IIT-JEE 2008 · Paper 2 · Chemistry Q49

ChemistrySingle correct+3 / -1Question 49 of 66

Both $\ce{[Ni(CO)4]}$ and $\ce{[Ni(CN)4]^{2-}}$ are diamagnetic. The hybridisations of nickel in these complexes, respectively, are

sp^3

sp^3

sp^3

dsp^2

dsp^2

sp^3

dsp^2

dsp^2

Answer and worked solution

Official answer: B

$\ce{[Ni(CO)4]}$ : Ni(0), $3d^84s^2$ ; strong-field CO pairs everything into $3d^{10}$ , and the empty 4s and 4p orbitals give $sp^3$ (tetrahedral), diamagnetic.
$\ce{[Ni(CN)4]^{2-}}$ : Ni(II), $d^8$ ; strong-field cyanide pairs the eight electrons into four d orbitals, leaving one 3d orbital for $dsp^2$ (square planar), diamagnetic.

(B)

Topics:Valence bond theory

IIT-JEE 2008 · Paper 2, Chemistry question 49