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

The problem has two parts. First, we need to identify which of the given elements (Sc, Mn, Co, Cu) has the highest enthalpy of atomisation. Second, we need to calculate the spin-only magnetic moment for the identified element in its +2 oxidation state.

Concept Used:

1. Enthalpy of Atomisation: This is the energy required to convert one mole of a substance in its standard state into gaseous atoms. For transition metals, it is a measure of the strength of the metallic bonding. The strength of metallic bonding depends on the number of unpaired electrons in the d-orbitals as well as the s-orbitals that can participate in bonding. A greater number of unpaired electrons generally leads to stronger interatomic interactions and thus a higher enthalpy of atomisation.

2. Spin-Only Magnetic Moment (\(\mu_s\)): The magnetic moment of an ion due to the spin of its unpaired electrons can be calculated using the formula:

\[ \mu_s = \sqrt{n(n+2)} \text{ BM} \]

where \(n\) is the number of unpaired electrons in the ion, and BM stands for Bohr Magneton, the unit of magnetic moment.

Step-by-Step Solution:

Step 1: Identify the element with the highest enthalpy of atomisation.

Let's analyze the electronic configurations and metallic bonding for each element:

• Scandium (Sc, Z=21): Electronic configuration is \([Ar] 3d^1 4s^2\). It has 3 valence electrons, leading to relatively weaker metallic bonding compared to other transition metals.
• Manganese (Mn, Z=25): Electronic configuration is \([Ar] 3d^5 4s^2\). Due to the very stable half-filled \(d^5\) configuration, the d-electrons are held more tightly and are less involved in metallic bonding. This results in weaker metallic bonds and an anomalously low enthalpy of atomisation for Mn.
• Cobalt (Co, Z=27): Electronic configuration is \([Ar] 3d^7 4s^2\). It has 3 unpaired electrons in the 3d subshell. These electrons, along with the 4s electrons, participate effectively in metallic bonding, leading to strong interatomic forces.
• Copper (Cu, Z=29): Electronic configuration is \([Ar] 3d^{10} 4s^1\). It has a completely filled d-orbital. Only the single 4s electron is primarily responsible for metallic bonding, which is relatively weak compared to elements in the middle of the series.

Comparing the options, Manganese (Mn) has a significant dip in atomisation enthalpy due to its stable electronic configuration. Scandium has fewer valence electrons. Copper has a filled d-shell. Cobalt (Co) has a high number of unpaired d-electrons that contribute to strong metallic bonding. Therefore, among the given choices, Cobalt (Co) has the highest enthalpy of atomisation.

Step 2: Determine the electronic configuration and the number of unpaired electrons in the \(Co^{2+}\) ion.

The neutral Cobalt atom (Co) has the configuration: \([Ar] 3d^7 4s^2\).

To form the \(Co^{2+}\) ion, two electrons are removed from the outermost shell, which is the \(n=4\) shell. So, the two 4s electrons are removed.

The electronic configuration of \(Co^{2+}\) is: \([Ar] 3d^7\).

Now, let's find the number of unpaired electrons (\(n\)) by filling the 3d orbitals according to Hund's rule of maximum multiplicity:

The \(3d^7\) configuration means there are 7 electrons in the 5 d-orbitals. The first 5 electrons will occupy each orbital singly, and the remaining 2 will pair up.

\[ \begin{array}{|c|c|c|c|c|} \hline \uparrow\downarrow & \uparrow\downarrow & \uparrow & \uparrow & \uparrow \\ \hline \end{array} \]

From the orbital diagram, we can see there are 3 unpaired electrons. So, \(n=3\).

Step 3: Calculate the spin-only magnetic moment (\(\mu_s\)).

Using the formula \(\mu_s = \sqrt{n(n+2)}\) with \(n=3\):

\[ \mu_s = \sqrt{3(3+2)} = \sqrt{3 \times 5} = \sqrt{15} \text{ BM} \]

Step 4: Find the numerical value and round to the nearest integer.

The value of \(\sqrt{15}\) is approximately 3.87.

\[ \mu_s \approx 3.87 \text{ BM} \]

Rounding this value to the nearest integer gives 4.

The spin only magnetic moment value of Cobalt in its +2 oxidation state is 4 BM (in nearest integer).

Answer 2

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)