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(D) $1$. The metal nitrate gives a golden yellow flame,identifying the metal as $Na$. Thus,the metal nitrate is $NaNO_3$.$2$. $2NaNO_3 \xrightarrow{\D

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) $1$. The metal nitrate gives a golden yellow flame,identifying the metal as $Na$. Thus,the metal nitrate is $NaNO_3$.$2$. $2NaNO_3 \xrightarrow{\Delta} 2NaNO_2 + O_2$. So,$A$ is $NaNO_2$.$3$. $NaNO_2$ reacts with thiourea $(NH_2CSNH_2)$ to form $NaSCN$ (Solution $B$). $SCN^-$ is a pseudohalide ion.$4$. $FeCl_3 + 3NaSCN \rightarrow Fe(SCN)_3 + 3NaCl$. The complex formed is $[Fe(SCN)_3(H_2O)_3]$ or $[Fe(SCN)(H_2O)_5]^{2+}$,which is blood red.$5$. $SCN^-$ is an ambidentate ligand and can show linkage isomerism (via $S$ or $N$ atom).$6$. In the complex $[Fe(SCN)(H_2O)_5]^{2+}$,$Fe$ is in $+3$ oxidation state $(3d^5)$. It has $5$ unpaired electrons. Magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \ B.M.$$7$. All statements $A$,$B$,and $C$ are correct. Therefore,the incorrect statement is 'None of these'.

Column-$1$ (Reaction)	Column-$2$ (Main product)
$A$. $4FeCl_3 + 3K_4[Fe(CN)_6] \rightarrow$	$I$. $Fe_4[Fe(CN)_6]_3$
$B$. $ZnCl_2 + 4NaOH \rightarrow$	$V$. $Na_2ZnO_2$
$C$. $2FeCl_3 + H_2S \rightarrow$	$II$. $FeCl_2$

Similar Questions

The equation which is balanced and represents the correct product$(s)$ is:

Which of the following change the colour of the aqueous solution of $FeCl_3$?

Match the following:
Column-$1$ (Reaction) Column-$2$ (Main product)
$A$. $4FeCl_3 + 3K_4[Fe(CN)_6] \rightarrow$ $I$. $Fe_4[Fe(CN)_6]_3$
$B$. $ZnCl_2 + 4NaOH \rightarrow$ $V$. $Na_2ZnO_2$
$C$. $2FeCl_3 + H_2S \rightarrow$ $II$. $FeCl_2$

$A$ blue colouration is not obtained when:

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