IIT JAM Chemistry Mock Test 1 - Chemistry MCQ

Correct Answer: D (12.18 K)

Explanation:
The relationship between Kp and Kc is given by the equation:

Kp = Kc(RT)^(Δn)

where Δn is the change in the number of moles of gas in the reaction, R is the ideal gas constant, and T is the temperature in Kelvin.

For the given equilibrium, N2O4(g) → 2NO2(g), the change in the number of moles of gas, Δn, is:

Δn = (moles of products) - (moles of reactants)
Δn = 2 - 1
Δn = 1

We know that Kp = Kc, so:

Kc(RT)^(Δn) = Kc

This equation simplifies to:
RT = 1
Let's consider the possibility that the constant R used in the question is given in a different unit, such as atm·L/(mol·K). In this case, R = 0.0821 atm·L/(mol·K). We can solve for T again:

T = 1/R
T = 1/0.0821
T ≈ 12.18 K

This value matches option D, so the correct answer is 12.18 K.

Correct answer is option D) All
Hydrophobic Amino Acids

• Phe (Phenylalanine)


• Try (Tryptophan)


• Ile (Isoleucine)

Answer:

• All of the mentioned amino acids (Phe, Try, and Ile) are hydrophobic.


 

The given reactions involve the partial hydrogenation of an alkyne.

Reaction A: H₂, Pd/BaSO₄​ (Lindlar's Catalyst)

Lindlar's catalyst is used to hydrogenate alkynes to cis-alkenes. It adds hydrogen to the alkyne in a syn addition manner, producing a cis-alkene. For the given compound, the alkyne would be converted to a cis-alkene.

Reaction B: Na, Liquid NH₃​ (Birch Reduction)

The Birch reduction reduces alkynes to trans-alkenes via an anti addition of hydrogen. For the given compound, the alkyne would be converted to a trans-alkene.

Given the molecular structure in the image, let's analyze the products of each reaction:

1. Lindlar's Catalyst (Reaction A):

   • Converts the alkyne to a cis-alkene.
   • The alkyne shown will be converted to a single cis-alkene product because the reaction is stereoselective.

2. Birch Reduction (Reaction B):

   • Converts the alkyne to a trans-alkene.
   • The alkyne shown will be converted to a single trans-alkene product because the reaction is stereoselective.

Conclusion:

Both reactions A and B give single products. Therefore, the correct answer is:

Option B: Both give single product

Correct answer is  option (D).
Decreasing Order of Stretching Frequency of C-O Bond (cm–1):
- IV) Mo(CO)3(PCl3)3
- II) Mo(CO)3[P(OPh)3]3
- III) Mo(CO)3(PMe3)3
- I) Mo(CO)3(NMe3)3

The decreasing order of stretching frequency of the C-O bond (cm–1) can be explained by considering the electron-donating or withdrawing ability of the ligands attached to the metal center. The higher the electron-donating ability of the ligands, the more electron density will be present on the metal center, which will weaken the C-O bond and result in a lower stretching frequency.

In this case, the order can be explained as follows:
- Mo(CO)3(PCl3)3 (IV): The PCl3 ligands are strong electron-withdrawing groups, which causes the C-O bond to be stronger and have a higher stretching frequency.
- Mo(CO)3[P(OPh)3]3 (II): The P(OPh)3 ligands are less electron-withdrawing than PCl3, resulting in a weaker C-O bond and a lower stretching frequency compared to IV.
- Mo(CO)3(PMe3)3 (III): The PMe3 ligands are electron-donating groups, which further weakens the C-O bond and lowers the stretching frequency compared to II.
- Mo(CO)3(NMe3)3 (I): The NMe3 ligands are strong electron-donating groups, which results in the weakest C-O bond and the lowest stretching frequency among all the complexes.

Therefore, the correct order is IV > II > III > I.

Correct answer is Option (C).

Explanation:

In the given molecule CN—CH2—CH2—CN, we can analyze the stable conformations based on the dihedral angle between the CN groups. The conformations are:

A: Gauche-form
- In this conformation, the dihedral angle between the CN groups is around 60°.
- The molecule will have a slight steric hindrance between the CN groups as they are close to each other.
- This form is not the most stable conformation.

B: Partially-eclipsed
- In this conformation, the dihedral angle between the CN groups is around 120°.
- The molecule will have a moderate steric hindrance between the CN groups.
- This form is less stable than the anti-form.

C: Anti-form
- In this conformation, the dihedral angle between the CN groups is 180°.
- This conformation has the least steric hindrance between the CN groups as they are farthest apart.
- This form is the most stable conformation.

D: Eclipsed-form
- In this conformation, the dihedral angle between the CN groups is 0°.
- The molecule will have maximum steric hindrance between the CN groups as they are directly overlapping.
- This form is the least stable conformation.

Hence, the most stable conformation for the given molecule CN—CH2—CH2—CN is the anti-form (Option C), where the dihedral angle between the CN groups is 180° and there is minimum steric hindrance.

Based on this, Option D is the most similar to the given energy equation for a particle in a 1D box, as it includes the correct placement of the 2m in the denominator along with the length of the box squared. The only difference is the absence of the quantum number n² and the π² term.

In bourdon tubes converts input pressure into displacement and displacement of the needle will be directly proportional to input pressure.