# Periodic Classification of Elements Class 10 Notes | Questions Chapter 5 | 2024

Last updated on July 14th, 2024 at 05:09 pm

## Periodic Classification of Elements Notes

Below are some of the very important NCERT Class 10 Science chapter 5 Periodic classification of elements notes and  Questions with Answers. These Class 10 Periodic classification of elements notes and questions have been prepared by expert teachers and subject experts based on the latest syllabus and pattern of term 2. Questions with Answers to help students understand the concept.

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All substances are made up of elements. At present, there are 118 elements known, out of which 98 are naturally occurring. In order to study the properties of all these elements separately, scientists felt the necessity to group elements having similar characteristics together.

Earlier Attempts at the Classification of Elements

Several attempts have been made to classify the elements according to their properties. Later, many classifications were tried. Some important of them are discussed below:

He arranged three elements with similar properties into groups which are known as triads and showed that when three elements in a triad were arranged in order of increasing atomic masses, the atomic mass of the middle element was roughly the average of atomic masses of other two elements. He could identify only three triads from the elements known at that time which are

Lr, Na, K, Ca, Sr, Ba, CI, Br, I

## Newland’s Law of Octaves

Newland arranged the known elements in order of increasing atomic masses and found that every eighth element has properties similar to that of the first. This law was applicable only upto calcium and he assumed that there were only 56 elements.

To fit elements into his table, he adjusted two

elements in the same slot and also put some unlike elements under the same column which have very different Properties than other elements.

## Mendeleev’s Periodic Table

According to this, the physical and chemical properties of the elements are periodic functions of their atomic masses, i.e. on arranging the elements in increasing order of their atomic masses, the similar properties were repeated after regular intervals.

He took the formulae of the hydrides and oxides formed by an element as one of the basic properties of an element for its classification. e.g. Hydride of carbon, CH4 as RH4 and its oxides, C02 as RO 9.

He then arranged 63 elements in the increasing order of their atomic masses and found that there was a periodic recurrence of elements with similar physical and chemical properties. He observed that elements with similar properties fall in the same vertical column. These vertical columns are called groups and horizontal rows of elements are called periods.

Features of Mendeleev’s Periodic Table

• It consists of 8 vertical columns, called groups and 6 horizontal rows, called periods.
• In every period, elements are arranged in increasing order of their atomic masses.
• He left gaps for the elements not discovered at that time and named such elements by prefixing a Sanskrit numerical Eka (one), divi (two) to the name of the preceding similar element in the same group. e.g. Eka-boron, Eka-alurniniurn, which after their discovery were named scandium, gallium.
• He also predicted the atomic masses and properties of several elements that were not known at that time.

Properties of Eka-aluminum and Gallium

Property      Eka-aluminum      Gallium

Atomic mass               68                69.7

Formula of oxide    E2O3            Ca2O3

Formula of chloride    ECI3            GaCI3

One of the strengths of Mendeleev’s periodic table was that, when noble gasses like helium, neon were discovered, they could be placed in a new group without disturbing the existing order.

Limitations of Mendeteev’s Periodic Table

• Elements with similar properties are kept in a same group
• Positions of hydrogen was not fixed in periodic table
• Elements with similar properties were kept in different groups
• Heavier elements were kept before the lighter elements
• Position of isotopes and isobars could not be explained

Mendeleev’s Periodic Table (Published in a German journal in 1872)

In the formula of oxide and hydrates at the top of the columns the letter R is used to represent any of the elements in the group

## Modern Periodic Table

In 1913, Henry Moseley showed that the atomic number of an element is a more fundamental property. On the basis of this, he modified Mendeleev’s periodic law as physical and chemical properties of the elements arc a periodic limction of their atomic number”. This is called modern periodic law.

When the elements are arranged in the increasing order of their atomic number, the obtained table is called the modern periodic table.

In this periodic table, hydrogen is kept at the top left corner because of its unique characteristics. The position of cobalt and nickel is also justified.

Features of Modern Periodic Table

This table has 18 vertical columns, known as groups and 7 horizontal rows, known as periods.

A few important features of the elements present in groups and periods are as follows

• The groups are not divided into sub-groups.
• The elements present in a group have the same number of valence electrons and valency.
• The number of shells increases as we go down the group.
• The elements present in a group have identical chemical properties and their physical properties like density, melting point vary gradually.
• Elements of a period have the same number of shells but they do not contain the same number of valence electrons. So, their chemical properties are also different.
• The number of valence shell electrons increases by one unit as the atomic number increases by one unit on moving from left  to right in a period.
• In this table, elements of group 13, 14, 15, 16 and 17 are called normal elements which includes metals, nonmetals and metalloids and elements of group 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12 are called transition elements.
• In this periodic table, elements from atomic number 58 to 71 called as lanthanides and elements from 91 to 103 called as actinides are kept out of the table.

Position of Elements in the Modern Periodic Table

For this, first of all write electronic configuration of the given element. Number of shells present in the electronic configuration shows the period number of that element. Number of valence electrons present in the electronic configuration show the group number of the element.

For example, Electronic configuration of the element with atomic number 19 is 2, 8, 8, 1, since it has four shells, thus it is element of fourth period. Due to presence of one electron in the last shell, its group number is l. After donating one electron, it acquires a stable configuration, hence its valency is also l.

Trends in Modern Periodic Table

(i) Valency. In a period, it increases with respect to hydrogen from 1 to 4 after that it decreases. On thc other hand with respect to oxygen, valency increases frorn 1 to 7. In a group, valency remains same as outer electronic configuration is same.

(ii) Atomic size. Atomic size decreases on moving from left to right in a period due to increase in nuclear charge. It increases down the group as new shells are being added.

iii) Metallic and non-metallic properties Effective nuclear charge acting on the valence shell electrons increases across a period and decreases down the group.

Therefore, metallic character decreases across a period and increases down a group. Non-metallic character, however, increases across a period and decreases down a group.

Metals like Na, Mg are present on left side of periodic table, whereas non-metals like S, Cl are present on right side of periodic table.

There are some metals which exhibits both the properties of metal and non-metals. These are called metalloids like Po, Te, Sb, etc.

(iv) Electronegativity. The electronegativity of the elements increases along a period, since the non-metallic character increases. Similarly, it decreases down the group, since the non-metallic character decreases.

(v) Nature of oxides: On moving from left to right in a period, due to increase in non-metallic character, basic nature of oxides decreases while acidic nature increases

On going down the group, the order is reversed.

1. What were the limitations of Dobereiner’s classification ?    (NCERT)

Ans. All the elements discovered at that time could not be classified into triads, only a limited number of elements could be arranged in such triads.

e.g. The three elements nitrogen (N), phosphorus (P) and  arsenic (As) have similar properties. Therefore, they should be regarded as forming a triad.

However, the actual mass of the middle element P (31.04) is  much lower than the average (44.454) of the atomic masses of nitrogen (14.4) and arsenic (74.94). Therefore, these three elements do not constitute a Dobereiner’s triad in spite of  their similar chemical properties.

2. Can the following groups of elements be classified as Dobereiner’s triad?

(i) Na, Si, cl    (ii) Be, Mg, Ca

Atomic mass of Be-9; Na-23; Mg-24; 5-28; cl-35; Ca-40

Ans.

(i) Na, Si and Cl have different properties, therefore, they   do not form Dobereiner’s triad even though the atomic mass of the middle atom (Si) is approximately the average of the atomic masses of Na and Cl, i.e. Na (23); Si (28); cl (35)

Atomic mass of Si = 23 + 35/2 =  29

(ii) Be, Mg and Ca have many similar properties and also the atomic mass of the middle element Mg is approximately the average of the atomic masses of Be and Ca, i.e.

Be (9); Mg (24); ca (40)

Atomic mass of Mg =  9+40/2 = 24.5

3. Did Dobereiner’s triads also exist in the columns of Newlands’ octaves? Compare and find out. (NCERT)

Ans.

Yes, Dobereiner’s triads also exist in the columns of Newlancls’ octaves, e.g. lithium (Li), sodium (Na) and potassium (K) constitute a Dobereiner’s triads.

Now, if we consider Li as the first element, then the eighth element from it is Na and if we consider Na as the first element, then  the eight element from it is K,

Similarly, Dobereiner’s triad consisting of the Clements beryllium (Be), magnesium (Mg) and calcium (Ca) is also included in the column of Newlands’ octaves.

Thus, Dobereiner’s triads are included in the columns of Newland’s octaves.

4. Elements have been arranged in the following sequence on the basis of their increasing atomic masses. F, Na, Mg, Al, Si, P, S, cl, Ar, K.

(i) Pick two sets of elements which have similar properties.

(ii) The given sequence represents which law of classification of elements ?

Ans.

(i) Here, the elements are arranged in the order of  increasing atomic masses, so according to Newlands’ law of octaves there is a repetition of every eighth element as compared to the given element; The two sets of elements which have similar properties are

Set I — F, cl

Set II — Na + K

F and Cl are the first and eighth elements in the above sequence, therefore, they have similar properties. Although Na and K have similar properties, they are   not related as first and eighth elements in the above   sequence.

(ii) The given sequence is according to Newlands’ law of octaves represented as

F Na Mg Al Si P S cl Ar K

5. What were the limitations of Newlands’ law of octaves? (NCERT)

Ans.

This law was applicable only upto calcium. After calcium, every eighth element did not possess the same properties similar to that of the first.

Newland assumed that there were only 56 elements that existed in nature and no more elements would be discovered in the future. But, later on, several new elements were discovered, whose properties did not fit into the law of octaves.

In order to fit elements into his table, Newlands adjusted two elements in the same slot and also put some unlike elements under the same column.

e.g. Cobalt and nickel are in the same slot and these are  placed in the same column as fluorine, chlorine and bromine which have very different properties than these elements. Iron, which resembles cobalt and nickel in properties, has been placed far away from these elements. Hence, Newlands’ law of octaves worked well with lighter elements only.

6. What were the criteria used by Mendeleev in creating his periodic table.    (NCERT)

Ans.

The criteria used by Mendeleev were:

(i) The arrangement of elements in increasing order of atomic masses.

(ii) Similarity in chemical properties of the elements.

7. In Mendeleev’s periodic the elements  were arranged in the increasing order of their atomic masses. However, Cobalt with atomic mass of 58.93 was placed before nickel having an atomic masses of 58.71 amu. Give reason for the same. NCERT (Exemplar)

Ans.

In Mendeleev’s periodic table, cobalt (Co) with a higher atomic mass of 58,93 u is placed before nickel (Ni) due to the

following reasons ;

(i) The properties of cobalt are similar to those of rhodium (Rh) and iridium (Ir) (same group) and

(ii) The properties of nickel are similar to those of palladium (Pd) and platinum (pt) (same group).

8. Write the formula of chlorides of Eka-silicon and Eka-aluminium, the elements predicted by Mendeleev.

Ans.

Eka-silicon is germanium (Ge), It lies in group 4 of the Mendeleev’s periodic table and thus, has a valency of 4.

The formula of its chloride is GeC14.

Eka-aluminum is gallium (Cya). It lies in group 3 of the Mendeleev’s periodic table and thus, has a valency of 3.

The formula of its chloride is GaCl3.

9. Write two main characteristics of Mendeleev’s periodic table and write the name of elements of the second period.

Ans.

Two main characteristics of Mendeleev’s periodic table are :

(i) It consists of 8 vertical columns, called groups and

6 horizontal rows, called period

(ii) In every period, elements are arranged in increasing order of their atomic masses.

The elements of the second period are lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine.

10. How it can be proved that the basic structure of the ‘modern periodic table’ is based on the electronic configuration of atoms of different elements? (CBSE 2019)

Ans.

Electronic configuration of an element decides its position in the modern periodic table.

If we take an example of sodium (Na), which has atomic number = 11, i.e. it’s electronic configuration = 29 8, 1 As Na contains 1 electron in its outermost shell, this means that it belongs to group I and sodium contains 3 shells so, it belongs to period number 3.

We can conclude that,

Group number = Number of valence electrons

(When valence electrons are 1 and 2) and group number = 10 + valence electrons

(When valence electrons are 3 and above)

Period number = Number of shells in which electrons are filled.

11. ‘Hydrogen occupies a unique position in the modern periodic table’. Justify the statement. (NCERT Exemplar)

Ans.

Hydrogen occupies a unique position in the modern periodic table due to the following reasons

(i) Both hydrogen and alkali metals have similar outer electronic configuration as both have one electron in  the valence shell. Therefore, some of the properties of hydrogen are similar to those of alkali metals and hence, it can be placed in group 1 alongwith alkali metals.

(ii) Both hydrogen and halogens have similar outer electronic configuration (both have one electron less than the nearest inert gas configuration). Therefore,   some of the properties of hydrogen are similar to those  of halogens and hence, it can be placed in group 17 alongwith halogens.

(iii) In some properties, it differs from both hydrogen and halogens, e.g. the oxide of hydrogen, i.e. H2O is neutral but the oxides of alkali metals (i.e. Na2O, K2O etc.) are basic while those of halogens (i.e. CI2O7, Br2O5, I2O5 etc.) are acidic.

12. (i) List any two distinguishing features between Mendeleev’s periodic table and the modern periodic table.

(ii) With the help of an example, explain Dobereiner’s Triads.

Ans.

Dobereiner arranged the elements with similar properties into groups having three elements each and named these groups as triads.

He showed that when the three elements in a triad Were arranged in the order of increasing atomic masses, the atomic mass of the middle element was roughly the average of the atomic masses of other two elements,

Elements        cl       Br        I

Atomic mass      35.5      80    127

Average atomic mass 35.5 + 127/2= 81.25 of first and that elements 2

(iii) State modern periodic law. (CBSE 2020)

Ans. Modern periodic law The physical and chemical  properties are a periodic function of their atomic number.

13. Write the formula of the product formed when the element A (atomic number 19) combines with the element B (atomic number 17).

Draw its electronic dot structure. What is the nature of the bond formed? (NCERT Exemplar)

Ans.

Atomic number of A = 19

Electronic configuration is 2, 8, 8, 1.

Hence, element A is metal potassium (K) and

Atomic number of B = 17.

Electronic configuration is 2, 8, 7.

It is a non-metal chlorine (Cl).

So, the electron dot structure of KCI is

The bond formed between K+ and Cl- is an ionic bond and formula of the product formed K+ Cl- or KCI.

(a) X has 12 protons and 12 electrons,

(b) Y has 12 protons and 10 electrons    (NCERT Exemplar)

Ans.

Since species X has 12 protons and 12 electrons, it is  electrically neutral. Since species Y has 12 protons and 10 electrons, therefore, it has two units of positive charge.

The electronic configuration of the two species are

Species X    Species Y

K    L    M           K     L

2    8     2            2     8

Since, species X has three shells while species Y has two shells, therefore, species Y has a smaller radius than species X.

15. Arrange the following elements in increasing order of their atomic radii.

(i) F and N    (ii) Cl, At, Br and 1 (NCERT Exemplar)

Ans.

(i) Atomic radii decrease along a period from left to right due to increase in nuclear charge. Li, Be, F and N belong to the same period. Thus, the atomic radii of Li, Be, F and N increases in the order:

F < N  < Be  <  Li

(ii) Atomic radii increase in a group from top to bottom due to the corresponding increase in the number of filled electronic shells. Cl, At, Br, I belong to the same group. Thus, atomic radii of Cl, At, Br and I increase in the order:

Cl < Br < I < At

16. An element X of group 15 exists as a diatomic molecule and combines with hydrogen at 773 K in presence of the catalyst to form a compound, ammonia which has a characteristic pungent smell.

(i) Identify the X. How many valence electrons does it have?

(ii) Draw the electron dot structure of the diatomic molecule of X. What type of bond is formed in it?

(iii) Draw the electron dot structure for ammonia and what type of bond is formed in it?

Ans.

Since, the element ‘X’ of group 15 exists as a diatomic molecule and combines with hydrogen at 773 K in presence of a catalyst to form ammonia which has a

characteristic smell, therefore, the element ‘X’ is nitrogen (N).

(i) The atomic number of nitrogen is 7. So, its electronic configuration is 2, 5. Thus, it has five valence electrons.

(ii) Nitrogen has 5 valence electrons. Therefore, it needs 3 more electrons to complete its octet.

To do so, it shares three Of its electrons with three electrons of the other nitrogen atom to form a diatomic molecule of N2 gas.

Thus, three covalent bonds are formed between two nitrogen atoms and each nitrogen atom is left with one lone pair of electrons.

In the NH3 molecule, there are three N—H single covalent bonds and one lone pair of electrons on the nitrogen atom.

17. A salt dissolved in water dissociates into cations and anions follows:

If both the ions consist of the same number of electrons and the weight of salt is 74.5, then identify the position of A and B in the periodic table.

Ans.

Since, both the ions consists of same number of  electrons and has +1 and -1 charges, hence the ions should belong to group 1A (cation, i.e. A+ ) and group VIl A (anion, i.e. B-).

Same number of electrons indicates that their electronic configuration is the same as that of a noble gas whose atomic number lies between that of the two elements A and B.

Dividing the molecular weight (which is sum of atomic masses of A and B), we get the rough idea about the atomic mass of the noble gas which is 74.5/2 = 37.25, i.e. nearest to argon (Ar – 40).

Hence, A is K (group IA, 4th period) and element B is Cl (group VIl A, 3rd period).

18. Three elements A, B and C have 3, 4 and 2 electrons respectively in their outermost shell. Give the group number to which they belong in the modern periodic table, Also, give their valencies.    (NCERT Exemplar)

Ans.

(i) Element A has 3 valence electrons, therefore, its valency is 3 and thus belongs to group 13 (3 + 10). As such, it could be any one of the following elements : B, Al, cya, In or Tl.

(ii) Element B has 4 valence electrons, therefore, its valency is 4 and it belongs to group 14 (4 + 10). The element B could be any one of the following : C, Si, Ge, Sn or Pb.

(iii) Element C has two valence electrons, therefore, its valency is 2 and it belongs to group 2. The element C could be any one of the following ; Be, Mg, Ca, Sr, Ba or Ra.

19. Based on the group valency of elements, write the molecular formula of the following compounds giving justification for each.

(i) Oxide of first group elements

(ii) Halide of the elements of group thirteen

(iii) Compounds formed when an element A of group 2 combines with an element, B of group seventeen. (CBSE 2019)

20.

From the elements Li, K, Mg, C, Al, S identify the

(i) Elements belonging to the same group.    (CBSE 2020)

(ii) Elements which have the tendency to lose two electrons.

(iii) Element which prefers sharing of electrons to complete its octet.

(iv) Most metallic element.

(v) Element that forms acidic oxide.

(vi) Element that belongs to group 13.

Ans. 19 and 20

(i) Elements belonging to the same group are Li and K as they both contain one electron in their outermost shell.

(ii) Element which has a tendency to lose two electrons is magnesium as it contains 2 electrons in its outermost shell.

(iii) Element which prefers the sharing of electrons to complete its octet is carbon due to its small size and strong C— C bond.

(iv) Most metallic element is potassium. Elements of group I are metallic in nature as they readily loose their one valence electron.

(v) Non-metals form acidic oxides. Among the given elements, S is a non-metal. Thus, it forms the most acidic oxide.

(vi) Aluminum belongs to group 13 as it contains 3 elements in its outermost shell.

21. The following table shows the position of five elements A, B, C, D and E in the modern periodic table.

(i) Which element is a metal with valency two?

(ii) Which element is least reactive?

(iii) Out of D and E which elements has a smaller atomic radius?

Ans.

(i) Element ‘D’ is a metal with valency two.

Because, the group number of an element having upto two valence electrons is equal to the number of valence electrons.

(ii) ‘C’ is the least reactive element. Because, it belongs to group 18. Group 18 (Noble gasses) are least reactive due zero valency of group 18 elements.

(iii) ‘E’ has a smaller atomic radius than ‘D’ because ‘E’ is on the right side of the modern periodic table. Across the period, atomic size/radius decreases on moving left to right.

This is due to an increase in nuclear charge which tends to pull the valence electrons closer to the nucleus and reduces the size of the atoms.

22.   (i) What term can be used Ibr the elements separating metal from non nonmetals and why?

(ii) Give the names of the  metalloids in the periodic table along with their atomic number.

(iii) In which groups of the periodic table are they located?

Ans.

(i) Metalloids as these elements show the properties of both the metals and non-metals.

(ii) The list of metalloids alongwith their atomic number is as follows:

B(5), Si(14), Ge(32), As(33), Sb(51), Te(52), Po(84)

(iii) These elements are located in groups 13, 14, 15 and 16.

23. A group of elements in the periodic table are given below (boron is the first member of the group and thallium is the last).

Boron, aluminum, gallium, indium, thallium

Answer the following question in relation to the above group of elements.

(i) Which element has the most metallic character?

(ii) Which element would be expected to have the highest electronegativity?

(iii) Will the elements in the group to the right of this boron group be more metallic or less metallic in character? Justify your answer.

Ans.

(i) Thallium has the most metallic character. Metallic character increases down in a group,

(ii) Boron has the highest electronegativity because electronegativity decreases down a group.

(iii) Less metallic in character, because on moving across a period, metallic nature decreases.

24. Use Mendeleev’s periodic table to predict the formula for the oxides of the following elements.

K, C, Al, Si and Ba.

Ans.

Oxygen is a number or group VI A in Mendeleevss periodic table, Its valency is 2. Similarly, the Valencies of all the elements given can be predicted from their respective group.

This can help in writing the formula of their oxides.

(i) Potassium (K) is a member of group IA. Its valency is 1. Therefore, the formula of their oxide is K2O.

(ii) Carbon (C) is a member of group IVA. Its valency is 4.

Therefore, the formula of its oxide is     or C2O4 or CO2

(iii) Aluminum (Al) belongs to group IIIA and its valency is 3. Therefore, the formula of its oxide is A12O3

(iv) Silicon (Si) is present in group IVA after carbon. Its valency is also 4. Therefore, the formula of its oxide is Si2O4 or Si02.

(v) Barium (Ba) belongs to group IIA and its valency is 2. Therefore, the formula of its oxide is Ba2O2 or BaO.

25. Compare and contrast the arrangement of elements in Mendeleev’s periodic table and the modern periodic table.    (NCERT)

Ans.

26. An element is placed in the 2nd group and 3rd period of the periodic table, burned in the presence of oxygen to form a basic oxide. (NCERT Exemplar)

(i) Identify the element.

(ii) Write the electronic configuration.

(iii) Write a balanced equation when it burns in the presence of air.

(iv) Write a balanced equation when this oxide is dissolved in water.

(v) Draw the electron dot structure for the formation of this oxide.

Ans.

27. Which elements has —

(i) two shells, both of which are completely filled with electrons ?

(ii) the electronic configuration 2, 8, 2?

(iii) a total of three shells, with four electrons in its valence shell?

(iv) a total of two shells, with three electrons in its valence shell?

(v) twice as many electrons in its second shell as in its first shell?    (NCERT)

Ans.

(i) Noble gasses are the elements which have completely. The noble gas with two shells (K, L) is Ne having atomic number and electronic eon figuration  both of the shells are completely filled.

(ii) Electronic configuration 2, 8, 2 suggests that the atomic number is 12 (2 + 8 + 2). Magnesium (Mg) has atomic number 12.

(iii) The element with three shells and four electrons in the valence shell will have electronic configuration K 2, L 8, M 4

The atomic number of this element is 14 (2+8+4) so it will belongs to group 14. Hence, it is silicon (Si).

(iv) Element with two shells and 3 electrons in the valence shell will exist in second period and will have the electronic configuration

The atomic number of this element will be 5 (2, 3). So, it will be boron (B).

(v) The element has two shells. We know that the first shell can have only 2 electrons, so according to the question there will be 4 electrons (double the number of electrons in the first shell) in the valence shell. The electronic configuration will be  K 2, L 4,    so the atomic number is 6. Hence, the element is carbon (C).

28. An element X (atomic number = 17 ) reacts with an element Y (atomic number = 20) to form a divalent halide.

(i) Where in the periodic table are elements X and Y placed?

(ii) Classify X and Y as metal(s), non-metal(s) or metalloid(s).

(iii) What will be the nature of oxide of element Y? Identify the nature of bonding in the compound formed.

(iv) Draw the electron dot structure of the divalent halide. (NCERT Exemplar)

Ans.

(i) The electronic configuration of element X with atomic number 17 is 2, 8, 7. Since, it has 7 valence electrons. Therefore, it lies in group 17 (10 + 7). Further, since in element X, the third shell is being filled, it lies in the third period.

In other words, X is chlorine.

The electronic configuration of element Y with atomic number 20 is 2, 8, 8, 2. Since it has 2 valence electrons, it lies in group 2. Further, since in element Y, the fourth shell is being filled, it lies in the 4th period. In other words, Y is calcium.

(ii) Since, element X (i.e. Cl) has seven electrons in the valence shell and needs one more electron to complete its octet. Therefore, it is non-metal.

Further, the element Y has two electrons in the valence shell that can be easily lost to achieve the stable electronic configuration of the nearest inert gas, therefore, it is a metal.

(iii) Since, element Y (i.e. Ca) is a metal, therefore, its oxide (i.e. CaO) must be in nature. Further metals and non-metals form ionic compounds, therefore, the nature of bonding in calcium oxide is ionic.

(iv) Electronic configuration of 20Ca = 2, 8, 8, 2 [valence electrons = 2], electronic configuration of 17Cl = 2, 8, 7 [valence electrons = 7]. The electron dot structure of divalent metal halide.

29. (i) Using the part of’ the periodic table given below answer the questions that follows:

(a) Na has physical and chemical properties similar to which element(s).

(b) Write the electronic configuration of N and P. Which one of these will be more electronegative and why?

(c) State a chemical property common to fluorine and chlorine.

(ii) The neutral atom of an element E consists of 12 electrons in its atoms.

(a) In which period and group is E placed?

(b) Name the element E.

(c) How many electrons does it needs to lose or gain to achieve noble gas configuration ?

(d) What will be the nature of oxide (acidic/basic) of E? Justify your answer.

(e) Write the formula of chloride of E.

Ans.

(i) (a) Lithium and potassium, due to the presence of the same number of valence electrons.

(b) N — 2,5       P — 2,8,5

N is a more electronegative element as  electronegativity decreases on moving down the  group.

(c) Both fluorine and chlorine form their hydrides on reacting with hydrogen.

H2+ F2 —  2HF

H2+ C12 — 2HCI

(ii) (a) The electronic configuration of E = K 2, L 8, M 2 Hence, it should be placed in 3rd period and group II A.

(b) The element E is magnesium (Mg).

(c) Mg loses 2 electrons to form noble gas (Ne) with configuration as K 2, L 8.

(d) Since, Mg is electropositive, it will form basic oxide.

(e) The formula of chloride is ECI2 or MgCI2.

### Case Based Questions

30. Read the following and answer the questions from (i) to (v) given below

Modern periodic law states that the physical and chemical properties of the elements are periodic functions of their atomic radius.

When these elements were arranged in the increasing order of their atomic number, the obtained table is called the modern periodic table.

Numerous forms of the periodic table have been devised from time to time. A new version, which is most convenient and widely used, is the modern periodic table. This table consists of 18 vertical columns called groups and 7 horizontal rows, known as periods.

The first period consists of two elements.

The subsequent consist of 8, 8, 18, 18 and 32 elements respectively. The seventh period is incomplete and like the sixth period would have a maximum of 32 electrons.

(i) An element belongs to group 17. It is present in the third period and its atomic number is 17. What is the atomic number of the element belonging to the stune group and present in the fifth period ?

(ii) Name the elements present in the first period of modern periodic table.

(iii) Atoms of different elements with the same number shells are placed in the same period. Explain.

(iv) What is the electronic configuration of the element present in the third period and belongs to group 15 ?

(v) How many elements are present in modern periodic table ?

Ans.

(i) The element whose atomic number is 17 and belongs to the third period is chlorine. If we go down the group, i.e. in the fifth period, the number of shells increases as 18e- are increased in consecutive periods.

Therefore, the atomic number of the element belonging to the same group and present in the fifth period is 17 + 18 + 18 = 53.

(ii) Hydrogen (H) and helium (He).

(iii) The number of valence shell electrons increases by one unit as the atomic number increases by one unit on moving from left to right in a period. Therefore, the atoms of difförent elements with the same number of shells are placed in the same period.

(iv) As the element belongs to group-15, i.e., it’s valence electrons are 5 and valency is 3, so it has the electronic configuration = 2, 8, 5.

(v) At present, 118 elements are known to us. All these have different properties. Out of these 118 elements, only 94 are naturally occurring.

31. Read the following and answer the questions from (i) to (v) given below

Valency is the combining capacity of an atom of an element to acquire noble gas configuration. It depends upon the number of valence electrons present in the outermost shell of its atom.

For the elements of group 1, 2, 13 and 14 valency = numbers of valence electrons(s), whereas for the elements of group 15 onwards valency = 8 — valence electrons.

The concept of valency is simple and rationalise the atomic composition of a large number of compounds. Yet, in many chemistry courses, it is sidelined in favor of electronic theories of bonding, which are more difficult.

When the theory of valency was devised, chemists thought that all compounds were molecular. We now know that many are non-molecular, i.e. they comprise a large number of atoms bound together in a continuous framework.

The theory can however be adapted to include non-molecular compounds.

The atomic number and valence of elements A, B, C, D and E are given in the below table.

Elements   Atomic number    Number of

valence electron

A                    3                          1

B                    9                          7

C                   17                         7

D                   20                         2

E              36                         0

(i) What is the valency of element B ?

(ii) If element C reacts With an element D, which type of compound they will form.

(iii) What is the name of element E ?

(iv) Element A belongs to which period in the modern periodic table ?

(v) Among A, C, D and E, which element belongs to group 17 ?

Ans.

(i) The atomic number of element B is 9 and the number of valence electrons is 7, i.e. it belongs to group 17.

Valency = 8 –  valence electrons

= 8 – 7= 1

Hence, the valency of B is 1.

(ii) Electronic configuration of element C is 2, 8, 7, i.e. it has 7 valence electrons and valency is -1 (as it requires 1 electron to complete its octet) and electronic   configuration of element D is 2, 8, 8, 2, i.e. it has 2 valence electrons and valency.

This forms a divalent compound (C2D).

(iii) The atomic number of element E is 36, so the  electronic configuration is [Ar]3d10 4s2 4p6. Therefore, this element belongs to 4th period as the last electron goes in 4th shell and has zero number of valence electrons.

Hence, the element is krypton.

(iv) The atomic number of element A is 3 and electronic  configuration is Is2 2s1. As last electron enters in 2nd shell, so, it belongs to period 2.

(v) The element that belongs to group 17 is ‘C’ because this group contains those elements which have 7 electrons in their outermost shell and element C also contains 7 valence electrons, Therefore, option (b) is correct.

Final Words

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