Coordination Compounds 2 Question 4

4. The crystal field stabilisation energy (CFSE) of $\left[Fe\left(H _2 O\right) _6\right] Cl _2$ and $K _2\left[NiCl _4\right]$, respectively, are

(2019 Main, 10 April II)

(a) $-0.4 \Delta _o$ and $-1.2 \Delta _t$

(b) $-0.4 \Delta _o$ and $-0.8 \Delta _t$

(c) $-2.4 \Delta _o$ and $-1.2 \Delta _t$

(d) $-0.6 \Delta _o$ and $-0.8 \Delta _t$

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Answer:

Correct Answer: 4. (b)

Solution:

  1. Key Idea: Crystal field stabilisation energy (CFSE) for octahedral complexes $=(-0.4 x+0.6 y) \Delta _o$

where, $x=$ number of electrons occupying $t _{2 g}$ orbital.

$y=$ number of electrons occupying $e _g$ orbital.

CFSE for tetrahedral complexes

$=(-0.6 x+0.4 y) \Delta _t$

where, $x=$ number of electrons occupying $e$ orbital.

$y=$ number of electrons occupying $t$ orbital.

In $\left[Fe\left(H _2 O\right) _6\right] Cl _2, H _2 O$ is a weak field ligand, so it is a high spin (outer orbital) octahedral complex of $Fe^{2+}$.

<img src=“https://temp-public-img-folder.s3.amazonaws.com/sathee.prutor.images/sathee_image/snip_images_QOhq0J_gUWFjY1IgivsFKMnl2AcFhkxZc1QB4JvYjj0_original_fullsize_png.jpg"width="300"/>

$\therefore CFSE=(-0.4 x+0.6 y) \Delta _o$

$ =[-0.4 \times 4+0.6 \times 2] \Delta _o=-0.4 \Delta _o $

In $K _2\left[NiCl _4\right], Cl^{-}$is a weak field ligand, so it is a high spin tetrahedral complex of $Ni^{2+}$.

<img src=“https://temp-public-img-folder.s3.amazonaws.com/sathee.prutor.images/sathee_image/cropped_2024_01_16_b4fdca9f34924034e8d8g-263_jpg_height_126_width_327_top_left_y_2092_top_left_x_528.jpg"width="260">

$\therefore CFSE=(-0.6 \times 4+0.4 \times 4) \Delta _t=-0.8 \Delta _t$