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H. 5 4 HH N H HH 2 H Н. С C C 1 Η 6 HH HH For parts A-G, provide a numerical answer in the blank (ex. 7). For parts H-J, answCan I please have answers to all the parts.

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Answer #1

a). Total no. of sigma bonds in the molecule are 30. (15 C-H+12 C-C+2 C-O+1C-N)

b.) Total no. of pi bonds in molecule are 6 (4 C=C+ 2 in C-tripple bond-N)

c). Total no. sp3 hybrid orbitals containing lone pairs is 1. i. e. O-atom labeled 1.

d). Total no. of sp2 hybrid orbitals containing lone pairs is 0. C-atoms have no lone pairs while N-atom in CN group does have a lone pair but it is sp hybridised.

e.) There are only H-atoms which undergoes bonding exclusively with unhybridised orbitals. Total no. is 15.

f.) Total 5 no. of atoms exclusively uses hybridised orbitals for bonding( 3 C-atoms from CH3 groups, 1 C-atom from CH2 group, 1 O-atom)

g.) All rest of the atoms in the molecule uses combination of hybridised and unhybridised orbitals for bonding, total no. is 10.

h.) No. It is impossible to rotate the bond between C and N atom as it is a tripple bond involving two pi bonds which are formed exclusivley from sidewise overlap of unhybridised orbitals. Hence no rotation without breaking the bonds.

i.) Yes, it can be rotated. It is a C-H sigma bond and can be rotated.

j.) Yes, it is drawn with maximum no. of atom in the same plane.

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