The reaction of $K_3[Fe(CN)_6]$ with freshly prepared $FeSO_4$ solution produces a dark blue precipitate called Turnbull's blue. The reaction of $K_4[Fe(CN)_6]$ with $FeSO_4$ solution in the complete absence of air produces a white precipitate $X$,which turns blue in air. Mixing the $FeSO_4$ solution with $NaNO_3$,followed by a slow addition of concentrated $H_2SO_4$ through the side of the test tube,produces a brown ring.
Precipitate $X$ is:
$(A)$ $Fe_4[Fe(CN)_6]_3$
$(B)$ $Fe[Fe(CN)_6]$
$(C)$ $K_2Fe[Fe(CN)_6]$
$(D)$ $KFe[Fe(CN)_6]$
Among the following,the brown ring is due to the formation of:
$(A)$ $[Fe(NO)_2(SO_4)_2]^{2-}$
$(B)$ $[Fe(NO)_2(H_2O)_4]^{3+}$
$(C)$ $[Fe(NO)_4(SO_4)_2]$
$(D)$ $[Fe(H_2O)_5(NO)]SO_4$

• A
  $C, D$
• B
  $C, B$
• C
  $A, D$
• D
  $D, C$