b (b)Electronic configuration of $_{16}{S^{32}} = 1{s^2},2{s^2},2{p^6},3{s^2},3{p^4}$. In the last orbit it has only $6$ electron. So it require $2$ electron to complete its octet, therefore it share $2$ electron with two hydrogen atom and forms $2$ covalent bond with it.
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