JEE MAIN - Chemistry (2020 - 8th January Evening Slot - No. 12)
The correct order of the calculated spin-only
magnetic moments of complexs (A) to (D) is:
(A) Ni(CO)4
(B) [Ni(H2O)6]Cl2
(C) Na2[Ni(CN)4]
(D) PdCl2(PPh3)2
(A) Ni(CO)4
(B) [Ni(H2O)6]Cl2
(C) Na2[Ni(CN)4]
(D) PdCl2(PPh3)2
(C) < (D) < (B) < (A)
(C) $$ \approx $$ (D) < (B) < (A)
(A) $$ \approx $$ (C) $$ \approx $$ (D) < (B)
(A) $$ \approx $$ (C) < (B) $$ \approx $$ (D)
Explanation
(A) Ni(CO)4
Ni = 3d84s2
CO is strong field ligand. So pairing of elections happens.
$$ \therefore $$ Number of unpaired electrons = 0
$$ \therefore $$ $$\mu $$spin = 0
(B) [Ni(H2O)6]Cl2
Ni+2 = 3d84s0
H2O is weak field ligand. So no pairing of electrons happens.
Number of unpaired electron = 2
$$ \therefore $$ $$\mu $$spin = $$\sqrt {n\left( {n + 2} \right)} $$ = $$\sqrt {2\left( {2 + 2} \right)} $$ = $$\sqrt 8 $$ B.M
(C) Na2[Ni(CN)4]
Ni+2 = 3d84s0
CN- is strong field ligand. So pairing of electrons happens.
Number of unpaired electron = 0
$$ \therefore $$ $$\mu $$spin = 0
(D) PdCl2(PPh3)2
Pd2+ = 4d8
This is dsp2 complex. And shape is square planar.
$$ \therefore $$ $$\mu $$spin = 0
Ni = 3d84s2
CO is strong field ligand. So pairing of elections happens.
$$ \therefore $$ Number of unpaired electrons = 0
$$ \therefore $$ $$\mu $$spin = 0
(B) [Ni(H2O)6]Cl2
Ni+2 = 3d84s0
H2O is weak field ligand. So no pairing of electrons happens.
Number of unpaired electron = 2
$$ \therefore $$ $$\mu $$spin = $$\sqrt {n\left( {n + 2} \right)} $$ = $$\sqrt {2\left( {2 + 2} \right)} $$ = $$\sqrt 8 $$ B.M
(C) Na2[Ni(CN)4]
Ni+2 = 3d84s0
CN- is strong field ligand. So pairing of electrons happens.
Number of unpaired electron = 0
$$ \therefore $$ $$\mu $$spin = 0
(D) PdCl2(PPh3)2
Pd2+ = 4d8
This is dsp2 complex. And shape is square planar.
$$ \therefore $$ $$\mu $$spin = 0
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