The pair having the same magnetic moment is :
[At. No. : Cr=24, Mn=25, Fe=26, Co=27]

 

  • Option 1)

     [Cr(H2O)6]2+ and [CoCl4]2−

     

  • Option 2)

     [Cr(H2O)6]2+ and [Fe(H2O)6]2+

     

  • Option 3)

     [Mn(H2O)6]2+ and [Cr(H2O)6]2+

     

  • Option 4)

     [CoCl4]2− and [Fe(H2O)6]2+

     

 

Answers (1)

As we learnt in 

Pauli Exclusion Principle -

“No two electrons in an atom can have the same set of four quantum number" .

-

 

 and

Hund’s Rule of Maximum Multiplicity -

"pairing of electrons in the orbitals belonging to the same subshell ( p,d or f ) does not take place until each orbital belonging to that subshell has got one electron each i.e.,It is singly occupied.”

-

 

 magnetic moment \mu = \sqrt {n \left ( n+2 \right )} where n is the number of unpaired electrons.

In given problem Cr^{2+} and Fe^{2+} has same number of unpaired electrons

Cr^{2+} => 3d^{4} four unpaired electrons

Fe^{2+} => 3d^{6} four unpaired electrons

Correct answer is 2

 

 


Option 1)

 [Cr(H2O)6]2+ and [CoCl4]2−

 

Incorrect option

Option 2)

 [Cr(H2O)6]2+ and [Fe(H2O)6]2+

 

Correct option

Option 3)

 [Mn(H2O)6]2+ and [Cr(H2O)6]2+

 

Incorrect option

Option 4)

 [CoCl4]2− and [Fe(H2O)6]2+

 

Incorrect option

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Knockout JEE Main January 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions