Question:medium
A transition metal 'M' among Sc, Ti, V , Cr, Mn and Fe has the highest second ionisation enthalpy. The spin only magnetic moment value of M+
ion is ______ BM (Near integer) (Given atomic number Sc : 21, Ti : 22, V : 23, Cr : 24, Mn : 25, Fe : 26)
A transition metal 'M' among Sc, Ti, V , Cr, Mn and Fe has the highest second ionisation enthalpy. The spin only magnetic moment value of M+
ion is ______ BM (Near integer) (Given atomic number Sc : 21, Ti : 22, V : 23, Cr : 24, Mn : 25, Fe : 26)
ion is ______ BM (Near integer) (Given atomic number Sc : 21, Ti : 22, V : 23, Cr : 24, Mn : 25, Fe : 26)
Updated On: Jan 13, 2026
Show Solution
Correct Answer: 6
Solution and Explanation
To identify the transition metal 'M' exhibiting the highest second ionization enthalpy, a comparison among Sc, Ti, V, Cr, Mn, and Fe is necessary. Generally, second ionization enthalpy escalates with increasing effective nuclear charge and diminished electron shielding. Within this group, Cr (Chromium) displays the highest second ionization enthalpy, attributable to its electronic configuration.
The electronic configuration of Cr is [Ar] 3d5 4s1. Following the removal of one electron, Cr transforms into Cr+, possessing the configuration [Ar] 3d5. The stability conferred by a half-filled d subshell in Cr+ results in a higher second ionization enthalpy.
Subsequently, the spin-only magnetic moment for a d5 configuration is computed via the formula:
µ = √n(n + 2) BM
Here, 'n' represents the count of unpaired electrons. For Cr+, n = 5. Consequently,
µ = √5(5 + 2) = √35 ≈ 5.92 BM.
Given the requirement for a near-integer value, this is rounded to 6 BM.
Thus, the spin-only magnetic moment of the M+ ion is 6 BM, aligning with the anticipated range of 6,6.
The electronic configuration of Cr is [Ar] 3d5 4s1. Following the removal of one electron, Cr transforms into Cr+, possessing the configuration [Ar] 3d5. The stability conferred by a half-filled d subshell in Cr+ results in a higher second ionization enthalpy.
Subsequently, the spin-only magnetic moment for a d5 configuration is computed via the formula:
µ = √n(n + 2) BM
Here, 'n' represents the count of unpaired electrons. For Cr+, n = 5. Consequently,
µ = √5(5 + 2) = √35 ≈ 5.92 BM.
Given the requirement for a near-integer value, this is rounded to 6 BM.
Thus, the spin-only magnetic moment of the M+ ion is 6 BM, aligning with the anticipated range of 6,6.
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