The reaction of $\mathrm{[PtCl_4]^{2-}}$ with two equivalents of $\mathrm{NH_3}$ produces

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A - cis-$\mathrm{[Pt(NH_3)_2Cl_2]}$. Trans effect: $\mathrm{Cl^-}$ has a greater trans effect than $\mathrm{NH_3}$, so the second $\mathrm{NH_3}$ replaces a Cl trans to Cl, giving cis-$\mathrm{[Pt(NH_3)_2Cl_2]}$ (cisplatin).

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B - trans-$\mathrm{[Pt(NH_3)_2Cl_2]}$

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C - both cis and trans-$\mathrm{[Pt(NH_3)_2Cl_2]}$

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D - trans-$\mathrm{[Pt(NH_3)_2Cl_4]}$

The reaction of \(\mathrm{[PtCl_4]^{2-}}\) with two equivalents of \(\mathrm{NH_3}\) produces

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