A high-spin aqueous M^2+ ion of a 3d metal has a spin-only magnetic moment of about 4.90 BM, and E deg(M^3+/M^2+) = -0.41 V. The spin-only magnetic moment (in BM) of the corresponding M^3+ ion is

JEE Advanced levelSingle correct optionDifficulty 3 of 3df072

A high-spin aqueous $M X^{2 +}$ ion of a 3d metal has a spin-only magnetic moment of about $4.90$ BM, and $E^{\circ} (M X^{3 +} / M X^{2 +}) = - 0.41 V$ . The spin-only magnetic moment (in BM) of the corresponding $M X^{3 +}$ ion is

μ = 4.90

BM means 4 unpaired electrons, i.e. a

d^{4}

d^{6}

ion (

Cr X^{2 +}

Mn X^{3 +}

-type

d^{4}

, or

Fe X^{2 +}

d^{6}

). A negative

E^{\circ} (M X^{3 +} / M X^{2 +})

means

M X^{2 +}

is a strong reductant, which fits

Cr X^{2 +}

(

E^{\circ} = - 0.41 V

) and not

Fe X^{2 +}

(

+ 0.77 V

Cr X^{3 +}

d^{3}

with 3 unpaired electrons, so

μ = 3 \times 5 = 3.87

BM.