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HPSC PGT Chemistry Mock Test - 2 - HPSC TGT/PGT MCQ


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30 Questions MCQ Test HPSC PGT Mock Test Series 2024 - HPSC PGT Chemistry Mock Test - 2

HPSC PGT Chemistry Mock Test - 2 for HPSC TGT/PGT 2024 is part of HPSC PGT Mock Test Series 2024 preparation. The HPSC PGT Chemistry Mock Test - 2 questions and answers have been prepared according to the HPSC TGT/PGT exam syllabus.The HPSC PGT Chemistry Mock Test - 2 MCQs are made for HPSC TGT/PGT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for HPSC PGT Chemistry Mock Test - 2 below.
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HPSC PGT Chemistry Mock Test - 2 - Question 1

Who is the custodian of the Indian Constitution?

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 1

The correct answer is The Supreme Court.

Key Points 

  • The Supreme Court of India is called the custodian of the Constitution in India.
  • The highest court in India, the Supreme Court is considered as the guardian of the Constitution.
  • The conflicts or quarrels of jurisdiction between the central government and the state governments or between the legislature and the executive are decided by the court.
  • It upholds and uplifts the rule of law and also ensures and protects citizens’ rights and liberties as given in the Constitution.
  • Therefore, the Supreme Court is also known as the Guardian of the Constitution.​

Additional Information 

  • The Indian constitution provides for a provision of the Supreme Court under Part V (The Union) and Chapter 6 (The Union Judiciary).
  • The Supreme Court of India is the highest judicial court and the final court of appeal under the Constitution of India, the highest constitutional court, with the power of judicial review.
  • India is a federal state and has a single and unified judicial system with a three-tier structure, i.e. Supreme Court, High Courts, and Subordinate Courts.
HPSC PGT Chemistry Mock Test - 2 - Question 2

Satyawart Kadian is associated with which of the following sports?

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 2

The correct answer is Wrestling.

Key Points

  • Satyawart Kadian is associated with wrestling sports.
    • He was born on 9th November 1993, in Rohtak, Haryana, India.  
  • The Achievements of Satyawart Kadian:
    • A bronze medal in the 100 kg category at the 2010 Youth Olympic Games, held in Singapore.
    • Earned a bronze medal in the 96 kg category at the World Junior Championships in 2013 that took place in Sofia.
    • Bagged silver and a bronze medal in the 97 kg category at the Commonwealth Games and Asian Championships of 2014.
    • Won the title at the 1st Rustam-e-International tournament which was held at Jammu in 2015.
    • Clinched a gold medal in the 97 kg category at the Commonwealth Championships of 2016, held in Singapore.
    • Claimed a bronze medal in the 97 kg category at the 2019 Asian Championships.

Additional Information

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HPSC PGT Chemistry Mock Test - 2 - Question 3

Recently who has been made the General Secretary of the Haryana Wrestling Association?

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 3

The correct answer is Rakesh Singh.

Key Points

  • Rakesh Koch -
    • He is a resident of Bahadurgarh, Haryana
    • He has been elected General Secretary of the Wrestling Association of Haryana.
    • He has been associated with sports for a long time and is also the director of the Model School located in Bahadurgarh.

Additional Information

  • ​Santosh Sharma -
    • This is the head constable of Panipat Police.
    • He has recently brought laurels to the state of Haryana by winning the gold medal in the World Police Sports Competition.
  • Preetpal Sangwan -
    • ​He is the Haryana police officer who has been awarded the Police Medal.
HPSC PGT Chemistry Mock Test - 2 - Question 4

If one of the electrons (1s2) of helium is taken in excited state then bond order of He2 is

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 4

The correct option is A increases by 1
He2 (4 electrons) : (σ1s2)(σ∗1s)2
Bond order = (2−2)/2 = 0
Electron is taken to next excited state that is (σ2s), then the electronic configuration is:
(σ1s2)(σ∗1s)2(σ2s)1
number of electrons in bonding molecular orbital = 3
and in anti-bonding molecular orbital = 1
Bond order =(3−1)/2=1
Thus, the bond order increases by 1 .

HPSC PGT Chemistry Mock Test - 2 - Question 5

If X is a nonmetal, its oxide X2O3 is expected to be a/an ______ oxide.​

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 5

Non metallic oxides are acidic.

HPSC PGT Chemistry Mock Test - 2 - Question 6

For a hypothetical H like atom which follows Bohr's model, some spectral lines were observed as shown. If it is known that line 'E' belongs to the visible region, then the lines possibly belonging to ultra violet region will be (n1 is not necessarily ground state)          

 [Assume for this atom, no spectral series shows overlaps with other series in the emmission spectrum]  

         

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 6

In the given figure if line 'E' is in visible region then line belonging to ultraviolet region will have more energy than 'E' i.e. line A

HPSC PGT Chemistry Mock Test - 2 - Question 7

Correct decreasing order of dipole moment of CH3F, CH3CI and CH3Br is

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 7




Bond length and (EN) oppose each other, with the larger halogens having longer bonds but weaker (EN). Thus CH3CI > CH3F > CH3Br.

HPSC PGT Chemistry Mock Test - 2 - Question 8

PH3 becomes spontaneously inflammable due to the presence of

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 8

Pure phoshine is not inflammable but due to contamination of P2H4 and P4 vapours, it becomes inflammabl.

HPSC PGT Chemistry Mock Test - 2 - Question 9

‘Hartree’ is the atomic unit of energy and is equal to

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 9

The potential energy of an electron in the first Bohr’s orbit in the H-atom is

This energy is defined as an atomic unit of energy called HARTREE.

HPSC PGT Chemistry Mock Test - 2 - Question 10

We can obtain ethylamine by Hoffmann bromamide reaction. The amide used in this reaction is:

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 10

The correct answer is option C
CH3​CH2​CONH2​ (A)⟶ CH3​ − CH2 ​− NH2​ (B)⟶ ​CH3​ − CH2 ​− OH
In the above sequence A & B respectively are Br2​/KOH and HNO2
The first step is Hoffmann bromamide degradation reaction in which an amide (propanamide) is converted to an amine  (ethylamine) containing one carbon atom less. The reagent A is bromine in presence of KOH. In the second step, aliphatic primary amine (ethyl amine) reacts with nitrous acid (reagent B) to form aliphatic primary alcohol (ethyl alcohol).
 

HPSC PGT Chemistry Mock Test - 2 - Question 11

An organic compound X (C5H12O) gives effervescence with NaH. X on treatment with acidic CrO3 solution turns solution to blue-green forming C5H10O2. Also X can be resolved into enantiomers. The correct statement regarding X is

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 11

X is a primary, chiral alcohol.

HPSC PGT Chemistry Mock Test - 2 - Question 12

 Enthalpy change for a cyclic process is

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 12

Enthalpy is a state function; which means change in enthalpy depends only on final and initial state. So, the change in enthalpy for a cyclic process is zero.

HPSC PGT Chemistry Mock Test - 2 - Question 13

Consider the following transformations.

Q. Which reaction sequence will bent bring about the above transformation?

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 13

On addition of H2/Pd/CaCO3
we get H-CH=C-CH2-CH2-CH3 
H-CH=C-CH2-CH2-CH3 + NBS/CCl4   →   H-CH=C-CHCl-CH2-CH3  (NBS adds Br to carbon at next alternate to double bond)
H-CH=C-CHCl-CH2-CH3 + KOH/EtOH  →  H2C=C-CH=CH-CH3
(by E2 elimination)

HPSC PGT Chemistry Mock Test - 2 - Question 14

For the reversible reaction, 

In a reaction vessel, [NO]= [O2]= 0.01 mol L-1 and [NO2]= 0.1 mol L-1 then above reaction is 

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 14

On substituting the values of conc. of NO, O2 and NO2 in given rate equation, we get a +ve (positive) value indicating that the reaction takes place in forward direction.

HPSC PGT Chemistry Mock Test - 2 - Question 15

Out of NH3, H2O and HF, which has the highest magnitude of Hydrogen bonding:

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 15

A small and highly electronegativity elements form a stronger hydrogen bond. The order of size of N , O and F is F < O < N .and so the order of strength of hydrogen bond is F > O > N.

Hence, electronegativity of F is the highest, therefore magnitude of positive charge on hydrogen and negative charge on F is the highest in HF and Hence, electrostatic attraction of H bonding is the strongest in HF.

HPSC PGT Chemistry Mock Test - 2 - Question 16

What is known as Auto-oxidation?

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 16
  • Autoxidation is any oxidation that occurs in presence of oxygen.
  • The term is usually used to describe the degradation of organic compounds in air (as a source of oxygen).
  • Autoxidation produces hydroperoxides and cyclic organic peroxides.
HPSC PGT Chemistry Mock Test - 2 - Question 17

Substances that alter the rate of a chemical reaction without being used up in a chemical reaction is known as

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 17

Catalyst only increases or decreases the activation energy of a reaction, It doesnot get used up in reaction.

HPSC PGT Chemistry Mock Test - 2 - Question 18

A chemical reaction [2A] + [2B] + [C] → product follows the rate equation : then order of reaction is -            [AIEEE-2002]

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 18

The correct answer is option C
The order of reaction with respect to A = 1
The order of reaction with respect to B =1
Total order of reaction = 1+1 = 2

HPSC PGT Chemistry Mock Test - 2 - Question 19

When sodium is dissolved in liquid ammonia, a solution of deep blue colour is obtained. The colour of the solution is due to

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 19

When sodium is dissolved in liquid ammonia, it donates electrons to the ammonia solvent, forming ammoniated electrons. These solvated electrons are responsible for the deep blue colour of the solution.

HPSC PGT Chemistry Mock Test - 2 - Question 20

Which of the following are all properties of nonmetals?

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 20

There are several ways to tell metals and nonmetals apart. Nonmetalsdon't have a metallic appearance. Unlike metals, they typically have lower melting and boiling points and tend not to conduct heat or electricity very well.

HPSC PGT Chemistry Mock Test - 2 - Question 21

Be can show coordination number four while other members show a value of six. This is because of:

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 21

The electronic configuration of Be is 1s22s2. We can see that there are only 4 orbitals in its valence shell( they are s and p). In other words, d orbital is absent. So, Be shows ccoordination no 4 and not 6 as like other members of its group.

HPSC PGT Chemistry Mock Test - 2 - Question 22

Which of the following has maximum boiling point?

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 22

Due to intermolecular H-bonding at two terminals boiling point of glycol is maximum.

HPSC PGT Chemistry Mock Test - 2 - Question 23

Which of the following types of octahedral complexes will exhibit geometrical isomerism (where, M = metal, a, b = achiral ligands)?

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 23

Octahedral complexes of formula [Ma4b2] in which the two ligands b may be oriented cis or trans to each other, e.g.

Geometrical isomers (cis and trans) of [Co(NH3)4CI2]+.

HPSC PGT Chemistry Mock Test - 2 - Question 24

What is the equilibrium constant for the reaction P4(s) + 5O2(g)  P4O10(s) :

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 24


HPSC PGT Chemistry Mock Test - 2 - Question 25

In the complete combustion of C2H6, 54 g of H2O is formed and 370 kcal of heat is evolved. Thus, ΔCH° of C2H6 is

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 25

54 gram of water is formed or three moles of water is formed.
C2H6 + (7/2)O2 → 2CO2 + 3H2O
So it indicates that only one mole of C2H6 is used in the combustion so the heat released is itself the value of heat of combustion of Ethane.

HPSC PGT Chemistry Mock Test - 2 - Question 26

The hydration energy of Mg2+ is -?

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 26

Hydration energy ∞ polarising power
Na+ < Mg2+ < Mg3+ < Be3+ < Al3+

HPSC PGT Chemistry Mock Test - 2 - Question 27

Entropy change when 2 moles of an ideal gas expands reversibly from an initial volume of 1 dm3 to a final volume of 10 dm3 at a constant temperature of 298 K is

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 27

∆S= 2.303nR × log(V2/V1)Here n=2, R=8.314, V2= 10, V1= 1

HPSC PGT Chemistry Mock Test - 2 - Question 28

Maximum number of electrons in a subshell with = 3 and = 4 is

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 28

In this question l=3 means the sub shell is f & n=4 means it is present in 4th orbit.
Finally it is a 4f sub shell.
The no of orbital in f sub shell =2l+1
                                                = 2*3+1
                                                =7
Each orbital can accommodate 2 electrons.
  7*2 = 14 electrons.

Maximum number of electrons in a sub shell with l=3& n=4 (4f) is 14electrons.

HPSC PGT Chemistry Mock Test - 2 - Question 29

If a is the degree of dissociation of Na2SO4, the vant Hoff's factor (i) used for calculating the molecular mass is–  

[AIEEE-2005]

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 29

The correct answer is option D
For Na2​SO4 ​: −i = n
Na2​SO4 ​⟶ 2Na+ + SO42−​​
                         n=2
i=1−α+nα
i=1−α+2α
i=1+2α

HPSC PGT Chemistry Mock Test - 2 - Question 30

How many grams of Mgl2 must be added to 250 mL of 0.0876 M Kl to produce a solution with [l-] = 0.10 M?

Detailed Solution for HPSC PGT Chemistry Mock Test - 2 - Question 30

 

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