Give the oxidation state,$d$-orbital occupation,and coordination number of the central metal ion in the following complexes:
$(i)$ $K_{3}[Co(C_{2}O_{4})_{3}]$
$(ii)$ $cis-[Cr(en)_{2}Cl_{2}]Cl$
$(iii)$ $(NH_{4})_{2}[CoF_{4}]$
$(iv)$ $[Mn(H_{2}O)_{6}]SO_{4}$

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(N/A) $(i)$ $K_{3}[Co(C_{2}O_{4})_{3}]$: Central metal is $Co$. Coordination number is $6$. Oxidation state: $x + 3(-2) = -3 \implies x = +3$. $Co^{3+}$ $(d^{6})$ is $t_{2g}^{6} e_{g}^{0}$.
$(ii)$ $cis-[Cr(en)_{2}Cl_{2}]Cl$: Central metal is $Cr$. Coordination number is $6$. Oxidation state: $x + 2(0) + 2(-1) = +1 \implies x = +3$. $Cr^{3+}$ $(d^{3})$ is $t_{2g}^{3} e_{g}^{0}$.
$(iii)$ $(NH_{4})_{2}[CoF_{4}]$: Central metal is $Co$. Coordination number is $4$. Oxidation state: $x + 4(-1) = -2 \implies x = +2$. $Co^{2+}$ $(d^{7})$ is $t_{2g}^{5} e_{g}^{2}$.
$(iv)$ $[Mn(H_{2}O)_{6}]SO_{4}$: Central metal is $Mn$. Coordination number is $6$. Oxidation state: $x + 6(0) = +2 \implies x = +2$. $Mn^{2+}$ $(d^{5})$ is $t_{2g}^{3} e_{g}^{2}$.

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Which of the following statements is correct?

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Select the incorrect statement.

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:
$(A)$ $[Fe(NO)_2(SO_4)_2]^{2-}$
$(B)$ $[Fe(NO)_2(H_2O)_4]^{3+}$
$(C)$ $[Fe(NO)_4(SO_4)_2]$
$(D)$ $[Fe(H_2O)_5(NO)]SO_4$

$0.02 \, \text{mole}$ of $[Co(NH_3)_5Br]Cl_2$ and $0.02 \, \text{mole}$ of $[Co(NH_3)_5Cl]SO_4$ are present in $200 \, \text{cc}$ of a solution $X$. The number of moles of the precipitates $Y$ and $Z$ that are formed when the solution $X$ is treated with excess silver nitrate and excess barium chloride are respectively:

When potassium iodide is added to an aqueous solution of potassium ferricyanide,a reversible reaction is observed in which a complex $P$ is formed. In a strong acidic medium,the equilibrium shifts completely towards $P$. Addition of zinc chloride to $P$ in a slightly acidic medium results in a sparingly soluble complex $Q$.
$(1)$ The number of moles of potassium iodide required to produce two moles of $P$ is. . .
$(2)$ The number of zinc ions present in the molecular formula of $Q$ is. . . .
Give the answer for $(1)$ and $(2)$.

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