Coordination Compounds 2 Question 31

33. $NiCl _2\{P\left(C _2 H _5\right) _2\left(C _6 H _5\right) \} _2$ exhibits temperature dependent magnetic behaviour (paramagnetic/diamagnetic) the coordination geometries of $Ni^{2+}$ in the paramagnetic and diamagnetic states respectively, are

(2012)

(a) tetrahedral and tetrahedral

(b) square planar and square planar

(c) tetrahedral and square planar

(d) square planar and tetrahedral

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Answer:

Correct Answer: 33. (c)

Solution:

  1. In the given complex, $NiCl _2\{P\left(C _2 H _5\right) _2\left(C _6 H _5\right) \} _2$ nickel is in $+2$ oxidation state and the ground state electronic configuration of $Ni^{2+}$ ions in free gaseous state is

<img src=“https://temp-public-img-folder.s3.amazonaws.com/sathee.prutor.images/sathee_image/cropped_2024_01_16_b4fdca9f34924034e8d8g-267_jpg_height_112_width_697_top_left_y_1758_top_left_x_1253.jpg"width="400">

For the given four coordinated complex to be paramagnetic, it must possess unpaired electrons in the valence shell. To satisfy this condition, four lone pairs from the four ligands occupies the four $s p^{3}$-hybrid orbitals as :

<img src=“https://temp-public-img-folder.s3.amazonaws.com/sathee.prutor.images/sathee_image/cropped_2024_01_16_b4fdca9f34924034e8d8g-267_jpg_height_123_width_675_top_left_y_2045_top_left_x_1281.jpg"width="400">

Therefore, geometry of paramagnetic complex must be tetrahedral. On the otherhand, for complex to be diamagnetic, there should not be any unpaired electrons in the valence shell.

This condition can be fulfilled by pairing electrons of $3 d$-orbitals against Hund’s rule as

<img src=“https://temp-public-img-folder.s3.amazonaws.com/sathee.prutor.images/sathee_image/cropped_2024_01_16_b4fdca9f34924034e8d8g-267_jpg_height_126_width_686_top_left_y_2385_top_left_x_1264.jpg"width="400">

The above electronic arrangement gives $d s p^{2}$-hybridisation and therefore, square planar geometry to the complex.