okay we are drawing bonds and i was wondering how i am supposed to know if you can make a double or triple bond
for example for NH2OH why cant i make the bond between N and O triple instead of leaving lone pairs..
thanks
Do that. Then count the electrons around H, O, and N.
I count two for each of the N-H bonds, 2 for the O-H bond, then for N I see 12. And that's too many if you are trying to obey the octet rule. Even if you aren't trying to obey the octet rule, N doesn't have the holes to accommodate 12 electrons. It has a
1s2 2s2 2p3 so it has "holes" for three bonds only (to fill the 2p3 orbitals and make 6. The 2 electrons in the 1s2 orbital are the lone pair around N if you draw the structure correctly.
You must write a Lewis structure that obeys the Octet Rule. You start by counting the number of valence electrons on the two central atoms, N(5), O(6), and the three H's (1 each). Electrons must be shared so that N and O end up with 8 electrons around them (shared and unshared total). Each hydrogen shares 2 electrons with O or N. Hydrogen is unique in that it does not go beyond 2 lectrons through sharing (no octet). We can't draw Lewis structures here but all bonds are single in H2N-O-H. The nitrogen has one unshared pair and the O two unshared pairs.
What I just described is determined by trial and error and it is the only way to obey the Octet Rule. The basic skill you need is writing Lewis structures.
okay i understand so writing the Lewis structures is just the math not the drawing?
Both.
No, I don't think so. It's the math, true enough, but it's also the drawing. We want to know that there are single bonds in H2NOH and that there are no double bonds or triple bonds. In other words, You can't place the electrons helter skelter (just anywhere) that looks good.
" ..It's the math, true enough, but it's also the drawing..". I think that is equivalent to my terse Both. The procedure for writing Lewis structures can't be fully demonstrated on a bulletin board posting like this one. Using some trial and error within the proper constraints, and producing a "helter skelter" electron dot pattern are not logically equivalent. My students usually can tell the difference.
When it comes to determining whether a double or triple bond can be formed between two atoms, you need to consider a few factors.
1. Valence electrons: First, determine the total number of valence electrons for each atom in the molecule. For NH2OH, nitrogen (N) has five valence electrons, oxygen (O) has six, and hydrogen (H) has one electron each.
2. Octet rule: The octet rule states that most atoms tend to achieve a stable electron configuration by having eight electrons in their outer shell (except for hydrogen, which only needs two electrons). This means that for NH2OH, nitrogen and oxygen will strive to have eight electrons in their outermost shell.
3. Electronegativity: Electronegativity is the ability of an atom to attract electrons towards it. In general, the higher the electronegativity difference between two atoms, the more likely they are to form a double or triple bond.
Now, applying these concepts to NH2OH:
- Starting with nitrogen (N), it has five valence electrons. To achieve an octet, it needs three more electrons. If N formed a triple bond with O, it would share three pairs of electrons, which would leave N with only one lone pair of electrons and not satisfy the octet rule. Therefore, N prefers to form a single bond with O and fulfill the octet rule by using the remaining electrons to form additional bonds or lone pairs.
- Oxygen (O), on the other hand, has six valence electrons and needs two more electrons to fulfill the octet rule. By forming a single bond with N and having one lone pair of electrons, O also achieves an octet.
So in the case of NH2OH, the arrangement of bonds (single or double) is determined by the octet rule and the requirement to satisfy the valence electron count for each atom. Since N needs more than two additional electrons to achieve an octet, a triple bond between N and O is not possible.