Correct Answer: Option C

Explanation

\(^{32} _{15}\)P \(\to\) 2\(^{0} _{-1}\)e + \(^{a} _{b}\)X

32 = 2(0) + a

a = 32

15 = 2(-1) + b

b = 15 + 2

b = 17

\(\therefore\) \(^{a} _{b}\)X = \(^{32} _{17}\)Cl

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