Coordination Compounds 2 Question 60

62. For the octahedral complexes of $Fe^{3+}$ in $SCN^{-}$ (thiocyanato-S) and in $CN^{-}$ ligand environments, the difference between the spin only magnetic moments in Bohr magnetons (when approximated to the nearest integer) is [atomic number of $Fe=26$ ]

(2015 Adv.)

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Answer:

Correct Answer: 62. $(4)$

Solution:

  1. When $S$ is donor atom of $SCN^{-}$, it produces weak ligand field and forms high spin complex as

$\left[Fe(SCN) _6\right]^{3-}: Fe^{3+}\left(3 d^{5}\right)=$

<img src=“https://temp-public-img-folder.s3.amazonaws.com/sathee.prutor.images/sathee_image/cropped_2024_01_16_b4fdca9f34924034e8d8g-271_jpg_height_118_width_577_top_left_y_1818_top_left_x_1307.jpg"width="400">

Spin only magnetic moment $\left(\mu _s\right)=\sqrt{5(5+2)} BM=\sqrt{35} $ $BM$

In case of $CN^{-}$ligand, carbon is the donor atom, it produces strong ligand field and forms low spin complex as

$\left[Fe(CN) _6\right]^{3-}: Fe^{3+}\left(3 d^{5}\right)$

<img src=“https://temp-public-img-folder.s3.amazonaws.com/sathee.prutor.images/sathee_image/cropped_2024_01_16_b4fdca9f34924034e8d8g-271_jpg_height_117_width_395_top_left_y_2137_top_left_x_1398.jpg"width="300">

Spin only magnetic moment $\left(\mu _s\right)=\sqrt{1(1+2)} BM=\sqrt{3} $ $BM$

Hence, difference in spin only magnetic moment

$ =\sqrt{35}-\sqrt{3} \approx 4$ $ BM $