Question

A magnetic moment of 1.73 BM will be shown by one among the following

• A. [Cu(NH₃)₄]²⁺
• B. [Ni(CN)₄]^2-
• C. TiCl₄
• D. [CoCl₆]^4-

Answer 1

The Correct Option is A

To determine which of the given complexes shows a magnetic moment of 1.73 BM (Bohr Magneton), we need to evaluate each option based on its electronic configuration and potential unpaired electrons.

1. [Cu(NH₃)₄]²⁺:

   • The oxidation state of Cu in this complex is +2.
   • Electronic configuration of Cu: [Ar] 3d¹⁰ 4s¹.
   • Upon losing two electrons (for Cu²⁺), the configuration becomes: [Ar] 3d⁹.
   • This configuration has one unpaired electron.
   • The magnetic moment (\mu) is given by the formula: \mu = \sqrt{n(n+2)}, where n is the number of unpaired electrons.
   • Substitute n=1 into the formula: \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 BM.

   Thus, this complex shows a magnetic moment of 1.73 BM.

2. [Ni(CN)₄]^2-:

   • The oxidation state of Ni in this complex is +2.
   • Electronic configuration of Ni: [Ar] 3d⁸ 4s².
   • For Ni²⁺: [Ar] 3d⁸.
   • In the presence of strong field ligands like CN^-, pairing of electrons occurs, resulting in no unpaired electrons.
   • This complex is diamagnetic with no unpaired electrons, hence zero magnetic moment.

3. TiCl₄:

   • The oxidation state of Ti is +4.
   • Electronic configuration of Ti: [Ar] 3d² 4s².
   • For Ti⁴⁺: [Ar] 3d⁰.
   • No unpaired electrons, hence zero magnetic moment.

4. [CoCl₆]^4-:

   • The oxidation state of Co is +2.
   • Electronic configuration of Co: [Ar] 3d⁷ 4s².
   • For Co²⁺: [Ar] 3d⁷.
   • There are 3 unpaired electrons, leading to a magnetic moment of \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 BM.

Comparing the options, [Cu(NH₃)₄]²⁺ is the correct answer as it shows a magnetic moment of 1.73 BM with one unpaired electron.