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IIT JAM Chemistry - MCQ Test 1 - Chemistry MCQ


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30 Questions MCQ Test - IIT JAM Chemistry - MCQ Test 1

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IIT JAM Chemistry - MCQ Test 1 - Question 1

The most stable canonical structure among all of above is:

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 1

The structure which has a maximum double bond is more stable.

IIT JAM Chemistry - MCQ Test 1 - Question 2

Strongest base among the following species is:

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 2

Nitrogen atom is more basic than the other two oxygen and carbon. Also, it has an available lone pair.

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*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 3

Compare relative stability of the following structures: 

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 3

Presence of negative charge on electronegative atom is more stable than that on a relatively less electronegative atom. If this is considered, negative charge on oxygen is more stable than that on nitrogen. Also, negative charge an nitrogen will be more stable than that on carbon. Hence, the order of stability is I > III > II.

IIT JAM Chemistry - MCQ Test 1 - Question 4

Strength of following bases decrease in the order? 

(I) Br   (II) F-      (III) NH2    (IV) CH3-

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 4
  1. Br⁻ (bromide ion): This is a halide ion. Halide ions become stronger bases as the size of the ion increases (i.e., as we move down the periodic table). Bromide is quite large, but it is not as large as iodide, making it a relatively weaker base compared to other larger halides.

  2. F⁻ (fluoride ion): Fluoride is a small ion with high electronegativity, which means it holds onto its electron pair tightly and is less inclined to share it to form a bond with a proton. Thus, in an aqueous solution, fluoride is a weaker base than larger halide ions.

  3. NH₂⁻ (amide ion): This ion has a nitrogen atom, which is less electronegative than fluorine, holding the negative charge. Nitrogen can share its electron pair more easily than fluorine. Also, the negative charge on the amide ion is not as effectively stabilized because nitrogen is less electronegative than fluorine.

  4. CH₃⁻ (methide ion): This ion has a carbon atom bearing the negative charge. Carbon is less electronegative than nitrogen and fluorine, so it is more ready to share its electron pair compared to NH₂⁻ and F⁻. However, the negative charge is not delocalized and is concentrated on a single carbon atom, making it a strong base.

Considering these factors, the order of increasing base strength from weakest to strongest should be:

F⁻ < Br⁻ < NH₂⁻ < CH₃⁻

This order takes into account that fluoride is very electronegative and holds onto its electron pair tightly, bromide is a larger halogen and can distribute its negative charge over a larger volume, the amide ion is less stable due to the higher energy of nitrogen-based anions, and the methide ion is a strong base due to carbon's low electronegativity and the localized negative charge.

IIT JAM Chemistry - MCQ Test 1 - Question 5

The ratio of van der Waal’s constant a and b, has the dimensions of:

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 5

The ratio of van der Waals’ constants a and b has the dimensions of atm L mole-1.

Hence B is the correct answer

IIT JAM Chemistry - MCQ Test 1 - Question 6

Find the corresponding subshell utilizing the information from graph. 

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 6

According to radial nodes formula:

Radial Nodes = n - l – 1

= 3 – 0 – 1

=2 nodes.
Therefore 3s.
Hence, A is the correct answer.

IIT JAM Chemistry - MCQ Test 1 - Question 7

Distribution of molecules with velocity is represented by curve as shown:

Velocity at point A is:

IIT JAM Chemistry - MCQ Test 1 - Question 8

Which have lowest bond angle?

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 8
  1. https://edurev.gumlet.io/ApplicationImages/Temp/94084588-6758-4742-94cd-660195a07a2e_lg.jpg

  2. https://edurev.gumlet.io/ApplicationImages/Temp/b2698cab-87a0-48c4-a6b8-5b9cc7056f9a_lg.jpg

  3. https://edurev.gumlet.io/ApplicationImages/Temp/0b2266c1-4fe5-49bd-b214-f323c8578b56_lg.jpg

  4. https://edurev.gumlet.io/ApplicationImages/Temp/215b7ea0-3771-4109-9f6e-50a7fd474502_lg.jpg

  5. PH3 because it follows Drago’s rule and hence there would be no or very low hybridization and hence pure p-orbital will be used. Therefore bond angle will be almost equal to 90 degree.

    Hence C is correct.

IIT JAM Chemistry - MCQ Test 1 - Question 9

Of the following acids, the one that is strongest is:

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 9

Higher the oxidation no. of the central atom, the stronger is the acid.
Here, Br=+7
Cl=+1
N=+3
P=+3
Thus, HBrO4 is the strongest acid amongst the given alternatives.

*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 10

Lattice energy (numerical value) of chloride of alkali metals is in order:

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 10
  • Lattice energy is the attraction force acting between the cation and anion.
  • The greater the size of cation, lower is the attractive force.
  • On moving down the group, size of the cation increases then,
    1. Attractive force decreases.
    2. Lattice energy decreases.
*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 11

Which of the following drawings is not a resonance structure of 1 -nitrocyclohexene:

IIT JAM Chemistry - MCQ Test 1 - Question 12

Which of the following is not resonating structure of each other?

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 12

Molecules in option (a) are linkage isomers of each other.

According to rules of Resonance, no change in position of atom should take place.

IIT JAM Chemistry - MCQ Test 1 - Question 13

Examine the following resonating structures of formic acid for their individual stability and then answer the question given below:

Which of the following arrangements gives the correct order of decreasing stability of the above-mentioned resonance contributors?

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 13

There are certain sets of rules that must be followed in order to check the stability of resonating structures.
These rules have a preference so we have to follow the order:

  • Uncharged species are more stable than charged species.
  • Charged species having more number of covalent bond are more stable.
  • The species in which octet rule is followed by every atom is more stable than the species in which octet rule is violated.
  • Species in which opposite charges are closer are more stable than species having same charges closer.
  • Species with negative charge on more electronegative atoms while positive charge on more electropositive atoms are more stable.

According to the above discussion, B is the correct answer.

IIT JAM Chemistry - MCQ Test 1 - Question 14

The correct stability order of the given resonating structures is:

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 14

Valence shell = 6e-
In II, octet is incomplete
E.N : O > N
The correct sequence is: III > I > II

IIT JAM Chemistry - MCQ Test 1 - Question 15

The correct order of stability for the given canonical structures is:

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 15

There are a certain set of rules that must be followed in order to check the stability of resonating structures.
These rules have a preference so we have to follow the order:

  • Uncharged species is more stable than charged species.
  • Charged species having more number of covalent bond are more stable.
  • The species in which the octet rule is followed by every atom is more stable than the species in which the octet rule is violated.
  • Species in which opposite charges are closer are more stable than species having same charges closer.
  • Species with negative charge on more electronegative atom while positive charge on more electropositive atom is more stable.

According to above discussion, B is the correct answer.

IIT JAM Chemistry - MCQ Test 1 - Question 16

The correct order of stability among the following canonical structure is: 

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 16

There are a certain set of rules that must be followed in order to check the stability of resonating structures.
These rules have a preference so we have to follow the order

  • Uncharged species is more stable than charged species.
  • Charged species having more number of covalent bond are more stable.
  • The species in which octet rule is followed by every atom is more stable than the species in which octet rule is violated.
  • Species in which opposite charges are closer are more stable than species having same charges closer.
  • Species with negative charge on more electronegative atom while positive charge on more electropositive atom is more stable.

According to above discussion, D is the correct answer.

IIT JAM Chemistry - MCQ Test 1 - Question 17

Which of the following is most basic?

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 17

Basicity is the tendency of an atom to donate their lone pair.

In case of  (a), (b),(d), lone pair of Nitrogen atom is in conjugation which is not available to be donated.

Hence (c) is more basic.

*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 18

If a particle in the box of length 'l' has wavelength, ψ = (l -x)x. (x is only variable)  What is its normalization constant?

IIT JAM Chemistry - MCQ Test 1 - Question 19

Match column I with column II and select the correct answers:

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 19
  • A critical point (or critical state) is the end point of a phase equilibrium curve. 
  • The Boyle temperature is the temperature at which a non ideal gas behaves most like an ideal gas.
  • The characteristic temperature below which a gas expands adiabatically into a region of low pressure through a porous plug with a fall in temperature is called inversion temperature (Ti).  Ti = 
  • The reduced temperature of a fluid is its actual temperature, divided by its critical temperature.
*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 20

The table indicates the value of van-der Waal’s constant ‘a’. The gas which can be liquefied most easily is:

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 20
  • The ease of liquification of a gas depends on their intermolecular force of attraction which in turn is measured in terms of Van Der Waals’ constant ‘a’.
  • Hence, higher the value of ‘a’, greater the intermolecular force of attraction, easier the liquification. In the present case, NH3 has highest ‘a’, can most easily be liquefied.
*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 21

rms velocity of hydrogen is √7 times the rms velocity of nitrogen. If T is the temperature of gas,then:

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 21

*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 22

Copper containing alloy weighing 0.3175g dissolved in an acid, and an excess of KI is added.
Estimation of copper in alloy is based on following reactions:

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 22

Iodometric determination of copper is based on the oxidation of iodides to iodine by copper (II) ions, which get reduced to Cu+.

Comparison of standard potentials for both half reactions (Cu2+/Cu+ E0=0.17 V,  I2/I- E0=0.54 V) suggests that it is iodine that should be acting as oxidizer. However, that's not the case, as Copper (I) Iodide (CuI) is very weakly soluble (Ksp = 10-12). That means concentration of Cu+ in the solution is very low and the standard potential of the half reaction Cu2+/Cu+ in the presence of iodides is much higher (around 0.88 V)

IIT JAM Chemistry - MCQ Test 1 - Question 23

If a particle has linear momentum at position , then its angular momentum is:

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 23

Angular momentum is equal to the cross product of the position vector and the linear mamentum.

J=r×p

(4i + 9j + 2k) × (3i - 3j + k)

= -15i-2j+39k

 

*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 24

Consider two molecules A & B.

∠α = ∠HCH; ∠β = ∠FCF

Which of following is true?

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 24

Though the substituting atoms are different in both cases but the hybridization of carbon is the same that is sp3, having the bond angle of 109′28″.

IIT JAM Chemistry - MCQ Test 1 - Question 25

Predict the shape of IF4+ molecule using VSEPR theory:

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 25

The electron geometry of IF4+ is trigonal bipyramidal. This is because it has five total electron groups. The central atom is bonded to four other atoms and has one lone pair of electrons. The lone pair of electrons will affect the molecular geometry of the compound as well as the bond angles.

*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 26

Which of the following is true for ionization energy:

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 26

The ionization energy of molecular nitrogen is 1503 kJ/mol, and that of atomic nitrogen is 1402 kJ/mol. We conclude that the energy of the electrons in molecular nitrogen is lower than that of the electrons in the separated atoms, so the molecule is bound.

*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 27

XeO2F2 has shape: 

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 27

XeO2F2 molecular geometry is originally said to be trigonal bipyramidal but due to the presence of lone pair on equatorial position, the actual shape will be see-saw. The repulsion between bond pair and lone pair of electrons will be more.

*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 28

What is the correct increasing order of bond lengths of bond indicated in following compund?

*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 29

Arrange the following in decreasing order of their bond angles: NH3, H2O , CH4

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 29

CH4 (109.28) > NH3(107) > H2O(104.5) 

IIT JAM Chemistry - MCQ Test 1 - Question 30

The ground state electronic configuration of valence shell electrons in a molecule A2 is written as• Hence bond order is_____________________.

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 30

Bond Order = 

Number of electrons in bonding MO = 8

Number of electrons in anti-bonding MO = 4

Bond order = (8 - 4)/2

Bond order = 4/2

Bond order = 2

 

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