Q.xii. Complete the following.

1. ICl$_3$ + H$_2$O $\longrightarrow$ ........ + ...... + ICl
2. I$_2$ + KClO$_3$ $\longrightarrow$ ....... + KIO$_2$
3. BrCl + H$_2$O $\longrightarrow$ ....... + HCl
4. Cl$_2$ + ClF$_3$ $\longrightarrow$ ........
5. H$_2$C = CH$_2$ + ICl $\longrightarrow$ .......
6. XeF$_4$ + SiO$_2$ $\longrightarrow$ ....... + SiF$_4$
7. XeF$_6$ + 6H$_2$O $\longrightarrow$ ........ + HF
8. XeOF$_4$ + H$_2$O $\longrightarrow$ ....... + HF

[!NOTE]
Two sub-items carry the book's own quirks, transcribed as printed: item 2 prints the product "KIO$_2$" where the chapter's own special reaction (section 7.12.2) gives KIO$_3$ (and only KIO$_3$ balances); item 7 prints "6H$_2$O ... + HF" while the chapter's own hydrolysis equation (section 7.13.2) is XeF$_6$ + 3H$_2$O $\longrightarrow$ XeO$_3$ + 6HF. The answer discloses both.

Item 1. 2ICl$_3$ + 3H$_2$O $\longrightarrow$ HIO$_3$ + 5HCl + ICl -- section 7.12.2's own XX'$_3$ hydrolysis (2 I, 6 Cl, 6 H, 3 O balance on both sides).

Item 2. I$_2$ + KClO$_3$, on heating, gives ICl + KIO$_3$ -- the chapter's own "special reaction for ICl" (section 7.12.2). The exercise prints the second product as "KIO$_2$", but only KIO$_3$ balances (3 O on each side); the printed KIO$_2$ is treated as a slip and disclosed, never silently used.

Item 3. BrCl + H$_2$O $\longrightarrow$ HOBr + HCl, exactly as in section 7.12.2's hydrolysis row.

Item 4. Cl$_2$ + ClF$_3$ $\longrightarrow$ 3ClF -- the halogen + interhalogen comproportionation, the exact chlorine analogue of the chapter's own worked example Br$_2$ + BrF$_3$ $\longrightarrow$ 3BrF (section 7.12.2, "reaction of halogen with interhalogen compounds"); 3 Cl and 3 F balance.

Item 5. H$_2$C=CH$_2$ + ICl $\longrightarrow$ ICH$_2$-CH$_2$Cl (iodine and chlorine adding across the double bond), as illustrated in section 7.12.2's addition-across-olefins row.

Item 6. 2XeF$_4$ + SiO$_2$ $\longrightarrow$ 2XeOF$_2$ + SiF$_4$ -- SiO$_2$ converts the xenon fluoride to its oxyfluoride, the direct XeF$_4$ analogue of the chapter's own 2XeOF$_4$ + SiO$_2$ $\longrightarrow$ 2XeO$_2$F$_2$ + SiF$_4$ (section 7.13.3), and XeOF$_2$ is exactly the oxyfluoride the chapter's own partial hydrolysis of XeF$_4$ produces (8 F, 2 Xe, 1 Si, 2 O balance). …

Matching each fill-in-the-blank reaction against the identical or closely analogous reaction already taught elsewhere in this same chapter, and flagging every case where the exercise's own printed react …

• Accepting the exercise's printed reactants at face value for items 4 and 6, producing an unbalanced or untaught reaction rather than che …