3014Si
3213Al
3217Cl
3216S
Correct answer is C
3215P → 20−1e + abX
32 = 2(0) + a
a = 32
15 = 2(-1) + b
b = 15 + 2
b = 17
∴ ^{a} _{b}X = ^{32} _{17}Cl
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