Please refer to Periodic Classification of Elements Class 10 Science Important Questions with solutions provided below. These questions and answers have been provided for Class 10 Science based on the latest syllabus and examination guidelines issued by CBSE, NCERT, and KVS. Students should learn these problem solutions as it will help them to gain more marks in examinations. We have provided Important Questions for Class 10 Science for all chapters in your book. These Board exam questions have been designed by expert teachers of Standard 10.
Class 10 Science Important Questions Periodic Classification of Elements
Very Short Answer:
Question. Write two reasons responsible for the late discovery of noble gases.
Answer : a. They are inert i.e., least reactive.
b. They are less abundant in nature except Argon.
Question. How many vertical columns are there in the modern periodic table and what are they called?
Answer : There are 18 vertical columns in Modern Periodic Table. They are called groups.
Question. List any two properties of elements belonging to the first group of modem periodic table.
Answer :
a. They have 1 valence electron.
b. Their valency is equal to 1.
Question. The atomic number of three elements A, B and C are 12, 18 and 20 respectively. State giving reason, which two elements will show similar properties.
Answer : A(12): 2, 8, 2, B(18): 2, 8,8, C(20): 2, 8, 8, 2
A and C belong to the same group because they have same number of valence electrons.
Question. Write the atomic numbers of two elements X and Y having electronic configurations 2, 8, 2 and 2, 8, 6, respectively.
Answer : X = 2 + 8 + 2 = 12
Y = 2 + 8 + 6 = 16
Question. Lithium, Sodium and potassium form a Dobereiner’s triad. The atomic masses of lithium and potassium are 7 and 39 respectively. Predict the atomic mass of sodium.
Answer : Atomic mass of Sodium
Question. Why was the system of classification of elements into triads not found suitable?
Answer : It is because all the elements discovered at that time could not be classified as Dobereneire’s triads.
Question. State the modern periodic law of classification of elements.
Answer : Modern Periodic Law: It states ‘properties of elements are a periodic function of their atomic number’.
Question. Out of the three elements P, Q and R having atomic numbers 11, 17 and 19 respectively, which two elements will show similar properties and why?
Answer : P(11): 2, 8, 1; Q(17): 2, 8, 7; R(19): 2, 8, 8, 1
P and R will show similar chemical properties because they have the same number of valence electrons.
Question. Write the formula which is used to determine the maximum number of electrons that a shell can accommodate.
Answer : 2n2 , where n is the shell number.
Question: Write the number of horizontal rows in the Modern Periodic Table. What are these rows called?
Answer: There are seven horizontal rows of elements
in the Modern periodic table which are known as periods.
Question: Why did Mendeleev have gaps in his periodic table?
Answer: Mendeleev placed elements with similar properties one below the other leaving gaps for elements which could match with other elements of that group but had not been discovered by that time.
Question: Write the number of vertical columns in the Modern Periodic Table. What are these columns called?
Answer: There are 18 vertical columns in the Modern periodic table which are called groups.
Question: Write any one di erence in the electronic configurations of group 1 and group 2 elements.
Answer: Group 1 elements have one electron in their outermost shell while group 2 elements have two electrons in their outermost shell.
Question: Write the atomic numbers of two elements ‘X’and ‘Y’ having electronic configurations 2, 8, 2 and 2, 8, 6 respectively.
Answer: Electronic configuration of X = 2, 8, 2
∴ Atomic number = 2 + 8 + 2 = 12 Similarly,
Electronic configuration of Y = 2, 8, 6
∴ Atomic number = 2 + 8 + 6 = 16
Question: 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.
Answer: Atomic number of A = 12
∴ Electronic configuration = 2, 8, 2
Similarly, for B(18) = 2, 8, 8
for C(20) = 2, 8, 8, 2
As elements A and C contain two valence electrons in their outermost shell (group-2) they will show similar properties.
Short Answer: I
Question: 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)?
Answer: Valency of an element is determined by the number of electrons present in its outermost shell. For elements having outermost electrons 1 to 4, valencies are equivalent to their respective valence electrons. For elements having outermost electrons 5 to 8, valency is calculated as; Valency = 8 – (Number of valence electrons) For element having atomic number = 9 Electronic configuration = 2, 7 Valency = 8 – 7 = 1
Question: Choose from the following :
6C, 8O, 10Ne, 11Na, 14Si
(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: The electronic configurations of the given elements are :
6C = 2, 4
8O = 2, 6
10 Ne = 2, 8
11Na = 2, 8, 1
14Si = 2, 8, 4
(i) 6C, 8O, 10Ne, all contain two shells hence, they belong to same period i.e., second period.
11Na, 14Si both contain three shells hence, they belong to third period.
(6C, 8O, 10Ne) ⇒ period number 2 (11Na, 14Si) ⇒ period number 3
(ii) 6C and 14Si belong to the same group as they both contain 4 electrons in their outermost shell.
Thus, 6C and 14Si belong to group 14.
Question: An element ‘X’ has atomic number 13 :
(a) Write its electronic 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: X has atomic number = 13
(a) Electronic configuration of X = 2, 8, 3
(b) As X contains 3 valence electrons in its outermost shell, it belongs to group 13.
(c) X is a metal as it contains 3 valence electrons which can be lost easily.
(d) Formula of X with bromine will be
Question: Choose from the following :
4Be, 9F, 19K, 20Ca
(i) The element having one electron in the outermost shell.
(ii) Two elements of the same group.
Answer: The electronic configurations of the given elements are :
4Be = 2, 2
9F = 2, 7
19K = 2, 8, 8, 1
20Ca = 2, 8, 8, 2
(i) Potassium (K) has one electron in its outermost shell.
(ii) Be and Ca have two electrons in their outermost shells hence, they belong to same group.
Question: An element has atomic number 13.
(a) What is the group and period number to which this element belongs?
(b) Is this element a metal or a non-metal?
Answer: Atomic number of element = 13 Thus, its electronic configuration = 2, 8, 3
(a) From the electronic configuration, it can be easily seen that there are 3 electrons in the outermost shell which indicates that it belongs to group number 10 + 3 = 13. Moreover, the element has 3 shells in which electrons are filled thus, it belongs to period number 3.
(b) As the element contains 3 valence electrons which can be easily lost thus, it is a metal.
Short Answer: II
Question: 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?
Answer: Electronic configuration of M is 2, 8, 2 which shows that it belongs to group 2 and period 3 of the Modern periodic table. As it has 2 valence electrons, so the valency of element M will be 2.The chemical formulae of the compounds formed will be M(NO3)2, MSO4, M3(PO4)2 As M has two valence electrons, it can easily loose these electrons to attain a noble gas configuration. Hence, M will form ionic compounds.
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) The 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: 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?
Answer: Period number of Na, Mg and Al = 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 (+4) 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: An element ‘X’ belongs to 3rd 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.
Answer: (a) As the element ‘X’ belongs to 3rd period so, it will have three energy shells. Moreover, it belongs
to 16th group, so it will have six valence electrons.
∴ Electronic configuration of X = 2, 8, 6 Thus, valence electrons = 6 and valency = 8 – 6 = 2
(b) Molecular formula of the compound formed when X reacts with hydrogen = H2X The electron dot structure is as :
(c) The element X is sulphur and it is a non-metal.
Question: An element ‘X’ has mass number 35 and number of neutrons 18. Write atomic number and electronic configuration of ‘X’. Also write group number, period number and valency of ‘X’.
Answer: Mass number of X = 35 Number of neutrons = 18
∴ Number of electrons = Number of protons = (Mass number – Number of neutrons) = 35 – 18 = 17
Number of electrons of X = Atomic number of X = 17 Thus, electronic configuration of X = 2, 8, 7
As it has 7 electrons in the outermost shell, so it belongs to 17th group. Moreover the electrons are present in three shells, so it belongs to 3rd period.
Valency of X = 8 – 7 = 1
Question: An element ‘X’ belongs to 3rd period and group 13 of the Modern Periodic Table.
(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: (a) As X belongs to group 13 so, it will have three valence electrons and valency of X will be 3.
(b) Atomic number of Y = 8
∴Electronic conffguration = 2, 6
Valency of Y = 8 – 6 = 2
Molecular formula of the compound when X reacts with element Y :
(c) As X belongs to 3rd period and group number 13, so it will be aluminium (Al).
For chlorine (17), electronic configuration = 2, 8, 7
∴ Valency of Cl = 8 – 7 = 1
∴ Formula of the compound :
Question: An element ‘X’ (atomic number 20) burns in the presence of oxygen to form a basic oxide.
(a) Identify the element and write its electronic confIguration.
(b) State its group number and period number in the Modern Periodic Table.
(c) Write a balanced chemical equation for the reaction when this oxide is dissolved in water.
Answer: An element ‘X’ (atomic number 20) burns in the presence of oxygen to form a basic oxide.
(a) Identify the element and write its electroniccalcium (Ca). Electronic configuration of Ca = 2, 8, 8, 2
(b) As calcium has two valence electrons in its outermost shell, so it belongs to group 2.
Moreover, it has four shells which indicates th at it belongs to period number 4.
(c) Calcium forms a basic oxide having the formula :
When calcium oxide is treated with water then calcium hydroxide is formed.
CaO+H2O→ Ca(OH)2
Calcium hydroxide
Question:The atomic number of an element X is 19.
(a) Write its electronic configuration.
(b) To which period of the Modern Periodic Table does it belong and what is its valency?
(c) If ‘X’ burns in oxygen to form its oxide, what will be its nature – acidic, basic or neutral?
(d) Write balanced chemical equation for the reaction when this oxide is dissolved in water.
Answer: Atomic number of X = 19
(a) Electronic configuration of X = 2, 8, 8, 1
(b) X has four shells so, the period number of X = 4. Moreover, it has one electron in its outermost shell, so the valency of X will be equal to one.
(c) Electronic configuration of X shows that it is a metal and metals form basic oxides.
(d) When oxide of X is dissolved in water then its hydroxide will be formed.
X2O + H2O →2XOH
Question: How does the tendency of the elements to loose electrons change in the Modern Periodic Table in (i) a group, (ii) a period and why?
Answer: (i) Tendency of the elements to loose electrons increases down the group. The reason being that at each succeeding element down a group, the number of shells increases. So, the distance of the valence shell from the nucleus increases due to which the e ective nuclear charge decreases on the last shell of electrons. So, it becomes easier for the atom to loose electrons.
(ii) Tendency of the elements to loose electrons decreases in a period from left to right.The reason being that as the electron enters to the same shell at each successive element so, the efiective nuclear charge on the valence shell electron increases, the attraction between the valence electrons and nucleus increases so, it becomes difficult to loose electrons.
Question: An element ‘X’ belongs to third period and second group of the Modern Periodic Table.
(a) Write its electronic configuration.
(b) Is it a metal or non-metal? Why?
(c) Write the formula of the compound formed when ‘X’ reacts with an element
(i) Y of electronic configuration 2, 6 and
(ii) Z of electronic configuration 2, 8, 7.
Answer:Third period indicates that it has three shells while group 2 indicates that it has two valence electrons in its outermost shell. Thus, X must be magnesium (Mg).
(a) Electronic configuration = 2, 8, 2
(b) As X has two valence electrons in its outermost shell which can be easily lost to form a noble gas configuration, so it will be a metal.
(c) (i) Electronic configuration of Y = 2, 6 Hence, valency of Y = 8 – 6 = 2 Formula of compound formed when X reacts with Y is
(ii) Electronic configuration of Z = 2, 8, 7 Hence, valency of Z = 8 – 7 = 1 Formula of compound formed when X reacts with Z is
Question: Na, Mg and Al are the elements of the same period of Modern Periodic Table having one, two and three valence electrons respectively. Which of these elements (i) has the largest atomic radius, (ii) isleast reactive? Justify your answer stating reason for each case.
Answer: Na, Mg and Al belong to same period of Modern periodic table.
Na Mg Al
Valence electrons 1 2 3
(i) Sodium (Na) will have the largest atomic radius because as we move from left to right in a period, atomic size decreases due to increase in effective nuclear charge which pulls the outermost electrons more closer to the nucleus.
(ii) Aluminium (Al) is least reactive because on moving from left to right in the periodic table the nuclear charge increases, so the valence electrons are pulled more closer to the nucleus.Therefore, the tendency to loose electrons decreases and hence, reactivity decreases.
Question: From the following elements :
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 and mention the nature of the compound formed by the union of 19K and element X (2, 8, 7).
Answer: The electronic configurations of the given elements are :
4Be = 2, 2
9F = 2, 7
19K = 2, 8, 8, 1
20Ca = 2, 8, 8, 2
(i) Potassium (K) has one electron in its outermost shell.
(ii) Be and Ca have two electrons in their outermost shells hence, they belong to same group .
Thus, formula of compound when K combines with X is
As K has one electron in its outermost shell, so it transfers this electron to outermost shell of X and hence, an ionic compound is formed.
Question: Write the number of periods the Modern Periodic Table has. 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 a group.
Answer: There are 7 periods in the Modern periodic table. As we move along the period from left to right then valency of the elements first increases from 1 to 4 and then decreases to 0. On moving from left to right in a period the metallic character of elements decreases as the electropositive character of elements decreases across the period. On moving down the group, the valency of the elements remains the same while atomic size increases. This is due to addition of new shell of electrons at every successive step.
Question: Two elements ‘P’ and ‘Q’ belong to the same period of the Modern 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 character.
(d) Their tendencies to loose electrons.
(e) The formula of their oxides.
(f) The formula of their chlorides.
Answer: The given characteristics can be tabulated as follows :
Question: 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.
Answer: Atomic number of the element = 16 Thus, electronic configuration = 2, 8, 6 Since, this element contains 3 shells hence, it belongs to period number 3.
As the element has 6 valence electrons, group number = 10 + 6 = 16 The valency of an element is determined by the number of electrons present in the outermost shell.
∴ Valency of the element = 8 – valence electrons = 8 – 6 = 2
Question: 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?
Answer: The electronic configuration of the given elements will be as follows :
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) Element D will have one electron in its outermost shell.
(b) Elements A and E will belong to same group as both of them have same electrons in their outermost shells.
(c) A and B belong to period number 2 (two shells).
(D) and E belong to period number 4 (four shells).
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?
Answer: (a) Atomic number of element X is 20 so, it is calcium (Ca). Electronic configuration of Ca = 2, 8, 8, 2
(b) 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.
(c) Calcium forms a basic oxide having the formula :
When calcium oxide is treated with water then calcium hydroxide is formed.
CaO+H2O →Ca(OH)2
Calcium hydroxide
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 configuation of X = 2, 8, 3
(b) Atomic number of Y = 17 Electronic configuation = 2, 8, 7 Valency of Y = 8 – 7 = 1
∴ Formula of compound formed when X reacts with Y :
(c) Al2O3 is amphoteric in nature i.e., acidic as well as basic oxide.
Question: In the following table, the positions 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 :
(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 B and C, whose atomic radius is bigger and why?
(v) Write the common name for the family
to which the elements D and F belong
Answer: (i) Element E will form only covalent compounds because it has 4 electrons in the outermost shell so, it can neither loose nor gain 4 electrons, hence E forms compounds by sharing of electrons.
(ii) Element B is a metal having valency 3 as it belongs to group 13.
(iii) C is a non-metal with valency (8 – 5 =) 3.
(iv) Out of B and C, B will be bigger in size because as we move along the period from left to right, the atomic radius decreases due to addition
of electrons in the same shell at each successive element. Hence, nucleus pulls electrons more towards the centre.
(v) D and F belong to group 18 and are called noble gases.
Question: Four elements P, Q, R and S belong to the third period of the Modern Periodic Table and have respectively 1, 3, 5 and 7 electrons in their outermost shells. Write the electronic configurations of Q and R and determine their valencies. Write the molecular formula of the compound formed when P and S combine.
Answer: P, Q, R and S all belong to 3rd period so, all of them will have 3 shells and the number of electrons in their outermost shell is 1, 3, 5 and 7 respectively.
∴ Electronic configuration of Q = 2, 8, 3 and its valency = 3
Similarly, electronic configuration of R = 2, 8, 5 and its valency = 8 – 5 = 3
Electronic configuration of P = 2, 8, 1
Thus, valency of P = 1
Electronic configuration of S = 2, 8, 7
Thus, valency of S = 8 – 7 = 1
Molecular formula of the compound :
Question: (a) Define the following terms :
(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 Modern Periodic Table?
Answer: (a) (i) Valency : It is defined as the combining capacity of the element which is determined by the number of valence electrons present in the
outermost shell of its atom.
(ii) Atomic size : It is defined as the distance between the centre of the nucleus and the outermost shell of an isolated atom.
(b) On moving from left to right in the period, the valency of elements increases from 1 to 4 and then decreases to 0. This is because the elements in a period do not have the same number of valence electrons hence, they do not show same valency.
The atomic size decreases on moving from left to right along a period due to increase in nuclear charge which tends to pull the electrons closer to the nucleus and reduces the size of the atom.
Question: Consider two elements X (atomic number 17) and Y (atomic number 20).
(i) Write the positions of these elements in the Modern Periodic Table giving justfication.
(ii) Write the formula of the compound formed by the combination of X and Y.
(iii) Draw the electron-dot structure of the compound formed and state the nature of the bond formed between the two elements.
Answer: Atomic number of X = 17
∴ Electronic configuration of X = 2, 8, 7 Atomic number of Y = 20
∴ Electronic configuration of Y = 2, 8, 8, 2
(i) From the electronic configurations, we can easily observe that X contains 3 shells so, it belongs to period 3 and it contains 7 electrons in the outermost shell so, it belongs to group-17. Similarly for Y, it has 4 shells which implies that it belongs to period 4 and Y contains two electrons in the outermost shell so, it belongs to group-2.
(ii) Valency of X = 1
Valency of Y = 2 Thus, formula of the compound formed will be
(iii) Electron dot structure of the compound will be
As two electrons present in the outermost shell of Y are donated to two di erent atoms of X thus, it will be an ionic bond (formed by the complete transfer of electrons).
Question: Consider two elements ‘A’ (Atomic number 17) and ‘B’ (Atomic number 19).
(i) Write the positions of these elements in the Modern 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.
Answer: Atomic number of A = 17
Electronic configuration of A = 2, 8, 7
Atomic number of B = 19
Electronic configuration of B = 2, 8, 8, 1
(i) From the electronic configuration of A, it can be easily observed that A contains three shells which indicates that it belongs to period 3.
Moreover, it has seven valence electrons in its outermost shell which indicates that it belongs to group 17. Similarly for B, it has 4 shells so, it belongs to
period 4 and it has one electron in outermost shell so, it belongs to group 1.
(ii) The molecular formula of the compound when A combines with B will be
As A contains 7 electrons in the outermost shell so, it is an electronegative element that is why A is placed after B.
(iii) The electron dot structure will be
The one electron present in the outermost shell of B gets transferred to the outermost shell of A and hence, ionic bond is formed.
Question: 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 :
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.
Answer: (i) Element E will form only covalent compounds because it has 4 electrons in the outermost shell so, it can neither loose nor gain 4 electrons, hence E forms compounds by sharing of electrons.
(ii) Element D is a metal having valency 3 as it belongs to group 13.
(iii) B is a non-metal with valency (8 – 5 =) 3.
(iv) Out of D and E, D will be bigger in size because as we move from left to right in a period there is addition of extra electron in the same shell due to which electrons are pulled more closer to the nucleus.
(v) C and F belong to group 18 and are called noble gases.
VERY SHORT ANSWER TYPE QUESTIONS
Question. In Modern Periodic Table what is common among all the elements in a group.
Answer : All elements in same group shows same valence electrons and same chemical properties.
Question. How many triads could Döbereiner identify from the existing elements then?
Answer : Döbernier could identify only three triads.
Question. An element ‘X’ belongs to II group and 2nd period. Write the atomic number and name of element.
Answer : K L ∴ Atomic Number = 4
2, 2 Element = Beryllium
Question. What is common among all the elements present in one period?
Answer : All the elements in same period show same number of shells e.g., all elements in period 3, show 3 electron shells each.
Question. What is the formula of oxide and hydride of Group I elements?
Answer : Oxide formula → R2O
Hydride formula → RH.
‘R’ represents element.
Question. What is atomic size?
Answer : The radius of an atom, i.e., the distance between the centre of the nucleus and the outermost shell of an atom is called atomic size.
The atomic radius is measured in picometre. (1 pm = 10–12 m)
Question. What happens to the size of atom as we move from left to right in a period.
Answer : The atomic size in a period decreases as we move from left to right.
Question. How do you think the tendency to lose electrons will change in a group?
Answer : Down the group, the effective nuclear charge experienced by valence electrons decreases, hence they can easily lose electrons.
Question. Give two limitations of Newlands’ law of Octaves.
Answer : Two limitations of Newlands’ law of octaves are:
(i) The law was applicable only upto calcium.
(ii) In order to fit elements into his table, Newlands adjusted two elements in wrong slot, and grouped unlike elements in same groups.
Question. How was the anomaly in arrangement of elements in the Mendeleev’s Periodic Table removed?
Answer : When elements were arranged in the increasing order of atomic number. The
anomalies of Mendeleev’s Periodic Table were removed.
Question. What are noble gases/inert gases?
Answer : The element which is inactive, does not react with any other element and it has its outermost shell completely filled are called inert gases or noble gases.
e.g., He, Ne, Ar, Xe.
Question. Fluorine (F) atomic number 9 and chlorine (Cl) atomic number = 17 are placed in group number17, what are the number of valence electrons present in them.
Answer : Fluorine atomic number 9 = 2, 7
Chlorine atomic number 17 = 2, 8, 7
Both of them show 7 valence electron.
Question. An element ‘A’ has atomic number 11, name the period and group number to which it belongs.
Answer : ‘A’ — atomic number = 11
Electronic configuration = K L M
2, 8, 1
∴ Period number = Shell No. = 3
Group number = Valence electron = 1
Question. An element ‘P’ belongs to group = 2 and period = 3, state whether it is a metal or nonmetal and nature of its oxides.
Answer : Group 2 = Metals
Nature of oxide = Basic oxide
Question. The electronic configuration of an atom is 2, 8, 7. Give its atomic number, nature of oxide.
Answer : Electronic configuration = 2, 8, 7
∴ Atomic number = 17
Nature of oxide = Acidic oxide
Question. State Modern Periodic Law.
Answer : ‘Properties of elements are a periodic function of their atomic number’.
Question. An element X belongs to group 17 and element Y belongs to group 1. What type of bond will they form?
Answer :
| X | Y | |
| Group Number Valency 1 | 17 (Non-metal) | 1 1 (Metal) |
Both of them will form ionic bond with the formula XY.
Question. The following elements belong to same period arrange them in order.
X Y Z
Atomic Radius → 231 262 242
Answer : Y Z X
262 242 231
The atomic size decreases in a period.
Question. What happens to the electropositive character of elements as we move from left to right of the period in the Periodic Table?
Answer : On moving from left to right in a period, the electropositive character decreases as the tendency to lose electrons decreases.
Question. What are halogens? Where are they located in the Periodic Table?
Answer : Halogens are those elements which react with metals to form salts. They are present in 17th group of the Periodic Table.
SHORT ANSWER TYPE QUESTIONS
Question. Why does the reactivity of metals increases and that of non-metals decreases as we move down the group?
Answer : Reactivity of metals depends on the tendency to lose electrons. If the atomic size increases, the valence electrons are easily removed thereby forming positive ion. In
case of metals the atomic size increases as we move down the table. Reactivity of non-metals depends on the tendency to gain electrons. As we move down the group, the tendency to gain electrons decreases because the atomic size increases, the nuclear force decreases.
Question. List the elements present in 2nd period. Write their atomic number and electronic configuration.
Answer : Elements → Li Be B C N O F Ne
Atomic No. → 3 4 5 6 7 8 9 10
Electronic
configuration → 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6 2, 7 2, 8
Question. For the following given elements predict the;
(a) Valency
(b) Period number
(c) Group number
Na (11), Al(13), Cl(17), K(19)
Answer :
| Element | Atomic Number | Valency | Period Number | Group Number | Electronic configuration |
| Na Al Cl K | 11 13 17 19 | 1 3 1 1 | 3 3 3 4 | 1 13 17 1 | 2, 8 ,1 2, 8, 3 2, 8, 7 2, 8, 8, 1 |
Question. Elements of group 1 are given below with their atomic number:
Li (3) Na (11) K (19)
(a) Give their atomic size.
(b) Reactivity.
Answer :
| Group 1 | Atomic Number | Electronic Configuration |
| Li Na K | 3 11 9 | 2 ,1 2 ,8 ,1 2 ,8 ,8 ,1 |
(a) The atomic size goes on increasing as new shell is added in each element as we move from top to bottom. So it is Li < Na < K.
(b) The reactivity increases as it is easy to lose electrons if the size of atom is big, the nuclear force decreases K is more reactive than Na and Li.
Question. Lithium, sodium potassium belong to same group called alkali metals. Why?
Answer : Lithium, sodium and potassium have same valence electron i.e. 1, hence they belong to same group. The group is called alkali metals group because all these elements form oxides which dissolve in water to form alkali.
Question. Carbon with atomic number 6 and silicon with atomic number 14 belong to same group although carbon is non-metal and silicon is semi-metal.
Answer : Carbon with atomic number 6, shows electronic configuration 2, 4. Silicon with atomic number 14 shows electronic configuration 2, 8, 4.
Both the elements have same valence electrons, hence they are placed in same group.
Question. What physical and chemical properties of elements were used by Mendeleev in creating his Periodic Table? List two observations which posed a challenge to Mendeleev’s Periodic Law.
Answer : The physical property used was the atomic mass of an element.
The chemical property used was the nature of oxide and hydride formed i.e. similarity in chemical properties were used by Mendeleev. The two observations that posed challenge in Mendeleev Periodic Law were:
(i) Arranging elements according to the increasing order of atomic mass could not be maintained. Chemical properties do not depend on atomic mass.
(ii) Isotopes were not given any place in the table as they have different atomic mass but same chemical properties.
Question. Table given below shows a part of the Periodic Table.
H He
Li Be B C N O F Ne
Na Mg Al Si P S Cl Ar
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 can readily lose electrons due to bigger size of atom.
(b) Mg has more number of protons than Na which attracts the electrons thereby reducing the size of Mg.
(c) Fluorine readily accepts/gains electrons to become F– ion due to its small atom size as compared to chlorine.
Question. The position of three elements A, B and C in the Periodic Table are shown below: (Img 71)
Giving reasons, explain the following:
(a) Element A is a non-metal.
(b) Element B has a larger atomic size than element C
(c) Element C has a valency of 1. (CBSE 2008 F)
Answer : (a) A is a non-metal because it can gain electrons easily, it has 7 valence electrons and form negative ions to become stable.
(b) The atomic number of B is less than C, it has less nuclear charge, less force of attraction between protons in the nucleus and valence electrons, hence its size is bigger than C.
(c) Element ‘C’ has 7 valence electrons, it can gain 1 electron to become stable so its valency is 1.
Question. The position of three elements A, B and C in the Periodic Table are shown below: (Img 71)
Giving reason, explain the following:
(a) Element A is a metal.
(b) Element C has a larger size than B.
(c) Element B has a valency of 3. (CBSE 2008 F)
Answer : (a) ‘A’ is a metal because its valence electron is 1, it can readily loose electron to became stable.
(b) Element ‘C’ belongs to 3rd period it has 3 shells whereas ‘B’ has only 2 shells, it belongs to 2nd period, distance between nucleus and valence electrons is more in C, hence its size is bigger than B.
(c) ‘B’ belongs to III group, has 3 valence electrons, it can loose 3 electrons to become stable hence its valency is 3.
Question. The position of 3 elements A, B and C in the Periodic Table is shown below: (Img 72)
Giving reasons, explain the following:
(a) Element A is a non-metal.
(b) Atom of element C has larger size than A.
(c) Element B has a valency of 1.
Answer : (a) A belongs to 7th group, has 7 valence electron, it can gain 1 electron to become stable. So it is a non-metal as it forms negative ion.
(b) ‘C’ has 3 shells ‘A’ has 2 shells so C is bigger than A.
(c) ‘B’ has one valence electron, it can loose this electron to become stable. So its valency is 1.
Question. The elements of the second period of the Periodic Table are given below:
Li Be B C N O F
(a) Give reason to explain why atomic radii decreases from Li to F.
(b) Identify the most
(i) Metallic and (ii) Non-metallic element.
Answer : (a) In a period on moving from left to right, the atomic number increases, the number of shells remains the same, nuclear charge increases the force of attraction of electrons towards the centre increases. The valence electrons are pulled at the centre, hence atomic radii decreases from Li to F.
(b) (i) Most metallic element → Li
(ii) Most non-metallic element → F.
Question. The elements of the third period of the Periodic Table are given below:
Group I II III IV V VI VII
Na Mg Al Si P S Cl
(a) Which atom is bigger — Na or Mg? Why?
(b) Identify the most
(i) Metallic and (ii) Non-metallic element in period 3.
Answer : (a) Na atom is bigger in size this is because as we move from Na to Cl, the atomic number goes on increasing and the nuclear charge also increases. It pulls/attract
the valence electrons at the centre and thus the atomic size decreases.
(b) (i) Most metallic — Na
(ii) Most non-metallic — Cl
Question. (a) What is meant by periodicity in properties of elements with reference to the Periodic Table?
(b) Why do all the elements of the same group have similar properties?
(c) How will the tendency to gain electrons change as we go from left to right across a period? Why?
Answer : (a) The repetition of same properties after definite interval is called periodicity in properties.
(b) All elements in group have same valence electrons.
(c) Tendency to gain electrons increases from L → R in the period because the atomic size goes on decreasing and nuclear charge increases, which can attract the nearby electron.
LONG ANSWER TYPE QUESTIONS
Question. Name the following elements
(a) Two shells, both of which are completely filled.
(b) Three shells with 2 valence electrons.
(c) Group 1, two shells.
(d) Group 17, period 3.
(e) Metal, with valency 3 group number 13 period 3.
Answer : (a) Two shells → K L
Filled → 2, 8
Atomic number → 10
∴ Element — Neon.
(b) Three shells → K, L, M
2 Valence electron → 2, 8, 2
Atomic number →12 ∴ Element = Magnesium
(c) Group 1
2 shells → K L
Electronic configuration → 2, 1
Atomic number → 3 ∴ Element = Lithium
(d) Group 17
Period → 3, K L M
Electronic configuration → 2, 8, 7
Atomic number → 17, ∴ Element = Chlorine
(e) Group → 13
Valency → 3
Period → 3 K L M
Electronic configuration → 2, 8, 3
∴ Atomic number → 13, ∴ Element = Aluminium.
Question. (a) What are ‘groups’ and ‘periods’ in the Periodic Table?
(b) Two elements M and N belong to groups I and II respectively and are in the same period of the Periodic Table. How do the following properties of M and N vary?
(i) Sizes of their atoms.
(ii) Their metallic characters.
(iii) Their valencies in forming oxides.
(iv) Molecular formulae of their chlorides. (CBSE 2009 F)
Answer : (a) The vertical column in the Periodic Table are called ‘groups’, the horizontal rows in the table are called periods.
(b) (i) ‘M’ and ‘N’ belong to the same period but group I and II respectively. N is smaller than M as the atomic size decreases on moving from left to right across the Periodic Table.
(ii) M is more metallic than ‘N’ because metallic character decreases from left to right as tendency to lose electrons decreases due to decrease in atomic size.
(iii) The valencies of M and N are 1 and 2 respectively, valency across the period first increases then decreases.
(iv) MCl and NCl2.
Question. On the baiss of the table of Mendeleev’s Periodic Table
(a) Name the element which is in
(i) I group and 3rd period. (ii) VII group and 2nd period.
(b) Suggest the formula for the following:
(i) Oxide of nitrogen (ii) Hydride of oxygen
(c) In group VIII of Periodic Table, why does cobalt with atomic mass 58.93 appear before nickel having atomic mass 58.71?
(d) Besides gallium, which two other elements have since been discovered for which Mendeleev had left gaps in his Periodic Table?
(e) Using atomic masses of Li, Na and K, find the average atomic mass of Li, and K and compare it with the atomic mass of Na. State the conclusion drawn from this activity.
Answer : (a) (i) Sodium (ii) Fluorine
(b) (i) N2O5 (ii) H2O
(c) Co resembles with Rh and Ir whereas ‘Ni’ resembles with Pd and Pt.
(d) Germanium and scandium
(e) Li Na K Average atomic mass of Na
6.939 22.99 39.102 = (6.939 + 39.102 )/2 = 23.0205
The atomic mass of Na is the average atomic mass of Li and K and these elements resemble with each other.
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 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 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 answers.
Answer : (a) Classification is done to study the properties of elements conveniently.
(b) Increasing order of atomic mass and similarity in chemical properties i.e. the nature of oxide and hydride formed.
(c) The gaps were left for undiscovered elements then.
(d) Noble gases were not invented at that time.
(e) Cl-35 and Cl-37 will be kept in the same slot as their chemical properties are same.
Question. (a) Which 2 criteria did Mendeleev use to classify the elements in his table.
(b) State Mendeleev’s Periodic law.
(c) Why could no fixed position be given to hydrogen in Mendeleev’s Periodic Table.
(d) How and why does the atomic size vary as you go:
(i) From left to right along a period
(ii) Down a group?
Answer : (a) (i) Increasing order of atomic mass and similarities in chemical properties of elements.
(ii) The formula of oxides and hydrides formed by elements.
(b) Mendeleev’s Periodic Law → Properties of elements are periodic functions of their atomic masses.
(c) Hydrogen had no fixed position in Mendeleev’s Periodic table because it resembles alkali metal by forming positive ions and resembles halogens by forming diatomic molecule.
(d) (i) Atomic size decreases from left to right, as the valence electrons are attracted by the nucleus due to increase in the nuclear force.
(ii) The atomic size increases from top to bottom in a group because the number
of shells keep on increasing therefore distance between nucleus and valence electrons increases.
Question. (a) Why did Mendeleev left gaps in his Periodic Table?
(b) State any 3 limitations of Mendeleev’s classification.
(c) How does electronic configuration of atoms change in period with increase in atomic number?
Answer : (a) Mendeleev left some gaps for undiscovered elements, because he predicted that there would be such elements which will fit in the gaps in future. He also predicted the properties of these elements.
(b) 3 limitations are:
(i) Position of hydrogen was not justified.
(ii) Increasing order of atomic mass could not be maintained.
(iii) Isotopes were not given separate place as they have different atomic mass.
(c) In a period the valence electrons goes on increasing from left to right as the number of shells is same.
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 compound formed 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.
e. (Image 172) is the formula of compound.
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.
