What are likely formulas for the following molecules?
a. CH₂Cl_?
b. BH_?
c. NI_?
d. SiCl_?

Question

What are likely formulas for the following molecules?

a. CH₂Cl_?

b. BH_?

c. NI_?

d. SiCl_?

Accepted Answer

Welcome back, everyone. Write the most probable formula 41 CHXF 32 B FX three NHX and four SI FX note that X in each case represents an unknown subscript. So we need to figure out what that subscript subscript is to find the correct chemical formula. Let's begin with compound one CHXF three. What we want to think of for part one is bonding preferences for each of these atoms. Let's begin with carbon. Recall that on our periodic table, carbon is located in group four A which corresponds to not only four valence electrons, but the fact that carbon is four more electrons short of a full valence shell. And so therefore, it prefers to make four bonds. We can also think of this in terms of group number corresponding to the number of bonds that the atom forms. Next, recall that typically carbon is going to be a central atom in compounds and it's typically bonded to other carbon atoms. But in this case, it's just going to need to make four bonds to any of the atoms within this compound being hydrogen or fluorine. So now let's consider hydrogen, which we should recall is in group one a on the periodic table corresponding to one bounce electron recall that hydrogen is typically going to be a terminal atom. And since it only has one valence electron, it's only capable of forming one bond. So next, we have fluorine recall that fluorine on our product table is located in group seven a our halogen group corresponding to seven valence electrons. This means that it's going to be one electron short from a full via show with eight electrons for an octet. And so therefore, recall that fluorine will be forming one bond typically in molecules with three lone pairs on itself. And typically fluorine is going to be a surrounding or terminal atom. With this understanding, we can see that having one bond and three loan pairs on flooring will give a full octet for its full valence shell. And we understand that this is important for fluorine to attain since fluorine is a period two element or period two atom. And so therefore, it should follow the octet rule in structures. And so what we can imagine is that we have our carbon atom, it needs four bonds. So we can make those, we can make three of those four bonds to fluorine. So we have three fluorine atoms bonded to carbon and then our fourth bond to carbon will now be to will be the hydrogen. Now recall that as we stated, our fluorine should each have three lone pairs for their full valence shell with directly attached seven valence electrons. And we can see from the structure that our formula is going to be CH which has a subscript of one. So we can say X is equal to one. Now recall that in formulas in chemical formulas, subscripts of one are implied. So we would just have CHF sub three and we can agree that this makes sense with the structure we've drawn. And so this would be our first answer. Now let's move on to part two where we have B FX. So we're gonna begin with the atom boron. Recall that boron on our product table is located in group three a corresponding to three bounce electrons. And based on its group number, it's going to form three bonds. Next. Again, we have fluorine as we stated before, located in group seven, a corresponding to seven valence electrons. Meaning yet again, it's one electron short from the full octet. And so just as before we would say that therefore, fluorine forms one bond with three lon pairs on itself. Now take note that in this case, since we understand that boron has three VMS electrons and forms three bonds, it's going to have three bonds to fluorine where we should recognize that this means that boron only has six electrons shared in covalent bonds because each line represents a bond with two electrons to between fluorine and boron. And so this means that boron is an exception to the octet rule. And even though it has six valence electrons or sorry six covalent shared bonding electrons, it's going to be forming a stable structure here. So it can violate the octet rule in this case. And that is because it only has three valence electrons which are directly attached to it being shared in covalent ponts. And just as before our fluorine atoms will have three lone pairs on themselves. So based on the structure we've drawn, we can confirm that our chemical formula is going to be BF sub three and so X is equal to three in this case. And this would be our second answer. BF three boron tri fluoride as our chemical formula. Next, we have part three which is for NHX. So beginning with our atom nitrogen, recall that nitrogen is located in group five A on the periodic table corresponding to five V electrons, meaning that nitrogen is three electrons short, I'm sorry. This says three electrons short from having a full valence shell or from having an octet. And so therefore, nitrogen will form three bonds and have one lone pair on itself. Next, we have hydrogen, which as we stated before is in group one a on the periodic table corresponding to one valence electron. And recall that because it only has one single valence electron, it can only form one bond and it is typically going to be a terminal at. So in this case, we have nitrogen as our central atom, we know that it needs to form three bonds. And so we will imply that it should have three covalent bonds to three different hydrogen atoms. And it still needs a lone pair. So we're gonna plug in that lone pair on nitrogen at the bottom. So based on this structure, we can understand that our chemical formula is going to be NH three. So in this case, x, our subscript for age is equal to three. So this is our third answer which we can recognize of the chemical formula for ammonia. Now moving on to part four of our prompt. So for part four, we have si F sub in which we begin with silicon recall that silicon is located in group four A on the periodic table corresponding to four bound electrons. And based on group number, we can say that silicon will form four bonds. Next, we have fluorine which we recall as force is located in group seven A on the periodic table corresponding to seven valence electrons. And also meaning that it's one electron short from having an octet. And so therefore, flooring will form one bond with three loan pairs on itself. So in this case, we can imagine that we have silicon as our central atom, it needs four bonds. So we would make four covalent bonds to four fluorine atoms where each floor atom will have its three lone pairs. And this is going to translate to the compound or chemical formula rather si F sub four, meaning that X is equal to four in this case. And so this is our compound silicon tetra fluoride. And this would be our fourth answer. So all of our answers highlighted in yellow are going to correspond to choice D in the multiple choice where for part one, we determined our chemical formula to be CHF sub three, which is our compound for boro or correction tri fluro methane. Then for part two, we determined our chemical formula to be BF sub three which corresponds to boron tri fluoride. For part three, we determined our formula to be NH sub three corresponding to our compound ammonia. And then for part four, we determined our formula to be si F four corresponding to silicon tetra fluoride as our fourth final answer. So I hope that this all made sense and let us know if you have any questions.

What are likely formulas for the following molecules?
a. CH₂Cl_?
b. BH_?
c. NI_?
d. SiCl_?

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What are likely formulas for the following molecules? a. CH2Cl? b. BH? c. NI?d. SiCl?

Verified video answer for a similar problem:

Key Concepts

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Common Chemical Notation

What are likely formulas for the following molecules?
a. CH₂Cl_?
b. BH_?
c. NI_?
d. SiCl_?