The difference between the electronegativity value of fluorine and sodium is more than 1.7 so the compound formed by them will be ionic in nature.
The difference between the electronegativity value of fluorine and sodium is more than 1.7 so the compound formed by them will be ionic in nature.
The correct option is A True
There are two parts in the statement that we need to confirm.
First, the electronegativity value for sodium is 0.93 and that for fluorine is 3.98. The difference between the electronegativity values is 3.05 which is greater than 1.7, therefore the first half of the statement given is true.
Second, we should know that if the difference of electronegativity values is greater than or equal to 1.7 the compound will be ionic in nature. If it is less than 1.7 the compound will be covalent in nature. Since the difference is more than 1.7 the compound formed will be ionic in nature.
Since both the parts of the statement are true the statement is true.
True
There are two parts in the statement that we need to confirm.
First, the electronegativity value for sodium is 0.93 and that for fluorine is 3.98. The difference between the electronegativity values is 3.05 which is greater than 1.7, therefore the first half of the statement given is true.
Second, we should know that if the difference of electronegativity values is greater than or equal to 1.7 the compound will be ionic in nature. If it is less than 1.7 the compound will be covalent in nature. Since the difference is more than 1.7 the compound formed will be ionic in nature.
Since both the parts of the statement are true the statement is true.
