Exams › JEE Advanced › Chemistry
Correct answer: (y - x) = (q - p)
Standard notation: ^A_y X means mass number A, atomic number y. Isobars (A)^y(X) and (B)^q(p... wait the notation xA^y is confusing. Let us interpret: nuclide 1 has atomic number x, mass number A (so neutrons = A-x), element symbol y is likely the symbol not the atomic number. Let me re-read: 'xA^y' — likely x=atomic number (Z), A=element symbol, y=mass number. Similarly pB^q: p=Z, q=mass number. mCⁿ: m=Z, n=mass number. Isobars xA^y and pB^q: same mass number -> y = q. Isotones xA^y and mCⁿ: same neutron number -> y-x = n-m. From (y-x)=(n-m) and isobar condition y=q: (q-x)=(n-m) -> n = q-x+m. That matches option B. Also (y-x)=(n-m) -> but option D says (y-x)=(q-p): since y=q (isobars), this becomes (q-x)=(q-p) -> x=p, which is option C. So from isobar: y=q. From isotone: y-x = n-m, i.e., neutrons equal: y-x = n-m. Option D: (y-x)=(q-p). Since y=q: q-x = q-p -> x=p. That would mean option C and D are equivalent and either true or false together. Since x and p are atomic numbers of different elements (A and B), x need not equal p. So D is not necessarily true. The isotone condition gives y-x = n-m, which means n = y-x+m = q-x+m (using y=q) -> option B is correct.