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

This problem involves two stages. Initially, it is required to determine which of the specified elements (Sc, Mn, Co, Cu) exhibits the highest enthalpy of atomisation. Subsequently, the spin-only magnetic moment for the selected element in its +2 oxidation state must be computed.

Concept Used:

1. Enthalpy of Atomisation: The energy necessary to convert one mole of a substance from its standard state into gaseous atoms is defined as its enthalpy of atomisation. For transition metals, this value quantifies the strength of metallic bonding. The metallic bond strength is influenced by the number of unpaired electrons in both the d-orbitals and s-orbitals available for bonding. A higher number of unpaired electrons typically correlates with stronger interatomic interactions and, consequently, a greater enthalpy of atomisation.

2. Spin-Only Magnetic Moment (\(\mu_s\)): The magnetic moment arising from the spin of unpaired electrons within an ion is calculated using the formula:

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

where \(n\) represents the count of unpaired electrons in the ion, and BM denotes Bohr Magneton, the standard unit for magnetic moment.

Step-by-Step Solution:

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

An examination of the electronic configurations and metallic bonding characteristics for each element is as follows:

• Scandium (Sc, Z=21): Electronic configuration: \([Ar] 3d^1 4s^2\). Possesses 3 valence electrons, leading to comparatively weaker metallic bonding relative to other transition metals.
• Manganese (Mn, Z=25): Electronic configuration: \([Ar] 3d^5 4s^2\). The highly stable half-filled \(d^5\) configuration results in d-electrons being more tightly bound and less involved in metallic bonding. This phenomenon weakens metallic bonds, causing an unusually low enthalpy of atomisation for Mn.
• Cobalt (Co, Z=27): Electronic configuration: \([Ar] 3d^7 4s^2\). Features 3 unpaired electrons within the 3d subshell. These electrons, along with the 4s electrons, actively participate in metallic bonding, establishing strong interatomic forces.
• Copper (Cu, Z=29): Electronic configuration: \([Ar] 3d^{10} 4s^1\). Exhibits a completely filled d-orbital. The single 4s electron is the primary contributor to metallic bonding, which is relatively weak compared to elements situated in the middle of the transition series.

Upon comparison, Manganese (Mn) shows a notable reduction in atomisation enthalpy due to its stable electronic configuration. Scandium has fewer valence electrons. Copper possesses a filled d-shell. Cobalt (Co), however, has a substantial number of unpaired d-electrons that contribute to robust metallic bonding. Consequently, among the provided options, Cobalt (Co) demonstrates the highest enthalpy of atomisation.

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

The electronic configuration of a neutral Cobalt atom (Co) is: \([Ar] 3d^7 4s^2\).

Formation of the \(Co^{2+}\) ion involves the removal of two electrons from the outermost shell, specifically the \(n=4\) shell. Thus, the two 4s electrons are removed.

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

The number of unpaired electrons (\(n\)) is determined by filling the 3d orbitals according to Hund's rule of maximum multiplicity:

A \(3d^7\) configuration implies 7 electrons distributed across 5 d-orbitals. The initial 5 electrons occupy each orbital singly, followed by pairing of the remaining 2 electrons.

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

The orbital diagram reveals 3 unpaired electrons. Therefore, \(n=3\).

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

Employing 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: Determine 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 yields 4.

The spin-only magnetic moment for Cobalt in its +2 oxidation state is approximately 4 BM (rounded to the nearest integer).