Is this structure correct for nitrate ion NO3- ?

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Aleksander Krol
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there's another structure for nitrate ions too where Nitrogen is boned with 4electrons of oxygen and that is considered to be the most stable structure of it. but i did it another way, and i'm not sure if this structure and the shape is valid.
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Explanation for the structure that I've drawn:
Nitrogen over here has 10 e- and it is bonded with 1 O- ion (7e-) and 2 Oxygen (6e-) (6e-) atoms, all having 8 electrons in their valence shell.
Also Nitrogen doesn't have any lone pair of electrons.
I used the lone pair formula to calculate the no. of lone pair of electrons in the central atom (this formula only works in the absence of Hydrogen), so in NO3- it is Nitrogen atom (central atom)
NO3- = 5+ 3(6) +1 = 24 e- so Nitrate ion has a total of 24e- in its valence.
Lp= [VE-8(n)]/2 (Lp -Lone pair) (VE- Valence Electrons) (n- no. atoms it is bonded with)
Lp for Nitrogen in Nitrate ion =[24-8(3)]÷2=0/2=0 so there are 0 lone pair of electrons in Nitrogen.
also x+3(-2)= -1
x-6= -1
x=6-1=5 so nitrogen has a charge of +5 in nitrate ion so it forms 5 bonds.
considering all of these calculations above, this structure seems right to me, so will i not be awarded any marks if i draw this structure in the examination for NO3- ion?
Last edited by Aleksander Krol; 1 month ago
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charco
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(Original post by Aleksander Krol)
there's another structure for nitrate ions too where Nitrogen is boned with 4electrons of oxygen and that is considered to be the most stable structure of it. but i did it another way, and i'm not sure if this structure and the shape is valid.
Name:  Screenshot_20210918-092155_Sketchbook.jpg
Views: 17
Size:  61.2 KB
Explanation for the structure that I've drawn:
Nitrogen over here has 10 e- and it is bonded with 1 O- ion (7e-) and 2 Oxygen (6e-) (6e-) atoms, all having 8 electrons in their valence shell.
Also Nitrogen doesn't have any lone pair of electrons.
I used the lone pair formula to calculate the no. of lone pair of electrons in the central atom (this formula only works in the absence of Hydrogen), so in NO3- it is Nitrogen atom (central atom)
NO3- = 5+ 3(6) +1 = 24 e- so Nitrate ion has a total of 24e- in its valence.
Lp= [VE-8(n)]/2 (Lp -Lone pair) (VE- Valence Electrons) (n- no. atoms it is bonded with)
Lp for Nitrogen in Nitrate ion =[24-8(3)]÷2=0/2=0 so there are 0 lone pair of electrons in Nitrogen.
also x+3(-2)= -1
x-6= -1
x=6-1=5 so nitrogen has a charge of +5 in nitrate ion so it forms 5 bonds.
considering all of these calculations above, this structure seems right to me, so will i not be awarded any marks if i draw this structure in the examination for NO3- ion?
No.
You cannot have 10 electrons in the outer shell.

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Aleksander Krol
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i finally understood the structure. if anyone has the same question like me or is perplexed about the structure of NO3- ion i'd recommend you to watch this video which i found extremely helpful. and i hope the same, that you do find it too.
https://youtu.be/D608NzB8Mr8
Last edited by Aleksander Krol; 1 month ago
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