# Q. If Hund's `2^(nd)` rule is violated then compare the magnetic moment of the following:- A) `Cr^+ = Fe^(2+) > Fe^(3+) > Cu^+` B) `Fe^(2+) > Cr^+ >Cu^+ >Fe^(3+)` C) `Fe^(3+) >...

Q. If Hund's `2^(nd)` rule is violated then compare the magnetic moment of the following:-

A) `Cr^+ = Fe^(2+) > Fe^(3+) > Cu^+`

B) `Fe^(2+) > Cr^+ >Cu^+ >Fe^(3+)`

C) `Fe^(3+) > Cr^+ >Fe^(2+) >Cu^+`

D) `Fe^(3+) = Cr^+>Fe^(2+)>Cu^+`

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The magnetic moment of an atom is due to spinning of electrons around their axes, with a small contribution from the orbiting of electrons around the nucleus. Magnetic moment due to fully filled orbitals is zero, as individual moments are cancelled out. For partially filled orbits, however, the magnetic measurements can be linked to molecular (rather electronic) properties.

Thus, the overall magnetic moment is given by,

`mu_(eff)= sqrt(L(L+1)+4S(S+1))` where, S is the spin angular momentum and L, the orbital angular momentum.

According to Hund’s second rule, for a given S, L has to be maximized, for lowest energy. Violation of Hund’s second rule implies that the other values of L have to be assumed. Keeping this in mind, the `mu_(eff)` values of the given ions can be computed as follows:

Species(elec. config.) L(violating 2nd rule) S mu_(eff) in B.M.

------------------- ----------------- ----- ------------

Cr^+ [Ar]3d^5 0 (half-filled) 5/2 5.9

Fe^(3+) [Ar]3d^5 0 (half-filled) 5/2 5.9

Fe^(2+)[Ar]3d^6 1 2 5.1

Fe^(2+)[Ar]3d^6 0 2 4.9

Fe^(2+)[Ar]3d^6 -1 2 4.9

Fe^(2+)[Ar]3d^6 -2 2 5.1

Cu^+ [Ar]3d^10 0 (fully filled) 0 0

---------------------------------------------------------

**Therefore, option D) is correct.**

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