The reaction of $K _3\left[ Fe ( CN )_6\right]$ with freshly prepared $FeSO _4$ solution produces a dark blue precipitate called Turnbull's blue. Reaction of $K _4\left[ Fe ( CN )_6\right]$ with the $FeSO _4$ solution in 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 $H _2 SO _4$ through the side of the test tube produces a brown ring.
Precipitate $X$ is
$(A)$ $Fe _4\left[ Fe ( CN )_6\right]_3$
$(B)$ $Fe \left[ Fe ( CN )_6\right]$
$(C)$ $K _2 Fe \left[ Fe ( CN )_6\right]$
$(D)$ $KFe \left[ Fe ( CN )_6\right]$
Among the following, the brown ring is due to the formation of
$(A)$ $\left[ Fe ( NO )_2\left( SO _4\right)_2\right]^{2-}$
$(B)$ $\left[ Fe ( NO )_2\left( H _2 O \right)_4\right]^{3+}$
$(C)$ $\left[ Fe ( NO )_4\left( SO _4\right)_2\right]$
$(D)$ $\left[ Fe ( NO )\left( H _2 O \right)_5\right]^{2+}$