The metal ions that have the calculated spin-only magnetic moment value of 4.9 B.M. are :
A. Cr²⁺
B. Fe²⁺
C. Fe³⁺
D. Co²⁺
E. Mn³⁺
Choose the correct answer from the options given below:

Solution

Chemistry ma’am · 3 steps

4.9 B.M. means exactly four unpaired electrons, so look for the d⁴ and d⁶ ions.

1. Turn 4.9 into n. The spin-only moment depends only on n, the number of unpaired electrons. Set it equal to 4.9 and solve for n.

   μ = √n(n+2) B.M. = 4.9 ⇒ n(n+2) = 24 ⇒ n = 4

2. Count unpaired electrons. Remove the 4s electrons first, then fill the five 3d orbitals one electron each before pairing (Hund's rule). A d⁴ ion has 4 unpaired electrons. A d⁶ ion also has 4, because only one orbital holds a pair.

   Ion	Configuration	Unpaired n	μ (B.M.)
   A. Cr²⁺	[Ar] 3d⁴	4	4.90
   B. Fe²⁺	[Ar] 3d⁶	4	4.90
   C. Fe³⁺	[Ar] 3d⁵	5	5.92
   D. Co²⁺	[Ar] 3d⁷	3	3.87
   E. Mn³⁺	[Ar] 3d⁴	4	4.90

3. So. Cr²⁺, Fe²⁺ and Mn³⁺ each have 4 unpaired electrons and a moment of 4.9 B.M. The answer is (b), A, B and E only.

The idea to keep. Learn the ladder: n = 1, 2, 3, 4, 5 gives 1.73, 2.83, 3.87, 4.90, 5.92 B.M. Then dⁿ and d¹⁰⁻ⁿ share the same unpaired count, so d⁴ and d⁶ both give 4.

Answer from the NTA final answer key. Solution by PrepWiser.