Question

‘X’, ‘Y’ and ‘Z’ belong to II-A group and obey the Dobereiner's Law of triads. Identify the true statement regarding X, Y and Z.

• A. $\mathrm{X}\to \mathrm{Ca},\mathrm{Y}\to \mathrm{Sr},\mathrm{Z}\to \mathrm{Ba}$, and $\mathrm{Ba}$ is a good reducing agent.
• B. $\mathrm{X}\to \mathrm{Mg},\mathrm{Y}\to \mathrm{Ca},\mathrm{Z}\to \mathrm{Sr}$, and $\mathrm{Ca}$ is a good reducing agent.
• C. $\mathrm{X}\to \mathrm{Mg},\mathrm{Y}\to \mathrm{Ca},\mathrm{Z}\to \mathrm{Sr}$, and $\mathrm{Mg}$ has low I.P. value.
• D. $\mathrm{X}\to \mathrm{Ca},\mathrm{Y}\to \mathrm{Sr},\mathrm{Z}\to \mathrm{Ba}$, and $\mathrm{Mg}$ has lower I.P. value than that of X, Y and Z.

Answer 1

The correct option is A

$\mathrm{X}\to \mathrm{Ca},\mathrm{Y}\to \mathrm{Sr},\mathrm{Z}\to \mathrm{Ba}$, and $\mathrm{Ba}$ is a good reducing agent.

Explanation for correct option: a)

1. According to Dobereiner's law of triads, when the elements are arranged in the increasing order of atomic masses the atomic mass of the central element will be approximately equal to the arithmetic mean of the atomic masses of the other two elements.
2. The reducing power of elements increases down the group because the tendency to lose electrons increases down the group.
3. Here the atomic number of strontium is $38$ which is the arithmetic mean of calcium of atomic number $20$ and barium of atomic number $56$.
4. Barium is a good reducing agent as compared to strontium and calcium.

Explanation for incorrect options:

b) Here the atomic number of calcium is not the arithmetic mean of the atomic number of magnesium and strontium. Hence it does not obey Dobereiner's Law of triads. But $\mathrm{Ca}$ is a good reducing agent.

c) The ionization potential of magnesium is found to be very high due to its completely filled electronic configuration and the given elements do not obey Dobereiner's Law of triads.

d) The ionization potential of magnesium is found to be very high as compared to $\mathrm{Ca},\mathrm{Sr}$ and $\mathrm{Ba}$ due its completely filled electronic configuration.

Hence option (a) is correct and the true statement regarding X, Y and Z is $\mathrm{X}\to \mathrm{Ca},\mathrm{Y}\to \mathrm{Sr},\mathrm{Z}\to \mathrm{Ba}$, and $\mathrm{Ba}$ is a good reducing agent.