IIT JAM Chemistry - MCQ Test 1 - Chemistry MCQ

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.

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.

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