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(C) $1$. The thermal decomposition of $B(OH)_3$ (orthoboric acid) at high temperature yields $B_2O_3$ and water vapor. Thus,$X = B_2O_3$ and $Z = H_2O

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$X = B_2O_3, Y = BCl_3, Z = HCl$

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(C) $1$. The thermal decomposition of $B(OH)_3$ (orthoboric acid) at high temperature yields $B_2O_3$ and water vapor. Thus,$X = B_2O_3$ and $Z = H_2O$ (though the question implies a gas,$H_2O$ vapor is released).$2$. $B_2O_3$ can be reduced to elemental $Boron$ using $Mg$ or $Fe$.$3$. $B_2O_3$ reacts with $PCl_5$ or other chlorinating agents to form $BCl_3$. $BCl_3$ reacts with $H_2O$ to regenerate $B(OH)_3$.$4$. $BCl_3$ reacts with $NH_3$ at high temperature to form inorganic benzene $(B_3N_3H_6)$. Thus,$Y = BCl_3$.$5$. Therefore,$X = B_2O_3, Y = BCl_3, Z = H_2O$ (Note: The provided options suggest $Z = HCl$ in option $C$,which is consistent with the reaction of $B_2O_3$ with $PCl_5$ or similar reagents to produce $BCl_3$ and $HCl$ gas).

On the basis of reactions given below identify $X, Y, Z$.
$Boron \xleftarrow{Mg \, or \, Fe} X \xleftarrow{\Delta} B(OH)_3 + \text{Gas-}Z$
$B(OH)_3 \xleftarrow{H_2O} Y \xrightarrow{2NH_3/\Delta} \text{Inorganic benzene}$

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On the basis of reactions given below identify $X, Y, Z$.$Boron \xleftarrow{Mg \, or \, Fe} X \xleftarrow{\Delta} B(OH)_3 + \text{Gas-}Z$$B(OH)_3 \xleftarrow{H_2O} Y \xrightarrow{2NH_3/\Delta} \text{Inorganic benzene}$