Question

Draw the molecular orbital diagram shown to determine which of the following is paramagnetic.
N22+
B22+
B22-
B2
C22-

Answer 1

Electronic configuration of nitrogen is as follows N (7): 1s'2s 2p Nitrogen has five valence electrons, two electrons in 2s orbital and remaining 3 are present in 2p orbitals. Thus, N, molecule has 10 valence electrons. In N,2ion, two electrons will be removed from N, molecule Thus, total number of valence electrons in the ion will be 8. Fill these electrons in the molecular orbitals that are shown in the following MO diagram by following Hund's rule. Hund's rule states that if two or more orbitals of equal energy are available, electrons will occupy them singly before pairing. MO diagram ofN2 is as follows: σ2px π2py 1-4-t 2 p -. . o2p, .. T2p n2p

σ2s 2s o2s Molecular orbitals of Atomic orbitals of N Atomic orbitals of N N22- ar orbitals have paired electrons in all the orbitals. Hence, N"is diamagnetic in nature

Write the electronic confoguraion of boron as shown below. B(5): 1s2 2s2 2p Boron has 3 valence electrons. In B2 molecule total 6 valence clectrons are present. In the formation of B2 2 valence electrons will be removed. Thus, B22+ ion has 4 valence electrons. In the formation of B2 2 valence electrons will be added to the molecular orbitals. Thus, Bion has 8 valence electrons Fill these electrons in the molecular orbitals that are shown in the following MO diagram by following Hunds's rule. Hund's rule states that if two or more orbitals of equal energy are available, electrons will occupy them singly before pairing. Draw the MO diagram for valence orbitals. MO diagram of B2 is as follows T2py t2pz σ2px σ2s 2s

Atomic orbitals of B Molecular orbitals of Atomic orbitals of B B2 The molecular orbitals have two unpaired electrons in π2P. and π2P. molecular orbitals. Presence of unpaired electrons makes it paramagnetic.

MO diagram of B2 is as follows: σ2px π2p n2pz σ2s Zs 2s ơ2s Molecular orbitals of Atomic orbitals of B Atomic orbitals of B The molecular orbitals have no unpaired electrons. Thus, the ion will be diamagnetic. Therefore, B,* is diamagnetic in nature.

MO diagram of B2is as follows: o2p T2py σ2px π2py Molecular orbitals of 4 2s Atomic orbitals of B Atomic orbitals of B Molecular orbitals of The molecular orbitals have no unpaired electrons. All the electrons are paired in the MOT. Hence, the ion is diamagnetic

C(6): 1s 2s2 2p2 Carbon has 4 valence electrons. In C2 molecule total 8 valence electrons are present. In the formation of C2 2 valence electrons will be added to the MO orbitals. Thus, C22ion has 10 valence electrons MO diagram of Cis as follows 6'2ps σ2px 2s 2s Atomic orbitals of C Molecular orbitals of Atomic orbitals of C 2- The molecular orbitals have no unpaired electrons. Therefore, C2 is diamagnetic