- ✓${H_2}Te$
- B${H_2}Se$
- C${H_2}O$
- D${H_2}S$
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$\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}$ $\xrightarrow[{{\text{H}}_{2}}{{\text{O}}_{2}},\bar{O}\text{H}]{\text{B}{{\text{H}}_{3}},\text{THF}}(P)$ $\xrightarrow[\text{C}{{\text{H}}_{2}}\text{C}{{\text{l}}_{2}}]{\text{ Pyridinium Chloro Chromate }(\text{PCC})}$ $(Q)$
$\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}$ $\xrightarrow[NaB{{H}_{4}}\,,\,H{{O}^{\Theta }}]{Hg\,{{(OAc)}_{2}}\,,\,{{H}_{2}}O}(R)$ $\xrightarrow[\text{C}{{\text{H}}_{2}}\text{C}{{\text{l}}_{2}}]{\text{ Pyridinium Chloro Chromate }(\text{PCC})}$ $(S)$

$(a)$ $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}\right]^{+}$
$(b)$ $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right) \mathrm{Cl}_{5}\right]^{-}$
$(c)$ $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}\left(\mathrm{NO}_{2}\right)\right]$
$(d)$ $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{ClBr}\right]^{2+}$
Nucleophile added in this reaction
