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An unknown element X can form a compound with the formula XF3. In which group on the periodic table would element X can be found?

1.) Group 1 2.) Group 2 3.) Group 13 4.) Group 14 This question doesn't make any sense to me :/ Can you please explain it?


Noticed you changed the choices by introducing groups 13 and 14. Well if that is the case, the group orders are following this table:

So here are the charges for elements in the following groups then:

Group 1: +1      Group 2: +2     Groups 3-12: Vary

Group 13: +3    Group 14: +4 or +2    

1. From my explanation previously, the only group with +3 charge for one of its elements is Group 13, so that is your answer. 

Group 15: -3    Group 16: -2        Group 17: -1

Rizul, could you please also explain what type of bonding this is? I'm pretty sure that changes as well as the answer but I'm not a chemistry person. 

*Since this is long, I have highlighted the answer at the bottom in green.

There are mainly two types of bondings going into high school chemistry (you will learn plenty more in advanced classes):

1. Covalent bonding: This is if you find a compound containing elements from groups 14 through 17 (nonmetals). The electrons in each compound are shared because the electronegativity difference between the elements is moderate.

2. Ionic bonding: This is if you find a compound containing an element from Groups 1-13 (which are metals) and an element from Groups 14-17 (nonmetals). The electronegativity difference is much greater here because chemistry is all about stability. Nonmetals are thirsty for electrons because they are a little shy of obtaining 8 electrons. On the other hand, metals want to lose electrons because they use much less energy by becoming stably by losing electrons to have 8 electrons in a lesser shell than try to obtain electrons to fill up their current occupied shell. Thus one element can only gain and another lose electrons in this relationship, sharing is not favorable.

In this problem, you may be confused first of how I know this is ionic and not covalent. 

Here are the covalent possibilities of flourine:


Well Flourine and any of the elements in group 17 need just one electron to make an octet (8 electrons in its current shell) since it has 7 outer electrons already (valence). Flourine can bond to hydrogen and share hydrogen's one and only electron to make an octet and hydrogen can share one of flourine's 7 electrons to fill its shell. Note that hydrogen is in the first group of periodic table and it is special because it only needs 2 electrons and not 8 to become stable. 


Flourine can also covalent bond with another flourine atom and both can share one electron from each to come up with 8 electrons at some moment in time.

The Bottomline Explanation/Answer:

We notice that in this particular problem there is just one X element but 3 flourine atoms. This doesn't fit with the covalent possibilities I've provided. So it has to be ionic since 3 Flourine atoms need 3 electrons to be shared for all three flourine atoms to have 8 electrons each and there is just one group that has elements with 3 valence electrons and this group is 13. You note that this group has metals and a Flourine atom is a nonmetal since it comes from group 17. Thus, this is an ionic bond. 

Thank you so much. That was very thorough. 

No problem, and the question was very valid for the OP. Happy to help anyway I can!

Your explanation is good for the question. But would like to add one point about the HF formation and its bond. Although the electro negativity difference between hydrogen and flourine is not significantly high enough to form an ionic bond and therefore it forms a covalent bond, the shared electron pair is closer to flourine atom than hydrogen and hence it is a polar covalent bond as small + and - charges surround hydrogen and flourine respectively.

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Rizul N. | UNC-CH Grad For Math and Science TutorUNC-CH Grad For Math and Science Tutor
4.8 4.8 (12 lesson ratings) (12)

XF3 is our compound. X is unknown

For the problem I am using the periodic table at this site,r:3,s:0,i:99

1. We know this compound has no charge because there is no + or - sign. 

2. flourine has -1 charge since it is in Group 7 and there are 3 of this element. So total minus charge is -1(3) = -3. 

Remember your charges, I have listed below for each element groupwise:

Group 1: +1           Group 2: +2     Group 3: +3     

Group 4: +4 or +2 (depends)

Group 7: -1       Group 6: -2          Group 5: -3    

3. Now going back to basic algebra, we know that neutral charge of compound = 0. We know we have -3 charge from flourine's side and we need a +3 charge to add to -3 to make the sum = 0. There is only one X, which means that only one element is offsetting the -3 charge. 

What element has +3 charge then? Look back at the Group charges and you will note that Group 3 has each of its elements possess +3 charge. 


What if question was find charge of X if compound was X2Cl2 ?

1. Compound is neutral because it doesn't have + or - next to it.

2. Chlorine (Cl) has -1 charge. There are 2 Chlorine atoms so total negative charge = 2(-1) = -2

3.  There should be +2 from X2 to offset -2 charge to make neutral compound. But we note that there are 2 X atoms. To find true charge of X, divide +2 by 2 (because of number of X atoms) and you get +1 which is real charge of X.

4. So +1 charge element comes from Group 1 in this case. 

Mykola V. | Math Tutor - Patient and ExperiencedMath Tutor - Patient and Experienced
4.9 4.9 (52 lesson ratings) (52)

F is the element Fluorine. It has 7 electons which means it needs one more to have a complete 2nd shell of 8 electrons. It gets this one more electron from elements in lower groups such as H. 

In our case we have element X forming a compound with 3 F's to make XF3. This means that our X element needs to give 3 electrons, one for each F element. If H is in group 1 and lends 1 electron, Mg is in group 2 and lends 2 electrons, then surely, one of the elements in group 3 will have 3 electrons to give to what ever it forms a compound with. So our answer choice is 3) Group 3. 


Oh and by the way we both were correct with Group 3 but the OP changed the choices and Group 3 became Group 13.