An element X decays into element Z by two-step process
     \[
  X \to Y + {}_2^4He \\
  Y \to Z + 2\overline e \\
 \]
A. X and Z are isobars
B. X and Y are isotopes
C. X and Z are isotones
D. X and Z are isotopes

Hint: Figure out the difference in the nucleons and electrons of X,Y and Z by the given information (assume that initially X has A nucleons and Z electrons). Then, by definition of isotopes, isotones and isobars compare X, Y and Z and choose the correct option.

Complete step by step answer:Let us assume that initially X had A nucleons (neutrons + protons) and Z number of electrons. Then, by comparing reactants and products in the first equation, we have,
     \[{}_Z^AX \to {}_{Z - 2}^{A - 4}Y + {}_2^4He\]
This occurs due to the fact that the number of nucleons and electrons before and after the reaction should be the same by law of conservation of matter and energy. Similarly, for second equation, we have,
     \[{}_{Z - 2}^{A - 4}Y \to {}_Z^{A - 4}Z + 2\overline {{e_{ - 1}}} \]
Now time for definitions:
Isotopes: Elements which have the same number of protons (=electrons) but different numbers of neutrons are called isotopes are called isotopes.
Same atomic number (Z), Different Mass number (A)
Isotones: Elements which have the same number of neutrons but different numbers of protons are called isotones.
Isobars: Elements which have the same number of nucleons but different numbers of electrons are called isobars. Same atomic mass (A), Different atomic number.
Examining the species of the radioactive reaction and the definitions we conclude that:
There is no relation between X and Y whatsoever.
X and Z have the same number of electrons i.e. same atomic number. Hence, X and Z are Isotones.
Y and Z have the same number of nucleons but different atomic numbers. Hence, Y and Z are Isobars.
The correct answer is thus option C.

Note:There is one important rule in chemistry which applies here which is ‘Always balance the equation’. Otherwise there will be a violation of the most fundamental law of conservation of matter and thus energy. So, checkout the missing nucleons and electrons and write the balanced equations. It will help in solving problems as well as in reducing errors.