Question

By constructing a molecular orbital picture for each of the following molecules, determine whether it is paramagnetic or diamagnetic.  2 B ;  2 O ; 2 2 N ; CO; 2 2 C

Answer 1

your complete question is -----> By constructing a molecular orbital picture for each of the following molecules, determine whether it is paramagnetic or diamagnetic ?

(a.) B2

(b.) C2

(c.) N2

(d.) NO

(e) CO

solution : we know, Paramagnetic materials, those with unpaired electrons, are attracted by magnetic fields whereas diamagnetic materials, those with no unpaired electrons, are weakly repelled by such fields.

now come to the question !

a. B2 : configuration of B2 = $\sigma(\mathrm{1s})^2, \sigma^*(\mathrm{1s})^2, \sigma(\mathrm{2s})^2, \sigma^*(\mathrm{2s})^2,\pi(\mathrm{2p}_x)^1 \approx\pi(\mathrm{2p}_y)^1$ it has two unpaired electrons one in each of its p orbitals, so, it is paramagnetic.

b. C2 : configuration of C2 =$\sigma(\mathrm{1s})^2, \sigma^*(\mathrm{1s})^2, \sigma(\mathrm{2s})^2, \sigma^*(\mathrm{2s})^2,\pi(\mathrm{2p}_x)^2 \approx\pi(\mathrm{2p}_y)^2$ it has paired electrons so, it is diamagnetic in nature.

c. O2 : configuration of O2 = $\sigma(\mathrm{1s})^2, \sigma^*(\mathrm{1s})^2, \sigma(\mathrm{2s})^2, \sigma^*(\mathrm{2s})^2,\sigma(\mathrm{2p}_z)^2,\pi(\mathrm{2p}_x)^1 \approx\pi(\mathrm{2p}_y)^1$ it has twounpaired electrons, one in each of its p* orbitals so, it is paramagnetic.

d. NO : configuration of NO =$\sigma(\mathrm{1s})^2, \sigma^*(\mathrm{1s})^2, \sigma(\mathrm{2s})^2, \sigma^*(\mathrm{2s})^2,\sigma(\mathrm{2p}_z)^2,\pi(\mathrm{2p}_x)^1 \approx\pi(\mathrm{2p}_y)^0$ it has an odd number of electrons and, therefore, must be paramagnetic.

e. CO : configuration of CO =$\sigma(\mathrm{1s})^2, \sigma^*(\mathrm{1s})^2, \sigma(\mathrm{2s})^2, \sigma^*(\mathrm{2s})^2,\sigma(\mathrm{2p}_z)^2,\pi(\mathrm{2p}_x)^2 \approx\pi(\mathrm{2p}_y)^2$ is diamagnetic because all of its electrons are paired.