Maharashtra Board Class 10 Science Part – 1 Chapter 2 Periodic Classification Of Elements Solution

Maharashtra Board Class 10 Science Part – 1 Solution Chapter 2 – Periodic Classification Of Elements

Balbharati Maharashtra Board Class 10 Science Part – 1 Solution Chapter 2: Periodic Classification Of Elements. Marathi or English Medium Students of Class 10 get here Periodic Classification Of Elements full Exercise 

 

Chapter: -2

(Periodic Classification Of Elements)

 

Exercise: –

 

1) Rearrange the columns 2 and 3 so as to match with the column 1.

 Column 1 Column 2   Column 3
1.       Triad a.Lightest and negatively charged particle in all the

atoms

1.Mendeleve
2.       Octave b.Concentrated mass and positive charge 2.Thomson
3.       Atomic number c.Average of the first and the third atomic mass 3.Newland
4.       Period d.Properties of the eighth element similar to the first 4.Rutherford
5.       Nucleus e.Positive charge on the nucleus 5.Dobernyer
6.       Electron f.Sequential change in molecular formulae 6.Mosley

 

Ans: –

1.Colulmn 1 2.Column 2 3.Column 3
1.       Triad c. Average of the first and the third atomic mass 5.Dobernyer
2.       Octave d. Properties of the eighth element similar to the first 3.Newland
3.       Atomic number e. Positive charge on the nucleus 6. Mosley
4.       Period f. sequential charge in molecular formula 1.Mendeleve
5.       Nucleus b. Concentrated mass and positive charge 4. Rutherford
6.       Electron a. Lightest and negatively charged particle in all the

atoms

2.Thomson

 

2) Choose the correct option and rewrite the statement.

The number of electrons in the outermost shell of alkali metals is……

i) 1 (ii) 2 (iii) 3 (iv) 7

Ans: -(i) 1.

 

b) Alkaline earth metals have valency 2. This means that their position in the modern periodic table is in ……

(i) Group 2 (ii) Group16  (iii) Period 2 (iv) d block.       

Ans: – (i) Group 2.

 

c) Molecular formula of the chloride of an element X is XCl. This compound is a solid having high melting point. Which of the following elements be present in

the same group as X.

i) Na (ii) Mg (iii) Al (iv) Si.

Ans: -(i) Na.

 

d) In which block of the modern periodic table are the non-metals found?

 (i) s-block (ii) p-block  (iii) d-block (iv) f-block.    

Ans: -(ii) p block.

 

3)  An element has its electron configuration as 2,8,2. Now answer the following questions.

a) What is the atomic number of this element?

b) What is the group of this element?

c ) To which period does this element belong?

With which of the following elements would this element resemble? (Atomic numbers are given in the brackets)

 N (7), Be (4), Ar (18), Cl (17)

Ans: – a) From the configure of this elements we can easily say that the atomic number is 12.

b) The group of this element is 2.

c)This element belongs to third period.

d) This element resembles to Be (4).

 

4) Write down the electronic configuration of the following elements from the given atomic numbers. Answer the following question with explanation.

a) 3Li, 14Si, 2He, 11Na, 15P Which of these elements belong to be period 3?

b) 1H, 7N, 20Ca, 16S, 4Be, 18ArWhich of these elements belong to the second group?

c) 7N, 6C, 8O, 5B, 13A1Which is the most electronegative element among these?

d) 4Be, 6C, 8O, 5B, 13A1.Which is the most electropositive element among these?

e) 11Na, 15P, 17C1, 14Si, 12MgWhich of these has largest atoms?

f) 19K, 3Li, 11Na, 4BeWhich of these atoms has smallest atomic radius?

g) 13A1, 14Si, 11Na, 12Mg, 16SWhich of the above elements has the highest metallic character?

h) 6C, 3Li, 9F, 7N, 8OWhich of the above elements has the highest non-metallic character?

Ans: – a) From electron configuration ofLi (2,1),Si (2,8,4), He (2), Na (2,8,1), P (2,8,5) we can say that Si, Na, P belongs to period 3.

b) From the electronic configuration of H (1), N (2,5), Ca (2,8,8,2), S (2,8,6), Be (2,2), Ar (2,8,8) it is seen that only Ca and Be has 2 electrons in their last shell, so these two belongs to group second.

c) When we move to left to right the electronegativity is increase in a period. From these element (N, C, B, O) oxygen lies at the right side of the period comparing these elements so O is more electronegative.

d) As we know that the electro positivity of an element is increase from top to bottom. From these elements all except Al lie on period and Al is lie on the next so it has more electropositive character.

e) In a period when we go from left to right due to increase in positive charge the electron stays closer to the nucleus so as Al lies on the left comparing this element it has more size.

f) The electronic configuration of the elements is K – (2,8,8,1), Li – (2,1), Na – (2,8,1), Be – (2,2). So from these configurations we see that only Be lies on group 2 and all other in group 1 so Be will have smallest radius.

g) In a period when we go from left to right the metallic character decrease so Al has more metallic character as comparison to others elements.

h) The non metallic character of an elements increase when we go left to right in a period. As fluorine (2,7) lies on the right side of the period in comparison to this element it shows more non metallic character.

 

5) Write the name and symbol of the element from the description.

a) The atom having the smallest size.

b.) The atom having the smallest atomic mass.

c) The most electronegative atom.

d) The noble gas with the smallest atomic radius.

e) The most reactive non-metal.

Ans: – a) The atom having smallest size is Helium.

b) The atom having smallest atomic mass is Hydrogen.

c) The most electronegative atom is fluorine (F).

d) The noble gas with the smallest atomic radius is He or helium.

e) The most reactive non metal is fluorine.

 

6) Write short notes.

a) Mendeleev’s periodic law.

b) Structure of the modern periodic table.

c) Position of isotopes in the Mendeleev’s and the modern periodic table.

Ans: –a) Mendeleev’s periodic law: – The physical and chemical property of an element in a periodic table will act according to the atomic mass of the elements. He states the row as period and column as group. The similar property showing elements will lie on the same period.

b) Modern periodic table consists of 18 group and 7 periods. The vertical line is called the group and the horizontal lines are called period. The element in the modern periodic table are according to the atomic number.

c) In modern periodic table the atoms arrange according to their atomic number. So in these case isotopes also be arranged in accordance to their atomic number. The isotope has similar atomic number but different mass number.

 

7) Write scientific reasons.

a) Atomic radius goes on decreasing while going from left to right in a period.

b) Metallic character goes on decreasing while going from left to right in a period.

c) Atomic radius goes on increasing down a group.

d) Elements belonging to the same group have the same valency.

e) The third period contains only eight elements even through the electron capacity of the third shell is 18.

Ans: – a) When we go from left to right in a period the atomic radius increase because of the increase of valance electron in the outer shell, the attraction towards the nucleus increase. So the atomic radius increase in a period for the increasing of atomic no.

b) In a period from left to right metallic character decrease because of the increase in the nuclear charge the valence electron strongly attached to nucleus and its difficult to pulled out.

c) In a group when we go from top to bottom we will see that the atomic radius increase because the new shell added while going into downward in a period.

d) In a group there are same electronic configuration in the last shell so all the elements in a group show same no of electron in last orbital, as the electron are same the valency must be same.

e) The third period has only 8 elements because other shell filled and there is no electron to become eighteen, and these electrons added to third shell and remaining added to the fourth period.

 

8) Write the names from the description.

a) The period with electrons in the shells K, L and M.

b) The group with valency zero.

c) The family of non-metals having valency one.

d) The family of metals having valency one.

e) The family of metals having valency two.

f) The metalloids in the second and third periods.

g) Non-metals in the third period.

h) Two elements having valency 4.

Ans: –a) period 3 has the electron of the shell K, L, M.

b) Group 18 has zero valency.

c) Halogen has the valency one in the family of non metals in periodic table.

d) The metal family having one valency is known as Alkali metal.

e) Alkaline earth metals has valency 2 in the metals family.

f) The metalloids Boron(B) in second period, and the metalloids Silicone (Si) in the third period.

g) Phosphorus(P),Sulphur(S), Chlorine(Cl), and Argon(Ar) are in the non metal lies on the third period.

h) Carbon (2,4) and silicone (2,8,4) have the 4 electron in their outer shell so there valency is four.

 

 

Here is your solution of Maharashtra Board Class 10 Science Part – 1 Chapter 2 Periodic Classification Of Elements

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Updated: December 1, 2021 — 12:53 pm

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