CBSE Class 12 Chemistry 2019 Outside Delhi Set 3 Solved Paper

Question : 7 of 13

Marks: +1, -0

(a) Although both

[\mathrm{NiCl}_4]^{2-}

and

[\mathrm{Ni}(\mathrm{CO})_4]

have

\text{sp}^3

hybridisation yet

[\mathrm{NiCl}_4]^{2-}

is paramagnetic and

[\mathrm{Ni}(\mathrm{CO})_4]

is diamagnetic. Give reason. (Atomic no. of

\mathrm{Ni}=28

)

(b) Write the electronic configuration of

d^5

on the basis of crystal field theory when.

(i)

\Delta_{o} < P

and

(ii)

\Delta_{o} > P

Solution:

(a) In

[\mathrm{NiCl}_4]^{2-}, \mathrm{Ni}

is in +2 oxidation state and each

\mathrm{Cl}^{-}

donates a pair of electron. So,

\mathrm{Cl}^{-}

acts as a weak ligand and does, not cause any forced pairing. Thus, electrons remain unpaired making it paramagnetic.

[\mathrm{Ni}(\mathrm{CO})_4], \mathrm{Ni}

is in zero oxidation state and

\mathrm{CO}

acts as a strong ligand causing forced pairing. Thus, no electron remains unpaired making it diamagnetic.

(b) (i)

t_{2g}^3 e_g^2

(ii)

t_{2g}^5 e_g^0

Go to Question:

Prev Question

Next Question