Solveeit Logo
Login

Question

Question: Which of the following pair will have effective magnetic moment equal?...

Which of the following pair will have effective magnetic moment equal?

A

Ti2+^{2+} and V2+^{2+}

B

Cr2+^{2+} and Fe2+^{2+}

C

Cr3+^{3+} and Mn2+^{2+}

D

V2+^{2+} and Sc3+^{3+}

Answer

Cr2+^{2+} and Fe2+^{2+}

Explanation

Solution

The effective magnetic moment (μeff\mu_{eff}) for transition metal ions is primarily due to the spin magnetic moment and is calculated using the formula:

μeff=n(n+2)\mu_{eff} = \sqrt{n(n+2)} BM (Bohr Magnetons)

where 'n' is the number of unpaired electrons. To have the same effective magnetic moment, the ions must have the same number of unpaired electrons.

  1. Electronic Configurations of Elements:

    • Sc (Z=21): [Ar] 3d1^1 4s2^2
    • Ti (Z=22): [Ar] 3d2^2 4s2^2
    • V (Z=23): [Ar] 3d3^3 4s2^2
    • Cr (Z=24): [Ar] 3d5^5 4s1^1 (exception)
    • Mn (Z=25): [Ar] 3d5^5 4s2^2
    • Fe (Z=26): [Ar] 3d6^6 4s2^2

  2. Electronic Configurations and Unpaired Electrons (n) for Ions:

    • Ti2+^{2+}: Formed by losing 2 electrons from 4s. [Ar] 3d2^2. Number of unpaired electrons (n) = 2.
    • V2+^{2+}: Formed by losing 2 electrons from 4s. [Ar] 3d3^3. Number of unpaired electrons (n) = 3.
    • Cr2+^{2+}: Formed by losing 1 electron from 4s and 1 from 3d. [Ar] 3d4^4. Number of unpaired electrons (n) = 4.
    • Fe2+^{2+}: Formed by losing 2 electrons from 4s. [Ar] 3d6^6. In 3d6^6, 4 electrons are unpaired (d-orbitals: ↑↓ ↑ ↑ ↑ ↑). Number of unpaired electrons (n) = 4.
    • Cr3+^{3+}: Formed by losing 1 electron from 4s and 2 from 3d. [Ar] 3d3^3. Number of unpaired electrons (n) = 3.
    • Mn2+^{2+}: Formed by losing 2 electrons from 4s. [Ar] 3d5^5. Number of unpaired electrons (n) = 5.
    • Sc3+^{3+}: Formed by losing 2 electrons from 4s and 1 from 3d. [Ar] 3d0^0. Number of unpaired electrons (n) = 0.

  3. Comparing Unpaired Electrons for Each Option:

    • (A) Ti2+^{2+} and V2+^{2+}: Ti2+^{2+} (n=2) and V2+^{2+} (n=3). Different number of unpaired electrons.
    • (B) Cr2+^{2+} and Fe2+^{2+}: Cr2+^{2+} (n=4) and Fe2+^{2+} (n=4). Same number of unpaired electrons.
    • (C) Cr3+^{3+} and Mn2+^{2+}: Cr3+^{3+} (n=3) and Mn2+^{2+} (n=5). Different number of unpaired electrons.
    • (D) V2+^{2+} and Sc3+^{3+}: V2+^{2+} (n=3) and Sc3+^{3+} (n=0). Different number of unpaired electrons.

Only the pair Cr2+^{2+} and Fe2+^{2+} have the same number of unpaired electrons (n=4), and therefore will have the same effective magnetic moment (μeff=4(4+2)=24\mu_{eff} = \sqrt{4(4+2)} = \sqrt{24} BM).