CBSE Class 10 Science Periodic Classification Of Elements Assignment Set 01

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 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.
Answer :  (A) Elements are classified to systematize their study and make the understanding of properties of elements and compounds simpler.
(B) The two main guiding factors for Mendeleev in the classification of the then known elements were:
(1) increasing atomic masses, and
(2) grouping together of elements having similar properties.
(C) These gaps were left for the elements which had not been discovered at that time. Mendeleev thought that these elements would be discovered later on in the future. The modern periodic table does not have any gaps because new elements were discovered later on, which were placed
(D) Noble gases were not known at that time. So, there was no group of noble gases in Mendeleev’s original periodic table.
(E) Both the isotopes of chlorine, Cl-35 and Cl-37, have the same atomic number of 17. Therefore, they have same chemical properties. Hence, both of them can be put at one place in the same group of the periodic table.

Question. The atomic number of an element is 20.
(A) Write its electronic configuration and determine its valency.
(B) Is it a metal or a non-metal?
(C) Write the formula of its chloride.
(D) Is it more reactive or less reactive than Mg (atomic number 12)?
Give reason for your answer. 
Answer :  (A) The electronic configuration of element having atomic number 20 is 2, 8, 8, 2, i.e., electrons in K, L, M and N shells are 2, 8, 8 and 2 respectively.
Valency of given element = 2
(B) It is a metal as it can lose electrons easily since it is electropositive.
(C) Let us denote the element by M. Since chlorine has valency 1, formula of the chloride of M is MCl₂.
(D) This element, M, is more reactive than Mg as Mg lies in 2nd Group, 3rd period and M lies in 2nd Group but 4th period of the periodic table. M is more electropositive than Mg as
the effective nuclear charge experienced by its valence electrons is less than that experienced by valence electrons of Mg because of increase in distance of valence electrons from the nucleus.

Question. An element ‘X' with electronic configuration (2, 8, 2) combines separately with two radicals, (NO₃)– and (SO₄)^2–.
(A) Is ‘X’ a metal or a non-metal ? Write the nature of its oxide.
(B) Write the formula of the compounds of ‘X’ formed by the combination of these radicals. Are these compounds covalent or electrovalent ? 
Answer :  (A) ‘X’ is a metal as it has 2 valence electrons. Its oxide is basic in nature.
(B) Valency of X is 2. The formula of compounds formed by X with (NO₃)^– and (SO₄)^2– :
a. X(NO₃)₂
b. XSO₄
These compounds are ionic (electrovalent) as these are formed by transfer of electrons

Question. An element ‘M’ has atomic number 12.
(A) Write its electronic configuration and valency.
(B) Is ‘M’ a metal or a non-metal? Give reason in support of your answer.
(C) Write the formula and nature (acidic/ basic) of the oxide of M. [17-311DC]
Answer :  Atomic number of element ‘M’ = 12
(A) Electronic configuration of M: (2, 8, 2), i.e.,k-shell = 2, L-shell = 8, M-shell = 2 electrons.
Valency of M = 2
(B) M is a metal as it is electropositive, since it has the tendency of losing electrons, thereby forming positive ion.
(C) Formula of the oxide of M: MO Nature of oxide: Basic

Question. How does the atomic radius of the elements change on going: (A) from left to right in a period, and (B) down a group in the Modern Periodic Table? (Give reasons) in support of your answer.
Answer :  (A) The atomic radius decreases on moving from left to right in a period in the Modern Periodic Table as the number of electrons and protons increases. Due to the large positive charge on the nucleus, electrons are pulled more strongly towards the nucleus. (B) The atomic radius increases on moving from top to bottom in a group, as new shells are being added, which increases the distance between the outermost electrons and the nucleus. So, although the nuclear charge increas, the atomic radius also increases

Question. a. Amongst the following elements identify the ones that would form anions:
K, O, Na, F, Ca, Cl, Mg
b. Write the electronic configuration of the anions identified above.
Answer : O, F, Cl will form anions.
O^2- (18): 2, 8
F^-(10): 2, 8
Cl^-(18): 2, 8, 8

Question. An element X belongs to 3rd period and group 17 of the periodic table. State its
(a) electronic configuration,
(b) valency. Justify your answer with a reason.
Answer : a. 2, 8, 7; because it has 3 shells as it belongs to 3rd period. Group 17 means 7 valence electrons.
b. Valency = 1
It can gain 1 electron to become stable i.e. to complete its octet.

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
Atomic mass of lithium
=      + Atomic mass of potassium/2
Na = 7 + 39/2 = 46/2 = 23

Question. An element ‘X’ has atomic number 13.
a. Write its electronic configuration.
b. State the group to which ‘X’ belongs to.
c. Is ‘X’ a metal or a non-metal?
d. Write the formula of its bromide.
Answer : a. X(13) : 2, 8, 3
b. It belongs to group 13.
c. It is a metal.
d. XBr₃ is the formula of its bromide.

Question. An element X has mass number 35 and the number of its neutrons is 18. Identify the group number, period and valency of element X’.
Answer : X has mass number 35, number of neutrons = 18 Atomic Number = 35 - 18 = 17
a. Electronic configuration: X = 2, 8, 7 It has 7 valence electrons. It belongs to group 17.
b. It has 3 shells, therefore it belongs to 3rd period.
c. It can gain 1 electron to become stable, so its valency is equal to 1.

Question. How can the valency of an element be determined if its electronic configuration is known? What will be valency of an element with atomic number 9?
Answer : Valency=Number of valence electrons in case of metals and metalloids. It is also equal to 8 — Number of valence electrons in case of non-metals.
F(9) has electronic configuration of 2, 7. It is a nonmetal. Its valency is equal to 1.

Question. Write the trend of atomic size and metallic character along a group and a period in modern periodic table.
Answer : Atomic size increases down the group and decreases along a period from left to right in the periodic table. Metallic character increases down the group and it decreases along a period from left to right.

Question. An element belongs to third period and second group of the periodic table:
a. State the number of valence electrons in it.
b. Is it a metal or a non-metal?
c. Name the element.
d. Write the formula of its oxide.
Answer : a. It has 2 valence electrons.
b. It is a metal
c. Magnesium.
d. MgO is the formula of its oxide.

Question. An element M has atomic number 12.
a. Write its electronic configuration.
b. State the group to which M belongs to.
c. Is M a metal or a non-metal?
d. Write the formula of its oxide.
Answer : a. 2, 8, 2,
b. Group 2,
c. Metal,
d. MO is formula of its oxide.

Question. Calcium is an element with atomic number 20.
a. Will it be a metal/non-metal?
b. What will be its valency?
c. What would be the formula of its chloride?
d. Will it be smaller / larger than K?
Answer : Ca(20): 2, 8, 8, 2
a. It will be a metal.
b. Its valency is equal to 2.
c. CaCl₂ is the formula of its chloride.
d. It will be smaller than K.

Question. The elements X, Y and Z having atomic numbers 11, 7 and 6 respectively react with oxygen to form their oxides.
a. Arrange these oxides in increasing order of their basic nature.
b. Give reason for your answer.
Answer : X(11) is sodium and it will form Na₂O, Y(7) is Nitrogen that will form N₂O₅, Z(6) is carbon which will form CO₂.
a. N₂O₅ < CO₂ < Na₂O
b. It is because non-metallic character increases along a period, therefore basic character of oxides decreases, and acidic nature increases from left to right across the period.

Question. An element M has atomic number 11.
a. Write its electronic configuration.
b. State the group to which M belongs to.
c. Is M a metal or a non-metal?
d. Write the formula of its chloride.
Answer : a. 2, 8, 1
b. M belongs to group 1.
c. It is a metal.
d. MCI is the formula of its chloride.

Question. List the anomalies of Mendeleev’s periodic table which were removed in Modem Periodic Table.
Answer : Co with higher atomic mass proceeds Ni with lower atomic mass. It was solved because Co has lower atomic number than Ni.
Isotopes should have been given different slots due to different atomic mass, but it is not possible due to same chemical properties. The problem was solved because isotopes have same atomic numbers.

Question. Why is lithium with atomic number 3 and potassium with atomic number 19 are placed in group one? What will be atomic number of the first two elements in the second group?
Answer : 
Group 1               Group 2
Li(3): 2,1             Be(4): 2, 2
K(19):2,8,8,1       Mg (12): 2, 8, 2
Li and K are placed in group 1 due to same number of valence electrons. In second group the atomic number of first two elements will be 4 and 12 respectively.

Question. a. Name the element with atomic number 17.
b. To which period does it belong to?
c. To which group does it belong to?
d. Write its electronic configuration.
Answer : a. Chlorine (17): 2, 8, 7
b. It belongs to 3rd period.
c. It belongs to group 17.
d. 2, 8, 7 is the electronic configuration of chlorine.

Question. Chlorine, Bromine and Iodine form Dobereiner’s triads. The atomic mass of Cl and I are 35.5 and 126.9 respectively. Predict the atomic mass of Bromine.
Answer :  Atomic Mass of Br
        =   Atomic mass of Cl + Atomic mass of I/2
        =  35.5 + 126.9 = 162.4/2 = 81.2
Electronic configuration: X = 2,8,7
It has 7 valence electrons. It belongs to group 17.
It has 3 shells, therefore it belongs to 3rd period.
It can gain 1 electron to become stable, so its valency is equal to 1.

Question. Choose from the following:
₆C, ₈O, ₁₀Ne, ₁₁Na, ₁₄S1
a. Elements that should be in the same period.
b. Elements that should be in the same group.
State the reason for your selection in each case.
Answer : a. ₆C(2, 4), ₈O(2, 6), ₁₀Ne(2, 8) belong to the same period i.e., 2nd period and ₁₁Na(2, 8, 1), ₁₄Si(2, 8, 4) belongs to the same period i.e., 3rd period. 
b. ₆C(2,4) and ₁₄Si(2,8,4) belongs to the same group -14 due to same number of valence electrons, which is equal to 4.

Question. What is meant by groups and periods in the Modern Periodic Table? Two elements ‘A’ and ‘B’ belong to group 1 and 2 respectively and are in the same period. How do the following
properties of ‘A’ and ‘B’ vary?
(A) Atomic size
(B) Metallic character
(C) Valencies in forming oxides
(D) Formula of their chlorides 
Answer :  Groups are the vertical Colums Periods are the horizontal rows.
(A) Atomic size: decreases as we go from left to right in a period hence atom of element A is bigger than B.
(B) Metallic Character: A is more metallic than B since metallic character decreases as we go along a period.
(C) Valancies in forming oxides: Valency of group 1 is 1 and valency of group 2 is 2.
(D) Formula of their chloride: ACl, BCl₂

Question. Write the names given to the vertical columns and horizontal rows in the Modern Periodic Table. How does the metallic character of elements vary on moving down a vertical column? How does the size of atomic radius vary on moving left to right in a horizontal row? Give reason in support of your answer in the above two cases. 
Answer :  The vertical columns in the Modern Periodic Table are known as "Groups" and horizontal rows are known as "Periods"The metallic character increases as we move from top to bottom in a group because the electropositive character of elements increases as the tendency of an atom to lose electrons increases. On moving from left to right in a period, the atomic radius decreases as the number of electrons and protons increases. Due to the large positive charge on the nucleus, electrons are pulled more strongly towards the nucleus.

Question. The electronic configuration of an element is 2, 8, 7.
(A) What is the group number and period number of this element in the Modern Periodic Table?
(B) Is this element a metal or a non-metal? Give reason to justify your answer. 
Answer : (A) Its electronic configuration is 2, 8, 7. Hence its atomic number is 17 and this element is chlorine. It belongs to 17th group and 3rd period in the modern periodic table
(B) This element is a non-metal because it gains the electron to complete their octet..

Question. (a)Name the element with atomic number 17.
(b)To which period does it belong?
(c) To which group does it belong?
(d)Write its electronic configuration.
Answer :  (a)Chlorine
(b) 3rd
(c) 17th
(d) 2,8,7

Question. An element X which is a yellow solid at room temperature shows catenation and allotropy. X forms two oxides which are also formed during the thermal decomposition of ferrous sulphate crystal and are major air pollutants.
i. Identify ‘X’.
ii. Write its electronic configuration.
iii. Write its balanced chemical equation for the thermal decomposition of ferrous sulphate crystals.
iv. What would be the nature (Acidic/Basic) of oxides formed?
v. Locate the position of the element in the periodic table.
Answer :  (i) ‘X’ is Sulphur
(ii) Electronic configuration of X is: 2,8,6
(iii) 2FeSO₄ —–→ Fe₂O₃+SO₂ +SO₃(iv)SO₂ and SO₃ are acidic oxides
(v) It belongs to group 16 and 3^rd period.

Case Study Based Questions

Question. Dobreiner triads and Newlands law of Octaves were early attempts at classifying elements into groups based on their properties. Dobreiner observed that groups of three elements(triads) could be formed in which all the elements shared similar physical and chemical properties. The identification of new elements made this model obsolete. Newly discovered elements did not fit into the triads. The British chemists John Newlands arranged then known elements in an ascending order of their atomic masses and observed that every 8th element had similar properties. On the basis of this observation, Newlands law of Octaves was formulated.
(a) What is the relationship in the atomic masses of elements of a Dobreiner triad?
(b) State Newlands law of Octaves.
(c) A and Bare the two elements having similar properties and obeying Newlands law of Octaves. How many elements are there in between A and B?
(d) Name the elements that form alkali metal triad.
Answer :  (a) In the Dobreiner triads of elements, the atomic weight of the middle element is the arithmetic means of the other two.
(b) When the elements with lower atomic masses were arranged in order of their increasing atomic masses, the properties of every eighth element were similar to those of the first one.
(c) There are six elements in between A and B.
(d) Lithium, Sodium and potassium form alkali metal triad 

Question. There are many observable patterns in the physical and chemical properties of elements as we descend in a group or move across a period in the periodic table. Such patterns are called Periodic trends. Major periodic trends include valency, atomic size, metallic and non-metallic properties.
(a) How does valency change on moving down in a group of modern periodic table?
(b) Why does atomic radius decrease on moving across a period in modern periodic table.
(c) how is metallic character related to the reactivity of metals.
(d) Discuss the variation of basic nature of oxides down the group.
Answer :  (a) On moving down a group, valency remains the same as all the elements of the group have same number of valence electrons
(b) On moving from left to right across a period, the atomic radius decreases due to increase in nuclear charge.
(c) Greater the metallic character, greater is the reactivity.
(d) Basic nature of oxides increases down the group due to increase in metallic character. 

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

Question : What is 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 : How does the tendency to lose electrons will change in a period.
Answer : The tendency to lose electrons will decrease across a period as the effective nuclear charge acting on the valence shell electrons increases.

Question : An element belongs to group 13 and period 3, name the element and give its valency.
Answer : The element is Aluminium.
The valency = 3

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

Question : What 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 : 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 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 : Name two alkali metals present in Group I.
Answer : Alkali metals are Li, Na, K.

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

HOTS: (High Order Thinking Skill) Questions with Answers

Question Q-1. X, Y and Z are the elements of a dobereiners triad. If the atomic mass of x is 7 and that of z is 39, what should be the atomic mass of y?
Answer: According to Dobereiner's Law of Triads, the atomic mass of the middle element of a triad is approximately the arithmetic mean of the atomic masses of the first and third elements.
Atomic mass of X = 7
Atomic mass of Z = 39
Atomic mass of Y = \(\frac{7 + 39}{2} = \frac{46}{2} = 23\)
In simple words: The atomic mass of the middle element is the average of the first and last elements.

Exam Tip: Remember Dobereiner's triad formula: \(\text{Middle mass} = \frac{\text{First mass} + \text{Third mass}}{2}\).

Question Q-2. A and B are the two elements having similar properties which obey Newlands law of octaves. How many elements are there in between A and B?
Answer: According to Newlands' Law of Octaves, when elements are arranged in increasing order of atomic masses, the properties of every eighth element are similar to the first. Therefore, if element A is the first element, there are 6 elements between A and B (making B the eighth element).
In simple words: There are 6 elements between A and B because every 8th element repeats properties like musical notes.

Exam Tip: In Newlands' Octaves, count the starting element as 1 and the repeating element as 8, meaning exactly 6 elements lie between them.

Question Q-3. The following is Newland's Octave Table. Observe it and answer the following questions:

i. Which of the element in 1st column has different properties from rest of the elements?
ii. Which of the elements resemble with each other in second column?
iii. Pick up odd element in second last column.
iv. Pick up elements which have similar properties in last column.
Answer:
(i) Co and Ni (placed in the first column along with fluorine, chlorine, and bromine, which are halogens, whereas Co and Ni are metals with entirely different properties).
(ii) Li, Na, K, Cu, Rb (alkali metals / group 1 elements like Li, Na, K share similarities, though Cu is grouped incorrectly here).
(iii) Mn (manganese is a transition metal, while N, P, As are non-metals / p-block elements in that column).
(iv) O, S, Se (oxygen, sulphur, and selenium belong to the same oxygen family / group 16).
In simple words: Cobalt and Nickel do not fit with halogens, while elements in vertical columns generally share similar chemical traits.

Exam Tip: Always look at group similarities and anomalies when evaluating Newlands' table limitations.

Question Q-4. In the Periodic Table given below, Lithium, carbon, oxygen and neon are placed in their correct positions and the positions of nine other elements are represented by letters. These letters are not the symbols for the elements?

By reference to the table, answer the following questions:
(a) Give the letter of the most reactive metal.
(b) Give the letter of the most reactive non-metal.
(c) Name the family of elements represented by L, Q, R, and T.
(d) Name one element in each case occurring in groups 2, 13 and 15
Answer:
(a) Z (As metallic character increases down a group, the element at the bottom of Group 1 is the most reactive metal).
(b) L (In the halogen group 17, reactivity decreases down the group, making the topmost halogen L the most reactive non-metal).
(c) Halogen family.
(d) Group 2: Magnesium (Mg), Group 13: Aluminium (Al), Group 15: Nitrogen (N) (or any other correct elements in those groups).
In simple words: Metals get more reactive going down, while non-metals get more reactive going up. Group 17 is the halogens.

Exam Tip: In modern periodic trends, reactivity of metals increases down the group, while reactivity of non-metals decreases down the group.

Question Q-5. Two elements X and Y have atomic numbers 12 and 16 respectively. Write the electronic configuration for these elements. To which period of the modern periodic table do these two elements belong? What type of bond will be formed between them and why?
Answer:
Electronic configuration of element X (Atomic number 12 - Magnesium): \(2, 8, 2\)
Electronic configuration of element Y (Atomic number 16 - Sulphur): \(2, 8, 6\)
Period number: Both elements have 3 shells, so they belong to the 3rd period.
Bond type: An ionic (electrovalent) bond is formed between X and Y because X is a metal (loses 2 electrons to achieve noble gas configuration) and Y is a non-metal (gains 2 electrons to achieve octet), resulting in complete transfer of electrons.
In simple words: Element X donates 2 electrons and Y accepts them, creating an ionic bond because opposites attract.

Exam Tip: Metals (1, 2, 3 valence electrons) combine with non-metals (5, 6, 7 valence electrons) via ionic bonds through complete electron transfer.

Question Q-6. An element X (2,8,2) combines separately with \(\text{NO}_3\), \((\text{SO}_4)^{2-}\), \((\text{PO}_4)^{3-}\) radicals. Write the formulae of the three compounds so formed. To which group of the periodic table does the element 'X' belong? Will it form covalent or ionic compound? Why?
Answer:
Formulae of compounds formed by element X (valency = 2):
1. With nitrate radical (\(\text{NO}_3^-\)): \(\text{X(NO}_3)_2\)
2. With sulphate radical (\(\text{SO}_4^{2-}\)): \(\text{XSO}_4\)
3. With phosphate radical (\(\text{PO}_4^{3-}\)): \(\text{X}_3(\text{PO}_4)_2\)
Group number: Element X belongs to Group 2 (since it has 2 valence electrons).
Nature of compound: It forms ionic compounds because it readily loses its 2 valence electrons to achieve the stable electronic configuration of the nearest noble gas (Neon).
In simple words: Use criss-cross valency rules for formulae. Since X has 2 valence electrons and is a metal, it forms ionic bonds.

Exam Tip: Always cross valencies correctly when writing chemical formulae with polyatomic radicals.

Question Q-7. The following table shows the position of six elements A, B, C, D, E and F in the periodic table.

Using the above table answer the following questions:
(a) Which element will form only covalent compounds?
(b) Which element is a metal with valency 2?
(c) Which element is a non-metal with valency of 3?
(d) Out of D and E, which one has a bigger atomic radius and why?
(e) Write a common name for the family of elements C and F.
Answer:
(a) E (Carbon family / group 14 element forms covalent bonds by sharing electrons).
(b) A and D (Both belong to group 2, hence they are divalent metals).
(c) B (Nitrogen family in group 15 has 5 valence electrons, so its valency is 8 - 5 = 3).
(d) D has a bigger atomic radius than E because atomic size decreases across a period from left to right due to increased nuclear charge pulling electrons closer.
(e) Noble gases (or Inert gases).
In simple words: Element E is carbon (covalent), A and D are group 2 metals, B has valency 3, D is larger than E because size shrinks across a period, and C and F are noble gases.

Exam Tip: Atomic size decreases across a period from left to right and increases down a group.

Question Q-8. The diagram below shows part of the Periodic Table

The position of three elements in the Periodic Table is shown:
i Write the atomic numbers of the elements.
ii Give the electronic distribution of the elements
iii Using these three elements as examples, describe the trend in chemical properties across the third period of the Periodic Table.
Answer:
(i) Atomic numbers: Sodium (Na) = 11, Chlorine (Cl) = 17, Argon (Ar) = 18.
(ii) Electronic distribution:
Sodium (Na) = \(2, 8, 1\)
Chlorine (Cl) = \(2, 8, 7\)
Argon (Ar) = \(2, 8, 8\)
(iii) Trend across the period: Metallic character and reducing character decrease across a period from left to right, while non-metallic and oxidizing character increase.
In simple words: Na is a reactive metal on the left, Cl is a reactive non-metal on the right, and Ar is an unreactive noble gas.

Exam Tip: Across a period, elements change from strongly metallic (left) to non-metallic (right) and end in noble gases.

Question Q-9.

(a) Write the formula of hydride and oxide of silicon
(b) Name the elements which is in
(i) II group and \(4^{\text{th}}\) period
(ii) VI group and \(3^{\text{rd}}\) period.
(c) Name the elements in group I which do not resemble with alkali metals
(d) In group VI why does Te with atomic mass 127.60 comes before I with atomic mass 126.90
Answer:
(a) Silicon is in group IV: Oxide = \(\text{SiO}_2\), Hydride = \(\text{SiH}_4\)
(b) (i) Calcium (\(\text{Ca}\)) and Zinc (\(\text{Zn}\))
(ii) Sulphur (\(\text{S}\))
(c) Hydrogen (\(\text{H}\))
(d) The sequence was inverted by Mendeleev so that elements with similar properties could be grouped together properly, ignoring strict atomic mass order.
In simple words: Silicon forms \(\text{SiO}_2\) and \(\text{SiH}_4\). Mendeleev placed Te before I to group similar elements correctly despite atomic weights.

Exam Tip: Mendeleev prioritized chemical similarities over strict atomic mass ordering when necessary.

Question Q-10. A metal M forms an oxide having the formula \(\text{M}_2\text{O}_3\). It belongs to \(3^{\text{rd}}\) period in the modern periodic table. Write the atomic number and valency of the metal.
Answer:
Valency of metal M = 3 (since oxygen has valency 2 and the formula is \(\text{M}_2\text{O}_3\)).
Since it belongs to the \(3^{\text{rd}}\) period, it has 3 electron shells (\(K, L, M\)). With a valency of 3, its electronic configuration is \(2, 8, 3\).
Atomic number = \(2 + 8 + 3 = 13\) (Element is Aluminium, \(\text{Al}\)).
In simple words: The formula \(\text{M}_2\text{O}_3\) shows M has valency 3. In period 3, its configuration is 2,8,3, giving atomic number 13.

Exam Tip: Total electrons in all filled shells plus valence electrons give the atomic number of the element.

PRACTICE QUESTIONS

Question Q-1. State the modern periodic law
Answer: Properties of elements are a periodic function of their atomic numbers.
In simple words: Elements show repeating properties when arranged by atomic number.

Exam Tip: Modern periodic law is based on atomic number, whereas Mendeleev's law was based on atomic mass.

Question Q-2. which of the two elements A = 2,8,1 B = 2,8,8,1 is more electropositive
Answer: Element B (\(2, 8, 8, 1\) - Potassium) is more electropositive than element A (\(2, 8, 1\) - Sodium) because electropositivity increases down a group as atomic size increases and valence electrons are lost more easily.
In simple words: B is larger, so its single outer electron is farther from the nucleus and lost more easily, making it more electropositive.

Exam Tip: Electropositivity (metallic character) increases down the group.

Question Q-3. How does the atomic size very in going from A) Left to right in a period B) Top to Bottom in a group
Answer:
A) Left to right in a period: Atomic size decreases because nuclear charge increases, pulling electrons closer.
B) Top to bottom in a group: Atomic size increases because new shells are added successively.
In simple words: Size shrinks across a period and grows down a group.

Exam Tip: Remember: Across = decreases; Down = increases.

Question Q-4. An element has atomic number 13. In which group and period it should be placed?
Answer:
Electronic configuration of atomic number 13 (Aluminium) = \(2, 8, 3\)
Period = 3 (since there are 3 shells)
Group = 13 (since it has 3 valence electrons, group number = \(10 + 3 = 13\))
In simple words: Configuration 2,8,3 means 3 shells (Period 3) and 3 valence electrons (Group 13).

Exam Tip: For elements with valence electrons > 2, add 10 to find the p-block group number.

Question Q-5. How many periods and groups are there in the long from of P.T?
Answer: There are 7 periods and 18 groups in the modern (long form) periodic table.
In simple words: The modern periodic table has 7 horizontal rows (periods) and 18 vertical columns (groups).

Question Q-6. Why does the size of the atoms progressively become smaller when we move from sodium (Na) to chlorine (cl) in the third period of the period table?
Answer: When moving from Sodium to Chlorine across the third period, the nuclear charge increases progressively (number of protons increases) while electrons are added to the same shell. This increased nuclear pull draws the electron cloud closer to the nucleus, reducing atomic size.
In simple words: More protons pull the same number of shells tighter together, making the atom smaller.

Exam Tip: Mention nuclear charge and shielding effect in your explanation.

Question Q-7. Give symbols for
A. A metal of group 2.
B. A metal of group 13.
C. Two non metals of group 16.
D. Most reactive non- metal of group 17.
Answer:
A. Magnesium (\(\text{Mg}\)) [or \(\text{Be}, \text{Ca}\)]
B. Aluminium (\(\text{Al}\))
C. Oxygen (\(\text{O}\)) and Sulphur (\(\text{S}\))
D. Fluorine (\(\text{F}\))
In simple words: Examples include Mg for group 2, Al for group 13, O and S for group 16, and F for group 17.

Exam Tip: Learn common element symbols for groups 1, 2, 13, 16, and 17.

Question Q-8. Explain Why-
1. All the elements of a group have similar chemical properties.
2. All the elements in a period have different chemical properties.
Answer:
1. Elements in the same group have the same number of valence electrons, which governs their chemical reactivity.
2. Elements in a period have different numbers of valence electrons, resulting in varying chemical properties.
In simple words: Same group means same outer electrons (similar properties); same period means different outer electrons (different properties).

Exam Tip: Chemical properties depend entirely on valence electrons.

Question Q-9. The atomic number of an element X is 17. Predict -
A. Its valency.
B. Nature of the elements.
C. Whether it is metal or non - metal.
D. Name of the element.
E. Relative size with respect to other members of its group.
Answer:
A. Valency = 1 (electronic configuration: \(2, 8, 7\); \(8 - 7 = 1\))
B. Nature: Highly reactive non-metal / halogen
C. Non-metal
D. Chlorine
E. It has a smaller size compared to elements below it in group 17 (since size increases down a group).
In simple words: Chlorine has 7 valence electrons, valency 1, acts as a non-metal, and is smaller than bromine or iodine.

Exam Tip: Elements with 5, 6, or 7 valence electrons are typically non-metals.

Question Q-10. The three elements predicted by mendeleev from the gaps in his periodic table were known as eka- boron, eka- aluminum, eka- silicon. What names were given to these elements when they were discovered later on?
Answer:
- Eka-boron was named Scandium (\(\text{Sc}\))
- Eka-aluminium was named Gallium (\(\text{Ga}\))
- Eka-silicon was named Germanium (\(\text{Ge}\))
In simple words: Mendeleev's predicted elements were later discovered as Scandium, Gallium, and Germanium.

Exam Tip: Memorize Mendeleev's predicted eka-elements and their modern discoveries.

Question Q-11. The atomic numbers of Nitrogen, Oxygen and fluorine are 7, 8, and 9 respectively. Write the electronic configuration of each element and answer the following:
(a) Out of N, O and F which is most electronegative and which one is least electronegative?
(b) What is the number of valence electrons of F?
(c) What is the valency of each of N, O and F?
Answer:
Electronic configurations:
- Nitrogen (7) = \(2, 5\)
- Oxygen (8) = \(2, 6\)
- Fluorine (9) = \(2, 7\)
(a) Most electronegative: Fluorine (F); Least electronegative: Nitrogen (N) (electronegativity increases across a period from left to right).
(b) Number of valence electrons of F = 7.
(c) Valencies: Nitrogen = 3, Oxygen = 2, Fluorine = 1.
In simple words: Fluorine is the most electronegative, has 7 valence electrons, and a valency of 1.

Exam Tip: Electronegativity increases across a period from left to right, making fluorine the most electronegative halogen.