, 7	7	1

(a)

	Group no.	Period no.
X	15	2
Y	16	2
Z	17	2

(b) X > Y > Z
X, Y, Z belong to the same period and size of the atom decreases on moving from left to right in a period.

(c) Formula of X when combined with Z

Question 41.
The position of eight elements in the Modern Periodic Table is given below where atomic numbers of elements are given in the parenthesis.  (2016 OD)

Period No.
2	Li (3)	Be (4)
3	Na (11)	Mg (12)
4	K (19)	Ca (20)
5	Rb (37)	Sr (38)

(i) Write the electronic configuration of Ca.
(it) Predict the number of valence electrons in Rb.
(iii) What is the number of shells in Sr?
(iv) Predict whether K is a metal or a non-metal.
(v) Which one of these elements has the largest atom in size?
(vi) Arrange Be, Ca, Mg and Rb in the increasing order of the size of their respective atoms.
Answer:
(i) Ca(20)
Electronic configuration: 2, 8, 8, 2

(ii) Li(3)

Thus Rb belongs to 1^st group
∴ Valence electrons in Rb = 1

(iii) Sr (38) belongs to 5^th period. Therefore, Sr has 5 shells.

(iv) K(19) belongs to 1^st group.
∴ Valence electrons in K(19) is 1. It has a high tendency to donate an electron and achieve a noble gas configuration. Hence, it is a metal.

(v) Largest atom = Rb

(vi) Increasing order of the atomic size
Be < Mg < Ca < Rb

Question 42.
An element ‘X’ belongs to 3^rd period and group 13 of the Modem Periodic Table.   (2016 OD)
(a) Determine the valence electrons and the valency of ‘X’.
(b) Molecular formula of the compound formed when ‘X’ reacts with an element ‘Y’ (atomic number = 8).
(c) Write the name and formula of the compound formed when ‘X’ combines with chlorine.
Answer:
X belongs to 3^rd period and the 13^th group.
(a) Since X belongs to 13^th group, therefore its valence electrons = 3.
∵ Valency of element X = 3

(b) When ‘X’ reacts with element ‘Y’:

∴ Molecular formula of compound formed is X₂Y₃.

(c) Electronic configuration = 2, 8, 3
∴ X is Aluminium (Al)
Atomic number of Chlorine = 17
Electronic configuration = 2, 8, 7
No. of valence electrons = 7
∴ Valency of Chlorine = 8 – 7 = 1

∴ Compound formed is Aluminium chloride and molecular formula of compound is AlCl₃.

Question 43.
Write the names given to the vertical columns and horizontal rows in the Modem 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 left to right in a horizontal row? Give reason in support of your answer in the above two cases. (2017 D)
Answer:
Vertical coloumns are called Groups.
The horizontal rows are called Periods.

Metallic character increases on going down in a group.
Reason. Ability to lose electrons increases on moving down the group due to increase in distance be¬tween the nucleus and the valence electrons. There is decrease in the attraction between the nucleus and the valence electrons.

Atomic radius decreases as we move from left to right in a period.
Reason. The nuclear charge increases on moving from left to right across a period resulting in in-crease in the attraction between the nucleus and the valence electrons.

Question 44.
An element P (atomic number 20) reacts with an element Q (atomic number 17) to form a compound. Answer the following questions giving the reason.
Write the position of P and Q in the Modern Periodic Table and the molecular formula of the compound formed when P reacts with Q.  (2017 D)
Answer:
Element P: Electronic configuration: 2, 8, 8, 2
So, element P belongs to 4^th period and group 2, as it has four shells and 2 valence electrons. Element Q: Electronic configuration: 2, 8, 7

So element Q belongs to 3^rd period and group 17, as it has three shells and seven valence electrons. Molecular formula of compound will be PQ₂. As valency of P is 2 and valency of Q is one.

Periodic Classification of Elements Class 10 Important Questions Long Answer Type

Question 1.
The elements of the third period of the Periodic Table are given below:

(a) Which atom is bigger, Na or Mg? Why?
(b) Identify the most (i) metallic and (ii) non-metallic element in Period 3.
(c) Which is more non-metallic, S or Cl?
(d) Which has higher atomic mass, A1 or Cl?  (2013 D)
Answer:
(a) Na is bigger than Mg because on moving from left to right in a period, the atomic number of elements increases which means that the number of protons and electrons in the atom increases. (The extra electrons being added to the same shell).

(b) (i) Most metallic element is Na. Most non-metallic element is Cl;
Because on moving from left to right in a period the nuclear charge increases thus the valence electrons are pulled in more strongly by the nucleus and it becomes more and more difficult for the atoms to lose electrons so tendency of atoms to lose electrons (i.e., metallic character) decreases on moving from left to right in a period.

On the other hand, due to increased nuclear charge, it becomes easier for the atoms to gain electrons. So the tendency to gain electrons (i.e. non-metallic character) increases on moving than left to right in a period.

(c) Cl is more non-metallic than S because on moving from left to right in a period, nuclear charge increases so the tendency to gain electrons increases (i.e., non-metallic character).

(d) Cl has higher atomic mass because on moving from left to right in a period, atomic number increases. Simultaneously, atomic mass increases.

Question 2.
Explain the trends in the Modern Periodic Table of various properties like valency, atomic size, metallic and non-metallic properties of the atoms of elements.  (2013 OD)
Answer:
(i) Valency. Elements belonging to same group have same number of valence electrons and thus same valency.
Valency in a particular period from left to right first increases as positive valency and then decreases as negative valency.
Example, In elements of 2^nd period:
Li has 1⁺ valency, then Be²⁺, Be³⁺, C⁴⁺ covalency, N^3- valency, then O^2- and F(^–) valency.

(ii) The atomic size or atomic radius increases as we move down in a group and it decreases as we move from left to right in a period. Atomic size increases down a group due to increase in the number of shells. Atomic size decreases along a period due to increase in the nuclear charge which tends to pull the electrons closer to the nucleus and reduces the size of the atom.

(iii) Metallic and non-metallic properties. In the modern periodic table metals are on the left side and non-metals are on the right side. A zigzag line of metalloids separates metals from non-metals. Metallic character decreases from left to right in a period and increases while going down in a group.

Non-metallic character increases from left to right in a period due to increase in the electro-negativity and this character decreases from top to bottom in a group due to decrease in the electro-negativity of atoms while going down in a group.

Question 3.
The position of three elements A, B and C in the Periodic Table are shown below:

Group →	I	II	III	IV	V	VI	VII
Period ↓
1	B
2							A
3						C

Giving reasons, explain the following:
(a) Element A is a non-metal.
(b) Atom of element C has a larger size than atom of element A.
(c) Element B has a valency of 1.  (2014 D)
Answer:
(a) Element A is a non-metal because it belongs to Group VII therefore it has 7 electrons in its outermost shell. It needs one electron to acquire nearest inert gas configuration and form a monovalent -ve ion.

(b) Atom of element C has a larger size than that of element A because C belongs to the 3^rd period therefore it has three shells whereas A belongs to the 2^nd period thus it has two shells.

(c) Element B has a valency of 1 because it belongs to Group I and has one valence electron. To complete its outermost shell or to acquire nearest noble gas configuration it loses one electron and forms a monovalent + ve ion.

Question 4.
(a) Why do we classify elements?
(b) What were the two criteria used by Mendeleev in creating his Periodic Table?
(c) Why did Mendeleev leave some gaps in his Periodic Table?
(d) In Mendeleev’s Periodic Table, why was there no mention of noble gases like Helium, Neon and Argon?
(e) Would you place the two isotopes of Chlorine, Cl-35 and Cl-37 in different slots because of their different atomic masses or in the same slot because their chemical properties are the same? Justify your answer. (2014 OD)
Answer:
(a) As different elements were being discovered, scientists gathered more information about the properties of these elements. It was observed that it was difficult to organise all the information or properties of these elements. So scientists started discovering some pattern in their properties to classify all the known elements to make their study easier.

(b) Atomic mass and similarity of chemical properties (compounds formed by elements with oxygen and hydrogen) were the two criteria used by Mendeleev in his Periodic Table.

(c) Mendeleev left some gaps in his Periodic Table as he predicted the existence of some elements that had not been discovered at that time.

(d) Noble gases like helium, neon, argon etc. were not mentioned in Mendeleev’s Periodic Table because these gases were discovered later as they are very inert and present in extremely low concentrations in our atmosphere. After the discovery of noble gases they could be placed in a new group without disturbing the existing order of the Periodic Table.

(e) The two isotopes of chlorine Cl-35 and Cl-37 would be placed in different slots of Mendeleev’s Periodic Table as both have different atomic masses. But the two isotopes have same chemical properties so would also be placed in the same slot. This is a contradiction. Thus position of isotopes of an element is not defined in Mendeleev’s Periodic Table.