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%

High-Yield Trend

Chapter Questions
4 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Molecular Orbital Theory

High-Yield Trend

Chapter Questions 4 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Chapter Questions
4 MCQs