Chemical Bonding And Molecular Structure Question 1
Question 1 - 24 January - Shift 2
What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species: $N_2: N_2^{+} ; O_2 ; O_2^{+}$?
(1) $0,1,2,1$
(2) $2,1,2,1$
(3) $0,1,0,1$
(4) $2,1,0,1$
Show Answer
Answer: (1)
Solution:
Formula: Energy level diagram
$N_2$ $\sigma 1s^{2} \sigma^{* } 1s^{2} \sigma 2s^{2} \sigma^{*} 2s^{2} \pi 2p_x^{2}=\pi 2p_y^{2} \frac{\sigma 2p_z^{2}}{HOMO}$
$N_2^{+}-\sigma 1s^{2} \sigma^{* } 1s^{2} \sigma^{} 2s^{2} \sigma^{*} 2s^{2} \pi 2p_x^{2}=\pi 2p_y^{2} \frac{\sigma 2p_z^{1}}{HOMO}$
$O_2-\sigma 1s^{2} \sigma^{* } 1s^{2} \sigma 2s^{2} \sigma^{* } 2s^{2} \sigma 2p_z^{2}$ $\pi 2 p_x^{2}=\pi 2 p_y^{2}$ $\pi^{} 2p_x^{1}=\pi^{} 2p_y^{1}$ (HOMO)
$O_2^{+}-\sigma 1s^{2} \sigma^{} 1s^{2} \sigma 2s^{2} \sigma^{} 2s^{2} \sigma 2p_z^{2} \pi 2p_x^{2}=\pi 2p_y^{2}$ $\pi^{} 2p_x^{1}=\pi^{} 2p_y^{0}$ (HOMO)
$N_2 \Rightarrow 0$ unpaired $e^{-}$in $HOMO$
$N_2^{+} \Rightarrow 1$ unpaired $e^{-}$in HOMO
$O_2 \Rightarrow 2$ unpaired $e^{-}$in HOMO
$O_2^{+} \Rightarrow 1$ unpaired $e^{-}$in $HOMO$
