Correct decreasing order of bond order of oxy | vedclass

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

Question

(A) The bond order of oxygen species is determined by the number of electrons in the molecular orbitals:$(I)$ $Na_2O_2$ contains the peroxide ion $O_2

Vedclass · Accepted Answer

$(II) > (IV) > (III) > (I)$

Vedclass · Answer

(A) The bond order of oxygen species is determined by the number of electrons in the molecular orbitals:$(I)$ $Na_2O_2$ contains the peroxide ion $O_2^{2-}$. The number of electrons is $18$. Bond order $= \frac{1}{2}(10 - 8) = 1.0$.$(II)$ $O_2[AsF_6]$ contains the dioxygenyl cation $O_2^+$. The number of electrons is $15$. Bond order $= \frac{1}{2}(10 - 5) = 2.5$.$(III)$ $CsO_2$ contains the superoxide ion $O_2^-$. The number of electrons is $17$. Bond order $= \frac{1}{2}(10 - 7) = 1.5$.$(IV)$ $O_2$ is the neutral dioxygen molecule. The number of electrons is $16$. Bond order $= \frac{1}{2}(10 - 6) = 2.0$.Comparing the bond orders: $2.5 (II) > 2.0 (IV) > 1.5 (III) > 1.0 (I)$.Thus,the correct decreasing order is $(II) > (IV) > (III) > (I)$.

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

Similar Questions

According to $MO$ theory,the bond orders for $O_2^{2-}$,$CO$,and $NO^{+}$ respectively,are:

According to molecular orbital theory,the number of unpaired electron$(s)$ in $O_{2}^{2-}$ is :

Which of the following has a three-electron bond in its structure?

$N_2$ and $O_2$ are converted to monocations $N_2^+$ and $O_2^+$ respectively. Which statement is wrong?

The number of antibonding electron pairs in $O_{2}^{2-}$ molecular ion on the basis of molecular orbital theory is (Atomic number of $O$ is $8$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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)$ $O_2$