JEE MAIN - Chemistry (2019 - 10th April Evening Slot - No. 4)
The crystal field stabilization energy (CFSE) of [Fe(H2O)6]Cl2 and K2[NiCl4] respectively, are :
– 0.4 $$\Delta $$0 and – 0.8 $$\Delta $$t
– 0.6 $$\Delta $$0 and – 0.8 $$\Delta $$t
– 2.4 $$\Delta $$0 and – 1.2 $$\Delta $$t
– 0.4 $$\Delta $$0 and – 1.2 $$\Delta $$t
Explanation
CN of [Fe(H2O)6]Cl2 = 6
For CN = 6, CFSE = $$\left[ {{3 \over 5} \times n - {2 \over 5} \times {n_1}} \right]{\Delta _0}$$
Here, n = number of electron in eg and
n1 = number of electron in t2g
Electronic configuration of Fe+2 according to crystal field theory = $$t_{2g}^4e_g^2$$
$$ \therefore $$ CFSE = $$\left[ {{3 \over 5} \times 2 - {2 \over 5} \times 4} \right]{\Delta _0}$$ = - 0.4 $${\Delta _0}$$
CN of K2[NiCl4] = 4
For CN = 4, CFSE = $$\left[ {{2 \over 5} \times n - {3 \over 5} \times {n_1}} \right]{\Delta _t}$$
Here, n = number of electron in t2g and
n1 = number of electron in eg
Electronic configuration of Ni+2 according to crystal field theory = $$e_g^4t_{2g}^4$$
$$ \therefore $$ CFSE = $$\left[ {{2 \over 5} \times 4 - {3 \over 5} \times 4} \right]{\Delta _t}$$ = - 0.8 $${\Delta _t}$$
For CN = 6, CFSE = $$\left[ {{3 \over 5} \times n - {2 \over 5} \times {n_1}} \right]{\Delta _0}$$
Here, n = number of electron in eg and
n1 = number of electron in t2g
Electronic configuration of Fe+2 according to crystal field theory = $$t_{2g}^4e_g^2$$
$$ \therefore $$ CFSE = $$\left[ {{3 \over 5} \times 2 - {2 \over 5} \times 4} \right]{\Delta _0}$$ = - 0.4 $${\Delta _0}$$
CN of K2[NiCl4] = 4
For CN = 4, CFSE = $$\left[ {{2 \over 5} \times n - {3 \over 5} \times {n_1}} \right]{\Delta _t}$$
Here, n = number of electron in t2g and
n1 = number of electron in eg
Electronic configuration of Ni+2 according to crystal field theory = $$e_g^4t_{2g}^4$$
$$ \therefore $$ CFSE = $$\left[ {{2 \over 5} \times 4 - {3 \over 5} \times 4} \right]{\Delta _t}$$ = - 0.8 $${\Delta _t}$$
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