IIT-JEE 2012 · Paper 2 · Chemistry Q21

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ChemistrySingle correct+3 / -1Question 21 of 60

$\ce{NiCl2\{P(C2H5)2(C6H5)\}2}$ exhibits temperature dependent magnetic behaviour (paramagnetic / diamagnetic). The coordination geometries of $\ce{Ni^{2+}}$ in the paramagnetic and diamagnetic states are respectively

tetrahedral and tetrahedral

square planar and square planar

tetrahedral and square planar

square planar and tetrahedral

Answer and worked solution

Official answer: C

Ni(II) is $d^8$ . With two phosphines and two chlorides (intermediate ligand field) the complex sits close to the tetrahedral/square-planar crossover, so both forms are accessible and interconvert with temperature.

Tetrahedral ( $sp^3$ ): two unpaired electrons, paramagnetic.
Square planar ( $dsp^2$ ): the $d_{x^2-y^2}$ orbital is empty and all eight electrons are paired, diamagnetic.

(C) tetrahedral (paramagnetic) and square planar (diamagnetic)

Topics:Valence bond theory Magnetic properties and colour

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IIT-JEE 2012 · Paper 2, Chemistry question 21