Molecular Orbital Theory

The question asks us to identify which species does not exist according to molecular orbital theory. Let's analyze each option using the molecular orbital theory to understand their existence.

1. :
   has a total of 3 electrons based on its electronic configuration.
   Molecular orbital electron configuration: .
   Bond Order = .
   Since the bond order is non-zero, can exist.
2. :
   has a total of 3 electrons.
   Molecular orbital electron configuration: .
   Bond Order = .
   Since the bond order is non-zero, can exist.
3. :
   has a total of 8 electrons based on its electronic configuration:
   Molecular orbital electron configuration: .
   Bond Order = .
   Since the bond order is zero, does not exist.
4. :
   has a total of 18 electrons.
   Molecular orbital electron configuration: .
   Bond Order = .
   Since the bond order is non-zero, can exist.

Hence, the species that does not exist is .

The correct answer is (C).
is anti-bonding molecular orbital
∼ is dipole moment
= bond order
is bonding molecular orbital.

1. Determine the electronic configuration and geometry for each species:

-
-
-
-
-
-

2. Count the paramagnetic species:

Paramagnetic species: .
Total = 4 species. Thus, the number of paramagnetic species is 4.

Paramagnetic species have unpaired electrons, determined by ligand field strength and electron configurations. Strong field ligands like CN and CO lead to low-spin complexes, while weak field ligands like Cl and H O result in high-spin complexes.

Using the formula for the radius of the nth orbit: For Li (\textit{Z} = 3, \textit{n} = 2): For Be (\textit{Z} = 4, \textit{n} = 3): The ratio of radii: Thus, the radius of the 3 orbit of Be is .

- Two lone pair one oxygen
- Molecule is ' v ' shaped
- Bond angle is less than 104.5∘(102∘)
- O.S. of 'O' is +2

The correct answer is 5

The orbitals and are considered axial orbitals. The -orbitals, such as and , have their lobes aligned along the x, y, and z axes, respectively.

These are termed as axial orbitals since they lie along the axis of symmetry of a molecule. The -orbitals, specifically and , also align with the axis of symmetry and are part of the axial set, which directly participate in bonding along the principal axis in a molecule.

Solution:
- The molecular orbital theory explains that bond order is given by the formula:
- has a bond order of 2 and is paramagnetic because it has two unpaired electrons.
- When undergoes reduction to , it gains two electrons, and its bond order decreases to 1.
- In the case of , it is paramagnetic with a bond order of 2. When loses an electron to become , it has no unpaired electrons and becomes diamagnetic. Furthermore, the bond order increases to 2.5.
Thus, the process satisfies both conditions of increasing bond order and changing from paramagnetic to diamagnetic.

The orbital angular momentum is given by: For an -orbital, . Substituting :

Final Answer: .

Step 1: Electronic Configuration of Fe²⁺

• The complex is K₄[Fe(CN)₆].
• In this complex, Fe is in the +2 oxidation state:

Step 2: Effect of Strong Field Ligand (CN⁻)

• CN⁻ is a strong field ligand as per the spectrochemical series.
• It causes pairing of electrons in the d-orbitals.
• The t_2g set of orbitals (lower energy) gets fully filled with 6 electrons, forming 3 pairs.

Step 3: Diagrammatic Representation

The electronic configuration in the d-orbitals is:

The t_2g set contains 6 electrons, forming 3 electron pairs.

Final Answer:

The number of electron pairs in the t_2g set is: 3.

• Homoleptic complexes are those in which the central metal ion is surrounded by only one type of ligand. For this case, the complex is [Co(H₂O)₆]²⁺.
• Co²⁺ has the electronic configuration [Ar]3d⁷. The ligand H₂O is a weak field ligand, meaning it does not cause a significant splitting of the d-orbitals.
• The t_2g and e_g orbitals are filled based on Hund’s rule of maximum multiplicity: t_2g⁵ e_g².
• This configuration results in one unpaired electron in the t_2g orbitals.

The molecular orbitals formed from 2s and 2p atomic orbitals are as follows:


Thus, the total number of molecular orbitals formed is 8.

To determine the truth of the given statements about molecular orbitals, we need to understand two concepts: bonding molecular orbitals (MOs) and antibonding molecular orbitals.

1. Statement I: A bonding MO has lower electron density above and below the inter-nuclear axis.
   • A bonding MO is formed by the lateral overlapping of p orbitals. In such MOs, the electron density is concentrated in two lobes above and below the plane of the inter-nuclear axis, not on the axis itself. The statement claims that there is "lower" electron density above and below, which contradicts how bonding MOs actually behave, where electron density is maximized in these regions. Thus, Statement I is false.
2. Statement II: The antibonding MO has a node between the nuclei.
   • A antibonding MO is also formed by the lateral overlapping of p orbitals, but in an out-of-phase manner. This results in a nodal plane between the nuclei where there is zero electron density, which is characteristic of antibonding molecular orbitals. Therefore, Statement II is true.

Based on the reasoning above:

• Statement I is false because the bonding MO should actually have increased electron density above and below the inter-nuclear axis.
• Statement II is true because the antibonding MO correctly has a node between the nuclei.

Thus, the correct answer is: Statement I is false but Statement II is true.

To determine the number of molecules with a bond order of 2, we first need to calculate the bond order for each given molecule. Bond order is given by the formula:

Bond Order = (Number of bonding electrons - Number of antibonding electrons) / 2

Let's apply this formula to each molecule:

Checking our results, and each have a bond order of 2. So, there are 2 molecules with a bond order of 2.

This matches the expected range of 2 to 2, confirming the calculated result is correct.

In molecular orbital theory, molecular orbitals form from the linear combination of atomic orbitals. When dealing with diatomic molecules, these combinations form bonding and antibonding orbitals.

The symbols and represent the wave functions of atomic orbitals from atoms A and B, respectively.

To form molecular orbitals, atomic orbitals can combine constructively or destructively:

• Constructive combination results in a bonding molecular orbital, which is generally represented as .
• Destructive combination results in an antibonding molecular orbital, denoted by , which is typically represented as .

The antibonding molecular orbital, , is formed when the wave functions interfere destructively, leading to a nodal plane between the nuclei where the electron density is low.

Given these explanations, the correct representation of is:

Correct Answer:

Let's rule out the other options:

• : This does not correspond to a standard form for molecular orbitals.
• : This would enhance constructive interference and does not represent an antibonding orbital.
• : This is typically the expression for a bonding orbital, not an antibonding one.

In a diatomic molecule, when combining atomic orbitals, both bonding and anti-bonding molecular orbitals are formed. For atomic orbitals 2s and 2p:

• Combination of 2s orbitals: When two 2s orbitals overlap, they form one bonding (σ_2s) and one anti-bonding (σ_2s^*) molecular orbital.
• Combination of 2p orbitals: The 2p orbitals split into three types: σ and two π orbitals. This results in bonding and anti-bonding combinations:
  • σ_2p orbitals: Form one σ_2p and one σ_2p^* orbital.
  • π_2p orbitals: Each set forms two degenerate π_2p and π_2p^* orbitals. Since there are two sets, this results in two bonding π_2p orbitals and two antibonding π_2p^* orbitals.

Total number of anti-bonding orbitals:

• From 2s: 1 σ_2s^* orbital.
• From 2p: 1 σ_2p^* and 2 π_2p^* orbitals.

Total anti-bonding orbitals = 1 (from 2s) + 1 (from σ_2p^*) + 2 (from 2 π_2p^*) = 4.
This total, 4, falls within the given range (4,4).

To calculate the total number of electrons in the molecular orbitals of , , and , we need to first understand the electronic configuration of molecular oxygen and its ions: : The electronic configuration is . Thus, there are 2 electrons in the orbitals. : Removing an electron gives , with 1 electron in the orbitals. : Adding an electron gives , with 3 electrons in the orbitals. Adding the electrons across these species: . The total number of electrons is 6, which is within the provided range.

The molecular orbitals formed from 2s and 2p atomic orbitals are as fo

The problem asks for the total number of molecular orbitals that are formed by the combination of the 2s and 2p atomic orbitals from the two atoms in a diatomic molecule.

Concept Used:

The formation of molecular orbitals (MOs) is explained by the Linear Combination of Atomic Orbitals (LCAO) theory. The fundamental principle of this theory is the conservation of orbitals, which states that the total number of molecular orbitals formed is always equal to the total number of atomic orbitals (AOs) that are combined.

Step-by-Step Solution:

Step 1: Identify the atomic orbitals contributed by the first atom.

For one atom, we are considering the atomic orbitals in the second principal energy level (n=2). These are:

• One 2s orbital.
• Three 2p orbitals (which are 2pₓ, 2pᵧ, and 2p₂).

So, the first atom contributes a total of atomic orbitals.

Step 2: Identify the atomic orbitals contributed by the second atom.

Since the molecule is diatomic, there is a second atom. This second atom also contributes its 2s and 2p atomic orbitals for bonding.

• One 2s orbital.
• Three 2p orbitals (2pₓ, 2pᵧ, and 2p₂).

The second atom also contributes a total of atomic orbitals.

Step 3: Calculate the total number of atomic orbitals being combined.

The total number of atomic orbitals from both atoms is the sum of the orbitals contributed by each atom.

So, a total of 8 atomic orbitals are combined.

Final Computation & Result:

Step 4: Apply the principle of conservation of orbitals to find the total number of molecular orbitals.

According to the LCAO theory, the number of molecular orbitals formed must be equal to the number of atomic orbitals that were combined.

These 8 molecular orbitals are typically designated as .

The total number of molecular orbitals formed is 8.

To determine which statements are true regarding H O, NH , and CH , let's analyze each statement one by one:

1. Statement (A): The central atoms of all the molecules are sp hybridized.
   • H O (Water) - The oxygen atom is sp hybridized, forming a bent shape due to two lone pairs.
   • NH (Ammonia) - The nitrogen atom is sp hybridized, forming a trigonal pyramidal shape due to one lone pair.
   • CH (Methane) - The carbon atom is sp hybridized, forming a tetrahedral shape with no lone pairs.
2. Statement (B): The H–O–H, H–N–H and H–C–H angles in the above molecules are 104.5°, 107.5° and 109.5° respectively.
   • The bond angle in H O is about 104.5° due to the repulsion between lone pairs.
   • The bond angle in NH is about 107.5° due to the presence of one lone pair.
   • The bond angle in CH is 109.5° as it is a perfect tetrahedral shape.
3. Statement (C): The increasing order of dipole moment is CH O.
   • CH (Methane) has no net dipole moment as it is symmetrical.
   • NH (Ammonia) has a dipole moment due to its trigonal pyramidal shape.
   • H O (Water) has the highest dipole moment due to its bent shape and strong hydrogen bonding.
4. Statement (D): Both H O and NH are Lewis acids and CH is a Lewis base.
   • Lewis Acid - An electron pair acceptor. H O and NH are not typically Lewis acids; rather, they can donate electrons.
   • Lewis Base - An electron pair donor. CH is not a Lewis base as it does not have available lone pairs to donate.
5. Statement (E): A solution of NH in H O is basic. In this solution NH and H O act as Lowry-Bronsted acid and base respectively.
   • In an aqueous solution, NH acts as a Bronsted-Lowry base as it accepts a proton from H O.
   • H O acts as a Bronsted-Lowry acid as it donates a proton to NH .

After evaluating each statement, the true statements are (A), (B), and (C). Therefore, the correct answer is: A, B, and C only.

Paramagnetism arises due to the presence of unpaired electrons in a molecule. Let’s analyze each molecule's electronic configuration to determine the number of paramagnetic molecules:

- O (Oxygen): Oxygen molecule has a total of 16 electrons. According to molecular orbital theory, the electronic configuration for O is: This shows that there are 2 unpaired electrons, making O paramagnetic.

- N (Nitrogen): Nitrogen molecule has a total of 14 electrons. Its molecular orbital configuration is: All electrons are paired in N , so it is diamagnetic.

- F (Fluorine): Fluorine molecule has a total of 18 electrons. The electronic configuration is: This configuration shows 2 unpaired electrons, making F paramagnetic.

- B (Boron): Boron molecule has a total of 10 electrons. Its electronic configuration is: This configuration shows 2 unpaired electrons, making B paramagnetic.

- Cl (Chlorine): Chlorine molecule has a total of 18 electrons. The electronic configuration is: This configuration shows 2 unpaired electrons, making Cl paramagnetic. Thus, the paramagnetic molecules are O , F , and B . Therefore, the number of paramagnetic molecules is 3.

To determine which linear combinations of atomic orbitals will lead to the formation of molecular orbitals in homonuclear diatomic molecules with the internuclear axis in the z-direction, we need to consider the symmetry and orientation of the atomic orbitals involved.

1. and : orbitals are aligned along the z-axis, while orbitals are aligned along the x-axis. Due to their differing orientations, these orbitals cannot effectively overlap to form a molecular orbital. Hence, this combination is not possible.
2. and : The orbital is spherically symmetric, while is oriented along the x-axis. The lack of proper orientation means they cannot combine to form a molecular orbital. Thus, this combination is not feasible.
3. and : Both these d-orbitals lie in the xy-plane and have symmetry that does not match the z-axis orientation necessary for forming bonds in diatomic molecules with the internuclear axis in the z-direction. Therefore, they cannot form molecular orbitals.
4. and : The spherical symmetry of and the z-orientation of allow them to combine effectively since the orientation along the z-axis is appropriate for bonding in homonuclear diatomic molecules. Thus, this combination will form molecular orbitals.
5. and : Similar to the point above, the orbital is in the z-direction, whereas the orbital lies in the xy-plane, making it inappropriate for forming molecular orbitals along the z-axis. Hence, this combination is not possible.

Therefore, the correct choice is "D only," as only the combination of and leads to molecular orbital formation in the context described.

Step 1: Determine bond order and magnetic nature. 

Step 2: Identify matching properties. 
O and N both have bond order 2.5 and are paramagnetic.