Chapter 10: Problem 123
Use the shortcut rules to assign an oxidation state to each atom: (a) \(\mathrm{NaPO}_{3}\) (b) \(\mathrm{B}(\mathrm{OH})_{3}\) (c) \(\mathrm{V}_{2} \mathrm{O}_{5}\) (d) \(\mathrm{K}_{2} \mathrm{TiF}_{6}\)
Short Answer
Expert verified
The oxidation states of each atom in the given compounds are:
(a) In \(\mathrm{NaPO}_{3}\), Na: +1, P: +5, O: -2
(b) In \(\mathrm{B}(\mathrm{OH})_{3}\), B: +3, O: -2, H: +1
(c) In \(\mathrm{V}_{2} \mathrm{O}_{5}\), V: +5, O: -2
(d) In \(\mathrm{K}_{2}\mathrm{TiF}_{6}\), K: +1, Ti: +4, F: -1
Step by step solution
01
(a) Finding the oxidation states in NaPO3)
In NaPO3, we have the elements Sodium (Na), Phosphorus (P), and Oxygen (O). Using the shortcut rules for assigning oxidation states:
1. Elemental oxygen typically has an oxidation state of -2.
2. Elemental sodium typically has an oxidation state of +1.
Now we can find the oxidation state of Phosphorus (P).
Let x be the oxidation state of P. Since the compound is neutral, the sum of all oxidation states must be zero. Therefore,
\(1 * (+1) + x + 3 * (-2) = 0\)
Solving for x,
\(x = 1 - 3*(-2) = 7\)
So, the oxidation states in \(\mathrm{NaPO}_{3}\) are:
Na: +1
P: +5
O: -2
02
(b) Finding the oxidation states in B(OH)3)
In B(OH)3, we have the elements Boron (B), Oxygen (O), and Hydrogen (H). Using the shortcut rules for assigning oxidation states:
1. Elemental hydrogen typically has an oxidation state of +1.
2. Elemental oxygen typically has an oxidation state of -2.
Now we can find the oxidation state of Boron (B).
Let x be the oxidation state of B. Since the compound is neutral, the sum of all oxidation states must be zero. Therefore,
\(x + 3 * (+1) + 3 * (-2) = 0\)
Solving for x,
\(x = 3*(-2) + 3*(1) = 3\)
So, the oxidation states in \(\mathrm{B}(\mathrm{OH})_{3}\) are:
B: +3
O: -2
H: +1
03
(c) Finding the oxidation states in V2O5)
In V2O5, we have the elements Vanadium (V) and Oxygen (O). Using the shortcut rules for assigning oxidation states:
1. Elemental oxygen typically has an oxidation state of -2.
Now we can find the oxidation state of Vanadium (V).
Let x be the oxidation state of V. Since the compound is neutral, the sum of all oxidation states must be zero. Therefore,
\(2 * x + 5 * (-2) = 0\)
Solving for x,
\(x = 5*(-2) / 2 = 5\)
So, the oxidation states in \(\mathrm{V}_{2}\mathrm{O}_{5}\) are:
V: +5
O: -2
04
(d) Finding the oxidation states in K2TiF6)
In K2TiF6, we have the elements Potassium (K), Titanium (Ti), and Fluorine (F). Using the shortcut rules for assigning oxidation states:
1. Elemental potassium typically has an oxidation state of +1.
2. Elemental fluorine typically has an oxidation state of -1.
Now we can find the oxidation state of Titanium (Ti).
Let x be the oxidation state of Ti. Since the compound is neutral, the sum of all oxidation states must be zero. Therefore,
\(2 * (+1) + x + 6 * (-1) = 0\)
Solving for x,
\(x = 2 - 6*(-1) = 8\)
So, the oxidation states in \(\mathrm{K}_{2}\mathrm{TiF}_{6}\) are:
K: +1
Ti: +4
F: -1
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Assigning Oxidation Numbers
Understanding how to assign oxidation numbers is crucial for analyzing redox reactions, which involve electron transfer between atoms. Oxidation numbers, often thought of as hypothetical charges, provide insight into the degree of oxidation or reduction of an atom within a compound.
Here's a simplified approach to assigning oxidation numbers:
Using this method makes it easier to balance redox reactions, predict reaction products, and understand electron flow during chemical reactions.
Here's a simplified approach to assigning oxidation numbers:
- For an atom in its elemental form, the oxidation number is always 0.
- For a monoatomic ion, the oxidation number is equal to the charge of the ion.
- Oxygen usually has an oxidation number of -2, except in peroxides where it's -1.
- Hydrogen is normally +1 when bonded to nonmetals and -1 when bonded to metals.
- The sum of all oxidation numbers in a neutral compound must be zero, while in a polyatomic ion, it must equal the ion's charge.
Using this method makes it easier to balance redox reactions, predict reaction products, and understand electron flow during chemical reactions.
Balancing Redox Reactions
Balancing redox (oxidation-reduction) reactions is an essential skill for thoroughly understanding chemical processes. Here's how to balance a redox reaction step by step:
- Separate the reaction into half-reactions, one for oxidation and another for reduction.
- Balance all elements except for hydrogen and oxygen.
- Balance oxygen by adding water (H2O).
- Balance hydrogen by adding hydrogen ions (H+).
- Balance charges by adding electrons (e-). The number of electrons should equal the difference in oxidation numbers from reactants to products.
- Equalize the number of electrons in both half-reactions by multiplying them with appropriate coefficients.
- Add the half-reactions back together, canceling out electrons and any other species that appear on both sides.
- Resolve any additional imbalances, including those involving charges, especially if the reaction occurs in basic solution - you may need to add hydroxide ions (OH-).
Chemical Compound Formulation
The formulation of a chemical compound entails a clear understanding of the chemical symbols, oxidation numbers, and valency, which is the combining power of an element. When formulating a compound, follow these guidelines:
- Identify the valency or oxidation number of each element.
- Apply the 'criss-cross' method where the valency or oxidation number of one ion becomes the subscript of the other, revealing how atoms combine to achieve stability.
- Adjust subscripts to the simplest whole number ratio to adhere to the Law of Multiple Proportions.
- Ensure that the total charge of the compounds balances to zero for neutral compounds or matches the net charge for ions.
Oxidation-Reduction Chemistry
Oxidation-reduction (redox) chemistry involves the transfer of electrons between chemical species. The substance that loses electrons undergoes oxidation, and the one that gains electrons undergoes reduction. This electron shift corresponds with a change in the oxidation numbers of the atoms involved.
Redox chemistry is fundamental in numerous biological processes, industrial applications, and even the production of energy in batteries. A few indicators that a redox reaction has occurred include:
Redox chemistry is fundamental in numerous biological processes, industrial applications, and even the production of energy in batteries. A few indicators that a redox reaction has occurred include:
- Change in color of the reactants or formation of a precipitate.
- Evolution of gas.
- Change in electrical conductivity.