QJEEM19Cha4bd: JEE Main 2019 Chemistry - Class 12 JEEM19Cha4bd Solution: The correct order of the spin-only magnetic moment of metal…

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 t_2g is less than e_g. As electrons always go to the lower energy level orbitals first, that is why electrons goes to the t_2g 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 e_g and t_2g orbital is very high.

(2) [Fe(CN)₆]^4– is an octahedral complex. And for octahedral complex Energy gap between e_g and t_2g orbital is called $$\mathrm{\Delta }$$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, $$\mathrm{\Delta }$$₀ is very high and for weak field ligand, $$\mathrm{\Delta }$$₀ is very low.

(5) For strong field ligand, $$\mathrm{\Delta }$$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, $$\mathrm{\Delta }$$₀ < P, so when electron gets energy, pairing of electrons does not happens as the energy required to enter into the e_g orbital is less than pairing energy. That is why electron go to e_g 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 e_g and t_2g 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⁺