The reaction of $\mathrm{SbCl_3}$, with 3 equivalents of EtMgBr yields compound X. Two equivalents of $\mathrm{SbI_3}$ react with one equivalent of X to give Y. In the solid state, Y has a 1D-polymeric structure in which each Sb is in a square pyramidal environment. Compounds X and Y respectively, are
(a)$\mathrm{SbEt_3}$, and $[\mathrm{Sb(Et)I_2}]_n$
(c)$\mathrm{SbEt_3}$, and $[\mathrm{SbEt_2Br_2}]_n$
(b)$\mathrm{Sb(Et_2)Cl}$ and $[\mathrm{Sb(Et)Cl}]_n$
(d)$\mathrm{Sb(Et)Br_2}$ and $[\mathrm{SbEt(I)(Br)}]_n$