Exams › JEE Advanced › Chemistry

Arrange the following ions in decreasing order of their magnetic moments: (A) V^+4, (B) Mn^+4, (C) Fe^+3, (D) Ni^+2.

1. B > C > A > D
2. C > D > B > A
3. C > B > D > A
4. A > D > C > B

Correct answer: C > B > D > A

Solution

V^+4: [Ar]3d¹ — 1 unpaired e, mu = sqrt(3) ≈ 1.73 BM. Mn^+4: [Ar]3d³ — 3 unpaired e, mu = sqrt(15) ≈ 3.87 BM. Fe^+3: [Ar]3d⁵ — 5 unpaired e (high spin), mu = sqrt(35) ≈ 5.92 BM. Ni^+2: [Ar]3d⁸ — 2 unpaired e, mu = sqrt(8) ≈ 2.83 BM. Order: Fe³+ > Mn⁴+ > Ni²+ > V⁴+, i.e., C > B > D > A.

Related JEE Advanced Chemistry questions

• The count of d-electrons in Fe2+ differs from the number of:
• Which sequence of reactions accurately depicts the chemical processes involving iron and its associated compounds?
• When H2S interacts with an acidic solution of K2Cr2O7, which of the following compounds is not produced?
• When an excess amount of sodium thiosulfate is combined with copper sulfate in water, the resulting product is:
• Among the given lanthanoid ions, which one exhibits diamagnetic behavior? (Atomic numbers: Ce = 58, Sm = 62, Eu = 63, Yb = 70)
• Which of these statements is accurate?

⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →