Question

1. Nitrogen is unique among the elements in forming 7 molecular oxides: (0) (iv) N2O NO N2O3 NO2 N204 N2O5 NO3 (vi) (vii) Three of these are paramagnetic, that is they do not have an even number of electrons so in their Lewis structure there is one unpaired electron. Draw Lewis structures for all the oxides and assign oxidation states to nitrogen and oxygen in each compound.

Answer 1

Answer a N20 Valence election in N-atonn = 5 Valence electron in o-aton = 6 EN IN = ŘEN-O = =8; oxidation state on N, N20 22+(-2) =o are =+2 r = +2/2 = t1 a =+1 oxidation state on Nidrogen = t1 oxidation Sted on oxygle = -2 (i) No NO :N::0: :N=0: oxidation state oxidation state on on N 0 paramagnetic unpaired electron. 12 – 2 = – ii) N2O3 O: IN. N:: 0 = = - = 0 0:

oxidation State on N N .03 20+ 3(-2) 20 2x - 6 =0 24 = to * +612 = +3 oxidation state on N=+3 oxidation state on o 2(+3) +34 =0 to +32 = 0 de = 4613 = - 2 oxidation steate on 0 = – 2 (iv) NO2 0 :N::0 = : Ö-N =Ö ON= paramagneltic unpained eleckton. oxication state on N, NO2 x + 2(-2) = 0 n - 4 =o n= t4 Oxidation State on N = +4. oxidation state or o=-2

N204 0:0 O=ze : :..Ñ:: 0 0:0 0:0: oxidation state on N N204 2 20+ 4 (-2) = 0 21 - 8 - 20 - to n = +8/2 = 14 oxidation & trade on N= 14 oxidation & take on o= -2 vi) Neos 0 No No :0: ZO 0=20 0:0: oxidation State on N, N2os 24+51-2) = 0 24 – lo no 2n = to n- +10/2 =ts oxidation oxidation & teate on State on N= ts 0 -- 2

NO3, N : -N- 5.0 IN lē :0 Paramag heltic due to unpaired electon. oxidation State on N, NO3 n+ 3(-2) = 0 n = +6 oxidation strate ono =-2 "Plecke rate it'