\(^{30} _{14}\)Si
\(^{32} _{13}\)Al
\(^{32} _{17}\)Cl
\(^{32} _{16}\)S
Correct answer is C
\(^{32} _{15}\)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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