Question

Please show work, and explain, thanks

Which of these Lewis structures would be closer to reality and why? :NEN-Ö: Ø N=N=5 (9) A because the negative charge ends up on the more electronegative atom. OB because the magnitude of formal charges is less. A because the magnitude of formal charges is less. O B because the octet rule is satisfied. OB because the negative charge ends up on the more electronegative atom. A because the octet rule is satisfied.

Answer 1

Option 1 is the correct answer.

I am attaching images for the explanation.

Answer Given, A :NEN-Ö ® N=N=0 let's find FC on each atom of (A), formal charge = (No of Valence e-)-( No of lone pair? (5x bond paire) Formal charge on Ni= (5 valence e-) - (2 lone pair) -(x 6 bond pair e-) - 5-2-3=0 Formal charge on N₂ = (5 valence e-) - (o lone paire want ou inton - 1 x 8 bond pair e-) d e 5 - 0 - 4 = +1 Formal charge on O = ( 6 Nalence e-) - (6 lone pary ( -4x2 Bond paire) = 6-6-1 = -1 Scanned with CamScanner

on each atom of 8 Now, Let's find formal charge V=N=0 formal charge on Ni (5 Valenu e-) - (4 Lone pars e) - (1 x 4 bond pais e-) = 5-4-2 = -1 formal charge on N = (5 valence e-)-(o lone pair) - 6 x 8 bond paix e = 5-0-4 = +1 formal charge on O = (6 valence &-) - (4 Lone paire) - (1 x 4 bond pair e-) .= 6- 4-2=0 Writing ② and ③ with formal charges INEN-Ö: ® NENE Ö (-ve charge + 6 yon Natom O +1 O Option 1- option (1) is the correct answer In negative charge is on oxygen atom and electronegativity of Oxygen is higher than Nitrogen- This means negative charge on oxygen atom is more stable than a megative charge on nitrogen atom. ho would be more closer to reality because the (cs negative change ends up on the more electronegative

atom " option @ and option ③ are incorrect because a and ② both have some magnitude of formal charge option @ and 6 are incorrect because each atom of ④ and 6 are satisfying the octet rule. Each atom has to electron count equal to 8. Note @ NEN ILP+B BP 2 +2x2 se- | (8) N = I BP (Bond part) = 20 ILP ( Lone Paus r) = 20 4 BP 31P+1 BP 4x2 3x2 +1+2 se- se- octect satistel N= 0 1 0 FT CO. 2LP+ 2BP 4BP 2BP +22P a 2x2 + 2x2 4x2 2x2 + 2x2 8e se octet satisfied otti option 6 is incorrect because in . negative charge is on 'N'atom. we know, electronegativity of Nitage is lesser than Oxygen. It means & is less stable than A. since A has negative change on oxygen atom, Scanned with CamScanner