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The number of unpaired electrons in a paramagnetic diatomic molecule of an element with atomic number 16 is :-
A
2
B
3
C
4
D
1
Detailed Solution
The element with atomic number 16 is sulphur. Its diatomic molecule is $S_2$.
Sulphur belongs to the same group as oxygen, so the molecular orbital configuration of $S_2$ in its valence shell is like that of $O_2$.
In $O_2$ the last two electrons enter the two degenerate antibonding orbitals $\pi^*2p_x$ and $\pi^*2p_y$ singly (Hund's rule).
Similarly, in $S_2$ the last two electrons occupy the two degenerate $\pi^*3p$ orbitals singly: $\pi^*3p_x^1 = \pi^*3p_y^1$.
So $S_2$ has two unpaired electrons according to molecular orbital theory, which makes it paramagnetic like $O_2$.
The number of unpaired electrons is 2.
Sulphur belongs to the same group as oxygen, so the molecular orbital configuration of $S_2$ in its valence shell is like that of $O_2$.
In $O_2$ the last two electrons enter the two degenerate antibonding orbitals $\pi^*2p_x$ and $\pi^*2p_y$ singly (Hund's rule).
Similarly, in $S_2$ the last two electrons occupy the two degenerate $\pi^*3p$ orbitals singly: $\pi^*3p_x^1 = \pi^*3p_y^1$.
So $S_2$ has two unpaired electrons according to molecular orbital theory, which makes it paramagnetic like $O_2$.
The number of unpaired electrons is 2.