Chemical Bonding And Molecular Structure

To solve the problem, we need to identify which group of elements in the periodic table form electron precise hydrides.

1. Understanding Electron Precise Hydrides:
Electron precise hydrides are those in which all valence electrons are involved in normal 2-center-2-electron (2c–2e) bonds. There are no electron deficiencies or lone pair electrons on the central atom in such compounds.

2. Analyze Each Group:

• Group 12: Elements like Zn, Cd, Hg form hydrides that are mostly ionic or metallic, not electron precise.
• Group 13: Forms electron deficient hydrides, e.g., (diborane), which have multi-center bonding — hence not electron precise.
• Group 14: Carbon, silicon, etc., form hydrides like , , etc., which are electron precise. All valence electrons are involved in standard covalent bonds.
• Group 16: Forms hydrides like , , etc., where lone pairs are present on the central atom — hence not electron precise.

Final Answer:
The correct answer is Group 14, as their hydrides are electron precise.

To solve the problem, we need to calculate the change in Gibbs free energy ( ) using the given internal energy change ( ) and entropy change ( ) at a constant temperature.

1. Recall the formula for Gibbs free energy change:
The relation is given by:

However, for a reaction involving gases, we use:
, and since ,
we modify the formula as:

2. Gather the given data:




From the reaction:
So, change in moles of gas,

3. Apply the formula:

Substituting the values:
(since we convert Cal to K.Cal)


Wait — this is too high for given options. Let’s re-express all values in **consistent units**:

4. Final Answer:
The value of is

To solve the problem, we need to identify which molecule, according to Molecular Orbital (MO) theory, contains only π (pi) bonds and no σ (sigma) bonds between atoms.

1. Understanding MO Theory for C₂:
According to MO theory, the molecular orbital configuration of diatomic carbon (C₂) is:

From the configuration above, we note:

• The bonding electrons in and contribute to π-bonds.
• There are no electrons in , meaning there is no σ-bond formed by the 2p orbitals.

2. Bond Order of C₂:
Bond order = ½ (Number of bonding electrons − Number of antibonding electrons)
= ½ (8 − 2) = 3

But this bond order includes 2 π bonds and one bond due to lower energy levels, but not from 2p σ overlap. Effectively, between the two carbon atoms, only π bonds contribute from 2p orbitals.

3. Comparison with Other Molecules:

• N₂: Has both σ and π bonds.
• O₂: Has σ and π bonds, and also unpaired electrons in π* orbitals.
• B₂: Has only π bonds but fewer electrons and a lower bond order than C₂.

Only C₂ meets the condition of having exclusively π bonds (no 2p σ bond).

Final Answer:
The molecule that contains only π bonds between the atoms is .

To solve the problem, we need to identify in which chemical change the hybridization of the central atom remains the same before and after the reaction.

1. Analyze Each Option:

• Option A:
  - In , carbon is hybridized.
  - In , carbon is hybridized.
  Hybridization changes → Incorrect
• Option B:
  - In , phosphorus is hybridized.
  - In , phosphorus is hybridized.
  Hybridization changes → Incorrect
• Option C:
  - In , boron is hybridized.
  - In , boron is hybridized.
  Hybridization changes → Incorrect
• Option D:
  - In , nitrogen is hybridized.
  - In , nitrogen is still hybridized (lone pair donates to form a coordinate bond).
  No change in hybridization → Correct

Final Answer:
The reaction shows no change in the hybridization of the central atom.

Finding the Temperature at Which the RMS Velocity of SO₂ Matches the Most Probable Velocity of O₂

Step 1: Formula for RMS Velocity of a Gas

The root mean square (RMS) velocity is given by the formula:

v_rms = √(3kT/m)

where:

k is the Boltzmann constant

T is the temperature in Kelvin

m is the molar mass of the gas

For sulfur dioxide (SO₂), at 400 K, the RMS velocity is given by:

v_rms,SO₂ = √(3kT_SO₂/m_SO₂)

Step 2: Formula for Most Probable Velocity of a Gas

The most probable velocity (v_mp) is given by:

v_mp = √(2kT/m)

For oxygen (O₂), the most probable velocity is given by:

v_mp,O₂ = √(2kT_O₂/m_O₂)

Step 3: Setting the RMS Velocity of SO₂ Equal to the Most Probable Velocity of O₂

We equate the two velocities:

√(3kT_SO₂/m_SO₂) = √(2kT_O₂/m_O₂)

Step 4: Simplifying the Equation

Canceling out k and squaring both sides:

(3T_SO₂/m_SO₂) = (2T_O₂/m_O₂)

(T_SO₂/T_O₂) = (2m_SO₂)/(3m_O₂)

Step 5: Using Molecular Masses

The molar masses are:

SO₂: 64 g/mol

O₂: 32 g/mol

Substituting:

(T_SO₂/T_O₂) = (2 × 64)/(3 × 32) = 128/96 = 4/3

Step 6: Solving for T_O₂

Given T_SO₂ = 400 K:

400/T_O₂ = 4/3

T_O₂ = (3 × 400) / 4 = 300 K

Final Answer:

The temperature at which the RMS velocity of SO₂ at 400 K matches the most probable velocity of O₂ is:

300 K

Step 1: Identify Molecule - In the reaction , the gas produced is . - Therefore, is .

Step 2: Structure of - has a bent (or V-shaped) structure due to the presence of a lone pair on the sulfur atom and two S–O double bonds. - The bond angle is approximately , which makes it an \emph{angular} molecule. Hence, the shape of molecule ( ) is (angular).

Step 1: Understanding the Reaction
- Urea ( ) hydrolyzes in the presence of water to form carbamic acid ( ), which is unstable and decomposes into ammonia ( ) and carbonic acid ( ).
- Here, X = Carbamic acid ( ) and Y = Carbon dioxide ( ).
Step 2: Determining Hybridization of Carbon in and
- Carbamic acid ( ) contains a carbonyl ( ) functional group, where the carbon is sp -hybridized due to one double bond and two single bonds.
- Carbon dioxide ( ) has a linear structure with two double bonds, making the carbon sp-hybridized.
Step 3: Identifying the Correct Option
- → Carbon is -hybridized.
- → Carbon is -hybridized.
- This corresponds to Option (1): . ✅

Step 1: Understanding Work Done in a PV Diagram
- Work done by or on a gas is given by the area under the PV curve.
- For an expansion process, work is done by the gas and is positive.
- For a compression process, work is done on the gas and is negative.

Step 2: Identifying the Correct Graph
- The given graph represents a process where pressure decreases as volume increases.
- The shaded area under the curve represents the work done on or by the gas.
- Among the given options, Option (2) correctly represents the work done, as the shading is correctly placed under the curve. Thus, the correct answer is .

Intramolecular hydrogen bonding in o-hydroxybenzaldehyde (A) results in a less rigid structure compared to p-hydroxybenzaldehyde (B), where intermolecular hydrogen bonding leads to a higher degree of molecular association and therefore a higher melting point.

Step 1: Identifying and

• When reacts with at , the product is (diborane).
• Diborane then reacts with water to form boric acid (denoted here as ) and hydrogen gas.

Step 2: Verifying the Statements

1. is an electron-deficient molecule: has 3-center-2-electron (banana) bonds and is electron-deficient. This is true.
2. In , a B--B bond is present: There is no direct B--B single bond in diborane; instead, there are bridging hydrogens. This is false.
3. is a weak tribasic acid: Boric acid ( ) is monobasic in water because it primarily acts via Lewis acidity (accepting ), not by losing 3 protons. This is false.

Step 1: Understanding the Hydrolysis of Xenon (VI) Fluoride
- undergoes complete hydrolysis to form Xenon trioxide ( ): - The molecular structure of is trigonal pyramidal.
Step 2: Counting and Bonds
- Each Xe–O bond contains one -bond.
- Each O has one -bond with Xe.
- Total bonds: bonds.