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Suppose that in an ionic compound, "m" represents a metal that could form more than one type of ion. in the formula mf2 , the charge of the m ion would be:

Respuesta :

premedchemist
F (Fluorine) is in column (group/family) VIIA, or the "halogens". When you see the halogens (Fluorine, Chlorine, Bromine, and Iodine) in combination with a metal, each halogen atom present will carry a -1 charge. We can see that the atom has no charge, so the metal must cancel out the negative charges brought by the two fluorine atoms.
(Charge on m) + 2*(charge on fluorine) = 0
(Charge on m) + 2*(-1) = 0
(Charge on m) - 2 = 0
Charge on m ion = +2

The charge of metal would be +2. Ionic compounds are neutral compounds. It is formed from positively charged ions (cations) and negatively charged ions (anions).

In MF2 is a neutral ionic compound. Therefore, net charge on MF2 is zero. F belongs to halogen family. It gains one electron to get noble gas configuration. So, fluorine has -1 charge.

Hence, charge on metal would be:

Charge of metal + 2*charge of fluorine =0

Charge of metal + 2*(-1) =0

Charge of metal - 2=0

Charge of metal= +2

Thus, we can conclude that, charge of metal in MF2 is +2. i.e., metal loses two electrons to attain noble gas configuration.

Learn more about of charge of ionic compounds here:

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