Mix Test

Result

Mix Test - 4

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Weekly Quiz Competition

Question 1
1 / -0

When electrons are trapped into the crystal lattice in place of anion vacancy, the defect in the crystal is known as

A
the Schottky defect

B
the Frenkel defect

C
dislocations

D
F-centre

Solution

When the electron is trapped in anion, valency of crystal lattice is referred to as F-centre (means colour).
Question 2
1 / -0

The pH of the solution containing 5mL of M/5 HCl and 10 mL of M/10 NaOH is

A
5

B
3

C
7

D
8

Solution
Question 3
1 / -0

The coagulating power of an electrolyte for arsenious sulphide decreases in the order:

A
Na⁺ > Al³⁺ > Ba²⁺

B

C

D

E

Solution

According to the Hardy Schulze rule, the coagulating power of an electrolyte is directly proportional to the charge on the active ion.
Since arsenius sulphide is a negatively charged sol, the cation of the electrolyte is the active ion.
Therefore, coagulating power of an electrolyte for arsenious sulphide decreases as: Al³⁺ > Ba²⁺ > Na⁺
Question 4
1 / -0

The correct order of hybridisation of the central atom in the following species?

NH₃, [PtCl₄]^2-, PCl₅ and BCl₃ is

A
dsp², dsp³, sp², sp³

B
sp³, dsp², dsp³, sp²

C
dsp², sp², sp³, dsp³

D
dsp², sp³, sp², sp³

Solution
Question 5
1 / -0

As compared to nitrogen, oxygen is

A
less electronegative and less reactive

B
more electronegative and less reactive

C
more electronegative and more reactive

D
less electronegative and more reactive

Solution

Electronegativity increases on moving from left to right in the periodic table. So, as compared to nitrogen, oxygen is more electronegative and more reactive.
Question 6
1 / -0

Which of the following conversions involves a change in both hybridisation and geometry?

A
NH₃ to NH₄⁺

B
AlCl₃to AlCl₄^-

C
H₂O to H₃O⁺

D
PCl₃ to PCl₄⁺

E
CH₄ to CCl₄

Solution

AlCl₃: sp² hybridisation, trigonal planar geometry
AlCl₄^-: sp³ hybridisation, tetrahedral geometry
Hence, option (2) is correct.

NH₃: sp³ hybridisation, trigonal pyramidal geometry
NH₄⁺: sp³ hybridisation, tetrahedral geometry
AlCl₃: sp² hybridisation, trigonal planar geometry
AlCl₄^-: sp³ hybridisation, tetrahedral geometry
PCl₃: sp³ hybridisation, trigonal pyramidal geometry
PCl₄⁺: sp³ hybridisation, tetrahedral geometry
Question 7
1 / -0

At low pressures, the compression factor of CO₂ having constant value of molar volume

A
increases with increase in temperature

B
decreases with increase in temperature

C
remains constant with change in temperature

D
is greater than one

Solution

At lower pressure
Hence, increasing temperature increases the value of Z.
Question 8
1 / -0

For the reaction

C₃H₈ (g) + 5O₂ (g) 3CO₂ (g) + 4H₂O ()

At constant temperature, – is

A
– RT

B
+ RT

C
– 3RT

D
+ 3RT

Solution

For the reaction

C₃H₈ (g) + 5O₂ (g) 3CO₂ (g) + 4H₂O (l)
Question 9
1 / -0

Which of the following is the strongest base?

OH^–, F^–,,

A

B

C
F^–

D
OH^–

Solution

The weakest acid amongst H₂O, HF, NH₃ and CH₄ is CH₄. Hence, the strongest base will be .
Question 10
1 / -0

Which of the following processes is used to convert vegetable oil to vegetable ghee?

A
Hydrogenation

B
Oxidation

C
Reduction

D
Decomposition

Solution

On commercial scale, vegetable oils are converted into vegetable ghee by hydrogenation process.
Vegetable oil + Hydrogen Vegetable ghee
Question 11
1 / -0

The product(s) of photochemical smog is/are:

1. Peroxyacetyl nitrate
2. Carbon monoxide
3. Ozone

A
Only 1

B
1 and 2

C
2 and 3

D
1 and 3

E
None of these

Solution

The products of photochemical smog are peroxyacetyl nitrate and ozone.

Formation of Photochemical Smog:

Step 1: People begin driving in the morning; nitrogen is burned or oxidised.
N₂+ O₂ 2NO ... (1)
The oxidation number of N₂ is 0. The nitrogen in NO has acquired an oxidation number of +2.

Step 2: After a few hours, NO combines with O₂, in another oxidation reaction.
2NO + O₂ 2NO₂... (2)
The nitrogen in NO has an oxidation number of +2. The nitrogen in NO₂ has an oxidation number of +4.

Step 3: Nitrogen dioxide absorbs light energy, resulting in a reduction reaction.
NO₂ NO + O ... (3)
The nitrogen in NO₂ has an oxidation number of +4 and the nitrogen in NO has an oxidation number of +2.

Step 4: In sunlight, atomic oxygen combines with oxygen gas to form ozone.
O + O₂ O₃... (4)

Step 5: Reaction is temperature and sunlight dependent.
O₃+ NO NO₂+ O

This self consuming cycle is disrupted when there are unburnt hydrocarbons present in the atmosphere.
NO and NO₂ can also react with the hydrocarbons instead of ozone to form other volatile compounds known as PAN (peroxyacetyl nitrate). The accumulation of ozone and volatile organic compounds along with the energy from the sun forms the brown, photochemical smog seen on hot, sunny days.
Question 12
1 / -0

The general formula B_nH_n+6 for boranes, corresponds to

A
closo-borane

B
nido-boranes

C
arachno-borane

D
hypho-borane

E
None of these

Solution

Arachno-borane has the structural formula- B_nH_n+6.
Question 13
1 / -0

In the reaction 2X + Y X₂Y, the reactant X disappears at

A
half the rate as that of the disappearance of Y

B
the same rate as that of the disappearance of Y

C
the same rate as that of the appearance of X₂Y

D
twice the rate as that of the appearance of X₂Y

Solution
Question 14
1 / -0

A compound of Xe and F found to have 53.5% Xe, the oxidation number of Xe in this compound is (at.wt. of Xe = 131, F = 19)

A
- 6

B
+ 6

C
- 4

D
+ 4

Solution

Number of gram atoms of Xe and F are 53.5/131 = 0.4 and 46.5/20 respectively. Xe and F are present in the ratio 1 : 6 (XeF₆). Oxidation number of Xe is + 6.
Question 15
1 / -0

Nitration of aniline in strongly acidic medium results in the formation of m-nitroaniline as a major product. This is because

A
amino group acts as meta orienting during electrophilic substitution reaction

B
in strongly acidic conditions, aniline is present as anilinium ion

C
nitration of aniline is a nucleophilic substitution reaction in strongly acidic medium

D
nitro group always goes to the meta position, irrespective of the substituents

E
None of these

Solution

In strongly acidic medium, aniline is protonated to give anilinium ion, which is electron withdrawing and meta-directing.

For this reason, aniline, on nitration, gives a substantial amount of m-nitroaniline.