Dobereiner’s Triads: Explanation, Examples, and Limitations

Collegedunia Team logo

Collegedunia Team

Content Curator

In chemistry, triads refer to a group of three elements that share similar chemical properties. The atomic mass of the middle element in a triad is approximately equal to the mean of the atomic masses of the first and last elements in the triad.

  • Examples of such triads are Calcium-strontium-barium, Chlorine-bromine-iodine, and Sulphur-selenium-tellurium.
  • These triads were identified by J.W Dobereiner, a German chemist in the year between 1817 and 1829.
  • Therefore, these triads are known as Dobereiner’s Triads.
  • It is also considered to be the earlier atomic-weight classification of various elements.
Key Terms: Triads, Dobereiner’s Triads, Law of Dobereiner’s Triads, Examples of Dobereiner’s Triads, Atomic mass, Chemical properties of elements, Average

What are Dobereiner’s Triads?

[Click Here for Sample Questions]

Johann Wolfgang Dobereiner, a German chemist, discovered the Dobereiner triads, which are groupings of elements having similar properties.

  • He also observed that the groups of three elements could be formed, where all the elements have the same physical and chemical properties.
  • This grouping of three elements is known as a Dobereiner’s Triads.

Law of Triads

According to Dobereiner's law of triads, the atomic masses of the first and third elements in a triad will be approximately equal to the atomic mass of the second element.

  • Dobereiner’s first triad consisted of the alkaline earth metals that were strontium-barium-calcium that he discovered in 1817.
  • After this, he identified three more triads in 1829.
Dobereiner's Triads
Dobereiner's Triads

Also check: 


Examples of Dobereiner’s Triads

[Click Here for Sample Questions]

Examples of a few triads are given below:

Triad 1

The first triad consisted of alkali metals like Lithium, Sodium, and Potassium.

Triad Atomic Masses
Lithium 6.94
Sodium 22.99
Potassium 39.1

The mean of the atomic masses of Lithium and Potassium is 23.02, which is almost identical to the atomic mass of sodium.

Triad 2

The second triad consisted of Calcium, Strontium, and Barium.

Triad Atomic Masses
Calcium 40.1
Strontium 87.6
Barium 137.6

The mean of the atomic masses of Calcium and Barium is 88.7, which is almost identical to the atomic mass of Strontium.

Triad 3

This triad consisted of halogens like Chlorine, Bromine, and Iodine.

Triad Atomic Masses
Chlorine 35.4
Bromine 79.9
Iodine

126.9

The mean of the atomic masses of Iodine and Chlorine is 81.12, which is almost identical to the atomic mass of Bromine.

Triad 4

The fourth triad consisted of elements like Sulfur, Selenium, and Tellurium.

Triad Atomic Masses
Sulfur 32.1
Selenium 78.9
Tellurium 127.6

The mean of the atomic masses of Selenium and Tellurium is 79.85, which is almost identical to the atomic mass of Selenium.

Triad 5

This triad consisted of elements like Cobalt, Iron, and Nickel.

Triad Atomic Masses
Iron 55.8
Cobalt 58.9
Nickel 58.7

The mean of the atomic masses of Nickel and Iron is 57.3, which is almost identical to the atomic mass of Cobalt.


Limitations of Dobereiner's Triads

[Click Here for Sample Questions]

The following are the limitations of Dobereiner's Triads

  • As more elements were discovered, it became clear that this grouping method wasn't universally applicable.
  • Triads only considered a small number of elements and didn't account for the full range of periodic properties.
  • Only 5 of Dobereiner's triads have been identified.
  • Even a few known elements did not fit into any of the triads.

Also check: 


Things To Remember

  • A triad is a group of three elements that show similar chemical properties.
  • These triads were identified by a German chemist J.W Dobereiner.
  • According to the law of Dobereiner’s triads, the atomic mass of the middle element is approximately equal to the mean of the atomic masses of the first and last elements.
  • Calcium-strontium-barium, Chlorine-bromine-iodine, and Sulphur-selenium-tellurium are examples of Dobereiner’s triads.
  • Only 5 of Dobereiner's triads have been identified yet.

Sample Questions

Ques. Explain Dobereiner’s triad with examples. (2 Marks)

Ans. Dobereiner’s triad states that, in sets of three chemical elements, the atomic mass of one element will be equal to the mean of the atomic mass of two other elements.

Examples: Sulfur-selenium-tellurium; iodine-bromine-chlorine; calcium-strontium-barium.

Ques. What is the law of triads? (2 Marks)

Ans. According to the law of triads, the arithmetic mean of the atomic masses of the first and third elements in a triad is approximately equal to the atomic mass of the second element.

Ques. Why was Dobereiner’s triad discarded? (1 Mark)

Ans. Dobereiner's triad was discarded because it failed to arrange all known elements in triads.

Ques. If A, B, and C are considered three elements of Dobereiner’s triad, calculate the Atomic Mass of element B. The Atomic Mass of A is 7 and C is 39. (2 Marks)

Ans. Given

  • Atomic mass of A = 7
  • Atomic mass of C = 39

According to Dobereiner's law of triads, the average of the atomic masses of element A and element C is approximately equal to the atomic mass of element B

Therefore, the atomic mass of element B = (7 + 39)/2 = 23

Ques. The three elements A, B, and C have similar properties and also have atomic masses X, Y, and Z respectively. Now, the mass of Y is approximately equal to the average mass of X and Z. This arrangement of elements will be called? (1 Mark)
A) Modern triad
B) Dobereiner’s Triad
C) Mendeleev triad
D) None of these

Ans. The correct answer is B. Dobereiner’s Triad.

Explanation: Dobereiner has found the theory that the atomic mass of one atom will be equal to the average of the other two atoms, in a group of three. This specific arrangement is called Dobereiner’s Triad.

Ques. If in Dobereiner's triad, the two members of a group are chlorine and iodine, then what will be the third member of this triad? (2 Marks)
A) Calcium
B) Sodium
C) Bromine
D) Fluorine

Ans. The correct option is C. Bromine.

The atomic mass of the two elements chlorine and iodine are 35.5 and 127.

According to Dobereiner’s Triad, the atomic mass of the third element = (35.5+127)/2 = 81.25

So, 81.25 is the atomic mass of the element Bromine.

Ques. Why did Dobereiner’s Triad become obsolete? (2 Marks)

Ans. Dobereiner’s Triad became obsolete as,

  • The elements of the triad did not have similar chemical or physical properties.
  • Also, the triad did not seem to be true for the newly discovered elements.

Ques. Which elements are not a Dobereiner’s Triad? (1 Mark)
A) Li, Na, K
B) Cl, Br, I
C) Be, Mg, Cr
D) Ca, Sr, Ba

Ans. The correct answer is C. Be, Mg, Cr

Explanation: The mean of atomic masses of Be and Cr is approximately equal to the atomic mass of Mg.

Ques. If in the Dobereiner triads, the two members are phosphorus and antimony, what is the third member of this triad? (2 Marks)
A) Arsenic
B) Iodine
C) Calcium
D) Sulphur

Ans. The correct answer is A. Arsenic

Explanation: According to the Dobereiner triad, the atomic mass of the third element can be found using the arithmetic mean of the two elements i.e. phosphorous and antimony.

We have, the atomic mass of phosphorous is 31 and the atomic mass of antimony is 121.75.

Therefore the atomic mass of the third element = (31+121.75)/2 =76.37

So, the atomic mass of the element is 76.37 and the element is Arsenic.

Ques. Which of the following sets of atomic masses do not match Dobereiner's triad? (1 Mark)
A) 20, 38, 56
B) 54, 95, 183
C) 20, 38, 56
D) 47, 91, 178

Ans. The correct options are B and D

Explanation: These two options do not match Dobereiner’s Triad Law i.e. the mean of the atomic masses of the two matches with the atomic masses of the third element. Hence, only the rest two options A and C obey Dobereiner’s Triad Law.

For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates


Also Read:

CBSE CLASS XII Related Questions

  • 1.
    Write mechanism of acid dehydration of ethanol to ethene.


      • 2.
        For decomposition of $H_2O_2$ by $I^-$: Step I: $H_2O_2 + I^- \rightarrow H_2O + IO^-$ (slow). Step II: $H_2O_2 + IO^- \rightarrow H_2O + I^- + O_2$ (fast). (a) Write rate law. (b) Determine order w.r.t. $H_2O_2$ and $I^-$ and overall order. (c) Molecularity of Step II.


          • 3.
            What are reducing sugars?


              • 4.
                Under what condition can a bimolecular reaction become kinetically first order?


                  • 5.
                    Which isomer of $C_4H_9Br$ is most reactive towards $S_N1$ reaction?


                      • 6.
                        Explain: (i) Presence of carbonyl group in glucose. (ii) Presence of five $-$OH groups attached to different carbon atoms.

                          CBSE CLASS XII Previous Year Papers

                          Comments


                          No Comments To Show