Question:
If the bond order of CO is \(x\), the bond order of \(O_2^{2-}\) ion is
If the bond order of CO is \(x\), the bond order of \(O_2^{2-}\) ion is
Show Hint
Using molecular orbital theory:
\[
\text{Bond Order}
=
\frac{\text{Bonding electrons}-\text{Antibonding electrons}}{2}
\]
CO has bond order \(3\), while \(O_2^{2-}\) has bond order \(1\).
Updated On: Jul 18, 2026
- \(x\)
- \(x/2\)
- \(x/3\)
- \(2x\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Calculate the bond order of CO.
Carbon monoxide is isoelectronic with \(N_2\). It contains \[ 14 \] electrons.
Using molecular orbital theory, \[ \text{Bond Order} =\frac{N_b-N_a}{2} \] For CO, \[ \text{Bond Order}=3 \] Hence, \[ x=3 \]
Step 2: Calculate the bond order of \(O_2^{2-}\).
Oxygen molecule has \[ 16 \] electrons. Therefore, \[ O_2^{2-} \] contains \[ 18 \] electrons.
Its molecular orbital configuration is similar to \(F_2\).
Number of bonding electrons: \[ N_b=10 \] Number of antibonding electrons: \[ N_a=6 \] Therefore, \[ \text{Bond Order} = \frac{10-6}{2} = 2 \] Wait, for \(O_2^{2-}\): \[ N_b=10,\qquad N_a=8 \] Thus, \[ \text{Bond Order} = \frac{10-8}{2} = 1 \]
Step 3: Express bond order in terms of \(x\).
Since \[ x=3 \] and \[ \text{Bond Order of }O_2^{2-}=1, \] we get \[ 1=\frac{3}{3} \] Therefore, \[ \text{Bond Order of }O_2^{2-} = \frac{x}{3} \]
Step 4: Verify the option.
The obtained expression matches option (3).
Step 5: Final conclusion.
Hence, \[ \boxed{\frac{x}{3}} \] is the correct answer.
Carbon monoxide is isoelectronic with \(N_2\). It contains \[ 14 \] electrons.
Using molecular orbital theory, \[ \text{Bond Order} =\frac{N_b-N_a}{2} \] For CO, \[ \text{Bond Order}=3 \] Hence, \[ x=3 \]
Step 2: Calculate the bond order of \(O_2^{2-}\).
Oxygen molecule has \[ 16 \] electrons. Therefore, \[ O_2^{2-} \] contains \[ 18 \] electrons.
Its molecular orbital configuration is similar to \(F_2\).
Number of bonding electrons: \[ N_b=10 \] Number of antibonding electrons: \[ N_a=6 \] Therefore, \[ \text{Bond Order} = \frac{10-6}{2} = 2 \] Wait, for \(O_2^{2-}\): \[ N_b=10,\qquad N_a=8 \] Thus, \[ \text{Bond Order} = \frac{10-8}{2} = 1 \]
Step 3: Express bond order in terms of \(x\).
Since \[ x=3 \] and \[ \text{Bond Order of }O_2^{2-}=1, \] we get \[ 1=\frac{3}{3} \] Therefore, \[ \text{Bond Order of }O_2^{2-} = \frac{x}{3} \]
Step 4: Verify the option.
The obtained expression matches option (3).
Step 5: Final conclusion.
Hence, \[ \boxed{\frac{x}{3}} \] is the correct answer.
Was this answer helpful?
0
0
Top AP EAPCET Chemistry Questions
- The four quantum numbers of the valence electron of potassium are :
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
- In which of the following the election gain enthalpy of elements is correctly arranged?
- The radii of $ \text{F,}{{\text{F}}^{-}}\text{O} $ and $ {{\text{O}}^{2-}} $ are in the order
- Which of the following has the highest first ionization potential?
View More Questions
Top AP EAPCET Molecular Stability Questions
- Arrange the following free radicals in the correct order of their stability:
- Arrange the following species in the decreasing order of their bond orders:
- Arrange the following in the correct order of their bond orders.
\[ \mathrm{N_2}\quad (I), \qquad \mathrm{O_2}\quad (II), \qquad \mathrm{O_2^{+}}\quad (III), \qquad \mathrm{O_2^{-}}\quad (IV) \] - The bond order values of anions of sodium peroxide and potassium superoxide are respectively:
- As per molecular orbital theory, the pair of molecules which do not exist is
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions