Chemistry Test 228

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Chemistry Test 228

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Weekly Quiz Competition

Question 1
4 / -1

Given below are two statements :

Statement (I) : A spectral line will be observed for a 2p_x → 2p_y transition.

Statement (II) : 2p_x and 2p_y are degenerate orbitals.

In the light of the above statements, choose the correct answer from the options given below :

A
Both Statement-I and Statement-II are true.

B
Both Statement-I and Statement-II are false.

C
Statement-I is true but Statement-II is false.

D
Statement-I is false but Statement-II is true.

Solution

Statement (I): A spectral line will be observed for a 2px → 2py transition.

This statement is false. A transition between 2px and 2py orbitals does not result in a spectral line because these orbitals are degenerate (having the same energy level) and such transitions do not involve the emission or absorption of light.

Statement (II): 2px and 2py are degenerate orbitals.

This statement is true. The 2px and 2py orbitals are degenerate, meaning they have the same energy level in the absence of an external field. All three 2p orbitals (2px, 2py, 2pz) in an atom are degenerate.

Considering the above analysis:

Correct Option: 4) Statement-I is false but Statement-II is true.
Question 2
4 / -1

Chlorine undergoes disproportionation in alkaline medium as shown below:

a CI₂(g) + b OH^-(aq) → c CIO^-(aq) + d CI^-(aq) + e H₂O(l)

The values of a, b, c and d in a balanced redox reaction are respectively:

A
1, 2, 1 and 1

B
2, 2, 1 and 3

C
3, 4, 4 and 2

D
2, 4, 1 and 3
Solution
- Step 1: Oxidation and Reduction Half-Reactions
  
  Reduction:
  
  Cl₂ → Cl⁻ + e⁻
  
  Oxidation:
  
  Cl₂ + 2OH⁻ → ClO⁻ + H₂O + 2e⁻
- Step 2: Combine the Half-Reactions
  
  2Cl₂ → 2Cl⁻ + 2e⁻
  
  Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O
  
  Multiply the reduction reaction by 2 to balance the electrons:
  
  Add the reduction and oxidation reactions:
- Step 3: Determine Coefficients
  
  a = 1 (Cl₂), b = 2 (OH⁻), c = 1 (ClO⁻), d = 1 (Cl⁻)
  
  From the balanced reaction:
Correct Answer: 1) 1, 2, 1, and 1
Question 3
4 / -1

Consider the following elements In, Tl, Al, Pb, Sn and Ge.

The most stable oxidation states of elements with highest and lowest first ionisation enthalpies, respectively, are

A
+2 and +3

B
+4 and +3

C
+4 and +1

D
+1 and +4
Solution
Given elements: In, Tl, Al, Pb, Sn, Ge.
- First ionisation enthalpies (IE₁) of the elements (trends generally observed in the periodic table):
  
  Highest IE₁: Ge (Germanium)
  
  Lowest IE₁: In (Indium)
- Most stable oxidation states of these elements:
  
  Ge (Germanium): +4
  
  In (Indium): +3
Therefore, the correct answer is +4 and +3
Question 4
4 / -1

Among the following:

The set which represents aromatic species is

A
I, II and III

B
III, IV and V

C
II and III

D
I, II and IV
Solution
- In the given problem, we need to determine which of the species (I, II, III, IV, and V) are aromatic.
- Looking at the structures of each, we analyze their conjugation and the number of π-electrons:
  
  Structure I: A benzene ring with alternating single and double bonds, possessing 6 π-electrons. According to Hückel's rule (4n+2), it is aromatic.
  
  Structure II: A naphthalene with 10 π-electrons. It follows the 4n+2 rule, so it is aromatic.
  
  Structure III: A 5-membered ring with 4 π-electrons. This doesnot fits the criteria for aromaticity and is aromatic, but based on the options, this structure is not considered aromatic in this case.
  
  Structure IV: This structure has 6 π-electrons and fits Hückel's rule, making it aromatic as well.
  
  Structure V: Does not follow the 4n+2 rule, making it non-aromatic.
Therefore, the aromatic species are I, II, and IV.
Question 5
4 / -1

The dipole moments of CCl₄, CHCl₃ and CH₄ are in the order:

A
CH₄ < CCl₄ < CHCl₃

B
CHCl₃ < CH₄ = CCl₄

C
CH₄ = CCl₄ < CHCl₃

D
CCl₄ < CH₄ < CHCl₃
Solution
- Tetrahedral geometry:
  
  CCl₄: Symmetrical structure with four Cl atoms. All dipole moments cancel out, so ({μ_net= 0).
  
  CH₄: Symmetrical structure with four H atoms. All dipole moments cancel out, so (μ_net = 0).
  
  CHCl₃: Asymmetrical structure with three Cl atoms and one H atom. Dipoles do not cancel out completely, so (μ_net > 0).
Order of dipole moments:
- CH₄ and CCl₄: Both have zero dipole moment.
- CHCl₃: Has a net dipole moment greater than zero.
Thus, the correct order is: 3) CH₄ = CCl₄ < CHCl₃
Question 6
4 / -1

The mechanism involved in the preparation of glycol from 1,2-dihaloethane using aqueous Na₂CO₃ is

A
SN¹ attack by OH^-

B
SN²attack by Br^-

C
SN² attack by OH^-

D
SN¹ attack by Br^-
Solution
In the preparation of glycol from 1,2-dihaloethane using aqueous Na₂CO₃:
- The mechanism involves:
  
  1,2-Dihaloethane (Br-CH2-CH2-Br) has two bromine atoms that can act as leaving groups.
  
  The hydroxide ion (OH-) from the aqueous Na2CO3 acts as the nucleophile.
  
  In the SN2 mechanism, the nucleophile (OH-) attacks the carbon atom bonded to the bromine atom from the opposite side, leading to the simultaneous displacement of the bromine atom.
  
  This results in the formation of a glycol (HO-CH2-CH2-OH) with inversion of configuration at the carbon centers.
Therefore, the mechanism involved in the preparation of glycol from 1,2-dihaloethane using aqueous Na₂CO₃ is SN² attack by OH-.
Question 7
4 / -1

Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction is spontaneous at boiling point of water. Choose the correct option.

A
Both ΔH and ΔS are (+ve)

B
ΔH is (–ve) but ΔS is (+ve)

C
ΔH is (+ve) but ΔS is (–ve)

D
Both ΔH and ΔS are (–ve)
Solution
- The reaction is non-spontaneous at the freezing point of water (273 K), meaning:
  
  ΔG = ΔH - TΔS > 0 at 273 K
- The reaction is spontaneous at the boiling point of water (373 K), meaning:
  
  ΔG = ΔH - TΔS < 0 at 373 K
- For an endothermic reaction to be non-spontaneous at low temperature and spontaneous at high temperature:
  
  ΔH must be positive (since the reaction is endothermic).
  
  ΔS must also be positive to ensure that increasing the temperature (T) makes the TΔS term large enough to make ΔG negative at higher temperatures.
Therefore, the correct answer is (1): Both ΔH and ΔS are positive.
Question 8
4 / -1

Choose the correct answer from the options given below :

A
A - IV, B - III, C - II, D - I

B
A - III, B - IV, C - II, D - I

C
A - III, B - IV, C - I, D - II

D
A - II, B - I, C - IV, D - III

Solution
Question 9
4 / -1

The Molarity (M) of an aqueous solution containing 5.85 g of NaCl in 500 mL water is :

(Given : Molar Mass Na : 23 and Cl : 35.5 gmol^–1)

A
20

B
0.2

C
2

D
4
Solution
- Mass of NaCl = 5.85 g
- Volume of water = 500 mL = 0.5 L
- Molar Mass of Na (Na) = 23 g/mol
- Molar Mass of Cl (Cl) = 35.5 g/mol
1. Calculate the molar mass of NaCl:
  
  Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
2. Calculate the number of moles of NaCl:
  
  Number of moles = mass of NaCl / molar mass of NaCl
  = 5.85 g / 58.5 g/mol
  = 0.1 moles
3. Convert the volume of the solution from mL to L:
  
  Volume = 500 mL = 0.5 L
4. Calculate the molarity (M):
  
  Molarity (M) = number of moles of solute / volume of solution in liters
  = 0.1 moles / 0.5 L
  = 0.2 M
The correct answer is: 0.2
Question 10
4 / -1

The correct statement amongst the following is :

A
The term 'standard state' implies that the temperature is 0°C

B
The standard state of pure gas is the pure gas at a pressure of 1 bar and temperature 273 K

C

D
Solution
1. Statement 1: Incorrect — Standard state does not mean temperature is fixed at 0°C.
2. Statement 2: Incorrect — Standard state requires 1 bar pressure, but temperature is not fixed at 273 K.
3. Statement 3: Incorrect — O(g) is not the most stable form of oxygen; Δ_fH° ≠ 0 for atomic oxygen.
4. Statement 4: Correct — O₂(g) in its diatomic form is the standard state of oxygen, so Δ_fH° = 0 at any specified temperature, including 500 K.
Therefore, the correct answer is— Δ_fH°₅₀₀ is zero for O₂(g).