Which of the following species has a bond ord | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The bond order is calculated using the formula: $	ext{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$

Vedclass · Accepted Answer

$O_{2}^{-}$

Vedclass · Answer

(B) The bond order is calculated using the formula: $	ext{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.For $O_{2}^{+}$ (total $15$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Bond order $= \frac{10 - 5}{2} = 2.5$.For $O_{2}^{-}$ (total $17$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. Bond order $= \frac{10 - 7}{2} = 1.5$.For $O_{2}^{2-}$ (total $18$ electrons): Bond order $= \frac{10 - 8}{2} = 1.0$.For $O_{2}$ (total $16$ electrons): Bond order $= \frac{10 - 6}{2} = 2.0$.Thus,$O_{2}^{-}$ has a bond order of $1.5$.

Which of the following species has a bond order of $1.5$?

Similar Questions

The $H.O.M.O.$ (Highest Occupied Molecular Orbital) of the $CO$ molecule is:

The correct sequence of bond order is

Which of the following is paramagnetic and contains a three-electron bond in its structure?

Calculate the bond order for $NO, NO^+, CN, CN^-$ and $CO$.

How many ions of the following have a bond order of $2.5 ?$
$N_2^{-} ; NO^{-} ; C_2^{-} ; N_2^{+} ; C_2^{2-} ; CN^{+}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module