JEE MAIN - Chemistry (2019 - 8th April Morning Slot - No. 15)

The correct order of the spin-only magnetic moment of metal ions in the following low-spin complexes, [V(CN)6]4–,[Fe(CN)6]4–, [Ru(NH3)6]3+, and [Cr(NH3)6]2+ , is :
V2+ > Ru3+ > Cr2+ > Fe2+
V2+ > Cr2+ > Ru3+ > Fe2+
Cr2+ > V2+ > Ru3+ > Fe2+
Cr2+ > Ru3+ > Fe2+ > V2+

Explanation


Here number of unpaired electrons = 3

$$ \therefore $$ Spin only magnetic moment ($$\mu $$) = $$\sqrt {3\left( {3 + 2} \right)} = \sqrt {15} $$ B.M

Note : Energy of t2g is less than eg. As electrons always go to the lower energy level orbitals first, that is why electrons goes to the t2g orbital first.
Here number of unpaired electrons = 0

$$ \therefore $$ Spin only magnetic moment ($$\mu $$) = 0 B.M

Note : (1) As CN- is a strong field ligand so Energy gap between eg and t2g orbital is very high.

(2) [Fe(CN)6]4– is an octahedral complex. And for octahedral complex Energy gap between eg and t2g orbital is called $$\Delta $$0 or Crystal Field splitting Energy.

(3) Energy required to pair up the electron in same orbital is called Pairing Energy(P).

(4) For strong field ligand, $$\Delta $$0 is very high and for weak field ligand, $$\Delta $$0 is very low.

(5) For strong field ligand, $$\Delta $$0 > P, so when electron gets energy, pairing of electrons happens as for pairing of electrons very low energy requied.

(6) For weak field ligand, $$\Delta $$0 < P, so when electron gets energy, pairing of electrons does not happens as the energy required to enter into the eg orbital is less than pairing energy. That is why electron go to eg orbital first.

(7) As CN- is a strong field ligand so pairing of electron occurs.
Here number of unpaired electrons = 1

$$ \therefore $$ Spin only magnetic moment ($$\mu $$) = $$\sqrt {1\left( {1 + 2} \right)} = \sqrt {3} $$ B.M

Note : (1) In [Ru(NH3)6]3+ complex, NH3 is a strong field ligand so Energy gap between eg and t2g orbital is very high. That is why pairing of electrons occurs.
Here number of unpaired electrons = 2

$$ \therefore $$ Spin only magnetic moment ($$\mu $$) = $$\sqrt {2\left( {2 + 2} \right)} = \sqrt {8} $$ B.M

$$ \therefore $$ Correct order of the spin-only magnetic moment of metal ions

V2+ > Cr2+ > Ru3+ > Fe2+

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