Question 1 - Past Year - 3 Mark Questions - Chapter 5 Class 10 - Periodic Classification of Elements

Last updated at April 16, 2024 by

Study the data of the following three categories A. B and C.

 

Answer

Category

Name of the element and Atomic Mass

Checking triad

A

  • Li (Atomic mass = 7)
  • Na (Atomic mass = 23)
  • K (Atomic mass = 39)

Mean of first and last = (7 + 39)/2 = 23

Atomic mass of middle = 23

Since they are the same , it is a dobereiner’s triad.

B

  • N (Atomic mass = 14)
  • P (Atomic mass = 31)
  • As (Atomic mass = 74)

Mean of first and last = (14 + 74)/2 = 44

Atomic mass of middle = 31

Since they are not the same , it is not a dobereiner’s triad.

C

  • B (Atomic mass = 10.8)
  • Al (Atomic mass = 27)
  • Ga (Atomic mass = 69.7)

Mean of first and last = (10.8 + 69.7)/2 = 40.25

Atomic mass of middle = 27

Since they are not the same , it is not a dobereiner’s triad.

Therefore, category 3 is a dobereiner’s triad.

Why did Mendeleev place elements of category  A, B and C in three different groups? 

  • Mendeleev arranged the elements according to  
  1. increasing atomic masses   
  2. The elements having similar chemical properties were placed together 
  • The elements of category A, B and C are similar among themselves ( have similar chemical properties ) but each category is different from each other.

Therefore, category A,B and C were placed in different groups.

 

Is Newland law of octaves applicable to all the three categories ? Give reason to justify your answer.

  • In newlands law of octave, the first and eighth elements show similar properties . 
  • This rule is not valid for elements after Ca in the periodic table. Therefore, the newland’s law of octave is not applicable for all the three categories.
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