I Draw the geometrical isomers of complex - Co en 2 Cl 2-ii On the basis of crystal field theory- write the electronic configuration for d 4 ion if -0- P-iii - NiCl 4-2- is paramagnetic- while - N i CO 4- is diamagnetic- though both are tetrahedral- Why- Atomic number of Ni -28

(i) Geometrical isomers of [Co(en)_2 Cl_2 ]^+ : (ii)If Δ_0 gt; P, it becomes more energetically favourable for the fourth electron to occupy a t_2g orbital ...

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Standard XII

Chemistry

CFSE Comparison

i Draw the ge...

Question

(i) Draw the geometrical isomers of complex $[C o (e n)_{2} C l_{2}]^{+}$ .

(ii) On the basis of crystal field theory, write the electronic configuration for d⁴ ion if $Δ_{0} > P$ .

(iii) $[N i C l_{4}]^{2 -}$ is paramagnetic, while $[N i (C O)_{4}]$ is diamagnetic, though both are tetrahedral. Why? (Atomic number of Ni = 28)

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Solution

(i) Geometrical isomers of $[C o (e n)_{2} C l_{2}]^{+}$ :

(ii)If $Δ_{0} > P$ , it becomes more energetically favourable for the fourth electron to occupy a $t_{2 g}$ orbital with configuration te.

(iii) In $[N i C l_{4}]^{2 -}$ , Ni is in the +2 state. $C l^{-}$ is a ligand which is a weak field ligand which does not cause pairing of unpaired 3d electrons. Hence, it is paramagnetic.

In $[N i (C O)_{4}]$ , Ni has 0 oxidation state. CO is a strong field ligand, which causes pairing of unpaired 3d electrons. No unpaired electrons are present in this case. So, it is diamagnetic.

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(i) Draw the geometrical isomers of complex [Co(en)2Cl2]+ . (ii) On the basis of crystal field theory, write the electronic configuration for d4 ion if Δ0>P. (iii) [NiCl4]2− is paramagnetic, while [Ni(CO)4] is diamagnetic, though both are tetrahedral. Why? (Atomic number of Ni = 28)