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 \(\left(\mathrm{FeO}_{4}^{2-}\right), \mathrm{Cl}^{-},\) and water. (b) Potassium hexacyanomanganate(II) reacts with \(\mathrm{K}\) metal to form \(\mathrm{K}_{6}\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]\) (c) Heating sodium superoxide \(\left(\mathrm{NaO}_{2}\right)\) with \(\mathrm{Co}_{3} \mathrm{O}_{4}\) produces \(\mathrm{Na}_{4} \mathrm{CoO}_{4}\) and \(\mathrm{O}_{2}\) gas. (d) Vanadium(III) chloride reacts with Na metal under a CO atmosphere to produce \(\mathrm{Na}\left[\mathrm{V}(\mathrm{CO})_{6}\right]\) and \(\mathrm{NaCl} .\) (e) Barium peroxide reacts with nickel(II) ions in basic solution to produce \(\mathrm{BaNiO}_{3}\) (f) Bubbling CO through a basic solution of cobalt(II) ion produces \(\left[\mathrm{Co}(\mathrm{CO})_{4}\right]^{-}, \mathrm{CO}_{3}^{2-},\) and water. (g) Heating cesium tetrafluorocuprate(II) with \(\mathrm{F}_{2}\) gas under pressure gives \(\mathrm{Cs}_{2} \mathrm{CuF}_{6}\) (h) Heating tantalum(V) chloride with Na metal produces \(\mathrm{NaCl}\) and \(\mathrm{Ta}_{6} \mathrm{Cl}_{15},\) in which half of the \(\mathrm{Ta}\) is in the +2 state. (i) Potassium tetracyanonickelate(II) reacts with hydrazine \(\left(\mathrm{N}_{2} \mathrm{H}_{4}\right)\) in basic solution to form \(\mathrm{K}_{4}\left[\mathrm{Ni}_{2}(\mathrm{CN})_{6}\right]\) and \(\mathrm{N}_{2}\) gas.

Step by step solution

01

- Iron(III) ion with hypochlorite ion

Write the balanced equation: Fe^{3+} + ClO^- + 4 OH^- → FeO_4^{2-} + Cl^- + 2 H_2O. Now, calculate the oxidation state of Fe in FeO_4^{2-}. Let it be x: 1Fe_x + 4(-2) = -2 → x - 8 = -2 → x = 6. Oxidation state of Fe is +6.

02

- Potassium hexacyanomanganate(II) with K metal

Write the balanced equation: K_4[Mn(CN)_6] + 2 K → K_6[Mn(CN)_6]. Oxidation state of Mn remains +2, as it is present in [Mn(CN)_6]^{4-}.

03

- Heating sodium superoxide with Co_3O_4

Write the balanced equation: 4 NaO_2 + Co_3O_4 → Na_4CoO_4 + 2 O_2. Calculate oxidation state of Co in Na_4CoO_4: 4(+1) + Co + 4(-2) = 0 → 4 - 8 + Co = 0 → Co = +4. Oxidation state of Co is +4.

04

- Vanadium(III) chloride with Na in CO atmosphere

Write the balanced equation: 2 VCl_3 + 6 Na + 6 CO → 2 Na[V(CO)_6] + 6 NaCl. V in Na[V(CO)_6] has +1 oxidation state.

05

- Barium peroxide with nickel(II) ions

Write the balanced equation: BaO_2 + Ni^{2+} + 2 OH^- → BaNiO_3 + H_2O. Determine oxidation state of Ni in BaNiO_3: Ba(+2), O_3(-6) → Ni is +4. Oxidation state of Ni is +4.

06

- Cobalt(II) ion with CO in basic solution

Write the balanced equation: Co^{2+} + 4 CO + 2 OH^- → [Co(CO)_4]^{-} + CO_3^{2-} + H_2O. For [Co(CO)_4]^{-}, oxidation state of Co is -1.

07

- Cesium tetrafluorocuprate(II) with F_2 gas

Write the balanced equation: Cs_2[CuF_4] + F_2 → Cs_2[CuF_6]. Oxidation state of Cu in Cs_2 [CuF_6] is +4.

08

- Heating TaCl_5 with Na metal

Write the balanced equation: 6 TaCl_5 + 13 Na → NaCl + Ta_6Cl_{15}. To find oxidation state: sum of oxidation states in Ta_6Cl_{15} is zero. Given half of the Ta is in +2 state; overall oxidation state is: 6Ta_x + 15(-1) = 0 → 6Ta_x - 15 = 0 → 6Ta_x = 15 → 4Ta_(+2) + 2Ta_( x) = 15; x = +3. Thus for mixed ox. states avg. is +2.5.

09

- Potassium tetracyanonickelate(II) with hydrazine

Write the balanced equation: 2 K_2[Ni(CN)_4] + N_2H_4 + 4OH^- → K_4[Ni_2(CN)_6] + N_2 + 4H_2O. Ni in K_4[Ni_2(CN)_6] is +2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are integral to many chemical processes. In these reactions, one substance loses electrons (oxidation), while another gains electrons (reduction). This transfer of electrons results in a change in the oxidation states of the involved elements.
For example, when Iron(III) ion (Fe³⁺) reacts with hypochlorite ion (ClO^-), the iron undergoes a change in oxidation state. Fe³⁺ is reduced to FeO₄^2-, where its oxidation state becomes +6, while the ClO^- ion is oxidized to Cl^-.
Redox reactions are essential in various applications, from energy production in batteries to biochemical processes in living organisms.

Balanced Chemical Equations

A balanced chemical equation ensures that the same number of each type of atom exists on both sides of the reaction. This reflects the conservation of mass principle, where matter is neither created nor destroyed.
For instance, the balanced equation for the reaction between potassium hexacyanomanganate(II) and potassium metal is:
K₄[Mn(CN)₆] + 2 K → K₆[Mn(CN)₆]
Here, both sides contain the same number of potassium, manganese, and cyanide ions. Balancing equations is crucial as it accurately represents the stoichiometry of the reaction, helping chemists understand the proportions of reactants and products involved.

Transition Metal Chemistry

Transition metals, found in the d-block of the periodic table, exhibit unique properties such as variable oxidation states, complex formation, and catalytic activity. These properties are due to their partially filled d-orbitals.
An example is cobalt (Co), which can exist in oxidation states ranging from -1 to +4 in compounds. In the reaction of cobalt(II) ion with carbon monoxide in a basic solution, cobalt forms the complex [Co(CO)₄]^- with an oxidation state of -1.
Understanding the behavior of transition metals is crucial in fields like industrial catalysis, materials science, and bioinorganic chemistry.

Oxidation State Determination

Determining the oxidation state of a transition metal in a compound involves understanding its bonding with other atoms. The oxidation state is the hypothetical charge the atom would have if all bonds were ionic.
For example, in barium nickelate (BaNiO₃), the oxidation state of nickel (Ni) can be calculated as follows:
Ba is +2, each oxygen (O) is -2, thus in the compound (Ba²⁺NiO₃)
2 + Ni + 3(-2) = 0 → Ni - 6 + 2 = 0 → Ni = +4.
This procedure helps chemists predict the reactivity and properties of compounds. Mastering oxidation state determination is essential for understanding redox reactions and transition metal chemistry.