Triad
(1)Mg (24.3), Ca (40.1), Sr (87.6)
(2)S (32.1), Se (79,0), Te (127.6)
(3)Be (9.0), Mg (24.3), Ca (40.1)
Find the atomic mass of middle element by calculation
Weather Dobereiner's triads or not? Why?
Answers
Answer:
given in ncert but still solving
Explanation:
mg+Sr/2
S+Te/2
so on
Answer:
1) Mg ( 24.3 ), Ca ( 40.1 ), Sr ( 87.6 ) is not a Dobereiner's triad.
2) S ( 32.1 ), Se ( 79.0 ), Te ( 127.6 ) is a Dobereiner's triad.
3) Be ( 9.0 ), Mg ( 24.3 ), Ca ( 40.1 ) is a Dobereiner's triad.
Explanation:
We have given some groups of elements.
We have to identify Dobereiner's triads from those groups.
1. Mg ( 24.3 ), Ca ( 40.1 ), Sr ( 87.6 )
Atomic mass of first element ( Mg ) = a = 24.3
Atomic mass of second element ( Ca ) = b = 40.1
Atomic mass of third element ( Sr ) = c = 87.6
We know that,
Atomic mass of middle element is equal to the mean of other two elements in a Dobereiner's triad.
∴ Mean = a + c / 2
⇒ Mean = 24.3 + 87.6 / 2
⇒ Mean = 111.9 / 2
⇒ Mean = 55.95
Here, the mean atomic mass of other two elements is not equal to the middle element.
∴ Mg ( 24.3 ), Ca ( 40.1 ), Sr ( 87.6 ) is not a Dobereiner's triad.
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2. S ( 32.1 ), Se ( 79.0 ), Te ( 127.6 )
Atomic mass of first element ( S ) = a = 32.1
Atomic mass of second element ( Se ) = b = 79.0
Atomic mass of third element ( Te ) = c = 127.6
Now, we know that,
Mean = a + c / 2
⇒ Mean = 32.1 + 127.6 / 2
⇒ Mean = 159.7 / 2
⇒ Mean = 79.85
∴ Mean ≈ b
Here, the mean atomic mass of other two elements is approximately equal to the atomic mass of the middle element.
∴ S ( 32.1 ), Se ( 79.0 ), Te ( 127.6 ) is a Dobereiner's triad.
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3. Be ( 9.0 ), Mg ( 24.3 ), Ca ( 40.1 )
Atomic mass of first element ( Be ) = a = 9.0
Atomic mass of second element ( Mg ) = b = 24.3
Atomic mass of third element ( Ca ) = 40.1
Now, we know that,
Mean = a + c / 2
⇒ Mean = 9.0 + 40.1 / 2
⇒ Mean = 49.1 / 2
⇒ Mean = 24.55
Here, the mean atomic mass of the other two elements is approximately equal to the atomic mass of middle element.
∴ Be ( 9.0 ), Mg ( 24.3 ), Ca ( 40.1 ) is a Dobereiner's triad.