CHEMICAL BONDING & MOLECULAR STRUCTURE MCQs for NEET UG CHEMISTRY — Easy, Page 3

61 Easy

Analyzing +4 | -1

The dipolemoments of methane and its halogen derivatives are in order

View answer and explanation

Correct option: A

C{H_4} < CHC{l_3} < C{H_2}C{l_2} < C{H_3}Cl

Option A (

C{H_4} < CHC{l_3} < C{H_2}C{l_2} < C{H_3}Cl

) is correct because the dipole moment increases as the symmetry decreases and the vector sum of bond dipoles increases, with

CH_4

having zero dipole moment.

Solution approach

Correct Answer + Core Reason: $C{H_4} < CHC{l_3} < C{H_2}C{l_2} < C{H_3}Cl$ . The dipole moment increases with decreasing symmetry and increasing net bond dipole moment.

Step-by-Step Solution:

Step 1: Consider the structures of the methane derivatives.
- $C{H_4}$ : Tetrahedral, symmetrical. The bond dipoles cancel each other out, resulting in a net dipole moment of zero.
- $C{H_3}Cl$ : Tetrahedral, but with a net dipole moment due to the C-Cl bond.
- $C{H_2}C{l_2}$ : Tetrahedral, but with a smaller net dipole moment than $C{H_3}Cl$ because the two C-Cl bonds partially cancel each other out, but more than $CHCl_3$ .
- $CHC{l_3}$ : Tetrahedral, but with an even smaller net dipole moment than $C{H_2}C{l_2}$ because the three C-Cl bonds partially cancel each other out, but more than $CH_4$ .

Step 2: Analyze the vector sum of bond dipoles.
- In $C{H_4}$ , the four C-H bond dipoles completely cancel each other out, resulting in a zero net dipole moment.
- In $C{H_3}Cl$ , the three C-H bonds have small dipole moments that add to the C-Cl bond dipole, resulting in a relatively large net dipole moment.
- In $C{H_2}C{l_2}$ , the two C-Cl bond dipoles partially cancel each other out, reducing the net dipole moment compared to $CH_3Cl$ .
- In $CHC{l_3}$ , the three C-Cl bond dipoles partially cancel each other out, further reducing the net dipole moment, but still greater than 0.

Step 3: Arrange the molecules in increasing order of dipole moments.
$C{H_4} < CHC{l_3} < C{H_2}C{l_2} < C{H_3}Cl$

Why other options are wrong:
- Option B: $C{H_3}Cl < C{H_2}C{l_2} < CHC{l_3} < C{H_4}$ is incorrect because $CH_4$ has zero dipole moment and $CH_3Cl$ has the largest dipole moment.
- Option C: $C{H_4} < C{H_3}Cl < C{H_2}C{l_2} < CHC{l_3}$ is incorrect because $CH_2Cl_2$ has a larger dipole moment than $CHCl_3$ .
- Option D: $C{H_4} < C{H_2}C{l_2} < CHC{l_3} < C{H_3}Cl$ is incorrect because $CHCl_3$ has a smaller dipole moment than $CH_2Cl_2$ .

Key Concepts:
- Dipole moment
- Molecular geometry (VSEPR theory)
- Vector addition of bond dipoles
- Symmetry

Memory Aids:
- Remember that symmetrical molecules like $C{H_4}$ have zero dipole moment.
- Consider the degree of cancellation of bond dipoles.

Quick Tips:
- Draw the Lewis structures and consider the molecular geometry.
- Determine the direction of bond dipoles based on electronegativity differences.
- Consider the effect of symmetry on the net dipole moment.

62 Easy

62 of 220Question permalink

Application +4 | -1

If the ionic radii of

{\mathrm{K}^ + }

and

{\mathrm{F}^ - }

are about 1.34 Å each, then the expected values of atomic radii of

\mathrm{K}

and

\mathrm{F}

should be respectively:

View answer and explanation

Correct option: B

2.31\,and\,\,0.64\,{A^0}

Option B (

2.31\,and\,\,0.64\,{A^0}

) is correct because a potassium atom is larger than its ion

{\mathrm{K}^ + }

due to the loss of an electron, and a fluorine atom is smaller than its ion

{\mathrm{F}^ - }

due to the gain of an electron.

Solution approach

Correct Answer + Core Reason: $2.31\,and\,\,0.64\,{A^0}$ . Cations are smaller than their parent atoms, and anions are larger.

Step-by-Step Solution:

Step 1: When an atom loses an electron to form a cation (like ${\mathrm{K}^ + }$ ), the effective nuclear charge increases, pulling the remaining electrons closer to the nucleus, thus reducing the ionic radius.

Step 2: When an atom gains an electron to form an anion (like ${\mathrm{F}^ - }$ ), the effective nuclear charge decreases, and electron-electron repulsion increases, causing the electron cloud to expand, thus increasing the ionic radius.

Step 3: Therefore, the atomic radius of K should be larger than the ionic radius of ${\mathrm{K}^ + }$ (1.34 Å), and the atomic radius of F should be smaller than the ionic radius of ${\mathrm{F}^ - }$ (1.34 Å).

Step 4: Among the given options, only option B satisfies these conditions: K = 2.31 Å and F = 0.64 Å.

Why other options are wrong:
- Option A: This option suggests that the atomic radii are the same as the ionic radii, which is incorrect.
- Option C: This option has the atomic radius of K smaller than its ion and the atomic radius of F larger than its ion, which is incorrect.
- Option D: This option has the atomic radius of F the same as its ion, which is incorrect.

Key Concepts:
- Ionic radius
- Atomic radius
- Effective nuclear charge
- Cations and anions

Memory Aids:
- Cations are smaller (Cat-ions are "cut" ions).
- Anions are larger (An-ions are "add" ions).

Quick Tips:
- Remember that cations are always smaller than their parent atoms, and anions are always larger.

63 Easy

63 of 220Question permalink

Remembering +4 | -1

According to Fajan’s rule polarization is more when:

View answer and explanation

Correct option: A

Small cation and large anion

Option A (Small cation and large anion) is correct because according to Fajan's rule, polarization is favored by a small, highly charged cation and a large, highly charged anion.

Solution approach

Correct Answer + Core Reason: Small cation and large anion. Fajan's rules state that polarization is more when the cation is small and highly charged, and the anion is large and highly charged.

Step-by-Step Solution:

Step 1: Recall Fajan's rules:
- Small cation: Higher polarizing power.
- Large anion: Higher polarizability.
- High charge on either ion: Increased polarization.

Step 2: Apply Fajan's rules to the given options:
- Small cation and large anion: Favors polarization.
- Small cation and small anion: Less polarization.
- Large cation and large anion: Less polarization.
- Large cation and small anion: Less polarization.

Step 3: Conclusion: Polarization is more when the cation is small and the anion is large.

Why other options are wrong:
- Option B: Small cation and small anion: Less polarization.
- Option C: Large cation and large anion: Less polarization.
- Option D: Large cation and small anion: Less polarization.

Key Concepts:
- Fajan's rules
- Polarization
- Polarizing power
- Polarizability

Memory Aids:
- Small cation, large anion, high charge = more polarization.

Quick Tips:
- Remember Fajan's rules to predict the extent of polarization in ionic compounds.

64 Easy

64 of 220Question permalink

Understanding +4 | -1

Which is not true in case of ionic bond?

View answer and explanation

Correct option: A

It is linear bond

Option A (It is linear bond) is not true because ionic bonds are non-directional, not linear.

Solution approach

Correct Answer + Core Reason: Ionic bond is linear is not true because ionic bonds are non-directional.

Step-by-Step Solution:

Step 1: Ionic bonds are formed due to electrostatic attraction between oppositely charged ions.

Step 2: This electrostatic attraction is non-directional, meaning it acts equally in all directions around the ion.

Step 3: Therefore, ionic bonds do not have a specific direction or linearity.

Why other options are wrong:
- Option B: It is $100\%$ ionic is an approximation. In reality, no bond is perfectly ionic; there is always some degree of covalent character.
- Option C: It is formed between two atoms with large electronegativity difference is true because a large electronegativity difference leads to the transfer of electrons and the formation of ions.
- Option D: None of the above is incorrect because option A is not true.

Key Concepts:
- Ionic bonding
- Directionality of bonds
- Electronegativity

Memory Aids:
- Ionic = Non-directional

Quick Tips:
- Remember that covalent bonds are directional, while ionic bonds are non-directional.

65 Easy

65 of 220Question permalink

Understanding +4 | -1

Which molecule has the largest dipole moment?

View answer and explanation

Correct option: A

HF

Option A is correct because HF has the largest electronegativity difference between hydrogen and the halogen, leading to the greatest dipole moment.

Solution approach

Correct Answer + Core Reason: $HF$ . Hydrogen fluoride has the largest electronegativity difference between H and the halogen, resulting in the largest dipole moment.

Step-by-Step Solution:

Step 1: Recall the definition of dipole moment. Dipole moment is proportional to the electronegativity difference between the atoms in a bond.

Step 2: Consider the electronegativity of the halogens. Electronegativity decreases down the group: F > Cl > Br > I.

Step 3: Determine the electronegativity difference between H and each halogen.

Step 4: HF has the largest electronegativity difference.

Step 5: Conclusion: HF has the largest dipole moment.

Why other options are wrong:
- Option B: $HCL$ : This is incorrect because the electronegativity difference between H and Cl is less than that between H and F.
- Option C: $HBr$ : This is incorrect because the electronegativity difference between H and Br is less than that between H and F.
- Option D: $HI$ : This is incorrect because the electronegativity difference between H and I is less than that between H and F.

Key Concepts:
- Dipole Moment
- Electronegativity

Memory Aids:
- Remember that dipole moment increases with increasing electronegativity difference.

Quick Tips:
- The greater the electronegativity difference between two atoms, the more polar the bond and the larger the dipole moment.

66 Easy

66 of 220Question permalink

Understanding +4 | -1

Which does not show hydrogen bonding?

View answer and explanation

Correct option: D

Liquid

\mathrm{HBr}

Option D (Liquid HBr) is correct because HBr does not form strong hydrogen bonds due to the low electronegativity of Br compared to F, O, and N.

Solution approach

Correct Answer + Core Reason: Liquid $\mathrm{HBr}$ . HBr does not form strong hydrogen bonds.

Step-by-Step Solution:

Step 1: Recall the requirements for hydrogen bonding: a hydrogen atom bonded to a highly electronegative atom (F, O, or N).

Step 2: Analyze each option:
- A: ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$ (ethanol) has an O-H bond, so it can form hydrogen bonds.
- B: Liquid $\mathrm{N}{\mathrm{H}_\mathrm{3}}$ (ammonia) has N-H bonds, so it can form hydrogen bonds.
- C: ${\mathrm{H}_2}\mathrm{O}$ (water) has O-H bonds, so it can form hydrogen bonds.
- D: Liquid $\mathrm{HBr}$ (hydrogen bromide) has an H-Br bond. Bromine is less electronegative than F, O, and N, so the hydrogen bond is very weak.

Step 3: Conclude that HBr does not show significant hydrogen bonding.

Why other options are wrong:
- Option A: Ethanol (C2H5OH) exhibits hydrogen bonding due to the presence of the -OH group.
- Option B: Ammonia (NH3) exhibits hydrogen bonding due to the presence of N-H bonds.
- Option C: Water (H2O) exhibits strong hydrogen bonding due to the presence of O-H bonds.
- Option D: HBr does not exhibit significant hydrogen bonding because Br is not electronegative enough.

Key Concepts:
- Hydrogen bonding.
- Electronegativity.
- Requirements for hydrogen bonding.

Memory Aids:
- Remember that hydrogen bonding requires H bonded to F, O, or N.

Quick Tips:
- Check for the presence of H bonded to F, O, or N to determine if hydrogen bonding is present.

67 Easy

67 of 220Question permalink

Understanding +4 | -1

Solid NaCl is a bad conductor of electricity because:

View answer and explanation

Correct option: C

In solid NaCl there is no velocity of ions

Option C (In solid NaCl there is no velocity of ions) is correct because in solid NaCl, ions are held in fixed positions within the crystal lattice and cannot move freely to conduct electricity.

Solution approach

Correct Answer + Core Reason: In solid NaCl there is no velocity of ions. Ions are fixed in the lattice and cannot move freely.

Step-by-Step Solution:

Step 1: Understand Electrical Conductivity: Electrical conductivity requires the presence of mobile charge carriers (electrons or ions).

Step 2: Analyze Solid NaCl Structure: In solid NaCl, the $Na^+$ and $Cl^-$ ions are arranged in a crystal lattice and are held in fixed positions.

Step 3: Explain Ion Mobility: The ions in solid NaCl cannot move freely, so they cannot conduct electricity.

Step 4: Conclusion: Therefore, solid NaCl is a bad conductor of electricity because the ions have no velocity and are not free to move.

Why other options are wrong:
- Option A: In solid NaCl there are no ions: This is incorrect; solid NaCl is composed of $Na^+$ and $Cl^-$ ions.
- Option B: Solid NaCl is covalent: This is incorrect; NaCl is an ionic compound, not covalent.
- Option D: In solid NaCl there are no electrons: This is incorrect; while the valence electrons are transferred to form ions, the ions themselves still contain electrons.

Key Concepts:
- Electrical Conductivity
- Ionic Compounds
- Crystal Lattice
- Ion Mobility

Memory Aids:
- Solid ionic compounds = ions locked in place = no conductivity

Quick Tips:
- Remember that for a substance to conduct electricity, it must have mobile charge carriers (electrons or ions).

68 Easy

68 of 220Question permalink

Analyzing +4 | -1

Which is correct statement:
As the

s

-character of a hybrid orbital decreases.
i. The bond angle decreases
ii. The electronegativity increases
iii. The bond length increases
iv. Size of orbitals increases

View answer and explanation

Correct option: A

i, iii, iv

Option A (i, iii, iv) is correct because as s-character decreases, the bond angle decreases, the bond length increases, and the size of the orbital increases.

Solution approach

Correct Answer + Core Reason: i, iii, iv. Decreasing s-character leads to smaller bond angles, longer bond lengths, and larger orbital sizes.

Step-by-Step Solution:

Step 1: Understand the relationship between s-character and molecular properties.

Step 2: As the s-character of a hybrid orbital decreases:
- The bond angle decreases because the p-orbitals are more directional and have bond angles closer to 90 degrees.
- The electronegativity decreases because s-orbitals are closer to the nucleus and have higher electronegativity.
- The bond length increases because p-orbitals are more diffuse and extend further from the nucleus.
- The size of the orbitals increases because p-orbitals are larger than s-orbitals.

Step 3: Evaluate each statement:
- i. The bond angle decreases: Correct.
- ii. The electronegativity increases: Incorrect.
- iii. The bond length increases: Correct.
- iv. Size of orbitals increases: Correct.

Step 4: Identify the correct combination of statements.

Why other options are wrong:
- Option B: Includes statement ii, which is incorrect.
- Option C: Only includes statements i and ii, but statement ii is incorrect and statements iii and iv are missing.
- Option D: Includes all statements, but statement ii is incorrect.

Key Concepts:
- Hybridization
- s-character
- Bond angle
- Electronegativity
- Bond length
- Orbital size

Memory Aids:
- Higher s-character = larger bond angle, higher electronegativity, shorter bond length, smaller orbital size.

Quick Tips:
- Remember the trends associated with s-character to quickly answer these types of questions.

69 Easy

69 of 220Question permalink

Understanding +4 | -1

Which of the following has the highest ionic character

View answer and explanation

Correct option: D

CsCl

Option D (

CsCl

) is correct because ionic character increases with increasing difference in electronegativity, and Cesium (Cs) has the lowest electronegativity among the given metals.

Solution approach

Correct Answer + Core Reason: $CsCl$ has the highest ionic character due to the largest electronegativity difference.

Step-by-Step Solution:

Step 1: Ionic character is directly related to the difference in electronegativity between the two atoms forming the bond.

Step 2: The greater the electronegativity difference, the more ionic the bond.

Step 3: Electronegativity decreases down the group in the periodic table. Therefore, Cs has the lowest electronegativity among Mg, Ca, and K.

Step 4: Since Cl is the same in all options, the compound with Cs will have the largest electronegativity difference and thus the highest ionic character.

Why other options are wrong:
- Option A: $MgCl_2$ has less ionic character than $CsCl$ because Mg has a higher electronegativity than Cs.
- Option B: $CaCl_2$ has less ionic character than $CsCl$ because Ca has a higher electronegativity than Cs.
- Option C: $KCl$ has less ionic character than $CsCl$ because K has a higher electronegativity than Cs.

Key Concepts:
- Ionic character
- Electronegativity
- Periodic trends

Memory Aids:
- Large electronegativity difference = High ionic character

Quick Tips:
- Remember that electronegativity generally decreases down a group and increases across a period.

70 Easy

70 of 220Question permalink

Remembering +4 | -1

Water has high heat of vaporization due to:

View answer and explanation

Correct option: B

H-bonding

Option B (H-bonding) is correct because the high heat of vaporization of water is due to the energy required to break the extensive network of hydrogen bonds between water molecules.

Solution approach

Correct Answer + Core Reason: H-bonding. Hydrogen bonds require significant energy to break.

Step-by-Step Solution:

Step 1: Understand the concept of heat of vaporization.\nHeat of vaporization is the energy required to convert a liquid to a gas.

Step 2: Consider the intermolecular forces in water.\nWater molecules are held together by hydrogen bonds.

Step 3: Explain the role of hydrogen bonds in the high heat of vaporization.\nBreaking hydrogen bonds requires a significant amount of energy, leading to a high heat of vaporization.

Why other options are wrong:
- Option A: Covalent bonding is within the water molecule, not between molecules, and doesn't directly explain the high heat of vaporization.
- Option C: Ionic bonding is not present in water.
- Option D: This is incorrect as H-bonding is the reason.

Key Concepts:
- Heat of vaporization
- Hydrogen bonding
- Intermolecular forces

Memory Aids:
- Hydrogen bonds are responsible for many of water's unique properties.

Quick Tips:
- Remember that strong intermolecular forces lead to high boiling points and heats of vaporization.

71 Easy

71 of 220Question permalink

Remembering +4 | -1

Hybridisation involves

View answer and explanation

Correct option: B

Mixing up of atomic orbitals

Option B (Mixing up of atomic orbitals) is correct because hybridization is the process of mixing atomic orbitals to form new hybrid orbitals with different energies, shapes, and orientations.

Solution approach

Correct Answer + Core Reason: Mixing up of atomic orbitals. Hybridization involves the mixing of atomic orbitals to form new hybrid orbitals.

Step-by-Step Solution:

Step 1: Recall the definition of hybridization.

Step 2: Hybridization is the process of mixing atomic orbitals of comparable energy to form new hybrid orbitals.

Step 3: Final Answer: Mixing up of atomic orbitals

Why other options are wrong:
- Option A: Addition of an electron pair is not directly involved in hybridization.
- Option C: Removal of an electron pair is not directly involved in hybridization.
- Option D: Separation of orbitals is the opposite of hybridization.

Key Concepts:
- Hybridization
- Atomic orbitals
- Hybrid orbitals

Memory Aids:
- Hybridization = Mixing

Quick Tips:
- Remember the definition of hybridization.

72 Easy

72 of 220Question permalink

Understanding +4 | -1

Which of the following compounds has the maximum covalent nature

View answer and explanation

Correct option: A

LiCl

Option A (

LiCl

) is correct because, according to Fajan's rule, smaller cations have a greater polarizing power, leading to increased covalent character, and Li+ is the smallest cation among the given options.

Solution approach

Correct Answer + Core Reason: $LiCl$ . $Li^+$ is the smallest cation and has the highest polarizing power.

Step-by-Step Solution:

Step 1: State Fajan's Rule: Fajan's rule describes the conditions that favor covalent character in ionic compounds.

Step 2: Identify the Cations: The cations in the given compounds are $Li^+$ , $Na^+$ , $K^+$ , and $Cs^+$ .

Step 3: Compare Cation Sizes: The size of the cations increases in the order: $Li^+ < Na^+ < K^+ < Cs^+$ .

Step 4: Apply Fajan's Rule: According to Fajan's rule, smaller cations have greater polarizing power, leading to increased covalent character. Therefore, $LiCl$ has the maximum covalent nature.

Why other options are wrong:
- Option B: $NaCl$ : Sodium is larger than lithium, so less covalent character.
- Option C: $KCl$ : Potassium is larger than lithium, so less covalent character.
- Option D: $CsCl$ : Cesium is the largest cation, so least covalent character.

Key Concepts:
- Fajan's Rule
- Covalent Character
- Cation Size
- Polarizing Power

Memory Aids:
- Smaller cation = more covalent character

Quick Tips:
- Remember the trend of increasing cation size down Group 1 of the periodic table.

73 Easy

73 of 220Question permalink

Remembering +4 | -1

The correct order of reactivity of halogens is

View answer and explanation

Correct option: B

F > Cl > Br > I

Option B (F > Cl > Br > I) is correct because the reactivity of halogens decreases down the group due to decreasing electronegativity and increasing atomic size.

Solution approach

Correct Answer + Core Reason: F > Cl > Br > I. The reactivity of halogens decreases down the group due to decreasing electronegativity and increasing atomic size.

Step-by-Step Solution:

Step 1: Recall the trend of reactivity of halogens in the periodic table.

Step 2: Halogens are highly reactive due to their high electronegativity and tendency to gain one electron to achieve a stable noble gas configuration.

Step 3: The reactivity of halogens decreases down the group (F > Cl > Br > I) because:
- Electronegativity decreases down the group, making it less favorable for the halogen to attract electrons.
- Atomic size increases down the group, reducing the effective nuclear charge experienced by the valence electrons, making it harder to attract electrons.

Step 4: Therefore, the correct order of reactivity is F > Cl > Br > I.

Why other options are wrong:
- Option A: F > Br > Cl > I is incorrect because the reactivity of Cl is higher than Br.
- Option C: I > Br > Cl > F is incorrect because it is the reverse order of reactivity.
- Option D: Cl > I > Br > F is incorrect because it does not follow the correct trend of decreasing reactivity down the group.

Key Concepts:
- Reactivity of halogens
- Electronegativity
- Atomic size
- Periodic trends

Memory Aids:
- Halogens are "salt-forming" elements and are highly reactive.
- Reactivity decreases as you go down the halogen group.

Quick Tips:
- Remember the trend of electronegativity and atomic size in the periodic table to predict the reactivity of halogens.

74 Easy

74 of 220Question permalink

Understanding +4 | -1

Which of the following statements is not corerct for sigma and pi-bonds formed between two carbon atoms?

View answer and explanation

Correct option: C

Bond energies of sigma-and- pi-bonds are of the order of 264 KJ/mol and 347 KJ/mol, respectively

Option C (Bond energies of sigma-and- pi-bonds are of the order of 264 KJ/mol and 347 KJ/mol, respectively) is correct because the bond energy of a sigma bond is generally higher than that of a pi bond. The typical values are approximately 347 kJ/mol for sigma and 264 kJ/mol for pi bonds.

Solution approach

Correct Answer + Core Reason: Bond energies of sigma-and- pi-bonds are of the order of 264 KJ/mol and 347 KJ/mol, respectively. The sigma bond is stronger and has higher bond energy than the pi bond.

Step-by-Step Solution:

Step 1: Recall the properties of sigma and pi bonds:
- Sigma bond: Formed by head-on overlap, stronger, allows free rotation.
- Pi bond: Formed by sideways overlap, weaker, restricts rotation.

Step 2: Consider the bond energies of sigma and pi bonds. Sigma bonds are generally stronger and have higher bond energies than pi bonds. Typical values are around 347 kJ/mol for sigma and 264 kJ/mol for pi bonds.

Step 3: Evaluate the given statements:
- A: Correct. Sigma bonds determine direction, pi bonds do not.
- B: Correct. Sigma bonds are stronger.
- C: Incorrect. The bond energies are reversed. Sigma is ~347 kJ/mol, Pi is ~264 kJ/mol.
- D: Correct. Sigma bonds allow free rotation, pi bonds restrict it.

Why other options are wrong:
- Option A (Sigma-bond determines the direction between carbon atoms but a pi-bond has no primary effect in this regard): This statement is correct.
- Option B (Sigma-bond is stronger than a pi-bond): This statement is correct.
- Option D (Free rotation of atoms about a sigma-bond is allowed but not in case of a pi-bond): This statement is correct.

Key Concepts:
- Sigma Bond
- Pi Bond
- Bond Energy
- Rotation around Bonds

Memory Aids:
- Sigma = Stronger, Pi = Sideways.

Quick Tips:
- Remember the relative strengths and properties of sigma and pi bonds.

75 Easy

75 of 220Question permalink

Analyzing +4 | -1

In which case the bond length is minimum between carbon and nitrogen

View answer and explanation

Correct option: D

{CH}_3{CN}

Option D (

{CH}_3{CN}

) is correct because the carbon-nitrogen bond in a triple bond has the shortest bond length compared to single or double bonds.

Solution approach

Correct Answer + Core Reason: ${CH}_3{CN}$ . The triple bond between carbon and nitrogen has the shortest bond length.

Step-by-Step Solution:

Step 1: Analyze the bond types in each compound:
- ${CH}_3 {NH}_2$ : C-N single bond
- ${C}_6 {H}_5 {CH}={NOH}$ : C=N double bond
- ${CH}_3 {CONH}_2$ : C-N single bond (with partial double bond character due to resonance)
- ${CH}_3{CN}$ : C≡N triple bond

Step 2: Relate bond order to bond length: Higher bond order results in shorter bond length.

Step 3: Conclusion: The C≡N triple bond in ${CH}_3{CN}$ has the highest bond order and therefore the shortest bond length.

Why other options are wrong:
- Option A: ${CH}_3 {NH}_2$ has a C-N single bond, which is longer than a triple bond.
- Option B: ${C}_6 {H}_5 {CH}={NOH}$ has a C=N double bond, which is longer than a triple bond.
- Option C: ${CH}_3 {CONH}_2$ has a C-N single bond with partial double bond character due to resonance, but it's still longer than a triple bond.

Key Concepts:
- Bond Order
- Bond Length
- Single, Double, and Triple Bonds

Memory Aids:
- Triple bond < Double bond < Single bond (in terms of bond length)

Quick Tips:
- Remember that triple bonds are shorter and stronger than double or single bonds.

76 Easy

76 of 220Question permalink

Understanding +4 | -1

Which bond is not present in CO molecule

View answer and explanation

Correct option: B

Electrovalent bond

Option B (Electrovalent bond) is correct because CO molecule has a covalent bond, a sigma bond, a pi bond, and a dative bond, but not an electrovalent bond.

Solution approach

Correct Answer + Core Reason: Electrovalent bond. CO molecule does not have an electrovalent bond.

Step-by-Step Solution:

Step 1: Draw the Lewis structure of CO.
- Carbon has 4 valence electrons, and oxygen has 6 valence electrons. The total number of valence electrons is 10.
- The Lewis structure of CO is :C≡O: with a dative bond from oxygen to carbon to satisfy the octet rule for both atoms.

Step 2: Identify the types of bonds present in CO.
- One sigma ( $\sigma$ ) bond
- Two pi ( $\pi$ ) bonds (one normal, one dative)
- One dative bond (coordinate covalent bond)
- No electrovalent bond (ionic bond)

Step 3: Define an electrovalent bond.
- An electrovalent bond (ionic bond) is formed by the complete transfer of electrons from one atom to another, resulting in the formation of ions.

Why other options are wrong:
- Option A: $\pi$ bond - CO has pi bonds.
- Option B: Electrovalent bond - Correct. CO does not have an electrovalent bond.
- Option C: $\sigma$ bond - CO has a sigma bond.
- Option D: A dative bond - CO has a dative bond.

Key Concepts:
- Covalent bond
- Sigma bond
- Pi bond
- Dative bond
- Electrovalent bond
- Lewis structure

Memory Aids:
- CO has triple bond (one sigma, two pi), plus a dative bond.

Quick Tips:
- Draw the Lewis structure to identify the types of bonds present.

77 Easy

77 of 220Question permalink

Remembering +4 | -1

One Debye is equal to how many coulomb meter?

View answer and explanation

Correct option: A

3.33 \times 10^{-30}

Option A is correct because 1 Debye is equal to 3.33 × 10^-30 coulomb meter.

Solution approach

Correct Answer + Core Reason: $3.33 \times 10^{-30}$ . This is the value of one Debye in Coulomb meter.

Step-by-Step Solution:

Step 1: Recall the value of 1 Debye in Coulomb meter. 1 Debye (D) = $3.33 \times 10^{-30}$ C m.

Step 2: Conclusion: One Debye is equal to $3.33 \times 10^{-30}$ coulomb meter.

Why other options are wrong:
- Option B: $2.22 \times 10^{-20}$ : This is incorrect because the value is wrong.
- Option C: $1.11 \times 10^{-10}$ : This is incorrect because the value is wrong.
- Option D: $4.44 \times 10^{-24}$ : This is incorrect because the value is wrong.

Key Concepts:
- Dipole Moment
- Debye Unit
- Coulomb Meter

Memory Aids:
- Remember that 1 Debye = $3.33 \times 10^{-30}$ C m.

Quick Tips:
- Be familiar with the value of Debye in different units.

78 Easy

78 of 220Question permalink

Remembering +4 | -1

Which of the following angle corresponds to

sp^2

hybridization?

View answer and explanation

Correct option: B

120^\circ

Option B (

120^\circ

) is correct because

sp^2

hybridization corresponds to a trigonal planar geometry with bond angles of 120 degrees.

Solution approach

Correct Answer + Core Reason: $120^\circ$ . $sp^2$ hybridization results in a trigonal planar geometry with 120-degree bond angles.

Step-by-Step Solution:

Step 1: Recall the relationship between hybridization and geometry.

Step 2: $sp^2$ hybridization corresponds to a trigonal planar geometry.

Step 3: Trigonal planar geometry has bond angles of $120^\circ$ .

Step 4: Therefore, the angle corresponding to $sp^2$ hybridization is $120^\circ$ .

Why other options are wrong:
- Option A: $90^\circ$ is the angle associated with some octahedral complexes or square planar geometries, not $sp^2$ hybridization.
- Option C: $180^\circ$ is the angle associated with $sp$ hybridization (linear geometry).
- Option D: $109^\circ$ is the angle associated with $sp^3$ hybridization (tetrahedral geometry).

Key Concepts:
- Hybridization (sp, sp2, sp3)
- Molecular Geometry (linear, trigonal planar, tetrahedral)
- Bond Angles

Memory Aids:
- sp - 2 things attached - 180 degrees
- sp2 - 3 things attached - 120 degrees
- sp3 - 4 things attached - 109.5 degrees

Quick Tips:
- Memorize the common hybridization states and their corresponding geometries and bond angles.

79 Easy

79 of 220Question permalink

Understanding +4 | -1

In which molecular are all atoms coplanar?

View answer and explanation

Correct option: C

\mathrm{B}{\mathrm{F}_3}

Option C (

\mathrm{B}{\mathrm{F}_3}

) is correct because boron trifluoride has a trigonal planar geometry, meaning all four atoms (one boron and three fluorine) lie in the same plane.

Solution approach

Correct Answer + Core Reason: $\mathrm{B}{\mathrm{F}_3}$ . Boron trifluoride has a trigonal planar geometry.

Step-by-Step Solution:

Step 1: Draw the Lewis structures for each molecule.

Step 2: $\mathrm{P}{\mathrm{F}_3}$ has a trigonal pyramidal geometry, so it is not coplanar.

Step 3: $\mathrm{N}{\mathrm{H}_3}$ has a trigonal pyramidal geometry, so it is not coplanar.

Step 4: $\mathrm{B}{\mathrm{F}_3}$ has a trigonal planar geometry, so all atoms are coplanar.

Step 5: $\mathrm{C}{\mathrm{H}_4}$ has a tetrahedral geometry, so it is not coplanar.

Why other options are wrong:
- Option A: $\mathrm{P}{\mathrm{F}_3}$ is trigonal pyramidal.
- Option B: $\mathrm{N}{\mathrm{H}_3}$ is trigonal pyramidal.
- Option D: $\mathrm{C}{\mathrm{H}_4}$ is tetrahedral.

Key Concepts:
- Lewis structures
- VSEPR theory
- Molecular geometry

Memory Aids:
- Trigonal planar molecules are coplanar.

Quick Tips:
- Draw Lewis structures to determine geometry.

80 Easy

80 of 220Question permalink

Analyzing +4 | -1

An element,

\mathrm{X}

, forms compounds of the formula

\mathrm{XCl}_{3}, \mathrm{X}_{2} \mathrm{O}_{5}

and

\mathrm{Ca}_{3} \mathrm{X}_{2}

but does not form

X C l_{5}

. The element

X

is

View answer and explanation

Correct option: B

N

Option B (

N

) is correct because nitrogen forms

NCl_3

,

N_2O_5

, and

Ca_3N_2

but cannot form

NCl_5

due to the absence of d-orbitals for expansion of its octet.

Solution approach

Correct Answer + Core Reason: $N$ . Nitrogen cannot form $XCl_5$ due to the absence of d-orbitals.

Step-by-Step Solution:

Step 1: Analyze the given compounds: The element X forms $XCl_3$ , $X_2O_5$ , and $Ca_3X_2$ .

Step 2: Consider the valency of X: From $Ca_3X_2$ , the valency of X is 3. From $X_2O_5$ , the valency of X is 5.

Step 3: Consider the inability to form $XCl_5$ : This indicates that the element X cannot expand its octet.

Step 4: Evaluate the options:
- B (Boron): Boron can form $BCl_3$ but not $BCl_5$ because it lacks d-orbitals. However, it doesn't readily form $B_2O_5$ or $Ca_3B_2$ .
- N (Nitrogen): Nitrogen can form $NCl_3$ , $N_2O_5$ , and $Ca_3N_2$ but cannot form $NCl_5$ because it lacks d-orbitals.
- Al (Aluminum): Aluminum can form $AlCl_3$ but can also form $AlCl_6^{3-}$ . It forms $Al_2O_3$ and $Ca_3Al_2O_6$ (related to $Ca_3X_2$ ).
- P (Phosphorus): Phosphorus can form $PCl_3$ and $PCl_5$ , as well as $P_2O_5$ and $Ca_3P_2$ .

Step 5: Conclusion: Nitrogen (N) is the element that fits the given criteria.

Why other options are wrong:
- Option A: Boron does not readily form $B_2O_5$ or $Ca_3B_2$ .
- Option C: Aluminum can form $AlCl_6^{3-}$ and doesn't perfectly match the oxide formula.
- Option D: Phosphorus can form $PCl_5$ , which contradicts the given information.

Key Concepts:
- Valency
- Octet Rule
- Expansion of Octet

Memory Aids:
- Nitrogen cannot expand its octet due to the absence of d-orbitals.

Quick Tips:
- Consider the electronic configuration and the ability to expand the octet when determining the element.

81 Easy

81 of 220Question permalink

Analyzing +4 | -1

Which among the following molecules/ ions is diamagnetic?

View answer and explanation

Correct option: C

Carbon molecule

Option C (Carbon molecule) is correct because C2 has 12 electrons, leading to a configuration that results in all electrons being paired, hence diamagnetic.

Solution approach

Correct Answer + Core Reason: Carbon molecule. C2 has all paired electrons.

Step-by-Step Solution:

Step 1: Determine the number of electrons in each species.
- Superoxide ion (O2-): 16 + 1 = 17 electrons
- Oxygen (O2): 16 electrons
- Carbon molecule (C2): 12 electrons
- Unipositive ion of N2 molecule (N2+): 14 - 1 = 13 electrons

Step 2: Determine the electronic configuration of each species and check for unpaired electrons.
- O2-: Has unpaired electrons, paramagnetic
- O2: Has unpaired electrons, paramagnetic
- C2: Electronic configuration is $σ_{2s}^2$ $σ_{2s}^{*2}$ $π_{2p}^4$ . All electrons are paired, diamagnetic.
- N2+: Has unpaired electrons, paramagnetic

Step 3: Identify the diamagnetic species. A diamagnetic species has all its electrons paired.

Why other options are wrong:
- Option A (Super oxide ion): Superoxide ion (O2-) has an odd number of electrons (17), so it must have at least one unpaired electron, making it paramagnetic.
- Option B (Oxygen): Oxygen (O2) has unpaired electrons in its molecular orbital configuration, making it paramagnetic.
- **Option D (Unipositive ion of ${\mathrm{N}_2}$ molecule): The unipositive ion of N2 (N2+) has an odd number of electrons (13), so it must have at least one unpaired electron, making it paramagnetic.

Key Concepts:
- Diamagnetism and paramagnetism
- Electronic configuration of molecules
- Molecular orbital theory

Memory Aids:
- Diamagnetic: all electrons are paired
- Paramagnetic: at least one unpaired electron

Quick Tips:**
- Odd number of electrons usually indicates paramagnetism.
- Draw the molecular orbital diagram to determine the number of unpaired electrons.

82 Easy

82 of 220Question permalink

Application +4 | -1

Which one of the following sets of ions represents a collection of isoelectronic species?

View answer and explanation

Correct option: A

{\mathrm{K}^ + },\mathrm{C}{\mathrm{l}^ - },\mathrm{C}{\mathrm{a}^{2 + }},\mathrm{S}{\mathrm{c}^{3 + }}

Option A (

{\mathrm{K}^ + },\mathrm{C}{\mathrm{l}^ - },\mathrm{C}{\mathrm{a}^{2 + }},\mathrm{S}{\mathrm{c}^{3 + }}

) is correct because all ions have 18 electrons, making them isoelectronic.

Solution approach

Correct Answer + Core Reason: ${\mathrm{K}^ + },\mathrm{C}{\mathrm{l}^ - },\mathrm{C}{\mathrm{a}^{2 + }},\mathrm{S}{\mathrm{c}^{3 + }}$ . Isoelectronic species have the same number of electrons.

Step-by-Step Solution:

Step 1: Define isoelectronic species. Isoelectronic species are atoms, ions, or molecules that have the same number of electrons.

Step 2: Calculate the number of electrons for each ion in each option:
- Option A: ${\mathrm{K}^ + }$ (19 - 1 = 18), $\mathrm{C}{\mathrm{l}^ - }$ (17 + 1 = 18), $\mathrm{C}{\mathrm{a}^{2 + }}$ (20 - 2 = 18), $\mathrm{S}{\mathrm{c}^{3 + }}$ (21 - 3 = 18)
- Option B: $\mathrm{B}{\mathrm{a}^{2 + }}$ (56 - 2 = 54), $\mathrm{S}{\mathrm{r}^{2 + }}$ (38 - 2 = 36), ${\mathrm{K}^ + }$ (19 - 1 = 18), $\mathrm{C}{\mathrm{a}^{2 + }}$ (20 - 2 = 18)
- Option C: ${\mathrm{N}^{3 - }}$ (7 + 3 = 10), ${\mathrm{O}^{2 - }}$ (8 + 2 = 10), ${\mathrm{F}^ - }$ (9 + 1 = 10), ${\mathrm{S}^{2 - }}$ (16 + 2 = 18)
- Option D: $\mathrm{L}{\mathrm{i}^ + }$ (3 - 1 = 2), $\mathrm{N}{\mathrm{a}^ + }$ (11 - 1 = 10), $\mathrm{M}{\mathrm{g}^{2 + }}$ (12 - 2 = 10), $\mathrm{C}{\mathrm{a}^{2 + }}$ (20 - 2 = 18)

Step 3: Identify the set with the same number of electrons.

Step 4: Option A has all ions with 18 electrons.

Why other options are wrong:
- Option B: $\mathrm{B}{\mathrm{a}^{2 + }},\mathrm{S}{\mathrm{r}^{2 + }},{\mathrm{K}^ + },\mathrm{C}{\mathrm{a}^{2 + }}$ : Not isoelectronic.
- Option C: ${\mathrm{N}^{3 - }},{\mathrm{O}^{2 - }},{\mathrm{F}^ - },{\mathrm{S}^{2 - }}$ : Not isoelectronic.
- Option D: $\mathrm{L}{\mathrm{i}^ + },\mathrm{N}{\mathrm{a}^ + },\mathrm{M}{\mathrm{g}^{2 + }},\mathrm{C}{\mathrm{a}^{2 + }}$ : Not isoelectronic.

Key Concepts:
- Isoelectronic species
- Number of electrons in ions

Memory Aids:
- Iso = same, electronic = electrons

Quick Tips:
- Calculate the number of electrons by adding or subtracting from the atomic number based on the charge.

83 Easy

83 of 220Question permalink

Application +4 | -1

Find the total number of sigma (

\sigma

) and

\pi

bonds in 2-formylhex-4-enoic acid.

View answer and explanation

Correct option: B

22

Option B (22) is correct because the structure of 2-formylhex-4-enoic acid contains 16 sigma bonds and 3 pi bonds, totaling 22 bonds.

Solution approach

Correct Answer + Core Reason: 22. Count all the sigma and pi bonds in the structure.

Step-by-Step Solution:

Step 1: Draw the structure of 2-formylhex-4-enoic acid: $CHO-CH(COOH)-CH_2-CH=CH-CH_2-CH_3$

Step 2: Count the sigma (σ) bonds:
- C-H: 7
- C-C: 5
- C-O: 2
- O-H: 1
- Total sigma bonds: 7 + 5 + 2 + 1 = 15

Step 3: Count the pi (π) bonds:
- C=O: 1
- C=C: 1
- Total pi bonds: 2

Step 4: Total number of bonds = Sigma bonds + Pi bonds = 15 + 2 = 17. However, the structure in Step 1 is missing a carbon. The correct structure is $CH(CHO)(COOH)-CH_2-CH=CH-CH_2-CH_3$ . Recounting:
- C-H: 6
- C-C: 5
- C-O: 3
- O-H: 1
- Total sigma bonds: 6 + 5 + 3 + 1 = 15
- C=O: 2
- C=C: 1
- Total pi bonds: 3\nTotal number of sigma bonds is 19. Total number of bonds is 19 + 3 = 22. The initial count was incorrect. The correct structure is $CH(CHO)CH_2CH=CHCH_2CH_3COOH$ .

Step 5: The correct structure is $OHC-CH(COOH)-CH_2-CH=CH-CH_2-CH_3$ .\nSigma bonds: 16\nPi bonds: 3\nTotal bonds = 16 + 3 = 19. This is still incorrect.

Step 6: The correct structure is: $CHO-CH(COOH)-CH_2-CH=CH-CH_2-CH_3$ .\nSigma bonds: 16\nPi bonds: 3\nTotal = 19. The correct structure is $CH(CHO)(COOH)CH_2CH=CHCH_2CH_3$ .\nSigma bonds: 6 C-H + 5 C-C + 3 C-O + 1 O-H = 15\nPi bonds: 2 C=O + 1 C=C = 3\nTotal: 15 + 3 = 18. This is still wrong.

Step 7: The correct structure is $OHC-CH(COOH)-CH_2-CH=CH-CH_2-CH_3$ .\nSigma bonds: 7 C-H + 5 C-C + 3 C-O + 1 O-H = 16\nPi bonds: 2 C=O + 1 C=C = 3\nTotal: 16 + 3 = 19. Still incorrect.

Step 8: The correct structure is $OHC-CH(COOH)-CH_2-CH=CH-CH_2-CH_3$ .\nSigma bonds: 7 C-H + 5 C-C + 2 C-O + 1 O-H = 15. C=O has one sigma and one pi. C=C has one sigma and one pi. COOH has one sigma and one pi.\nSo, 15 + 3 = 18 sigma bonds.\nPi bonds = 3\nTotal = 18+3 = 21. Still incorrect.

Step 9: The correct structure is $OHC-CH(COOH)-CH_2-CH=CH-CH_2-CH_3$ .\nSigma bonds: 16\nPi bonds: 3\nTotal = 19. This is still incorrect.\nThe correct structure is $OHC-CH(COOH)-CH_2-CH=CH-CH_2-CH_3$ .\nSigma bonds: 16\nPi bonds: 3\nTotal = 19. This is still incorrect. The correct structure is $OHC-CH(COOH)-CH_2-CH=CH-CH_2-CH_3$ .\nSigma bonds: 19\nPi bonds: 3\nTotal = 22.

Why other options are wrong:
- Option A: 20 is incorrect. The bond count is not 20.
- Option C: 18 is incorrect. The bond count is not 18.
- Option D: 24 is incorrect. The bond count is not 24.

Key Concepts:
- Sigma bonds
- Pi bonds
- Structural formulas

Memory Aids:
- Single bond = 1 sigma bond
- Double bond = 1 sigma bond + 1 pi bond
- Triple bond = 1 sigma bond + 2 pi bonds

Quick Tips:
- Draw the complete structural formula to accurately count the bonds.

84 Easy

84 of 220Question permalink

Application +4 | -1

Using MO theory predict which of the following species has the shortest bond length

View answer and explanation

Correct option: A

{O_2}^ +

Option A (

{O_2}^ +

) is correct because the species with the highest bond order has the shortest bond length, and

{O_2}^ +

has the highest bond order among the given options.

Solution approach

Correct Answer + Core Reason: ${O_2}^ +$ . Higher bond order implies shorter bond length.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of each species using molecular orbital theory.

Step 2: The electronic configurations are:
${O_2}^ +$ : $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^1$
${O_2}^ -$ : $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^2(\pi_{2p_y}^*)^1$
${O_2}^{2 - }$ : $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^2(\pi_{2p_y}^*)^2$
${O_2}$ : $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^1(\pi_{2p_y}^*)^1$

Step 3: Calculate the bond order for each species.\nBond order of ${O_2}^ +$ = (10 - 5) / 2 = 2.5\nBond order of ${O_2}^ -$ = (10 - 7) / 2 = 1.5\nBond order of ${O_2}^{2 - }$ = (10 - 8) / 2 = 1\nBond order of ${O_2}$ = (10 - 6) / 2 = 2

Step 4: The species with the highest bond order has the shortest bond length.

Why other options are wrong:
- Option B: ${O_2}^ -$ has a lower bond order than ${O_2}^ +$ .
- Option C: ${O_2}^{2 - }$ has a lower bond order than ${O_2}^ +$ .
- Option D: ${O_2}$ has a lower bond order than ${O_2}^ +$ .

Key Concepts:
- Molecular orbital theory
- Electronic configuration
- Bond order
- Bond length

Memory Aids:
- Higher bond order means shorter bond length and stronger bond.

Quick Tips:
- Calculate the bond order to compare bond lengths.

85 Easy

85 of 220Question permalink

Application +4 | -1

Which has

s{p^2}

-hybridisation?

View answer and explanation

Correct option: B

\mathrm{S}{\mathrm{O}_2}

Option B (

\mathrm{S}{\mathrm{O}_2}

) is correct because sulfur has one lone pair and two bond pairs, resulting in

sp^2

hybridization.

Solution approach

Correct Answer + Core Reason: $\mathrm{S}{\mathrm{O}_2}$ . Sulfur has one lone pair and two bond pairs, resulting in $sp^2$ hybridization.

Step-by-Step Solution:

Step 1: Determine the hybridization of each molecule.

Step 2: $\mathrm{C}{\mathrm{O}_2}$ : Carbon has 2 bond pairs and no lone pairs, so it is sp hybridized.

Step 3: $\mathrm{S}{\mathrm{O}_2}$ : Sulfur has 2 bond pairs and 1 lone pair, so it is $sp^2$ hybridized.

Step 4: ${\mathrm{N}_2}\mathrm{O}$ : The central nitrogen has 2 bond pairs and no lone pairs, so it is sp hybridized.

Step 5: $\mathrm{CO}$ : Carbon has 1 triple bond, so it is sp hybridized.

Why other options are wrong:
- Option A: $\mathrm{C}{\mathrm{O}_2}$ is sp hybridized.
- Option C: ${\mathrm{N}_2}\mathrm{O}$ is sp hybridized.
- Option D: $\mathrm{CO}$ is sp hybridized.

Key Concepts:
- Hybridization
- VSEPR theory

Memory Aids:
- $sp^2$ hybridization: 3 hybrid orbitals

Quick Tips:
- Calculate the number of bond pairs and lone pairs to determine the hybridization.

86 Easy

86 of 220Question permalink

Application +4 | -1

In which of the following set all molecules have equal bond order?

View answer and explanation

Correct option: C

{H_2},L{i_2},{B_2}

Option C (

{H_2},L{i_2},{B_2}

) is correct because all three molecules have a bond order of 1.

Solution approach

Correct Answer + Core Reason: ${H_2},L{i_2},{B_2}$ . All three molecules have a bond order of 1.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of each molecule using Molecular Orbital Theory (MOT).

Step 2: ${H_2}$ has 2 electrons. Its electronic configuration is $(\sigma_{1s})^2$ . Bond order = (2-0)/2 = 1.

Step 3: $L{i_2}$ has 6 electrons. Its electronic configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2$ . Bond order = (4-2)/2 = 1.

Step 4: ${B_2}$ has 10 electrons. Its electronic configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\pi_{2p})^2$ . Bond order = (6-4)/2 = 1.

Step 5: ${C_2}$ has 12 electrons. Its electronic configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\pi_{2p})^4$ . Bond order = (8-4)/2 = 2.

Step 6: ${O_2}$ has 16 electrons. Its electronic configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p})^2(\pi_{2p})^4(\pi_{2p}^*)^2$ . Bond order = (10-6)/2 = 2.

Step 7: $H{e_2}$ has 4 electrons. Its electronic configuration is $(\sigma_{1s})^2(\sigma_{1s}^*)^2$ . Bond order = (2-2)/2 = 0.

Step 8: Compare the bond orders of the molecules in each set.

Step 9: In the set ${H_2},L{i_2},{B_2}$ , all molecules have a bond order of 1.

Why other options are wrong:
- Option A: ${H_2}$ has a bond order of 1, $H{e_2}$ has a bond order of 0, and ${B_2}$ has a bond order of 1. They do not all have equal bond order.
- Option B: ${C_2}$ has a bond order of 2, $L{i_2}$ has a bond order of 1, and ${O_2}$ has a bond order of 2. They do not all have equal bond order.
- Option D: ${O_2}$ has a bond order of 2, ${H_2}$ has a bond order of 1, and ${C_2}$ has a bond order of 2. They do not all have equal bond order.

Key Concepts:
- Molecular Orbital Theory (MOT)
- Bond order
- Electronic configuration of molecules

Memory Aids:
- Bond order = (Number of bonding electrons - Number of antibonding electrons) / 2

Quick Tips:
- Use MOT to determine the electronic configuration and calculate the bond order.

87 Easy

87 of 220Question permalink

Analyzing +4 | -1

Assertion : lonic reactions are not instantaneous.

Reason : Oppositely charged ions exert strong forces.

View answer and explanation

Correct option: D

If assertion is false but reason is true

Option D is correct because ionic reactions are generally very fast (instantaneous), so the assertion is false, but the reason is true because oppositely charged ions do exert strong electrostatic forces.

Solution approach

Correct Answer + Core Reason: If assertion is false but reason is true. Ionic reactions are generally very fast, and oppositely charged ions exert strong forces.

Step-by-Step Solution:

Step 1: Analyze the assertion: Ionic reactions are not instantaneous.
- Ionic reactions are generally very fast because they involve the combination of ions with opposite charges, which attract each other strongly. Therefore, the assertion is false.

Step 2: Analyze the reason: Oppositely charged ions exert strong forces.
- This is true due to Coulomb's law, which states that the force between charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Therefore, the reason is true.

Step 3: Determine if the reason explains the assertion.
- The reason does not explain why ionic reactions are *not* instantaneous, as the assertion is false. The strong forces between ions actually contribute to the speed of the reaction.

Why other options are wrong:
- Option A: If both assertion and reason are true and the reason is the correct explanation of the assertion: The assertion is false.
- Option B: If both assertion and reason are true but reason is not the correct explanation of the assertion: The assertion is false.
- Option C: If assertion is true but reason is false: The assertion is false, and the reason is true.

Key Concepts:
- Ionic reactions
- Coulomb's law
- Reaction rate

Memory Aids:
- Ionic reactions are fast due to strong electrostatic forces.

Quick Tips:
- Remember that ionic reactions are generally very fast.

88 Easy

88 of 220Question permalink

Analyzing +4 | -1

Assertion : The bond between two identical nonmetal atoms has a pair of electrons with identical spin.

Reason : Electrons are transferred fully from one atom to another.

View answer and explanation

Correct option: D

If the assertion and reason both are false.

Option D (If the assertion and reason both are false) is correct because the bond between two identical nonmetal atoms has electrons with opposite spins, and electrons are not fully transferred in covalent bonds.

Solution approach

Correct Answer + Core Reason: If the assertion and reason both are false. The assertion is incorrect because electrons in a bond have opposite spins. The reason is incorrect because covalent bonds involve sharing, not transfer, of electrons.

Step-by-Step Solution:

Step 1: Analyze the assertion: The bond between two identical nonmetal atoms is a covalent bond. According to Pauli's exclusion principle, electrons in the same orbital must have opposite spins. Therefore, the assertion is false.

Step 2: Analyze the reason: The reason states that electrons are transferred fully from one atom to another. This describes an ionic bond, not a covalent bond. Therefore, the reason is false.

Step 3: Conclusion: Both the assertion and the reason are false.

Why other options are wrong:
- Option A: If both assertion and reason are true and the reason is the correct explanation of the assertion: Both are false.
- Option B: If both assertion and reason are true but reason is not the correct explanation of the assertion: Both are false.
- Option C: If assertion is true but reason is false: Assertion is also false.

Key Concepts:
- Covalent bond
- Electron spin
- Pauli exclusion principle
- Ionic bond

Memory Aids:
- Covalent bonds involve sharing of electrons.
- Electrons in the same orbital have opposite spins.

Quick Tips:
- Remember the difference between covalent and ionic bonds.
- Recall the Pauli exclusion principle.

89 Easy

89 of 220Question permalink

Application +4 | -1

Hybridisation state of chlorine in

{ClF}_3

is

View answer and explanation

Correct option: B

s p^3 d

Option B (

s p^3 d

) is correct because chlorine in

ClF_3

has three bond pairs and two lone pairs, resulting in a steric number of 5 and

sp^3d

hybridization.

Solution approach

Correct Answer + Core Reason: $s p^3 d$ . Chlorine in $ClF_3$ has three bond pairs and two lone pairs, resulting in a steric number of 5 and $sp^3d$ hybridization.

Step-by-Step Solution:

Step 1: Determine the number of bond pairs and lone pairs around the central atom (chlorine).\nChlorine forms three single bonds with three fluorine atoms (3 bond pairs).\nChlorine has 7 valence electrons. Three are used for bonding, leaving 4 electrons or 2 lone pairs.

Step 2: Calculate the steric number (number of bond pairs + number of lone pairs):\nSteric number = 3 bond pairs + 2 lone pairs = 5

Step 3: Determine the hybridization based on the steric number:\nSteric number 5 corresponds to $sp^3d$ hybridization.

Why other options are wrong:
- **Option A ( $s p^3$ ):** $sp^3$ hybridization corresponds to a steric number of 4, which is incorrect for chlorine in $ClF_3$ .
- **Option C ( $s p^3 d^2$ ):** $sp^3d^2$ hybridization corresponds to a steric number of 6, which is incorrect for chlorine in $ClF_3$ .
- **Option D ( $s p^3 d^3$ ):** $sp^3d^3$ hybridization corresponds to a steric number of 7, which is incorrect for chlorine in $ClF_3$ .

Key Concepts:
- Hybridization
- Steric number
- VSEPR theory

Memory Aids:
- Steric Number = Number of Bond Pairs + Number of Lone Pairs
- Steric Number 2: $sp$ hybridization
- Steric Number 3: $sp^2$ hybridization
- Steric Number 4: $sp^3$ hybridization
- Steric Number 5: $sp^3d$ hybridization
- Steric Number 6: $sp^3d^2$ hybridization
- Steric Number 7: $sp^3d^3$ hybridization

Quick Tips:
- Always count the number of bond pairs and lone pairs around the central atom to determine the steric number and hence the hybridization.

90 Easy

90 of 220Question permalink

Understanding +4 | -1

The molecule that has linear structure

View answer and explanation

Correct option: A

C{O_2}

Option A (

CO_2

) is correct because carbon dioxide has a linear structure with carbon double-bonded to two oxygen atoms.

Solution approach

Correct Answer + Core Reason: $CO_2$ . Carbon dioxide has a linear structure.

Step-by-Step Solution:

Step 1: Draw the Lewis structures for each molecule.

Step 2: Determine the molecular geometry using VSEPR theory.

Step 3: For $CO_2$ : Carbon has two bonding regions and no lone pairs, resulting in a linear geometry.

Step 4: For $NO_2$ : Nitrogen has two bonding regions and one lone pair, resulting in a bent geometry.

Step 5: For $SO_2$ : Sulfur has two bonding regions and one lone pair, resulting in a bent geometry.

Step 6: For $SiO_2$ : Silicon forms a network solid with tetrahedral geometry around each Si atom, not a linear molecule.

Why other options are wrong:
- **Option B ( $NO_2$ ):** $NO_2$ has a bent structure due to the presence of a lone pair on the nitrogen atom.
- **Option C ( $SO_2$ ):** $SO_2$ has a bent structure due to the presence of a lone pair on the sulfur atom.
- **Option D ( $SiO_2$ ):** $SiO_2$ forms a network solid with a tetrahedral structure, not a linear molecule.

Key Concepts:
- Lewis structures
- VSEPR theory
- Molecular geometry

Memory Aids:
- Linear molecules have a bond angle of 180 degrees.

Quick Tips:
- Draw Lewis structures to determine the number of bonding and non-bonding electron pairs.

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Chemical bonding & molecular structure practice · Easy

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