Molecular Orbital Theory

For

O%20_%7B2%7D

molecule, electronic configuration is

%5Csigma%5C%2C%201%20s%5E%7B2%7D%2C%20%5Coverset%7B*%7D%7B%5Csigma%7D%201%20s%5E%7B2%7D%2C%20%5Csigma%202%20s%5E%7B2%7D%2C%20%5Coverset%7B*%7D%7B%5Csigma%7D%202%20s%5E%7B2%7D%2C%20%5Csigma%202%20p_%7Bz%7D%5E%7B2%7D%2C%20%5Cpi%202%20p_%7Bx%7D%5E%7B2%7D%20%5Capprox%20%5Cpi%202%20p_%7By%7D%5E%7B2%7D%2C%20%5Coverset%7B*%7D%7B%5Cpi%20%7D2%20p_%7Bx%7D%5E%7B1%7D%3D%5Coverset%7B*%7D%7B%5Cpi%7D%202%20p_%7By%7D%5E%7B1%7D

%5Ctherefore

Bond order

%3D%5Cfrac%7B10-6%7D%7B2%7D%3D2

For

O%20_%7B2%7D%5E%7B%2B%7D

molecule, an electron is removed from

%5Coverset%7B*%7D%7B%5Cpi%20%7D2%20p_%7By%7D

orbital.

%5Ctherefore

Bond order

%3D%5Cfrac%7B10-5%7D%7B2%7D%3D2.5

For

O%20_%7B2%7D%5E%7B-%7D

molecule, an electron is added to

%5Coverset%7B*%7D%7B%5Cpi%7D%202%20p_%7Bx%7D%5E%7B1%7D

orbital.

%5Ctherefore

Bond order

%3D%5Cfrac%7B10-7%7D%7B2%7D%3D1.5

For

O%20_%7B2%7D%5E%7B2-%7D

molecule, two electrons are added to

%5Coverset%7B*%7D%7B%5Cpi%7D%202%20p_%7Bx%7D%5E%7B1%7D

and

%5Coverset%7B*%7D%7B%5Cpi%7D%202%20p_%7By%7D%5E%7B1%7D

orbitals.

%5Ctherefore

Bond order

%3D%5Cfrac%7B10-8%7D%7B2%7D%3D1

So, the order is :

$O _{2}^{2-}< \,O _{2}^{-}

To determine the percentage of s-character in the hybrid orbitals, we must analyze the type of hybridization for each molecule.
1. For $%5Ctext%7BNO%7D_2%5E%2B$ : $%5Ctext%7BNO%7D_2%5E%2B$ has a linear geometry with nitrogen having 2 bonding pairs and no lone pairs of electrons, implying $sp$ hybridization. In $sp$ hybridization, the s-character is 50%.
2. For $%5Ctext%7BNO%7D_3%5E-$ : $%5Ctext%7BNO%7D_3%5E-$ has a trigonal planar geometry with nitrogen having 3 bonding pairs and no lone pairs, implying $sp%5E2$ hybridization. In $sp%5E2$ hybridization, the s-character is 33.3%.
3. For $%5Ctext%7BNH%7D_4%5E%2B$ : $%5Ctext%7BNH%7D_4%5E%2B$ has a tetrahedral geometry with nitrogen having 4 bonding pairs and no lone pairs, implying $sp%5E3$ hybridization. In $sp%5E3$ hybridization, the s-character is 25%.
Thus, the percentage of s-character in the hybrid orbitals of nitrogen in $%5Ctext%7BNO%7D_2%5E%2B$ , $%5Ctext%7BNO%7D_3%5E-$ , and $%5Ctext%7BNH%7D_4%5E%2B$ are 50%, 33.3%, and 25%, respectively.

The correct option is (D): 50%, 33.3%, 25%

About Molecular Orbital Theory - KCET

Molecular Orbital Theory is a vital chapter for KCET aspirants. Mastering the concepts covered in this chapter is essential for securing a top rank.

By rigorously practicing the previous year questions associated with this chapter, you can identify high-yield topics, understand the examiner's perspective, and boost your confidence during the actual exam.

Frequently Asked Questions

Why focus on Molecular Orbital Theory PYQs?

Analyzing PYQs for this specific chapter reveals the most frequently tested concepts and the typical complexity of questions, allowing you to tailor your study plan efficiently.

How to best use this analysis?

Review the topic breakdown to see which sub-topics within Molecular Orbital Theory carry the most weight. Then, tackle the questions iteratively to solidify your understanding.

High-Yield Trend

Chapter Questions
4 MCQs

Official Solution

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About Molecular Orbital Theory - KCET

Frequently Asked Questions

Why focus on Molecular Orbital Theory PYQs?

How to best use this analysis?

Molecular Orbital Theory

High-Yield Trend

Chapter Questions 4 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

About Molecular Orbital Theory - KCET

Frequently Asked Questions

Why focus on Molecular Orbital Theory PYQs?

How to best use this analysis?

Chapter Questions
4 MCQs