Chemistry Test 226

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Chemistry Test 226

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Weekly Quiz Competition

Question 1
4 / -1

A mixture of Al and Zn weighing 1.67 g was completely dissolved in acid, and the evolved 1.69 litres of hydrogen gas was measured at 273 K and one atmosphere pressure. The respective masses of Al and Zn in the original mixture were

A
1.25 g, 0.42 g

B
0.42 g, 1.25 g

C
1.25 g, 4.2 g

D
12.5 g, 4.2 g

Solution

Let 'a' g and 'b 'g be the weights of Al and Zn in the mixture.

a + b = 1.67 … (1)

M_eq of Al + M_eq of Z_n = Meq of H₂

From equations (1) and (2)

a = 1.25 g Al

b = 0.42 g Zn
Question 2
4 / -1

In which of the following sets of quantum numbers will an electron have the highest energy?

A
4, 0, 0, 1/2

B
3, 2, 2, 1/2

C
3, 1, 1, 1/2

D
2, 1, 1, 1/2

Solution

The orbital with highest value of (n + l) has the highest energy.

For set, 3, 2, 2, 1/2.

Value of n + l is 3 + 2 = 5

It is the highest among given options, hence electron will have the highest energy for this set of quantum numbers.
Question 3
4 / -1

In aqueous solution, the ionisation constants for carbonic acid are K₁ = 4.2 × 10^-7 and K₂ = 4.8 × 10^-11.

Select the correct statement for a saturated 0.034 M solution of carbonic acid.

A
The concentration of CO₃²- is 0.034.

B
The concentration of CO₃^2- is greater than that of HCO₃^-.

C
The concentrations of H⁺ and HCO₃^- are approximately equal.

D
The concentration of H⁺ is double that of CO₃^2-.

Solution

Second dissociation constant is much smaller than the first one. Just a small fraction of total HCO^-₃ formed will undergo second stage of ionisation. Hence, in saturated solution,
Question 4
4 / -1

In which of the following arrangements is the order not according to the property indicated against it?

A
Al³⁺ < Mg² < Na⁺ < F^- - Increasing ionic size

B
B < C < N

C
I < Br < F < Cl - Increasing electron gain enthalpy (with negative sign)

D
Li < Na < K < Rb - Increasing metallic radius

Solution

As we move from left to right across a period, ionisation enthalpy increases with increasing atomic number. So, the order of increasing ionisation enthalpy should be B < C < N < O.

But, N (1s² 2s² 2p³) has a stable half filled electron configuration. So, ionisation enthalpy of nitrogen is greater than oxygen. So, the correct order of increasing first ionisation enthalpy is B < C < O < N.

Hence, the order indicated in the option (2) is correct.
Question 5
4 / -1

The metallic lustre exhibited by sodium is explained by

A
diffusion of sodium ions

B
oscillations of loose electrons

C
excitation of free protons

D
existence of body centered cubic lattices

Solution

The metallic lustre of sodium is due to oscillations of loose electrons. Metals are lustrous because of the free electrons they have. The free electrons can move freely in the metal, causing any light incident on them to get reflected back.
Question 6
4 / -1

What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?

A

B
and H2O are formed.

C

D

Solution

Dilute nitric acid converts chromate into dichromate and H₂O.

2K₂CrO₄ + 2HNO₃ → K₂Cr₂O₇ + 2KNO₃ + H₂O

⇒ 2CrO₄⁻² ⇌ H⁺ + Cr₂O₇²⁻ + H₂O

Acidic medium shifts the equilibrium in the direction of the dichromate ion.
Question 7
4 / -1

Which of the following is not aromatic?

A
Cyclopentadienyl cation

B
Cyclopentatrienyl cation

C
Cyclopentadienyl anion

D
Antracene

Solution

Cyclopentadienyl cation is not aromatic as it contains only 4 pi electrons. So, it does not follow (4n + 2) pi electron rule.
Question 8
4 / -1

Which of the following statements is correct?

A
The addition of H₂ to cis-2, 3-dibromobutene produces meso-2, 3-dibromobutane.

B
The addition of H₂ to trans-2, 3-dibromobutene produces meso-2, 3-dibromobutane.

C
The hydroxylation of cis-2, 3-dimethylbutene with Baeyer's reagent produces racemic-2, 3-dihydroxybutane.

D
The hydroxylation of trans-2, 3-dimethylbutene with Baeyer's reagent produces meso-2, 3-dihydroxybutane.

Solution

(1) Hydrogenation of an alkene is a heterogeneous process which occurs on the surface of a solid catalyst (PtO₂ or Pd/C). The addition occurs with syn stereochemistry (both hydrogens add from the same side). For cis isomer, the product is a meso compound.

(3) The addition of two hydroxyl groups involves syn addition (on the same side of a double bond). The result is same as that of hydrogenation. The cis isomer produces meso compound.

(4) The addition of two hydroxyl groups involves syn addition (on the same side of a double bond). The result is same as that of hydrogenation. The trans isomer produces racemic mixture.
Question 9
4 / -1

A

B

C

D

Solution
Question 10
4 / -1

Which of the following has the composition of Schweizer reagent?

A
[Ag(NH₃)₂]Cl

B
[Ag(NH₃)₄]Cl

C
[Cu(NH₃)₄(H₂O)₂](OH)₂

D
[Cu(NH₃)₃](OH)₂

Solution

Schweizer reagent is the chemical complex tetraamminediaquacopper dihydroxide, [Cu(NH₃)₄(H₂O)₂](OH)₂. It is prepared by precipitating copper (II) hydroxide from an aqueous solution of copper sulphate using sodium hydroxide or ammonia, then dissolving the precipitate in a solution of ammonia.