JEE Main 2014 · 19 Apr · Shift 1 · Chemistry Q46

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ChemistrySingle correct+4 / -1Question 46 of 90

Nickel (Z = 28) combines with a uninegative monodentate ligand to form a diamagnetic complex $\ce{[NiL4]^2-}$ . The hybridisation involved and the number of unpaired electrons present in the complex are respectively :

sp^3

, two

dsp^2

, zero

dsp^2

, one

sp^3

, zero

Answer and worked solution

Official answer: B

$\ce{Ni^2+}$ is $d^8$ . A diamagnetic four-coordinate $d^8$ complex is square planar, using $dsp^2$ hybrids with all eight d electrons paired: zero unpaired.

Topics:Valence bond theory Magnetic properties and colour

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JEE Main 2014 · 19 Apr · Shift 1, Chemistry question 46