Q121CP - Chemistry: Molecular Nature Of Matter And Change

Question

In many species, a transition metal has an unusually high or low oxidation state. Write balanced equations for the following and find the oxidation state of the transition metal in the product: 

(a) Iron (III) ion reacts with hypochlorite ion in basic solution to form ferrate ion $\mathbf{\left(}{\mathbf{FeO}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\mathbf{\right)}\mathbf{,}\mathbf{ }{\mathbf{Cl}}^{\mathbf{-}}$and water.

(b) Potassium hexacyanomanganate(II) reacts with the K metal to form ${\mathbf{K}}_{\mathbf{6}}\mathbf{\left[}\mathbf{Mn}{\mathbf{\left(}\mathbf{CN}\mathbf{\right)}}_{\mathbf{6}}\mathbf{\right]}$.

(c) Heating sodium superoxide $\left({\mathrm{NaO}}_2\right)$ with ${\mathbf{Co}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}$produces ${\mathbf{Na}}_{\mathbf{4}}{\mathbf{CoO}}_{\mathbf{4}}$ and ${\mathbf{O}}_{\mathbf{2}}$ gas.

(d) Vanadium(III) chloride reacts with Na metal under a CO atmosphere to produce $\mathbf{Na}\left[V{\left(\mathrm{CO}\right)}_6\right]$ and NaCl.

(e) Barium peroxide reacts with nickel(II) ions in a basic solution to produce ${\mathbf{BaNiO}}_{\mathbf{3}}$

(f) Bubbling CO through a basic solution of cobalt(II) ion produces ${\left[\mathrm{Co}{\left(\mathrm{CO}\right)}_4\right]}^{-}\mathbf{,}\mathbf{ }{{\mathbf{CO}}_{\mathbf{3}}}^{\mathbf{2}\mathbf{-}}$and water.

(g) Heating caesium tetrafluorocuprate(II) with ${\mathbf{F}}_{\mathbf{2}}$ gas under pressure gives ${\mathbf{Cs}}_{\mathbf{2}}{\mathbf{CuFe}}_{\mathbf{6}}$.

(h) Heating tantalum(V) chloride with Na metal produces NaCl and ${\mathbf{Ta}}_{\mathbf{6}}{\mathbf{Cl}}_{\mathbf{15}}$, in which half of the Ta is in the +2 state.

(i) Potassium tetracyanonickelate(II) reacts with hydrazine $\left(N_2{H}_4\right)$ in basic solution to form ${\mathbf{K}}_{\mathbf{4}}\left[{\mathrm{Ni}}_2{\left(\mathrm{CN}\right)}_6\right]$ and ${\mathbf{N}}_{\mathbf{2}}$ gas.

 

Step-by-Step Solution

Answer

(a)$4{\mathrm{OH}}^{-}+{\mathrm{Fe}}^{3+}+2{\mathrm{ClO}}^{-}+{\mathrm{H}}_2\mathrm{O}\to {{\mathrm{FeO}}_4}^{-}+2{\mathrm{Cl}}^{-}$; Charge of is Fe is 7+

(b)${\mathrm{K}}_4\left[\mathrm{Mn}{\left(\mathrm{CN}\right)}_6\right]+2\mathrm{K}\to {\mathrm{K}}_6\left[\mathrm{Mn}{\left(\mathrm{CN}\right)}_6\right]$; Charge of Mn is 4+

(c) $12{\mathrm{NaO}}_2+{\mathrm{Co}}_3{\mathrm{O}}_4\to 3{\mathrm{Na}}_4{\mathrm{CoO}}_4+8{\mathrm{O}}_2$; Charge of Co is 4+

(d)${\mathrm{VCl}}_3+4\mathrm{Na}+6\mathrm{CO}\to \mathrm{Na}\left[\mathrm{V}{\left(\mathrm{CO}\right)}_6\right]+3\mathrm{NaCl}$; Charge of V is 5+

(e)${\mathrm{Ni}}^{2+}+6\mathrm{OH}+{{\mathrm{O}}_2}^{2-}\to {\left[{\mathrm{NiO}}_3\right]}^{2-}+2{\mathrm{O}}^{2-}+3{\mathrm{H}}_2\mathrm{O}$; Charge of Ni is 4+

(f) $11\mathrm{CO}+2{\mathrm{Co}}^{2+}+12{\mathrm{OH}}^{-}\to 3{{\mathrm{CO}}_3}^{2-}+2{\left[\mathrm{Co}{\left(\mathrm{CO}\right)}_4\right]}^{-}+6{\mathrm{H}}_2\mathrm{O}$; Charge of Co is 1-

(g)${\mathrm{Cs}}_2\left[{\mathrm{CuF}}_4\right]+{\mathrm{F}}_2\to {\mathrm{Cs}}_2\left[{\mathrm{CuF}}_6\right]$ ; Charge of Cs is 2+

(h) $6{\mathrm{TaCl}}_5+15\mathrm{Na}\to {\mathrm{Ta}}_6{\mathrm{Cl}}_{15}+15\mathrm{NaCl}$; Charge of Ta is 3+

(i)  ${\mathrm{N}}_2{\mathrm{H}}_4+4{\mathrm{OH}}^{-}+4{\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}\to {\mathrm{N}}_2+2{\left[{\mathrm{Ni}}_2{\left(\mathrm{CN}\right)}_6\right]}^{4-}+4{\mathrm{CN}}^{-}+4{\mathrm{H}}_2\mathrm{O}$; Charge of Ni is 2+

Step 1: Given Data

The given information shows the rule of the naming of coordination compounds.

Step 2: Concept Introduction

A coordination complex consists of a central atom or ion, which is typically metallic and is termed the coordination center, and a surrounding array of bound molecules or ions, that are successively called ligands or complexing agents.

Step 3: (a) Balanced equation when iron reacts with hydrochloride ion

${\mathrm{Fe}}^{3+}+3{\mathrm{ClO}}^{-}\to {{\mathrm{FeO}}_4}^{-}+{\mathrm{Cl}}^{-}+{\mathrm{H}}_2\mathrm{O} \left(\mathrm{Basic}\right)$

1. Identify the half-reactions

$$\begin{gathered} \mathrm{OHR}: {\mathrm{Fe}}^{3+}\to {\mathrm{FeO}}_4^{-} \\ \mathrm{RHR}: {\mathrm{ClO}}^{-}\to {\mathrm{Cl}}^{-} \end{gathered}$$

2. Balance the electron and non-oxygen atoms

$$\begin{gathered} \mathrm{OHR}: {\mathrm{Fe}}^{3+}\to {\mathrm{FeO}}_4^{-}+4{\mathrm{e}}^{-} \\ \mathrm{RHR}: {\mathrm{ClO}}^{-}+2{\mathrm{e}}^{-}\to {\mathrm{Cl}}^{-} \end{gathered}$$

3. Balance the charges with hydroxide ion

$$\begin{gathered} \mathrm{OHR}: 8{\mathrm{OH}}^{-}+{\mathrm{Fe}}^{3+}\to {\mathrm{FeO}}_4^{-}+4{\mathrm{e}}^{-} \\ \mathrm{RHR}: {\mathrm{ClO}}^{-}+2{\mathrm{e}}^{-}\to {\mathrm{Cl}}^{-}+2{\mathrm{OH}}^{-} \end{gathered}$$

4. Balance the oxygen atoms by adding the water to another side

$$\begin{gathered} \mathrm{OHR}: 8{\mathrm{OH}}^{-}+{\mathrm{Fe}}^{3+}\to {\mathrm{FeO}}_4^{-}+4{\mathrm{e}}^{-}+{\mathrm{H}}_2\mathrm{O} \\ \mathrm{RHR}: {\mathrm{ClO}}^{-}+2{\mathrm{e}}^{-}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{Cl}}^{-}+2{\mathrm{OH}}^{-} \end{gathered}$$

5. Multiply the half-reactions to  get the electrons least common multiple and then cancel the respective atoms from both sides,

$$\begin{gathered} \mathrm{OHR}: 8{\mathrm{OH}}^{-}+{\mathrm{Fe}}^{3+}\to {\mathrm{FeO}}_4^{-}+4{\mathrm{e}}^{-}+{\mathrm{H}}_2\mathrm{O} \\ \mathrm{RHR}: {\mathrm{ClO}}^{-}+2{\mathrm{e}}^{-}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{Cl}}^{-}+2{\mathrm{OH}}^{-} \\ \mathrm{Overall} \mathrm{Reaction} \\ 4{\mathrm{OH}}^{-}+{\mathrm{Fe}}^{3+}+2{\mathrm{ClO}}^{-}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{FeO}}_4^{-}+2{\mathrm{Cl}}^{-} \end{gathered}$$

6. Get the oxidation state of the metal

${\mathrm{FeO}}_4^{-} \mathrm{has} \mathrm{a} \mathrm{charge} \mathrm{of} 1-$

Oxygen has a charge of 2-

thus the charge of Fe will be 8-1=7+

The charge of Fe is 7+

Step 4: (b) Balanced equation when potassium hexacynoate reacts with potassium metal

$$\begin{gathered} {\mathrm{K}}_4\left[\mathrm{Mn}{\left(\mathrm{CN}\right)}_6\right]+\mathrm{K}\to {\mathrm{K}}_6\left[\mathrm{Mn}{\left(\mathrm{CN}\right)}_6\right] \\ 1. \mathrm{Balance} \mathrm{the} \mathrm{atoms} \mathrm{involved} \\ {\mathrm{K}}_4\left[\mathrm{Mn}{\left(\mathrm{CN}\right)}_6\right]+2\mathrm{K}\to {\mathrm{K}}_6\left[\mathrm{Mn}{\left(\mathrm{CN}\right)}_6\right] \\ 2.\mathrm{Get} \mathrm{the} \mathrm{charge} \mathrm{of} \mathrm{oxidation} \mathrm{state} \mathrm{of} \mathrm{the} \mathrm{metal} \\ {\mathrm{K}}_6\left[\mathrm{Mn}{\left(\mathrm{CN}\right)}_6\right]: \left[\mathrm{Mn}{\left(\mathrm{CN}\right)}_6\right] \mathrm{has} \mathrm{a} \mathrm{charge} \mathrm{of} 6^{-} \\ \mathrm{CN} \mathrm{has} \mathrm{a} \mathrm{charge} \mathrm{of} -1 \\ 6\times \left(-1\right)=-6 \\ -6=-6+\mathrm{charge} \mathrm{of} \mathrm{Mn} \\ \mathrm{The} \mathrm{charge} \mathrm{of} \mathrm{Mn} \mathrm{is} 0. \end{gathered}$$

Step 3: (c) Balanced equation when sodium superoxide react

$$\begin{gathered} {\mathrm{NaO}}_2+{\mathrm{Co}}_3{\mathrm{O}}_4\to {\mathrm{Na}}_4{\mathrm{CoO}}_4+{\mathrm{O}}_2 \\ 1. \mathrm{Balance} \mathrm{the} \mathrm{atoms} \mathrm{involved} \\ 12{\mathrm{NaO}}_2+{\mathrm{Co}}_3{\mathrm{O}}_4\to 3{\mathrm{Na}}_4{\mathrm{CoO}}_4+8{\mathrm{O}}_2 \\ 2. \mathrm{Get} \mathrm{the} \mathrm{charge} \mathrm{of} \mathrm{the} \mathrm{oxidation} \mathrm{state} \\ {\mathrm{Na}}_4{\mathrm{CoO}}_4: \left[{\mathrm{CoO}}_4\right] \mathrm{has} \mathrm{a} \mathrm{charge} \mathrm{of} 4- \\ \mathrm{O} \mathrm{has} \mathrm{a} \mathrm{charge} \mathrm{of} 2- \\ 4\times -2=-8 \\ -4=-8+\mathrm{Charge} \mathrm{of} \mathrm{Co} \\ \mathrm{Charge} \mathrm{of} \mathrm{Co} \mathrm{is} 4+. \end{gathered}$$

Step 6: (d) Balanced equation when vanadium chloride reacts with Na metal

$$\begin{gathered} {\mathrm{VCl}}_3+\mathrm{Na}+\mathrm{CO}\to \mathrm{Na}\left[\mathrm{V}{\left(\mathrm{CO}\right)}_6\right]+\mathrm{NaCl} \\ 1. \mathrm{Balance} \mathrm{the} \mathrm{atoms} \mathrm{involved} \\ {\mathrm{VCl}}_3+4\mathrm{Na}+6\mathrm{CO}\to \mathrm{Na}\left[\mathrm{V}{\left(\mathrm{CO}\right)}_6\right]+3\mathrm{NaCl} \\ 2. \mathrm{Get} \mathrm{the} \mathrm{charge} \mathrm{of} \mathrm{the} \mathrm{oxidation} \mathrm{state} \mathrm{of} \mathrm{the} \mathrm{metal} \\ \mathrm{Na}\left[\mathrm{V}{\left(\mathrm{CO}\right)}_6\right]: \left[\mathrm{V}{\left(\mathrm{CO}\right)}_6\right] \mathrm{has} \mathrm{the} \mathrm{charge} \mathrm{of} 1- \\ \mathrm{CO} \mathrm{has} \mathrm{charge} 0 \\ \mathrm{Charge} \mathrm{of} \mathrm{V} \mathrm{is} 1-. \end{gathered}$$

Step 7: (e) Balanced equation when barium peroxide reacts with nickel

$$\begin{gathered} {\mathrm{BaO}}_2+{\mathrm{Ni}}^{2+}\to {\mathrm{BaNiO}}_3 \left(\mathrm{Basic}\right) \\ 1. \mathrm{Identify} \mathrm{the} \mathrm{half} \mathrm{reactions} \\ \mathrm{OHR}: {\mathrm{Ni}}^{2+}\to {\left[{\mathrm{NiO}}_3\right]}^{2-} \\ \mathrm{RHR}: {\mathrm{O}}_2^{2-}\to 2{\mathrm{O}}^{2-} \\ 2. \mathrm{Balance} \mathrm{the} \mathrm{electrons} \mathrm{and} \mathrm{non} \mathrm{O} \mathrm{atoms} \\ \mathrm{OHR}: {\mathrm{Ni}}^{2+}\to {\left[{\mathrm{NiO}}_3\right]}^{2-}+2{\mathrm{e}}^{-} \\ \mathrm{RHR}: {\mathrm{O}}_2^{2-}+2{\mathrm{e}}^{-}\to 2{\mathrm{O}}^{2-} \\ 3. \mathrm{Balance} \mathrm{the} \mathrm{charges} \mathrm{with} {\mathrm{OH}}^{-} \\ \mathrm{OHR}: {\mathrm{Ni}}^{2+}+6{\mathrm{OH}}^{-}\to {\left[{\mathrm{NiO}}_3\right]}^{2-}+2{\mathrm{e}}^{-} \\ \mathrm{RHR}: {\mathrm{O}}_2^{2-}+2{\mathrm{e}}^{-}\to 2{\mathrm{O}}^{2-} \\ 4. \mathrm{Balance} \mathrm{the} \mathrm{O} \mathrm{by} \mathrm{adding} {\mathrm{H}}_2\mathrm{O} \\ \mathrm{OHR}: {\mathrm{Ni}}^{2+}+6{\mathrm{OH}}^{-}\to {\left[{\mathrm{NiO}}_3\right]}^{2-}+2{\mathrm{e}}^{-}+3{\mathrm{H}}_2\mathrm{O} \\ \mathrm{RHR}: {\mathrm{O}}_2^{2-}+2{\mathrm{e}}^{-}\to 2{\mathrm{O}}^{2-} \\ 5. \mathrm{Combine} \mathrm{the} \mathrm{half} \mathrm{reactions} \\ \mathrm{OHR}: {\mathrm{Ni}}^{2+}+6{\mathrm{OH}}^{-}\to {\left[{\mathrm{NiO}}_3\right]}^{2-}+2{\mathrm{e}}^{-}+3{\mathrm{H}}_2\mathrm{O} \\ \mathrm{RHR}: {\mathrm{O}}_2^{2-}+2{\mathrm{e}}^{-}\to 2{\mathrm{O}}^{2-} \\ \mathrm{Overall} \mathrm{Reaction}: \\ {\mathrm{Ni}}^{2+}+6{\mathrm{OH}}^{-}+{\mathrm{O}}_2^{2-}\to {\left[{\mathrm{NiO}}_3\right]}^{2-}+2{\mathrm{O}}^{2-}+3{\mathrm{H}}_2\mathrm{O} \\ 6. \mathrm{Get} \mathrm{the} \mathrm{charge} \mathrm{of} \mathrm{the} \mathrm{oxidation} \mathrm{state} \mathrm{of} \mathrm{the} \mathrm{metal} \\ {\left[{\mathrm{NiO}}_3\right]}^{2-} \mathrm{has} \mathrm{a} \mathrm{charge} \mathrm{of} 2- \\ \mathrm{Thus}, \\ 3\times \left(2-\right)+\mathrm{Charge} \mathrm{of} \mathrm{Ni}=2- \\ \mathrm{Charge} \mathrm{of} \mathrm{Ni} =4+ \\ \mathrm{Charge} \mathrm{of} \mathrm{Ni} \mathrm{is} 4+ \end{gathered}$$

Step 8: (f) Balanced equation when CO reacts with Cobalt

$$\begin{gathered} \mathrm{CO}+{\mathrm{Co}}^{2+}\to {\left[\mathrm{Co}{\left(\mathrm{CO}\right)}_4\right]}^{-}+{\mathrm{CO}}_3^{2-}+{\mathrm{H}}_2\mathrm{O} \\ 1. \mathrm{Identify} \mathrm{the} \mathrm{half} \mathrm{reactions} \\ \mathrm{OHR}: \mathrm{CO}+{\mathrm{Co}}^{2+}\to {\left[\mathrm{Co}{\left(\mathrm{CO}\right)}_4\right]}^{-} \\ \mathrm{RHR}:\hspace{0.17em}\mathrm{CO}\to {\mathrm{CO}}_3^{2-} \\ 2. \mathrm{Balance} \mathrm{the} \mathrm{electrons} \mathrm{and} \mathrm{non} \mathrm{O} \mathrm{atoms} \\ \mathrm{OHR}: 4\mathrm{CO}+{\mathrm{Co}}^{2+}+3{\mathrm{e}}^{-}\to {\left[\mathrm{Co}{\left(\mathrm{CO}\right)}_4\right]}^{-} \\ \mathrm{RHR}:\hspace{0.17em}\mathrm{CO}\to {\mathrm{CO}}_3^{2-}+2{\mathrm{e}}^{-} \\ 3. \mathrm{Balacne} \mathrm{the} \mathrm{charges} \mathrm{with} {\mathrm{OH}}^{-} \\ \mathrm{OHR}: 4\mathrm{CO}+{\mathrm{Co}}^{2+}+3{\mathrm{e}}^{-}\to {\left[\mathrm{Co}{\left(\mathrm{CO}\right)}_4\right]}^{-} \\ \mathrm{RHR}:\hspace{0.17em}\mathrm{CO}+4{\mathrm{OH}}^{-}\to {\mathrm{CO}}_3^{2-}+2{\mathrm{e}}^{-} \\ 4. \mathrm{Balance} \mathrm{the} \mathrm{O} \mathrm{atoms} \mathrm{by} \mathrm{adding} {\mathrm{H}}_2\mathrm{O} \\ \mathrm{OHR}: 4\mathrm{CO}+{\mathrm{Co}}^{2+}+3{\mathrm{e}}^{-}\to {\left[\mathrm{Co}{\left(\mathrm{CO}\right)}_4\right]}^{-} \\ \mathrm{RHR}:\hspace{0.17em}\mathrm{CO}+4{\mathrm{OH}}^{-}\to {\mathrm{CO}}_3^{2-}+2{\mathrm{e}}^{-}+{\mathrm{H}}_2\mathrm{O} \\ 5. \mathrm{Multiply} \mathrm{the} \mathrm{half} \mathrm{reactions} \\ 2\times \mathrm{OHR}: 8\mathrm{CO}+2{\mathrm{Co}}^{2+}+6{\mathrm{e}}^{-}\to 2{\left[\mathrm{Co}{\left(\mathrm{CO}\right)}_4\right]}^{-} \\ 3\times \mathrm{RHR}:\hspace{0.17em}3\mathrm{CO}+12{\mathrm{OH}}^{-}\to 3{\mathrm{CO}}_3^{2-}+6{\mathrm{e}}^{-}+3{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Overall} \mathrm{Reaction} \\ 8\mathrm{CO}+2{\mathrm{Co}}^{2+}+3\mathrm{CO}+12{\mathrm{OH}}^{-}\to 2{\left[\mathrm{Co}{\left(\mathrm{CO}\right)}_4\right]}^{-}+3{\mathrm{CO}}_3^{2-}+3{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

$$\begin{gathered} 6. \mathrm{Get} \mathrm{the} \mathrm{charge} \mathrm{of} \mathrm{the} \mathrm{oxidation} \mathrm{state} \mathrm{of} \mathrm{the} \mathrm{metal} \\ \left[\mathrm{Co}{\left(\mathrm{CO}\right)}_4\right] \mathrm{has} \mathrm{a} \mathrm{charge} \mathrm{of} 1- \\ \mathrm{Co} \mathrm{has} \mathrm{the} \mathrm{charge} \mathrm{of} 0 \\ \mathrm{Thus} \mathrm{the} \mathrm{charge} \mathrm{of} \mathrm{Co} \mathrm{is} 1- \end{gathered}$$

Step 9: (g) Balanced equation when Cesium tetra fluoro cuprate react

$$\begin{gathered} \mathrm{Cs}\left[{\mathrm{CuF}}_4\right]+{\mathrm{F}}_2\to \mathrm{Cs}\left[{\mathrm{CuF}}_6\right] \\ 1. \mathrm{Balance} \mathrm{the} \mathrm{atoms} \mathrm{involved} \\ \mathrm{Cs}\left[{\mathrm{CuF}}_4\right]+{\mathrm{F}}_2\to \mathrm{Cs}\left[{\mathrm{CuF}}_6\right] \\ 2. \mathrm{Get} \mathrm{the} \mathrm{charge} \mathrm{of} \mathrm{the} \mathrm{oxidation} \mathrm{state} \mathrm{of} \mathrm{the} \mathrm{metal} \\ \left[{\mathrm{CuF}}_6\right] \mathrm{has} \mathrm{a} \mathrm{charge} \mathrm{of} 4- \\ \mathrm{F} \mathrm{has} \mathrm{a} \mathrm{charge} \mathrm{of} 1- \\ 6\times \left(-1\right)+\mathrm{Charge} \mathrm{of} \mathrm{Cu}=4- \\ \mathrm{Charge} \mathrm{of} \mathrm{Cu}=2+ \\ \mathrm{Thus} \mathrm{the} \mathrm{charge} \mathrm{of} \mathrm{Cu} \mathrm{is} 2+ \end{gathered}$$

Step 10: (h) Balanced equation when tantalum chloride react

$$\begin{gathered} {\mathrm{TaCl}}_5+\mathrm{Na}\to \mathrm{NaCl}+{\mathrm{Ta}}_6{\mathrm{Cl}}_{15} \\ 1. \mathrm{Balance} \mathrm{the} \mathrm{atoms} \mathrm{involved} \\ 6{\mathrm{TaCl}}_5+15\mathrm{Na}\to 15\mathrm{NaCl}+{\mathrm{Ta}}_6{\mathrm{Cl}}_{15} \\ 2. \mathrm{Get} \mathrm{the} \mathrm{charge} \mathrm{of} \mathrm{the} \mathrm{oxidation} \mathrm{state} \mathrm{of} \mathrm{the} \mathrm{metal} \\ {\mathrm{Ta}}_6{\mathrm{Cl}}_{15} \mathrm{has} \mathrm{the} \mathrm{charge} \mathrm{of} 0 \\ \mathrm{Cl} \mathrm{has} \mathrm{a} \mathrm{charge} \mathrm{of} 1- \\ \mathrm{Half} \mathrm{of} \mathrm{Ta} \mathrm{has} \mathrm{a} \mathrm{charge} \mathrm{of} 2+ \\ \mathrm{Half} \mathrm{of} \mathrm{Ta} \mathrm{has} \mathrm{a} \mathrm{charge} \mathrm{of} 3+ \end{gathered}$$

Step 11: (i) Balanced equation when Potassium cynaonickelate react

$$\begin{gathered} {\mathrm{K}}_2\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]+{\mathrm{N}}_2{\mathrm{H}}_4\to {\mathrm{K}}_4\left[{\mathrm{Ni}}_2{\left(\mathrm{CN}\right)}_6\right]+{\mathrm{N}}_2 \left(\mathrm{Basic}\right) \\ 1. \mathrm{Identify} \mathrm{the} \mathrm{half} \mathrm{reactions} \\ \mathrm{OHR}: {\mathrm{N}}_2{\mathrm{H}}_4\to {\mathrm{N}}_2 \\ \mathrm{RHR}: {\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}\to {\left[{\mathrm{Ni}}_2{\left(\mathrm{CN}\right)}_6\right]}^{4-} \\ 2. \mathrm{Balance} \mathrm{the} \mathrm{electrons} \mathrm{and} \mathrm{non} \mathrm{O} \mathrm{atoms} \\ \mathrm{OHR}: {\mathrm{N}}_2{\mathrm{H}}_4\to {\mathrm{N}}_2+4{\mathrm{e}}^{-} \\ \mathrm{RHR}: {\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}+2{\mathrm{e}}^{-}\to {\left[{\mathrm{Ni}}_2{\left(\mathrm{CN}\right)}_6\right]}^{4-}+2{\mathrm{CN}}^{-} \\ 3. \mathrm{Balance} \mathrm{the} \mathrm{charges} \mathrm{with} {\mathrm{OH}}^{-} \\ \mathrm{OHR}: {\mathrm{N}}_2{\mathrm{H}}_4+4{\mathrm{OH}}^{-}\to {\mathrm{N}}_2+4{\mathrm{e}}^{-} \\ \mathrm{RHR}: {\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}+2{\mathrm{e}}^{-}\to {\left[{\mathrm{Ni}}_2{\left(\mathrm{CN}\right)}_6\right]}^{4-}+2{\mathrm{CN}}^{-} \\ 4. \mathrm{Balance} \mathrm{the} \mathrm{O} \mathrm{atoms} \mathrm{with} {\mathrm{H}}_2\mathrm{O} \\ \mathrm{OHR}: {\mathrm{N}}_2{\mathrm{H}}_4+4{\mathrm{OH}}^{-}\to {\mathrm{N}}_2+4{\mathrm{e}}^{-}+4{\mathrm{H}}_2\mathrm{O} \\ \mathrm{RHR}: {\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}+2{\mathrm{e}}^{-}\to {\left[{\mathrm{Ni}}_2{\left(\mathrm{CN}\right)}_6\right]}^{4-}+2{\mathrm{CN}}^{-} \\ 5. \mathrm{Multiply} \mathrm{the} \mathrm{half} \mathrm{reactions} \mathrm{with} \mathrm{get} \mathrm{least} \mathrm{common} \mathrm{multiple} \mathrm{and} \mathrm{then} \mathrm{combine} \mathrm{them} \mathrm{to} \mathrm{get}, \\ \mathrm{OHR}: {\mathrm{N}}_2{\mathrm{H}}_4+4{\mathrm{OH}}^{-}\to {\mathrm{N}}_2+4{\mathrm{e}}^{-}+4{\mathrm{H}}_2\mathrm{O} \\ 2\times \mathrm{RHR}: 2{\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}+4{\mathrm{e}}^{-}\to 2{\left[{\mathrm{Ni}}_2{\left(\mathrm{CN}\right)}_6\right]}^{4-}+4{\mathrm{CN}}^{-} \\ \mathrm{Overeall} \mathrm{Reaction} \\ 2{\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}+{\mathrm{N}}_2{\mathrm{H}}_4+4{\mathrm{OH}}^{-}\to {\mathrm{N}}_2+4{\mathrm{H}}_2\mathrm{O}+2{\left[{\mathrm{Ni}}_2{\left(\mathrm{CN}\right)}_6\right]}^{4-}+4{\mathrm{CN}}^{-} \\ 6. \mathrm{Charge} \mathrm{of} \mathrm{the} \mathrm{oxidation} \mathrm{state} \mathrm{of} \mathrm{the} \mathrm{metal} \\ {\left[{\mathrm{Ni}}_2{\left(\mathrm{CN}\right)}_6\right]}^{4-} \mathrm{has} \mathrm{a} \mathrm{charge} \mathrm{of} 4- \\ \mathrm{CN} \mathrm{has} \mathrm{a} \mathrm{charge} \mathrm{of} 1- \\ 6\times \left(-1\right)+2\times \mathrm{Charge} \mathrm{of} \mathrm{Ni}=4- \\ \mathrm{Charge} \mathrm{of} \mathrm{Ni}=1+ \\ \mathrm{Therefore} \mathrm{the} \mathrm{cahrge} \mathrm{of} \mathrm{Ni} \mathrm{is} 1+ \end{gathered}$$