Well, I think you mean [math] ext{nitrate ion}[/math], the which is [math]NO_{3}^{-}[/math] ….
And here we got [math]5+3×6+1=24• ext{valence electrons}[/math] to distribute across 4 centres … and thus we get a quaternized nitrogen atom with a formal positive charge, and TWO of the oxygen atoms have a formal negative charge…and thus …
[math]O=stackrel{+}N(-O^{-})_{2}[/math]
….you will have to add the 2 lone pairs on the neutral oxygen, and the three lone pairs on the anionic oxygens…nitrogen “owns” 4 valence electrons, and thus is associated with a formal positive charge….
Well, I think you mean [math]ext{nitrate ion}[/math], the which is [math]NO_{3}^{-}[/math] ….
And here we got [math]5+3×6+1=24•ext{valence electrons}[/math] to distribute across 4 centres … and thus we get a quaternized nitrogen atom with a formal positive charge, and TWO of the oxygen atoms have a formal negative charge…and thus …
[math]O=stackrel{+}N(-O^{-})_{2}[/math]
….you will have to add the 2 lone pairs on the neutral oxygen, and the three lone pairs on the anionic oxygens…nitrogen “owns” 4 valence electrons, and thus is associated with a formal positive charge….
If (NO4)(3-) existed (and I’m pretty sure it doesn’t), it would have a nitrogen with a +1 formal charge in the center of the ion, and four oxygen atoms each having a -1 formal charge. Every atom would then have a complete octet. Sorry about not being able to draw the Lewis structure here in Quora.
If (NO4)(3-) existed (and I’m pretty sure it doesn’t), it would have a nitrogen with a +1 formal charge in the center of the ion, and four oxygen atoms each having a -1 formal charge. Every atom would then have a complete octet. Sorry about not being able to draw the Lewis structure here in Quora.
Well, I think you mean [math] ext{nitrate ion}[/math], the which is [math]NO_{3}^{-}[/math] ….
And here we got [math]5+3×6+1=24• ext{valence electrons}[/math] to distribute across 4 centres … and thus we get a quaternized nitrogen atom with a formal positive charge, and TWO of the oxygen atoms have a formal negative charge…and thus …
[math]O=stackrel{+}N(-O^{-})_{2}[/math]
….you will have to add the 2 lone pairs on the neutral oxygen, and the three lone pairs on the anionic oxygens…nitrogen “owns” 4 valence electrons, and thus is associated with a formal positive charge….
Well, I think you mean [math]ext{nitrate ion}[/math], the which is [math]NO_{3}^{-}[/math] ….
And here we got [math]5+3×6+1=24•ext{valence electrons}[/math] to distribute across 4 centres … and thus we get a quaternized nitrogen atom with a formal positive charge, and TWO of the oxygen atoms have a formal negative charge…and thus …
[math]O=stackrel{+}N(-O^{-})_{2}[/math]
….you will have to add the 2 lone pairs on the neutral oxygen, and the three lone pairs on the anionic oxygens…nitrogen “owns” 4 valence electrons, and thus is associated with a formal positive charge….
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If (NO4)(3-) existed (and I’m pretty sure it doesn’t), it would have a nitrogen with a +1 formal charge in the center of the ion, and four oxygen atoms each having a -1 formal charge. Every atom would then have a complete octet. Sorry about not being able to draw the Lewis structure here in Quora.
If (NO4)(3-) existed (and I’m pretty sure it doesn’t), it would have a nitrogen with a +1 formal charge in the center of the ion, and four oxygen atoms each having a -1 formal charge. Every atom would then have a complete octet. Sorry about not being able to draw the Lewis structure here in Quora.
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