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No answer submission is required for this problem during this quiz. Please read the instructions and then go to the next problem 1. This question has no time limit. 2. You will upload your answer later with your other "Show-Your-Work" pages. After you take a picture of the problem below, type "OK" in the textbox and continue to the next problem. This problem is worth 10 points. Answer both part A and part B below. Resonance and Formal Charges Problem A. The compound N2O has three (3) resonance forms (N in the middle, N-N-O). 1. Draw the Lewis structures for all three resonance forms. 2. Assign formal charges to each atom in each resonance form. 3. Based on the formal charges, predict which resonance form is the most stable. B. Analysis of N20 shows the N-N bond is 112 pm and the N-O bond is 119 pm. Using the following bond length data, identify the resonance form that best represents the actual bonding in N20. N-N single 167 pm, N-N double 120 pm, N-N triple 110 pm, N-O single 147 pm, N-O double 115 pm, N-O triple 105 pm.

Answer 1

A) 1) Lewis structure for all three resonance :NEN -Ö: N=N=ö <>. nzö: (11) (11) 2. Formal charge on each atom in each resorence form FC = be - Boneling formal charge 1CBE ME) 2 N-Valamu e of free atom. NEJ Non-Bonding er on structure I Fic on 0 - 6 - (12BE + NE) = 6-( 42 * 2 +6)= 6-(1) = -) F.C. on Midden=5-CY₂x840) at F.C on terminal N=5-(1/2 x642)zo in- IN SN-01 Formal charge on on structure a F.O. on o = 6 1/2 x 4+4) = 6-6=0 f.c.on middle N = 5-(1/2 % 8+o)=t FC on terminal M=5-(12*4+2) = -1 ti N = M = 0 0 ..-1 O

F C structure TiT on 0 6-C1246+2). + on middle al = S-48eati on terminal N = 5-(2x246) = -2 FC F.C +) No ON Only two more dectronegade +1 #1 3. IN SM ษ์ - 0.! =N= No (1) (14) (11) Resonance structure I is most stable because it have two formal changes atoins have formal charges land 102 atom at formal charge and us electronegatin 'N' contain al charge. B.) Experimental show that N-N Bond lenght is 112pm, N-0 Bond lenght is 119 pm Actual Bond lenght of N-M in N20 - 112pm which lies bw N-M double bond lenght 720pm and N-N triple bond lenght Hopm can be possible Strudlung Resonance for I N = = 0; . This +1

Actual Bond lenght of N-0 is 19pm. in M20, which lies blw N-o single bonel lenght (147pm) and unght clispm). This can be possible for strucline N-o double sesongne +9 :MZN 0 best resonance by the مصطح forms analy sis bond lenght form is resonance m (I) MENO
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