IIT JAM Chemistry - MCQ Test 1

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

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

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.

Br- 2.96
F- 3.98
NH₂, The electronegativity of Nitrogen is 3.04 and Hydrogen is 2.2 So this difference makes the N-H bond polar.
CH₃ = 5.25

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

Hence B is the correct answer

According to radial nodes formula:

Radial Nodes = n - l – 1

= 3 – 0 – 1

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

1. 

2. 

3. 

4. 

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.

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.

• 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.

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

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

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.

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

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.

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.

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.

• 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).  T_{i =} 
• The reduced temperature of a fluid is its actual temperature, divided by its critical temperature.
  

• 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.

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 (K_sp = 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)

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

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″.

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.

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.

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.

CH₄ (109.28) > NH₃(107) > H₂O(104.5) 

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

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.

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.

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.

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.

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 .

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.

Percentage of an atom in any molecule can calculated by formula: n x (atomic weight/molecular weight) x 100

where n is total number of atomsin molecule

(i) 60g CH₃COOH

Atomic weight of carbon = 12

Molecular weight = 60

Number of atoms = 8

Percentage of atom =  n x (atomic weight/molecular weight) x 100

Percentage of atom = 8 x (12/60) x 100

Percentage of atom = 160

(ii) 3g HCHO

Atomic weight of carbon = 12

Molecular weight = 30

Number of atoms = 4

Percentage of atom =  n x (atomic weight/molecular weight) x 100

Percentage of atom = 4 x (12/30) x 100

Percentage of atom = 160

(iii) 60g NH₂CONH₂

Atomic weight of carbon = 12

Molecular weight = 62

Number of atoms = 8

Percentage of atom =  n x (atomic weight/molecular weight) x 100

Percentage of atom = 8 x (12/62) x 100

Percentage of atom = 154.2

4) 180g C₆H₁₂O₆

Atomic weight of carbon = 12

Molecular weight = 180

Number of atoms = 24

Percentage of atom =  n x (atomic weight/molecular weight) x 100

Percentage of atom = 24 x (12/180) x 100

Percentage of atom = 160

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)

• Nucleophile is a chemical species that donates an electron pair to form a chemical bond in relation to a reaction.
• All molecules or ions with a free pair of electrons or at least one pi bond can act as nucleophiles.
• Because nucleophiles donate electrons, they are by definition Lewis bases.

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.

ΔE = 13.6 [ 1/9 - 1/16]

ΔE = 13.6 [7/144]

ΔE = 13.6 x 0.048

ΔE = 0.66

Ideal gas equation, PV = nRT

Dividing both sides by nRT

⇒ PV/nRT = nRT/nRT

⇒ PV/nRT = 1

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 + 2H₂O 

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.

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. 

• 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.

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

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%

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