Question

The calculated spin only magnetic moments of [Cr(NH₃)₆]³⁺ and [CuF₆]³⁻ in B.M., respectively are ______.

[Atomic numbers of Cr and Cu are 24 and 29 respectively]

पर्याय

• 3.87 and 2.84

• 4.90 and 1.73

• 3.87 and 1.73

• 4.90 and 2.84

Answer 1

The calculated spin only magnetic moments of [Cr(NH₃)₆]³⁺ and [CuF₆]³⁻ in B.M., respectively are 3.87 and 2.84.

Explanation:

To calculate the spin-only magnetic moment for the given complexes [Cr(NH₃)₆]³⁺ and [CuF₆]³⁻, we use the formula for the spin-only magnetic moment:

`mu_"spin" = sqrt(n(n + 2)) BM`

For [Cr(NH₃)₆]³⁺

Cr³⁺ = [Ar] 3d⁵ 

It has 3 unpaired electrons.

`mu_"spin" = sqrt(3(3 + 2)) BM`

= `sqrt(3(5)) BM`

= `sqrt(15) BM`

= 3.87 BM

For [CuF₆]³⁻ 

Cu³⁺ = [Ar] 3d⁸ 

It has two unpaired electrons.

`mu_"spin" = sqrt(2(2 + 2)) BM`

= `sqrt(2(4)) BM`

= `sqrt(8) BM`

= 2.84 BM