Select the correct answer.
What is the formula of the compound vanadium(V) oxide? Use the list of polyatomic ions and the periodic table to help you answer.
A. $VO _2$
B. $VO _5$
C. $V _5 O _2$
D. $V _2 O _5$
E. $V _2 O _2$
Answer (2)
Vanadium(V) oxide consists of vanadium ions with a +5 charge and oxide ions with a -2 charge.
Balance the charges to achieve a neutral compound.
Two vanadium ions ( 2 × + 5 = + 10 ) and five oxide ions ( 5 × − 2 = − 10 ) are required.
The chemical formula is therefore V 2 O 5 .
Explanation
Understanding the Problem We are asked to determine the chemical formula of vanadium(V) oxide. Vanadium(V) indicates that vanadium has a +5 charge, and oxygen typically has a -2 charge. We need to find the smallest whole number ratio of vanadium and oxygen ions that results in a neutral compound.
Balancing the Charges To balance the charges, we need to find the least common multiple (LCM) of 5 and 2, which is 10. Therefore, we need a total positive charge of +10 and a total negative charge of -10. To get +10 from vanadium ions with a +5 charge each, we need 2 vanadium ions ( 2 × + 5 = + 10 ). To get -10 from oxide ions with a -2 charge each, we need 5 oxide ions ( 5 × − 2 = − 10 ).
Determining the Formula Thus, the chemical formula of vanadium(V) oxide is V 2 O 5 .
Final Answer The correct answer is D. V 2 O 5 .
Examples
Understanding chemical formulas like that of vanadium(V) oxide ( V 2 O 5 ) is crucial in various real-world applications. For instance, V 2 O 5 is used as a catalyst in the production of sulfuric acid, a vital chemical in many industrial processes, including fertilizer production and petroleum refining. Knowing the correct formula ensures that the right amount of each element is used, leading to efficient and safe chemical reactions. This knowledge also helps in predicting the properties of the compound and its behavior in different conditions, which is essential for materials science and chemical engineering.
The formula for vanadium(V) oxide is V 2 O 5 , which is derived from the +5 oxidation state of vanadium and the -2 oxidation state of oxygen. To balance the charges, we need 2 vanadium ions and 5 oxide ions. Thus, the correct answer is D. V 2 O 5 .
;
