Question

a)

Construct the framework of the molecule acrylonitrile (H2CCHCN) and assign o and/or t bond(s) to each bond placement.

b)

c)

Answer 1

which is not the case with the above atoms. Hence the order is- rasou as cras<0+ 25 < (+2p== A2py] <82p2 <ixtapa - Trpy J < 0*2p2 second season can be electronegativity, as we move from left to light in peciodic table it increases which increases the stability for signa (op) orbitals and hence it occurs fiest. lik B<c<N<OTE [ electronegatning order Let us consider the Molecular oviral diagrams for O2 and 12 palat 111 2 1

Molecular diagram for Hey- He atom configuration - 152; Het 15 212 MO's required as and ot 15 which have to accomodate 3 e (hent terse ) T K (aunbonding 15 TT ( bonding 6 Bond older = ras No. of eo in bonding cubital I No. of ee in anibonding albirere 2-1 = 1 I c It is stable because no. of elections in bonding orbital are more than in annbonding. a) It is a paramagnehe molecule due to presence of unboiled election in antibending * (45) orbital.

acrylonitule CMC CHON] fancak ! because н Alouber of Sigma bords 6 pi bonds 3 1 single bond = ir bend 1 double bond - 15 & 1 x bond 1 triple bond = 10+ 2 a bond coscopic date have bee periodic tanec molecules According to Molecular orbital theory, Antibonding Mo's :- * 25 *2pz at 2px 2py bonding Mo's - 525 & 2p2 apin tapy The energy levels of these molecular orbitals have been determined experimentally spectioscopic data for homo- - nudear diabomuc molecules e ccord cow elements of periodic table. The increasing order of energies of various redeculas orbitals for oz and his ras Cols 40 2550*25 < 1 2pz < (72pn = A2py] < [* 2px = x 2py I co*2p2 ramenes this sequence changes for the larges sired aboms like ₂, Bez, B2, C2, N2 because in the case of Ne, O F2, the rp is attracted a little more to nucleus hence shifts down in molecular orbital diagram.