(Yellow,ppt),T xleftarrow{{{K_2}Cr{O_4}}} X x | vedclass

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(B) $1$. $X$ gives a green flame test,which is characteristic of $Ba^{2+}$ ions.$2$. $X$ reacts with $K_2CrO_4$ to form a yellow precipitate $T$. $BaC

Vedclass · Accepted Answer

$BaS_2O_3$

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(B) $1$. $X$ gives a green flame test,which is characteristic of $Ba^{2+}$ ions.$2$. $X$ reacts with $K_2CrO_4$ to form a yellow precipitate $T$. $BaCrO_4$ is a yellow precipitate.$3$. $X$ reacts with $dil. HCl$ to form a yellow precipitate $Y$ and a pungent smelling gas $Z$.$4$. $BaS_2O_3 + 2HCl \rightarrow BaCl_2 + S(s) + SO_2(g) + H_2O$. However,$BaS_2O_3$ is not the correct fit for the yellow precipitate $Y$ in this context.$5$. Considering $PbS_2O_3 + 2HCl \rightarrow PbCl_2(s) + S(s) + SO_2(g) + H_2O$. $PbCl_2$ is white. $6$. Re-evaluating: $BaS_2O_3$ is the correct choice as $Ba^{2+}$ gives a green flame test. The yellow precipitate $T$ is $BaCrO_4$. The reaction with $HCl$ produces $SO_2$ (pungent gas) and the precipitate $Y$ is sulfur (often appearing yellow).

$(Yellow\,ppt)\,T \xleftarrow{{{K_2}Cr{O_4}}} X$ $\xrightarrow{{dil.\,HCl}} Y(Yellow\,ppt) + Z \uparrow (pungent\,smelling\,gas)$. If $X$ gives green flame test,then $X$ is:

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