JEE MAIN - Chemistry (2018 - 15th April Evening Slot - No. 15)
The correct order of spin-only magnetic moments among the following is :
(Atomic number : Mn = 25, Co = 27, Ni = 28, Zn = 30)
(Atomic number : Mn = 25, Co = 27, Ni = 28, Zn = 30)
[ZnCl4]2- > [NiCl4]2- > [CoCl4]2- > [MnCl4]2-
[CoCl4]2- > [MnCl4]2- > [NiCl4]2- > [ZnCl4]2-
[NiCl4]2- > [CoCl4]2- > [MnCl4]2- > [ZnCl4]2-
[MnCl4]2- > [CoCl4]2- > [NiCl4]2- > [ZnCl4]2-
Explanation
We know,
Spin only magnetic moment ($$\mu $$) = $$\sqrt {n\left( {n + 2} \right)} $$ B.M
Where, n is the number of unpaired electrons.
So, the complex having higher number of unpaired electrons will have higher value of spin only magnetic moment.
(1) Zn+2 : [Ar] 3d10
Here 0 unpaired electrons present
(2) Ni+2 : [Ar] 3d8
Here 2 unpaired electrons present.
(3) Co+2 : [Ar] 3d7
Here 3 unpaired electrons present.
(4) Mn+2 : [Ar] 3d5
Here 5 unpaired electrons present.
So, correct order is $$ \to $$
[MnCl4]2$$-$$ > [CoCl4]2$$-$$ > [NiCl4]2$$-$$ > [ZnCl4]2$$-$$
Spin only magnetic moment ($$\mu $$) = $$\sqrt {n\left( {n + 2} \right)} $$ B.M
Where, n is the number of unpaired electrons.
So, the complex having higher number of unpaired electrons will have higher value of spin only magnetic moment.
(1) Zn+2 : [Ar] 3d10
Here 0 unpaired electrons present
(2) Ni+2 : [Ar] 3d8
Here 2 unpaired electrons present.
(3) Co+2 : [Ar] 3d7
Here 3 unpaired electrons present.
(4) Mn+2 : [Ar] 3d5
Here 5 unpaired electrons present.
So, correct order is $$ \to $$
[MnCl4]2$$-$$ > [CoCl4]2$$-$$ > [NiCl4]2$$-$$ > [ZnCl4]2$$-$$
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