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)
Correct Answer: 6
Approach Solution - 1
The electronic configuration of Cr is [Ar] 3d5 4s1. Upon losing one electron, Cr becomes Cr+ with the configuration [Ar] 3d5. The presence of a half-filled d subshell in Cr+ offers additional stability, contributing to a higher second ionisation enthalpy.
Now, the spin-only magnetic moment for a d5 configuration is calculated using the formula:
µ = √n(n + 2) BM
Where n is the number of unpaired electrons. For Cr+, n = 5. Thus,
µ = √5(5 + 2) = √35 ≈ 5.92 BM.
Since the question asks for the value as a near integer, we round this to 6 BM.
Therefore, the spin-only magnetic moment of the M+ ion is 6 BM, which lies within the expected range of 6,6.
Approach Solution -2
To determine which transition metal 'M' has the highest second ionization enthalpy among Sc, Ti, V, Cr, Mn, and Fe, we need to analyze the electronic configurations and related properties. The second ionization enthalpy is the energy required to remove an electron from a singly charged cation (M+). It is typically high for elements where removing an electron significantly disrupts a stable configuration.
1. Determine the electronic configuration of each element:
- Sc (Z=21): [Ar] 3d1 4s2
- Ti (Z=22): [Ar] 3d2 4s2
- V (Z=23): [Ar] 3d3 4s2
- Cr (Z=24): [Ar] 3d5 4s1
- Mn (Z=25): [Ar] 3d5 4s2
- Fe (Z=26): [Ar] 3d6 4s2
2. Evaluate the second ionization: Removing the second electron from M+.
3. Cr ([Ar] 3d5) has a half-filled 3d subshell, which is notably stable. Removing one more electron to form Cr2+ disrupts this stability, resulting in high ionization enthalpy.
Answer: Cr has the highest second ionization enthalpy.
4. Find the spin-only magnetic moment of Cr+ (d5 configuration):
The magnetic moment (µ) is given by µ=√n(n+2), where n is the number of unpaired electrons.
- Cr+ has 5 unpaired electrons: 3d5
- µ=√5(5+2)=√35 ≈ 5.92 BM
5. Verification:
The calculated magnetic moment is approximately 6 BM, which is within the given range of 6,6.
Conclusively, the spin-only magnetic moment of Cr+ ion is approximately 6 BM.
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