Question

Which are correct? In molecular orbital (MO) energy level diagram for cyanide ion? The MO shows: (a) bond order = 2 (b) bonder order =3 ) cyanide ion is paramagnetic ) cyanide ion is! diamagnetic (e) HOMO is nu HOMO is og (6%) (2) Write down the diatomic molecules 2p orbitals for (a), (b), (6%) (3) What is the bond order of (a) OF(b) to which atom ( O or F ) the HOMO is polarized (c) which MO energy level is the HOMO ? (9%) (4) The ion NO can react with H * to form a chemical bond. (A) Which one is more likely to be formed. HNO or HON. (b) NO is easily to dimerize to N,04 with a trans configuration, which energy levels are involved in the dimerization: o *g (2p), *u (2p), oku (2p). *g (2p) (6%) (5) (i) Which sketch is correct for the interaction of 2 Pz orbitals to form a molecular orbital of (a) o and (b) o* (ID interaction between 2 Px orbitals to form (C) and (d) * molecular orbital (12%) (6) Use Drago's theory to calculate All for the reactions of (a) pyridine and BFs (b) pyridine and B(CH3)3 ? (E parameter for pyridine-1.17. BF3 = 9.88 and B(CH)3 - 6.14;C marameter for pyridine=6.40, BF; = 1.62 and B(CH3), = 1.70)

question 1 2 3 4 5 6

Answer 1

1) Electronic Configuration of Cyanide ion is σ_1s² > σ_1s^2* > σ_2s² >σ_2s^2* > π_2px² = π_2py² > σ_2pz² > π_2px^* = π_2py^* > σ_2pz^*

Bond Order = 1/2 [N_b -N_a] where N_b is number of electron in bonding molecular orbitals and N_a is number of electron in antibonding molecular orbitals.

Bond order of cyanide ion = 1/2 [10-4] =1/2[6] =3

2p 2p --HOMO 2p 2s N C CN MO diagram for CN

Based on molecular orbital diagram, Cyanide ion is diamagnetic due to no unpaired electron and HOMO of cyanide ion is σ_g

Option b, d, f are correct.

2) diatomic 2p orbitals of a) C₂

8 Py Pz Atomic Orbital of C

8-3 T2px 2py 80-8 O2pz 2py Molecular orbital of C2 8.8

2p 2p 2p C C 2p MO diagram for C2

b) diatomic 2p orbital of O₂

8 Px Ру Р. Atomic Orbital of O 8-3 2pz T2px T2py 8-8 T2px O 2pz T2py Molecular orbital of O 8.8

2p 2p 2p O б2р MO diagram for O2

σ_2p bonding orbital are gerade whereas π_2p bonding orbital are ungerade

σ_2p^* antibonding orbital are ungerade whereas π_2p^* antibonding orbital are gerade

3) a) Electronic Configuration of OF^- is σ_1s² > σ_1s^2* > σ_2s² >σ_2s^2* > π_2px² = π_2py² > σ_2pz² > π_2px^*2 = π_2py^*2 > σ_2pz^*2

Bond Order = 1/2 [N_b -N_a] where N_b is number of electron in bonding molecular orbitals and N_a is number of electron in antibonding molecular orbitals.

Bond order of OF^- = 1/2 [10-10] =1/2[0] =0

b) HOMO is polarized towards fluorine atoms due to electronegativity,

c) σ_2pz^* sigma antibonding energy levels is HOMO

O2p 2p 2p O2p O25 2s 2s OF MO diagram for OF

4) A) HOMO of NO^- have more N character and so hydrogen attack the N atom instead of O atom to form HNO.

H NO

O2p 2p 2p 2p 2s 2s O NO MO diagram for NO

B) NO^- is easily dimerize to N₂O₄^2- with a trans configuration using π^*_g (2p).

5) i) interaction 2p_z orbitals to form molecular orbital of

a) σ bonding orbitals Option d)

2pz
or
O2p

b) σ* antibonding orbitals Option a)

ос уре 2рz
or
Огр

ii) interaction 2p_x orbitals to form molecular orbital of

c) π bonding orbitals Option c)

8-3 2px
or
T2px

d) π* antibonding orbitals Option b)

8-8 2px
or
T2px

6) a) Using Drago's Theory,

-ΔH = E_aE_b + C_aC_b

Where ΔH is enthalpy, E is electrostatic parameter and C is covalent parameter

-ΔH = E_pyE_BF3 +C_pyC_BF3

= (1.17*9.88) + (6.40*1.62)

-ΔH =1.1916

ΔH = -1.1916

Exothermic reaction

b) Using Drago's Theory,

-ΔH = E_aE_b + C_aC_b

Where H is enthalpy, E is electrostatic parameter and C is covalent parameter

-ΔH = E_pyE_BMe3 +C_pyC_BMe3

= (1.17*6.14) + (6.40*1.70)

-ΔH =-3.6962

ΔH = 3.6962

Endothermic reaction