ExamHabit
Practise free

JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)

JEE Main 2023 Chemistry question on d- and f-Block Elements

The magnetic moment of a transition metal compound has been calculated to be 3.87 B.M. The metal ion is
  1. A.
    V2+\text{V}^{2+}
  2. B.
    Ti2+\text{Ti}^{2+}
  3. C.
    Cr2+\text{Cr}^{2+}
  4. D.
    Mn2+\text{Mn}^{2+}
Show answer and solution

Correct option: A

Solution

Using the spin-only formula μ = √(n(n+2)) B.M., a value of 3.87 B.M. corresponds to n = 3 unpaired electrons (√(3×5) = √15 ≈ 3.87).

V²⁺ has the configuration [Ar] 3d³, giving exactly 3 unpaired electrons. Ti²⁺ has 3d² (2 unpaired, μ ≈ 2.83), Cr²⁺ has 3d⁴ (4 unpaired, μ ≈ 4.90), and Mn²⁺ has 3d⁵ (5 unpaired, μ ≈ 5.92), none of which match 3.87 B.M.

[JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)]

Practise d- and f-Block Elements the way it is asked

A timed set of real JEE Main d- and f-Block Elements questions, graded as the exam grades. No sign-up.

Start a d- and f-Block Elements test

More: all JEE Main d- and f-Block Elements questions · JEE Main 2023 · Chemistry chapter list