The bond order in O_2^{2-} is | vedclass

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

Question

(D) Total number of electrons in $O_2^{2-} = 18$.The molecular orbital electronic configuration of $O_2^{2-}$ is:$\sigma 1s^2, \sigma^* 1s^2, \sigma 2

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) Total number of electrons in $O_2^{2-} = 18$.The molecular orbital electronic configuration of $O_2^{2-}$ is:$\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$.Bond order is calculated using the formula: $BO = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.Here,$N_b = 10$ and $N_a = 8$.Therefore,$BO = \frac{10 - 8}{2} = 1$.

The bond order in $O_2^{2-}$ is

Similar Questions

The bond order of $He_2^+$ molecule ion is

Which of the following is paramagnetic as well as it has a fractional bond order?

If $x-axis$ is the molecular axis and $p_z$ orbitals participate in bonding,then which one of the following will be the nodal plane present within the $\pi-$ bond of the $O_2$ molecule?

Correct decreasing order of bond order of oxygen species in $I$ to $IV$:
$(I)$ $Na_2O_2$
$(II)$ $O_2[AsF_6]$
$(III)$ $CsO_2$
$(IV)$ $O_2$

The sum of bond order values of $C_2$ and $O_2^{2+}$ is $x$,which is equal to the sum of bond order values of $a, b$ and $c$. What are $a, b$ and $c$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module