Question

which of the following complex show magnetic moment (spin only) = 5.91 BM

Answer 1

which of the following complex show magnetic moment (spin only) = 5.91 BM

1. [Ni(CO)₄]
2. [FeF6]³¯
3. [Fe(CN)6]³¯
4. [Cr(H₂O)6]³⁺

solution : The metal of coordination compound is generally from 3d transition series so that magnetic moment of such compound can be calculated using the formula,

$\quad \mu=\sqrt{n(n+2)}$ BM

where n is number of unpaired electrons in the central metal atom.

so first of all we have to find configuration of central metal atom of each complex and then using crystal field theory to find no of unpaired electrons.

1. [Ni(CO)₄] ⇒3d¹⁰ so no of unpaired electrons = 0

therefore μ = 0 BM

2. [FeF6]³¯ ⇒3d^5 no of unpaired electrons = 0

therefore μ = √5(5+2) = √35 = 5.91 BM

3. [Fe(CN)6]³¯ ⇒3d^5 no of unpaired electrons = 1

therefore μ = √1(1+2) = √3 ≈ 1.732 BM

4. [Cr(H₂O)6]³⁺ ⇒3d³ no of unpaired electrons = 5

therefore μ = √3(3+2) = √15 = 3.83 BM

Therefore the correct option is 2. [FeF6]³¯

Answer 2

Question :

Which of the following complex show magnetic moment (spin only) = 5.91 BM

1) [Ni(CO)₄]

2) [FeF6]³¯

3) [Fe(CN)6]³¯

4) [Cr(H₂O)6]³⁺

Theory :

The magnetic moment is determined by the no of unpaired electrons and is calculated by spin-only formula :

$\tt\:\mu=\sqrt{n(n+2)}$

Where, n is the no if unpaired electrons

Solution :

1) [Ni(CO)₄]

Here , oxidation state of Ni is 0

$\sf\:Ni=3d^8\:4s^2$

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No of unpaired electrons= 0

Thus , $\sf\:\mu=\sqrt{n(n+2)}$

$\sf=0\:BM$

2) [FeF6]³¯

Here , oxidation state of Fe is

x-6= -3

x=-3+6= 3

Thus , oxidation state is 3

$\sf\:Fe^{+3}=3d^5$

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No of unpaired electrons= 5

Thus , $\sf\:\mu=\sqrt{n(n+2)}$

$\sf=\sqrt{5(2+5)}$

$\sf=\sqrt{35}=5.91BM$

3) [Fe(CN)6]³¯

Here , oxidation state of Fe is

x-6= -3

x=-3+6= 3

Thus, oxidation state of Fe is 3

$\sf\:Fe^{+3}=3d^5$

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No of unpaired electrons= 1

Thus , $\sf\:\mu=\sqrt{n(n+2)}$

$\sf=\sqrt{1(1+2)}$

$\sf=\sqrt{3}=1.7BM$

4) [Cr(H₂O)6]³⁺

Here , Oxidation state of Cr is 3

$\sf\:Cr^3=3d^3$

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No of unpaired electrons= 3

Thus , $\sf\:\mu=\sqrt{n(n+2)}$

$\sf=\sqrt{3(3+2)}$

$\sf=\sqrt{15}=3.87\:BM$

Hence , The correct option is 2)

Magnetic moment of [FeF6]³¯ is 5.91 BM