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IIT JAM Chemistry - MCQ Test 1 - Chemistry MCQ


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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.

*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

 

*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 31

Which of these structures is practically not a valid resonance structure for formaldehyde?

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 31

The lowest energy form is I, because every atom has a complete octet. It is the major contributor.

II is lower in energy than III, because the negative charge is in the more electronegative O atom.

*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 32

Which of the following is an electrophile?

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 32

Electrophile is an electron pair acceptor. Electrophiles are positively charged or neutral species having vacant orbitals that are attracted to an electron rich centre. It participates in a chemical reaction by accepting an electron pair in order to bond to a nucleophile.

*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 33

Select the correct statement:

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 33

Benzoic acid is a stronger acid than phenol because the benzoate ion is stabilised by two equivalent resonance structures in which the negative charge is present at the more electronegative oxygen atom.

The conjugate base of phenol, a phenoxide ion, has non-equivalent resonance structures in which the negative charge is at the less electronegative carbon atom. Thus, the benzoate ion is more stable than phenoxide ion. Hence, benzoic acid is a stronger acid than phenol.

*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 34

An operator ‘A’ is defined as Which one of the following statement is true?

*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 35

Identify which of the following operator is Hermitian:

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 35

*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 36

Which of the following is/are true about zero point energy?

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 36

Zero-point energy (ZPE) is the lowest possible energy that a quantum mechanical system may have. Unlike in classical mechanics, quantum systems constantly fluctuate in their lowest energy state as described by the Heisenberg uncertainty principle.

 

*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 37

Which of the following is/are true about Uncertainty principle: 

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 37

A) As uncertainity principle tells us about the minimum uncertain value of position for a certain value of momentum or vice versa. Hence it doesn't put an upper but a lower bound.

B) The equation,  is true under all conditions because it is the law of nature if anything exist it should have a certain amount of uncertainity relation, however it could be small in the bigger objects but it exists.

C) Both position and momentum are inversely proportional to each other . 

D) If one of them become zero then the other should go infinity at the same time .

 

*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 38

Which of the following is not a disproportionation reaction?

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 38

A disproportionation reaction is when a multiatomic species whose pertinent element has a specific oxidation state gets oxidized and reduced in two separate half-reactions, yielding two other products containing the same pertinent element.

*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 39

Consider the following cases:

(I) 60g CH3COOH (II) 3g HCHO (III) 60g NH2CONH2 (IV) 180g C6H12O6

Percentage of carbon is identical in:

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 39



 

In formaldehyde (CH2O), the structure consists of one carbon atom (C), two hydrogen atoms (H), and one oxygen atom (O). The molecular formula indicates that there are two hydrogen atoms in the molecule, but only one of them is considered when discussing certain aspects, such as its hybridization or geometry.

The reason for considering only one hydrogen is related to the concept of formal charge and the fact that the hydrogen atom attached to the oxygen is part of a hydroxyl (OH) group, not a methyl (CH3) group.

*Multiple options can be correct
IIT JAM Chemistry - MCQ Test 1 - Question 40

In trigonal bipyramidal geometry, which of the following holds true?

Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 40

The axial bonds are longer than equatorial bonds because of greater repulsion from equatorial bonds. The 2-axial bonds are at 90º degree angle to 3-equatorial bonds while all equatorial bonds are at 120º angle to each other. So, none of the statements is correct hence option(d)

*Answer can only contain numeric values
IIT JAM Chemistry - MCQ Test 1 - Question 41

Total number of nucleophiles among following molecules is/are___________.
CH3O-, carbene, Nitrene, H2O , ROH, ZnCl2, NHR2, SnCl4-NH2 , Cl-


Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 41
  • Nucleophiles are molecules or ions that are attracted to electron-deficient sites and have a tendency to donate electrons. Let's identify the nucleophiles among the given molecules:

  • CH3O- (methoxide ion) - Nucleophile
  • Carbene - Nucleophile
  • Nitrene - Nucleophile
  • H2O - Nucleophile
  • ROH (alcohol) - Nucleophile
  • ZnCl2 - Not a nucleophile (does not have lone pairs to donate)
  • NHR2 - Nucleophile
  • SnCl4 - Not a nucleophile (does not have lone pairs to donate)
  • -NH2 (amide ion) - Nucleophile
  • Cl- - Nucleophile
  • So, the total number of nucleophiles among the given molecules is 8.

*Answer can only contain numeric values
IIT JAM Chemistry - MCQ Test 1 - Question 42

Total number of a-hydrogen in the given following compound is:


*Answer can only contain numeric values
IIT JAM Chemistry - MCQ Test 1 - Question 43

Number of carbenes among following:

  :HgCl2, :CCl2 ,     , :CCl3, :SnCl2, BeCl2, :CH2.


Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 43

A carbene is a molecule containing a neutral carbon atom with a valence of two and two unshared valence electrons. The general formula is R-(C:)-R' or R=C: where the R represent substituents or hydrogen atoms.

*Answer can only contain numeric values
IIT JAM Chemistry - MCQ Test 1 - Question 44

A particle having energy 66h2/8mL2 for a 3D box has degeneracy?


*Answer can only contain numeric values
IIT JAM Chemistry - MCQ Test 1 - Question 45

Find ΔE for states n = 3,4 in Bohr’s model of hydrogenic atoms (in eV):


Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 45

ΔE = 13.6 [ 1/9 - 1/16]

ΔE = 13.6 [7/144]

ΔE = 13.6 x 0.048

ΔE = 0.66

*Answer can only contain numeric values
IIT JAM Chemistry - MCQ Test 1 - Question 46

By what factor does the average velocity of a gaseous molecule increase, when temperature is doubled:


Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 46

*Answer can only contain numeric values
IIT JAM Chemistry - MCQ Test 1 - Question 47

What is the value of PV/nRT for ideal gas?


Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 47

Ideal gas equation, PV = nRT

Dividing both sides by nRT

⇒ PV/nRT = nRT/nRT

⇒ PV/nRT = 1

IIT JAM Chemistry - MCQ Test 1 - Question 48

Normality of a solution obtained by mixing 10 mL of  N/5 HCl & 30 mL of N/10 HCl is_______:


Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 48

*Answer can only contain numeric values
IIT JAM Chemistry - MCQ Test 1 - Question 49

Ca(OH)2 + H3PO4 → CaHPO4 + 2H2

Equivalent weight of H3PO4 in the above reaction is:


Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 49

H3PO4  is an acid and the equivalent mass of an acid is calculated by using the formula:

From the given reaction:

Ca(OH)2 + H3PO4 → CaHPO4 + 2H2

As, the replaceable hydrogen ions in acid is 2, by putting the values in above equation, we get:

Hence, The equivalent mass of H3PO4 for the given reaction is 49 g eq.

 

*Answer can only contain numeric values
IIT JAM Chemistry - MCQ Test 1 - Question 50

How many equivalents are there per mol of H2S in its oxidation to SO2?


Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 50

*Answer can only contain numeric values
IIT JAM Chemistry - MCQ Test 1 - Question 51

Total number of the molecules that violates the rules of resonance are__________________



Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 51

    

 

In the above structure Nitrogen is having 5 bonds which is not possible, hence these molecules are violating the rules of resonance.

 

In above molecule Oxygen is forming 4 bonds that is not possible, hence this molecules is also violating the rules of resonance. 

   

In both molcules electron pair is moving from Sulphur and Oxygen which already have positive charge. Also in case of 2nd molecule, Oxygen is forming 5 bonds which is again not possible. 

 

 

 

*Answer can only contain numeric values
IIT JAM Chemistry - MCQ Test 1 - Question 52

Total number of delocalized lone pairs (combined) in Adenine & Guanine:


*Answer can only contain numeric values
IIT JAM Chemistry - MCQ Test 1 - Question 53

Number of electrophiles in following:

BF3, +CH3, -NH2, I+, PCl5, NH3, OH2, SnCl4, ROR


Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 53
  • In organic chemistry, an electrophile is an electron pair acceptor. Electrophiles are positively charged or neutral species having vacant orbitals that are attracted to an electron rich centre.
  • It participates in a chemical reaction by accepting an electron pair in order to bond to a nucleophile. Because electrophiles accept electrons, they are Lewis acids. Most electrophiles are positively charged, have an atom that carries a partial positive charge, or have an atom that does not have an octet of electrons.
  • They appear to attract electrons as well and seem to behave as though they are partially empty. These partially empty substances thus require an electron rich center, and thus they are filled. Electrophiles can be observed as electron-sensitive or photo-sensitive.
*Answer can only contain numeric values
IIT JAM Chemistry - MCQ Test 1 - Question 54

Find the probability of finding an e- in a box of length ‘L’ in middle half of box. (in %)


*Answer can only contain numeric values
IIT JAM Chemistry - MCQ Test 1 - Question 55

At what temperature (in K), will the total KE of 0.30 mole of He be same as the total KE of 0.40 mole of Ar at 400 K?


Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 55

*Answer can only contain numeric values
IIT JAM Chemistry - MCQ Test 1 - Question 56

 What is the value of n2?

[Given is the wavefunction for a particle in 1-D box of length ‘L' If energy of ‘nth’ state is given by 


*Answer can only contain numeric values
IIT JAM Chemistry - MCQ Test 1 - Question 57

Number of molecules in 1L of water at 4°C is close to: (Give the multiple of NA)


Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 57

1 lit = 1000 cm3 = 1000 × 1 g = 1000 g = 1000/18 mol =55.5556 mol . It Contain 55.5556 N(A) number of molecule.

*Answer can only contain numeric values
IIT JAM Chemistry - MCQ Test 1 - Question 58

Consider AB2 molecule Calculate % s-character in A—B bond:


Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 58

From Bent’s Rule equation,  cosx = s/s-1

where x is an bond angle and s is the % s character in the bond, it is clear that if we increase bond angle, the %s Character will be maximum.

Here angle is 150,

⇒ Cosx = s/(s-1)

⇒ Cos 150° = s/(s-1)

⇒ -0.86 = s/(s-1)

⇒ -0.86s +0.86 = s

⇒ 0.86 = 1.86s

⇒ s = 0.86/1.86

⇒ s = 0.462

⇒ %s = 0.462 X 10

⇒ %s = 46.2%

*Answer can only contain numeric values
IIT JAM Chemistry - MCQ Test 1 - Question 59

Number of π-bonds in will be:


*Answer can only contain numeric values
IIT JAM Chemistry - MCQ Test 1 - Question 60

Concentration of 10% CH3COOH (w/v) in mol L-1 will be ________?


Detailed Solution for IIT JAM Chemistry - MCQ Test 1 - Question 60

M.W of acetic acid is 60 gm
10% (w/v) acetic acid means.
In 100 ml solution acetic acid is = 10 gm
so, in 100 ml solution acetic acid is = 10/60 mole.
so. in 1000ml solution acetic acid is = (0.167 × 1000)/100
= 1.67 mole / lit

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