D Block Elements

The correct answer is Option A) : paramagnetic behaviour

V = 3 unpaired electrons 
Cr = 4 unpaired electrons 
Mn = 5 unpaired electrons 
Fe = 4 unpaired electrons 
Hence, the order of paramagnetic behaviour should be 

Ionic size decreases from left to right in the same period 
Paramagnetic property is dependent on the number of unpaired electrons. As the unpaired electrons increase in number, the paramagnetic property also increases.

Discover More From Chapter: D and F Block Elements

When is acidified with dilute and exposed to , it gets reduced, turning green due to the formation of ions:

.

Here are the colors of the given compounds:

• : White
• : Yellow
• : Brown
• (Prussian blue): Blue

Conclusion:

The colors are as listed above, with Prussian blue being the most notable due to its distinct deep blue color.

The question asks about the stability of copper in its different oxidation states, Cu(I) and Cu(II). To determine which is more stable under standard conditions, we need to consider both the electronic configuration and some known empirical facts about copper salts.

Electronic Configuration:

• The electronic configuration of Cu(II) is 3d⁹.
• The electronic configuration of Cu(I) is 3d¹⁰.

Cu(II) has one less electron in the d-orbital compared to Cu(I). In terms of electronic configuration, a completely filled or half-filled d-orbital is generally more stable. Yet, in this case, the d⁹ configuration of Cu(II) may initially appear less stable compared to the d¹⁰ of Cu(I).

Stability of Copper Oxidation States:

The stability of different oxidation states of an element can be influenced by various factors, including lattice energy, solvation energy, and the Jahn-Teller effect:

• Jahn-Teller Effect: Cu(II)'s d⁹ configuration is subject to Jahn-Teller distortion, leading to stabilization.
• Coordination Chemistry: Cu(II) is more stable in aqueous solution due to the higher hydration energy compared to Cu(I).

Considering the above factors, Cu(II) is generally more stable than Cu(I) in aqueous solution, and the empirical evidence from copper salts often shows that Cu(II) compounds are more common and stable.

Conclusion:

The correct answer is: Cu(II) is more stable. Therefore, the assertion that "Cu(II) is more stable" is supported by the greater stability of Cu(II) due to factors like the Jahn-Teller effect and higher hydration energy.

To determine the pair of transition metal ions with the same number of unpaired electrons, we need to analyze the electronic configurations of the given ions.

1. First, let's find the electronic configuration for each ion in their ground state:
2. Vanadium (V) has an atomic number of 23:
   • Neutral Vanadium (V):
   • For : remove two electrons from 4s, giving . It has 3 unpaired electrons.
3. Cobalt (Co) has an atomic number of 27:
   • Neutral Cobalt (Co):
   • For : remove two electrons from 4s, giving . It has 3 unpaired electrons.
4. Titanium (Ti) has an atomic number of 22:
   • For (similar method as above): . It has 2 unpaired electrons.
   • For : . It has 1 unpaired electron.
5. Iron (Fe) has an atomic number of 26:
   • For : . It has 5 unpaired electrons.
6. Chromium (Cr) has an atomic number of 24:
   • For : . It has 4 unpaired electrons.
7. Manganese (Mn) has an atomic number of 25:
   • For : . It has 5 unpaired electrons.

By analyzing these configurations, and both have 3 unpaired electrons. Hence, the correct answer is:

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 ( ): The magnetic moment of an ion due to the spin of its unpaired electrons can be calculated using the formula:

where 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 . It has 3 valence electrons, leading to relatively weaker metallic bonding compared to other transition metals.
• Manganese (Mn, Z=25): Electronic configuration is . Due to the very stable half-filled 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 . 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 . 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 ion.

The neutral Cobalt atom (Co) has the configuration: .

To form the ion, two electrons are removed from the outermost shell, which is the shell. So, the two 4s electrons are removed.

The electronic configuration of is: .

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

The 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.

From the orbital diagram, we can see there are 3 unpaired electrons. So, .

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

Using the formula with :

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

The value of is approximately 3.87.

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).

To determine the correctness of the given statements, let's analyze each one in detail:

Statement I: CrO₃ is a stronger oxidizing agent than MoO₃

This statement is true. Chromium trioxide (CrO₃) is known to be a powerful oxidizing agent. It readily accepts electrons due to its high oxidation state of +6, which makes it an effective electron acceptor and oxidizing agent. In contrast, Molybdenum trioxide (MoO₃) is a less effective oxidizing agent in comparison. The difference in oxidizing strengths can be attributed to the electronic configuration and periodic trends; Cr is generally a stronger oxidizing agent than Mo.

Statement II: Cr(VI) is more stable than Mo(VI)

This statement is false. The stability of oxidation states tends to decrease as you move down a group in the periodic table. Chromium in the +6 oxidation state (Cr(VI)) is not as stable as Molybdenum in the +6 oxidation state (Mo(VI)). Molybdenum's higher atomic number and lower position in the group make Mo(VI) more stable compared to Cr(VI), which is more prone to reduction and disproportionation.

Conclusion:

Based on the above analysis, Statement I is true but Statement II is false. Therefore, the correct answer is: Statement I is true but Statement II is false.