Question

Which of the following molecule has a double bond?

• A. $H_2$
• B. $F_2$
• C. $C{l}_2$
• D. $O_2$

Answer 1

Answer: (D)

$O_2$

The electronic configuration of hydrogen is $1s^1$
Diatomic molecule of hydrogen $\left(H_2\right)$ is formed by s-s overlap leads to one $\sigma -bond$.
(b) The electronic configuration of fluorine is $1s^{2}2s^{2}2p^5$
Diatomic molecule of fluorine $\left(F_2\right)$ is formed by p - p overlap leads to one $\sigma -bond$.
(c) The electronic configuration of Chlorine is $1s^{2}2s^{2}2p^{6}3s^{2}3p^5$. Diatomic molecule of chlorine $\left(C{l}_2\right)$ is formed bu p - p overlap leads to one $\sigma -bond$.
(d) The electronic configuration of oxygen is $1s^{2}2s^{2}2p^4$.
Diatomic molecule of oxygen $\left(O_2\right)$ is formed by one $p_z-p_z$ overlap leads to one $\sigma -bond$ and side on overlap of $p_y-p_y$ leads to $\pi -bond$.
So, $O_2$ contains a double bond.