Numerical/Formula-based

1. TITLE

How to solve Bond Order Calculation in NEET — Step-by-step

2. PROBLEM TYPE INTRO 

This is a Molecular Orbital Theory (MOT) numerical. To solve this, you need the formula for Bond Order (B.O.) based on the number of bonding ($N_b$) and antibonding ($N_a$) electrons. A typical mistake students make is miscounting the total electrons when ions (positive or negative charges) are involved, or confusing the energy sequence of orbitals for molecules with more than 14 electrons.

3. MCQ 

Problem:
Calculate the bond order for the dioxygen cation ($O_2^+$).
Given: Atomic number of Oxygen ($Z$) = 8.

A) 1.5
B) 2.0
C) 2.5
D) 3.5

4. STEP-BY-STEP SOLUTION

SOLUTION

Given:
* Species: $O_2^+$
* Atomic number of Oxygen ($Z$): 8
* Total electrons in $O_2$ neutral molecule: $8 + 8 = 16$
* Charge: $+1$ (indicates loss of 1 electron)
* Total electrons in species: $16 - 1 = 15$ electrons

To Find:
* Bond Order (B.O.)

Formula:

$$\text{Bond Order} = \frac{1}{2}(N_b - N_a)$$


[Citation: 94]
Use this formula after determining the electronic configuration according to Molecular Orbital Theory.

 

Step 1: Write the Electronic Configuration
For molecules with more than 14 electrons (like $O_2, F_2$), the energy order is:

$\sigma 1s < \sigma^ 1s < \sigma 2s < \sigma^ 2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^ 2p_x = \pi^ 2p_y) < \sigma^* 2p_z$.

Filling 15 electrons:
1. $\sigma 1s^2$ (2)
2.  $\sigma^ 1s^2$ (2)
3. $\sigma 2s^2$ (2)
4. $\sigma^ 2s^2$ (2)
5. $\sigma 2p_z^2$ (2)
6. $\pi 2p_x^2 = \pi 2p_y^2$ (4)
7.  $\pi^ 2p_x^1$ (1)
(Total = 15 electrons)*

Step 2: Count Bonding and Antibonding Electrons
* Bonding electrons ($N_b$): Electrons in orbitals without a star ().
$N_b = 2 (\sigma 1s) + 2 (\sigma 2s) + 2 (\sigma 2p_z) + 4 (\pi 2p) = \mathbf{10}$
* Antibonding electrons ($N_a$): Electrons in orbitals with a star ().

$N_a = 2 (\sigma^ 1s) + 2 (\sigma^ 2s) + 1 (\pi^* 2p) = \mathbf{5}$

Step 3: Calculate Bond Order

$$B.O. = \frac{1}{2}(10 - 5)$$

 

$$B.O. = \frac{1}{2}(5)$$

 

$$B.O. = 2.5$$

 

Answer: 2.5 (Option C)

5. WHERE STUDENTS GO WRONG

  • ❌ Mistake 1: Ignoring the charge. Students often calculate for neutral $O_2$ (16 electrons), getting a B.O. of 2.0.
  • ❌ Mistake 2: Mixing up electron counts. Swapping $N_b$ and $N_a$ leads to a negative bond order, which indicates an unstable molecule.
  • ❌ Mistake 3: Wrong Orbital Sequence. For molecules with $\le 14$ electrons (like $N_2$), the $\sigma 2p_z$ energy is higher than $\pi 2p$. While this often yields the same B.O. number, it causes errors in predicting magnetic nature.

6. QUICK FORMULA REFERENCE

Formulas Used:

  1. Bond Order: $\text{B.O.} = \frac{1}{2}(N_b - N_a)$ — Use for all MOT calculations.
  2. Stability Check: If $N_b > N_a$, the molecule is stable.
  3. Shortcut (14-Electron Rule):
    • 14 electrons ($N_2$) $\rightarrow$ B.O. = 3.0
    • For every 1 electron added or removed, subtract 0.5 from the B.O.
    • 15 electrons ($O_2^+$) $\rightarrow$ $3.0 - 0.5 = 2.5$.

7. SIMILAR PROBLEM HINT

"If NEET changes the values, the method stays the same. Practice with: $O_2^-$ (Superoxide ion). It has 17 electrons ($16+1$). Using the shortcut: 14e (3.0) $\rightarrow$ 17e (diff of 3). $3.0 - 1.5 = 1.5$."

8. NCERT REFERENCE

Based on: NCERT Class 11, Chapter 4, Section 4.7.5 and Example of Oxygen Molecule ($O_2$).

9. CTA

More solved numericals → Chemical Bonding Complete Guide
Practice problems → Mock exams