Question

Among the following reactions , the impossible one is :

• A. ${}^{2}H{e}_4+$ ${}^{4}B{e}_9⟶$ ${}^0{n}_1+$ ${}^6{C}_{12}$
• B. ${}^{2}H{e}_4+$ ${}^7{N}_{14}⟶$ ${}^1{H}_1+$ ${}^8{O}_{17}$
• C. $4{(}^1{H}_1)⟶$ ${}^{2}H{e}_4+$ ${}^2{(}^{-1}{e}_0)$
• D. ${}^{3}L{i}_7+$ ${}^1{H}_1⟶$ ${}^{4}B{e}_8$

Answer 1

The correct option is C $4{(}^1{H}_1)⟶$ ${}^{2}H{e}_4+$ ${}^2{(}^{-1}{e}_0)$

For possible reactions :
Sum of mass numbers LHS should be equal to the sum of mass numbers on RHS of the equation.
As well as sum of Atomic numbers on LHS should be equal to the sum of Atomic numbers on RHS of the equation.
So, in option C we can see that on LHS mass number is 4 x 1 = 4 and Atomic number is 4 x 1 = 4. on RHS mass number is 2 + 2(-1) = 0 and Atomic number is 4 + 0 = 4.
So, mass number are different in LHS and RHS of the equation. So, this reaction is not possible.