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To determine the formula for barium fluoride, we first need to understand the chemical symbols and charges of the ions involved.
1. Barium (Ba) Ion:
- Barium is an alkaline earth metal and typically forms a +2 charge when it becomes an ion. Therefore, the barium ion is denoted as [tex]\( \text{Ba}^{2+} \)[/tex].
2. Fluoride (F) Ion:
- Fluorine is a halogen and typically forms a -1 charge when it becomes an ion. Therefore, the fluoride ion is denoted as [tex]\( \text{F}^{-} \)[/tex].
To form a neutral compound, the total positive charge must balance the total negative charge. Given that a barium ion has a charge of +2 and a fluoride ion has a charge of -1, we need two fluoride ions to balance the charge of one barium ion:
[tex]\[ \text{Ba}^{2+} + 2\text{F}^{-} \][/tex]
This results in the chemical formula:
[tex]\[ \text{BaF}_2 \][/tex]
Therefore, the formula for barium fluoride is [tex]\( \text{BaF}_2 \)[/tex].
Let's identify the correct choice from the options provided:
A. [tex]\( \text{Ba}_2\text{F} \)[/tex]
B. [tex]\( \text{Ba}_2\text{F}_2 \)[/tex]
C. [tex]\( \text{BaF} \)[/tex]
D. [tex]\( \text{BaF}_2 \)[/tex]
The correct answer is:
D. [tex]\( \text{BaF}_2 \)[/tex]
1. Barium (Ba) Ion:
- Barium is an alkaline earth metal and typically forms a +2 charge when it becomes an ion. Therefore, the barium ion is denoted as [tex]\( \text{Ba}^{2+} \)[/tex].
2. Fluoride (F) Ion:
- Fluorine is a halogen and typically forms a -1 charge when it becomes an ion. Therefore, the fluoride ion is denoted as [tex]\( \text{F}^{-} \)[/tex].
To form a neutral compound, the total positive charge must balance the total negative charge. Given that a barium ion has a charge of +2 and a fluoride ion has a charge of -1, we need two fluoride ions to balance the charge of one barium ion:
[tex]\[ \text{Ba}^{2+} + 2\text{F}^{-} \][/tex]
This results in the chemical formula:
[tex]\[ \text{BaF}_2 \][/tex]
Therefore, the formula for barium fluoride is [tex]\( \text{BaF}_2 \)[/tex].
Let's identify the correct choice from the options provided:
A. [tex]\( \text{Ba}_2\text{F} \)[/tex]
B. [tex]\( \text{Ba}_2\text{F}_2 \)[/tex]
C. [tex]\( \text{BaF} \)[/tex]
D. [tex]\( \text{BaF}_2 \)[/tex]
The correct answer is:
D. [tex]\( \text{BaF}_2 \)[/tex]
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