Question

The formation of a double bond in an oxygen molecule is explained below. Arrange the given points in a sequential order.
(a) 2p${}_y$ orbitals of each oxygen atom overlap laterally/sidewise to form a pi bond.
(b) Thus, a double bond between two oxygen atoms in which one p${}_z$ - p${}_z$ sigma bond and p${}_y$ - p${}_y$ $\pi$ bond is formed.
(c) All the three 2p orbitals are perpendicular to each other. Hence 2p${}_z$ orbitals of each oxygen atom overlap end to end to form a sigma bond.
(d) The electronic configuration of oxygen is $O$: 1s${}^2$2s${}^2$2p${}_x^1$2p${}_y^2$2p${}_z^1$

• A. $cadb$
• B. $dcab$
• C. $cdba$
• D. $cdab$

Answer 1

The correct option is B $dcab$

Formation of double bond in $O_2$

Electronic configuration of $O$ : $1s^{2}2s^{2}2p_x^{1}2p_y^{2}2p_z^1$. All $2p_z$ orbitals each of $O$ atoms overlap end to end as all the three $2p$ are perpendicular to each other to form $\sigma$ bond.

$2p_y$ of each $O$ atom overlap sideways to form a $\pi$ bond

Thus double bond between two $O$ atoms in which $p_z-p_z$ forms $\sigma$ bond and $p_y-p_y$ forms $\pi$ bond. $(O=O)$