JEE MAIN - Chemistry (2020 - 9th January Morning Slot - No. 4)
[Pd(F)(Cl)(Br)(I)]2– has n number of
geometrical isomers. Then, the spin-only
magnetic moment and crystal field stabilisation
energy [CFSE] of [Fe(CN)6]n–6, respectively,
are:
[Note : Ignore the pairing energy]
[Note : Ignore the pairing energy]
1.73 BM and –2.0 $$\Delta $$0
5.92 BM and 0
2.84 BM and –1.6 $$\Delta $$0
0 BM and –2.4 $$\Delta $$0
Explanation
Complex [Pd(F)(Cl)(Br)(I)]2– (square planar
geometry)-has 3 geometrical isomers.
$$ \therefore $$ n = 3
[Fe(CN)6]n–6 becomes [Fe(CN)6]3-
Here Oxidation state of Fe = +3
$$ \therefore $$ Fe+3 = [Ar]3d54s0
CN- is strong field ligand so it pairing of electrons happens.
E.C. according to CFT = ($$t_{2g}^5e{g^0}$$)
$$ \therefore $$ No. of unpaired e- = 1
Then Magnetic moment = $$\sqrt {n\left( {n + 2} \right)} $$ = $$\sqrt 3 $$ = 1.73 B.M
CFSE = [(–0.4 × 5) + (0.6 × 0)] $$\Delta $$0 = –2.0 $$\Delta $$0
$$ \therefore $$ n = 3
[Fe(CN)6]n–6 becomes [Fe(CN)6]3-
Here Oxidation state of Fe = +3
$$ \therefore $$ Fe+3 = [Ar]3d54s0
CN- is strong field ligand so it pairing of electrons happens.
E.C. according to CFT = ($$t_{2g}^5e{g^0}$$)
$$ \therefore $$ No. of unpaired e- = 1
Then Magnetic moment = $$\sqrt {n\left( {n + 2} \right)} $$ = $$\sqrt 3 $$ = 1.73 B.M
CFSE = [(–0.4 × 5) + (0.6 × 0)] $$\Delta $$0 = –2.0 $$\Delta $$0
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