Question

Among, Sc, Mn, Co and Cu, identify the element with highest enthalpy of atomisation. The spin only magnetic moment value of that element in its +2 oxidation state is _______BM (in nearest integer).

Hint:

The enthalpy of atomisation is related to the strength of metallic bonding, which often correlates with the number of unpaired d-electrons. The spin-only magnetic moment can be calculated using the formula \( \sqrt{n(n+2)} \) BM, where n is the number of unpaired electrons in the ion. Remember to determine the correct electronic configuration of the ion in its given oxidation state.

Answer 1

Correct Answer: 4

Enthalpy of atomisation is the enthalpy change when one mole of a substance is completely converted into gaseous atoms. It depends on the strength of the metallic bonds in the solid state. Stronger metallic bonds lead to higher enthalpy of atomisation. 
The strength of metallic bonds in transition metals is related to the number of unpaired d-electrons that can participate in bonding. Electronic configurations of the given elements: 
Sc (Z=21): [Ar] 3d\(^1\) 4s\(^2\) 
Mn (Z=25): [Ar] 3d\(^5\) 4s\(^2\) 
Co (Z=27): [Ar] 3d\(^7\) 4s\(^2\) 
Cu (Z=29): [Ar] 3d\(^{10}\) 4s\(^1\) 
Number of unpaired electrons in the ground state: 
Sc: 1 unpaired electron 
Mn: 5 unpaired electrons 
Co: 3 unpaired electrons 
Cu: 1 unpaired electron (due to stable fully filled d-orbital) 
However, enthalpy of atomisation also depends on other factors. 
From the given table: Enthalpy of Atomisation (kJ/mole): 
Sc: 326 
Mn: 281 
Co: 425 
Cu: 339 
The element with the highest enthalpy of atomisation is Co (425 kJ/mole). 
Now, we need to find the spin-only magnetic moment of Co in its +2 oxidation state (Co\(^{2+}\)). 
Electronic configuration of Co\(^{2+}\): [Ar] 3d\(^7\) 
To find the number of unpaired electrons in Co\(^{2+}\), we can use Hund's rule to fill the d orbitals: 
3d: \( \uparrow \downarrow \) \( \uparrow \downarrow \) \( \uparrow \) \( \uparrow \) \( \uparrow \) 
There are 3 unpaired electrons (n = 3). The spin-only magnetic moment (\( \mu_{spin-only} \)) is given by the formula: \[ \mu_{spin-only} = \sqrt{n(n+2)} \, \text{BM} \] where n is the number of unpaired electrons. 
For Co\(^{2+}\) (n = 3): \[ \mu_{spin-only} = \sqrt{3(3+2)} \, \text{BM} = \sqrt{3 \times 5} \, \text{BM} = \sqrt{15} \, \text{BM} \] \( \sqrt{15} \approx 3.87 \, \text{BM} \) The nearest integer to 3.87 is 4. 
Final Answer: (4)