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

1 Easy

Understanding +4 | -1

The example of the

p - p

-orbital overlapping is the formation of:

View answer and explanation

Correct option: B

\mathrm{C}{\mathrm{l}_2}

molecule

Option B (

\mathrm{C}{\mathrm{l}_2}

molecule) is correct because chlorine has the electronic configuration 1s² 2s² 2p⁶ 3s² 3p⁵, and the bond is formed by the overlap of p orbitals.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{l}_2}$ molecule. Chlorine forms a bond by the overlap of p orbitals.

Step-by-Step Solution:

Step 1: Write the electronic configuration of chlorine (Cl): 1s² 2s² 2p⁶ 3s² 3p⁵.

Step 2: Chlorine has 7 valence electrons in the 3p orbitals. It needs one more electron to complete its octet.

Step 3: The bond in Cl2 is formed by the overlap of the half-filled 3p orbitals of two chlorine atoms.

Step 4: This is a p-p overlap, forming a sigma bond.

Why other options are wrong:
- Option A: ${\mathrm{H}_2}$ molecule is incorrect because hydrogen forms a bond by s-s overlap.
- Option C: Hydrogen chloride is incorrect because it involves s-p overlap between hydrogen and chlorine.
- Option D: Hydrogen bromide molecule is incorrect because it involves s-p overlap between hydrogen and bromine.

Key Concepts:
- Overlapping of orbitals
- Electronic configuration
- Sigma bond

Memory Aids:
- Halogens form bonds by p-p overlap.

Quick Tips:
- Consider the electronic configuration to determine the type of orbital overlap.

2 Easy

2 of 220Question permalink

Understanding +4 | -1

Which of the following is correct?

View answer and explanation

Correct option: A

Decreases in bond length means increase in bond strength

Option A (Decreases in bond length means increase in bond strength) is correct because shorter bond lengths generally indicate stronger bonds due to increased electron density between the atoms.

Solution approach

Correct Answer + Core Reason: Decreases in bond length means increase in bond strength. Shorter bonds are generally stronger.

Step-by-Step Solution:

Step 1: Recall the relationship between bond length and bond strength. Shorter bond lengths generally correspond to stronger bonds.

Step 2: Consider the other options:
- Covalent radius decreases across a period (C is to the left of N, so it has a larger radius)
- Single bonds are weaker than double bonds.
- FeCl3 exists as a dimer (Fe2Cl6).

Why other options are wrong:
- Option B: Covalent radius of carbon is greater than that of nitrogen.
- Option C: Single bonds are weaker than double bonds.
- Option D: $\mathrm{Fe}$ ( $III$ ) chloride exists in the dimeric form $\mathrm{F}{\mathrm{e}_2}\mathrm{C}{\mathrm{l}_6}$ .

Key Concepts:
- Bond length
- Bond strength
- Covalent radius

Memory Aids:
- Remember that shorter bonds are stronger.

Quick Tips:
- Understand the trends in bond length and bond strength.

3 Easy

3 of 220Question permalink

Analyzing +4 | -1

Which has the lowest anion to cation size ratio?

View answer and explanation

Correct option: D

\mathrm{CsF}

Option D (CsF) is correct because Cs has the largest cation size and F has the smallest anion size among the given options, leading to the lowest ratio.

Solution approach

Correct Answer + Core Reason: CsF. The ratio is lowest when the cation is largest and the anion is smallest.

Step-by-Step Solution:

Step 1: Understand the question. We need to find the compound with the smallest anion to cation size ratio.

Step 2: Consider the trends in ionic size. Cations decrease in size across a period and increase down a group. Anions decrease in size across a period and increase down a group.

Step 3: Compare the cation sizes. Li < Na < Cs. Cs is the largest cation.

Step 4: Compare the anion sizes. F < I. F is the smallest anion.

Step 5: Determine the compound with the largest cation and smallest anion. This is CsF.

Step 6: Conclude that CsF has the lowest anion to cation size ratio.

Why other options are wrong:
- Option A: LiF has a small cation and a small anion, but the ratio is higher than CsF because Cs is much larger than Li.
- Option B: NaF has a small cation and a small anion, but the ratio is higher than CsF because Cs is much larger than Na.
- Option C: CsI has a large cation and a large anion, resulting in a higher anion to cation size ratio compared to CsF.

Key Concepts:
- Ionic size trends in the periodic table
- Cation and anion size
- Anion to cation size ratio

Memory Aids:
- Remember the trends: Down a group, size increases; across a period, size decreases.

Quick Tips:
- Always consider the positions of the elements in the periodic table to determine their relative sizes.

4 Easy

4 of 220Question permalink

Analyzing +4 | -1

Which of the following has been arranged in order of decreasing dipole moment

View answer and explanation

Correct option: A

C{H_3}Cl > C{H_3}F > C{H_3}Br > C{H_3}I

Option A (

C{H_3}Cl > C{H_3}F > C{H_3}Br > C{H_3}I

) is correct because the dipole moment depends on both electronegativity difference and bond length, and the product of these factors is largest for

CH_3Cl

.

Solution approach

Correct Answer + Core Reason: $C{H_3}Cl > C{H_3}F > C{H_3}Br > C{H_3}I$ . The dipole moment depends on both electronegativity difference and bond length.

Step-by-Step Solution:

Step 1: Consider the electronegativity differences between carbon and the halogens.
- Electronegativity: F > Cl > Br > I
- The electronegativity difference between C and F is the largest, followed by C and Cl, then C and Br, and finally C and I.

Step 2: Consider the bond lengths.
- Bond length increases as the size of the halogen increases: C-F < C-Cl < C-Br < C-I

Step 3: Dipole moment is the product of charge (related to electronegativity difference) and distance (bond length).
$\mu = q \times d$

Step 4: Analyze the trend.
- $C{H_3}F$ : Large electronegativity difference, but short bond length.
- $C{H_3}Cl$ : Moderate electronegativity difference and moderate bond length. The product is largest for $CH_3Cl$ .
- $C{H_3}Br$ : Smaller electronegativity difference and longer bond length.
- $C{H_3}I$ : Smallest electronegativity difference and longest bond length.

Step 5: The dipole moment is highest for $C{H_3}Cl$ because the product of electronegativity difference and bond length is the greatest. The dipole moment decreases as we go down the group from Cl to I, and F has a smaller dipole moment than Cl due to its shorter bond length.
$C{H_3}Cl > C{H_3}F > C{H_3}Br > C{H_3}I$

Why other options are wrong:
- Option B: $C{H_3}F > C{H_3}Cl > C{H_3}Br > C{H_3}I$ is incorrect because the dipole moment of $CH_3Cl$ is greater than $CH_3F$ .
- Option C: $C{H_3}Cl > C{H_3}Br > C{H_3}I > C{H_3}F$ is incorrect because the dipole moment of $CH_3F$ is greater than $CH_3Br$ and $CH_3I$ .
- Option D: $C{H_3}F > C{H_3}Cl > C{H_3}I > C{H_3}Br$ is incorrect because the dipole moment of $CH_3Cl$ is greater than $CH_3F$ .

Key Concepts:
- Dipole moment
- Electronegativity
- Bond length

Memory Aids:
- Remember that dipole moment is the product of charge and distance.

Quick Tips:
- Consider both electronegativity difference and bond length when comparing dipole moments.
- The trend is not always straightforward and may require careful consideration of both factors.

5 Easy

5 of 220Question permalink

Remembering +4 | -1

Valence bond theory of metallic bond was given by

View answer and explanation

Correct option: D

Pauling

Option D (Pauling) is correct because Linus Pauling is credited with developing the valence bond theory to explain metallic bonding.

Solution approach

Correct Answer + Core Reason: Pauling. Linus Pauling developed the valence bond theory of metallic bonding.

Step-by-Step Solution:

Step 1: Recall the scientists associated with different bonding theories.

Step 2: Valence bond theory of metallic bonding is attributed to Linus Pauling.

Why other options are wrong:
- Option A: Dalton is known for his atomic theory, not metallic bonding.
- Option B: Drudel proposed the Drude model, a classical model of electrical conduction in metals, but not the valence bond theory.
- Option C: Fajan is known for Fajan's rules, which describe the polarization of ions in ionic compounds, not metallic bonding.

Key Concepts:
- Metallic bonding
- Valence bond theory
- History of chemistry

Memory Aids:
- Pauling = Valence Bond Theory (for metallic bonds)

Quick Tips:
- Associate specific scientists with their key contributions to bonding theories.

6 Easy

6 of 220Question permalink

Analyzing +4 | -1

Species having same bond order are

View answer and explanation

Correct option: C

F_2^ + ,\,\,\,O_2^{ - 1}

Option C (

F_2^+ , O_2^{-1}

) is correct because both species have a bond order of 1.5 based on their molecular orbital configurations.

Solution approach

Correct Answer + Core Reason: $F_2^ + ,\\,O_2^{ - 1}$ . Both have a bond order of 1.5.

Step-by-Step Solution:

Step 1: Determine the molecular orbital configuration and bond order of $N_2$ .

Step 2: $N_2$ has 14 electrons. The molecular orbital configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)$ . Bond order = (10 - 4)/2 = 3.

Step 3: Determine the molecular orbital configuration and bond order of $N_2^-$ .

Step 4: $N_2^-$ has 15 electrons. The molecular orbital configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)$ . Bond order = (10 - 5)/2 = 2.5.

Step 5: Determine the molecular orbital configuration and bond order of $O_2$ .

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

Step 7: Determine the molecular orbital configuration and bond order of $O_2^{-2}$ .

Step 8: $O_2^{-2}$ has 18 electrons. The molecular orbital configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)$ . Bond order = (10 - 8)/2 = 1.

Step 9: Determine the molecular orbital configuration and bond order of $F_2^+$ .

Step 10: $F_2^+$ has 17 electrons. The molecular orbital configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)$ . Bond order = (10 - 7)/2 = 1.5.

Step 11: Determine the molecular orbital configuration and bond order of $O_2^{-1}$ .

Step 12: $O_2^{-1}$ has 17 electrons. The molecular orbital configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)$ . Bond order = (10 - 7)/2 = 1.5.

Step 13: Determine the molecular orbital configuration and bond order of $F_2^{+1}$ .

Step 14: $F_2^{+1}$ has 17 electrons. The molecular orbital configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)$ . Bond order = (10 - 7)/2 = 1.5.

Step 15: Determine the molecular orbital configuration and bond order of $O_2^{+1}$ .

Step 16: $O_2^{+1}$ has 15 electrons. The molecular orbital configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)$ . Bond order = (10 - 5)/2 = 2.5.

Why other options are wrong:
- Option A: $N_2$ has a bond order of 3, while $N_2^-$ has a bond order of 2.5.
- Option B: $O_2$ has a bond order of 2, while $O_2^{-2}$ has a bond order of 1.
- Option D: $F_2^{+1}$ has a bond order of 1.5, while $O_2^{+1}$ has a bond order of 2.5.

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

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

Quick Tips:
- Determine the molecular orbital configuration to calculate the bond order.

7 Easy

7 of 220Question permalink

Application +4 | -1

Molecule having non-zero value of dipole moment is

View answer and explanation

Correct option: B

{CHCl}_3

Option B (

{CHCl}_3

) is correct because chloroform has polar C-Cl bonds and an asymmetrical tetrahedral structure, resulting in a non-zero dipole moment, while the other molecules are symmetrical and non-polar.

Solution approach

Correct Answer + Core Reason: ${CHCl}_3$ . Asymmetrical structure and polar bonds lead to a non-zero dipole moment.

Step-by-Step Solution:

Step 1: Analyze the molecular structure of each compound.

Step 2: ${CCl}_4$ (Carbon tetrachloride) has a symmetrical tetrahedral shape. The C-Cl bonds are polar, but the bond dipoles cancel each other out due to the symmetry of the molecule, resulting in a net dipole moment of zero.

Step 3: ${CHCl}_3$ (Chloroform) has a tetrahedral shape with polar C-H and C-Cl bonds. The bond dipoles do not cancel out, resulting in a net dipole moment.

Step 4: ${BF}_3$ (Boron trifluoride) has a trigonal planar shape. The B-F bonds are polar, but the bond dipoles cancel each other out due to the symmetry of the molecule, resulting in a net dipole moment of zero.

Step 5: ${BCl}_3$ (Boron trichloride) has a trigonal planar shape. The B-Cl bonds are polar, but the bond dipoles cancel each other out due to the symmetry of the molecule, resulting in a net dipole moment of zero.

Why other options are wrong:
- **Option A ( ${CCl}_4$ ):** ${CCl}_4$ has a zero dipole moment due to its symmetrical tetrahedral shape.
- **Option C ( ${BF}_3$ ):** ${BF}_3$ has a zero dipole moment due to its symmetrical trigonal planar shape.
- **Option D ( ${BCl}_3$ ):** ${BCl}_3$ has a zero dipole moment due to its symmetrical trigonal planar shape.

Key Concepts:
- Dipole moment and molecular symmetry
- Electronegativity differences and bond polarity
- Molecular geometry and its effect on dipole moment

Memory Aids:
- Symmetrical molecules with identical bonds have zero dipole moment.

Quick Tips:
- Consider both bond polarity and molecular geometry when determining the net dipole moment.

8 Easy

8 of 220Question permalink

Analyzing +4 | -1

Which has the least bond angle

View answer and explanation

Correct option: C

{H}_2 {O}

Option C (

{H}_2 {O}

) is correct because its bent geometry and two lone pairs on the oxygen atom result in the smallest bond angle (approximately 104.5 degrees) compared to the other options.

Solution approach

Correct Answer + Core Reason: ${H}_2 {O}$ . The bent geometry and lone pairs result in the smallest bond angle.

Step-by-Step Solution:

Step 1: Determine the geometry of ${NH}_3$ : ${NH}_3$ has a trigonal pyramidal geometry with a bond angle of approximately 107 degrees.

Step 2: Determine the geometry of ${BeF}_2$ : ${BeF}_2$ has a linear geometry with a bond angle of 180 degrees.

Step 3: Determine the geometry of ${H}_2 {O}$ : ${H}_2 {O}$ has a bent geometry with a bond angle of approximately 104.5 degrees due to the two lone pairs on the oxygen atom.

Step 4: Determine the geometry of ${CH}_4$ : ${CH}_4$ has a tetrahedral geometry with a bond angle of 109.5 degrees.

Step 5: Conclusion: ${H}_2 {O}$ has the smallest bond angle (approximately 104.5 degrees).

Why other options are wrong:
- Option A: ${NH}_3$ has a bond angle of approximately 107 degrees, which is larger than that of ${H}_2 {O}$ .
- Option B: ${BeF}_2$ has a linear geometry with a bond angle of 180 degrees, which is much larger than that of ${H}_2 {O}$ .
- Option D: ${CH}_4$ has a tetrahedral geometry with a bond angle of 109.5 degrees, which is larger than that of ${H}_2 {O}$ .

Key Concepts:
- VSEPR Theory
- Molecular Geometry
- Bond Angle

Memory Aids:
- Lone pairs cause greater repulsion, decreasing bond angles.

Quick Tips:
- Use VSEPR theory to determine the geometry and bond angles of molecules, considering the effect of lone pairs.

9 Easy

9 of 220Question permalink

Remembering +4 | -1

Co-ordinate covalent bond is formed by

View answer and explanation

Correct option: C

donation of electrons

Option C (donation of electrons) is correct because a coordinate covalent bond is formed when one atom donates a pair of electrons to another atom.

Solution approach

Correct Answer + Core Reason: donation of electrons. A coordinate covalent bond is formed when one atom donates a pair of electrons to another atom.

Step-by-Step Solution:

Step 1: Define a coordinate covalent bond. A coordinate covalent bond (also called a dative bond) is a covalent bond in which both electrons shared in the bond come from the same atom.

Step 2: Consider the options:
- Transfer of electrons: This describes ionic bonding.
- Sharing of electrons: This describes regular covalent bonding.
- Donation of electrons: This describes coordinate covalent bonding.

Why other options are wrong:
- Option A (Transfer of electrons): This describes ionic bonding, not coordinate covalent bonding.
- Option B (Sharing of electrons): This describes regular covalent bonding, not coordinate covalent bonding.
- Option D (None of these): Coordinate covalent bonds are formed by donation of electrons.

Key Concepts:
- Coordinate Covalent Bond (Dative Bond)
- Covalent Bond
- Ionic Bond

Memory Aids:
- Dative = Donation.

Quick Tips:
- Remember that in coordinate covalent bonds, one atom provides both electrons for the bond.

10 Easy

10 of 220Question permalink

Remembering +4 | -1

The carbon atoms of graphite are

View answer and explanation

Correct option: A

sp^2

hybridised

Option A (

sp^2

hybridised) is correct because each carbon atom in graphite is bonded to three other carbon atoms, resulting in

sp^2

hybridization.

Solution approach

Correct Answer + Core Reason: $sp^2$ hybridised. Each carbon is bonded to three other carbons.

Step-by-Step Solution:

Step 1: Recall the structure of graphite.

Step 2: Each carbon atom in graphite is bonded to three other carbon atoms in a planar arrangement.

Step 3: This arrangement corresponds to $sp^2$ hybridization.

Step 4: Final Answer: $sp^2$ hybridised

Why other options are wrong:
- Option B: $sp^3$ hybridization corresponds to tetrahedral geometry (e.g., diamond).
- Option C: $sp$ hybridization corresponds to linear geometry.
- Option D: $dsp^2$ hybridization is not typical for carbon.

Key Concepts:
- Hybridization
- Graphite structure
- $sp^2$ hybridization

Memory Aids:
- Graphite: $sp^2$ (3 bonds)
- Diamond: $sp^3$ (4 bonds)

Quick Tips:
- Count the number of sigma bonds around the carbon atom.
- Relate the number of sigma bonds to the hybridization.

11 Easy

11 of 220Question permalink

Analyzing +4 | -1

Which among the following can have largest

C - O

bond length?

View answer and explanation

Correct option: D

{\left[ {Cr{{\left( {CO} \right)}_5}} \right]^{2 - }}

Option D (

{\left[ {Cr{{\left( {CO} \right)}_5}} \right]^{2 - }}

) is correct because the negative charge on the metal increases back-bonding to CO, weakening the C-O bond and thus increasing its length.

Solution approach

Correct Answer + Core Reason: ${\left[ {Cr{{\left( {CO} \right)}_5}} \right]^{2 - }}$ . The negative charge on the metal increases back-bonding to CO, weakening the C-O bond.

Step-by-Step Solution:

Step 1: Recall the concept of back-bonding in metal carbonyls.

Step 2: Back-bonding involves the donation of electron density from the filled d-orbitals of the metal to the empty $\pi^*$ antibonding orbitals of CO.

Step 3: Increased electron density in the $\pi^*$ antibonding orbitals of CO weakens the C-O bond and increases its length.

Step 4: The negative charge on the metal increases the extent of back-bonding.

Step 5: Compare the charges on the metal in each complex:
- $\left[ {Ni{{\left( {CO} \right)}_4}} \right]$ : Ni (0)
- $\left[ {Fe{{\left( {CO} \right)}_5}} \right]$ : Fe (0)
- ${\left[ {Co{{\left( {CO} \right)}_4}} \right]^ - }$ : Co (-1)
- ${\left[ {Cr{{\left( {CO} \right)}_5}} \right]^{2 - }}$ : Cr (-2)

Step 6: Since Cr has the most negative charge, it will have the greatest back-bonding and the longest C-O bond length.

Why other options are wrong:
- Option A: $\left[ {Ni{{\left( {CO} \right)}_4}} \right]$ : Less back-bonding compared to Cr(-2).
- Option B: $\left[ {Fe{{\left( {CO} \right)}_5}} \right]$ : Less back-bonding compared to Cr(-2).
- Option C: ${\left[ {Co{{\left( {CO} \right)}_4}} \right]^ - }$ : Less back-bonding compared to Cr(-2).

Key Concepts:
- Back-bonding in metal carbonyls
- Effect of charge on back-bonding
- Relationship between back-bonding and bond length

Memory Aids:
- More negative charge on the metal, more back-bonding, longer C-O bond.

Quick Tips:
- Consider the charge on the metal when comparing C-O bond lengths in metal carbonyls.

12 Easy

12 of 220Question permalink

Understanding +4 | -1

Which bond has the highest bond energy?

View answer and explanation

Correct option: C

Multiple bond

Option C (Multiple bond) is correct because multiple bonds (double or triple bonds) have higher bond energy than single bonds (sigma bonds) or coordinate bonds.

Solution approach

Correct Answer + Core Reason: Multiple bond. Multiple bonds have higher bond energy.

Step-by-Step Solution:

Step 1: Recall the definition of bond energy. It is the energy required to break a bond.

Step 2: Compare the bond energies of different types of bonds. Multiple bonds (double and triple bonds) are stronger than single bonds (sigma bonds) because they involve more shared electrons.

Step 3: Coordinate bonds are a type of covalent bond and are generally weaker than typical covalent bonds.

Why other options are wrong:
- Option A: Coordinate bonds are weaker than multiple bonds.
- Option B: Sigma bonds are single bonds and are weaker than multiple bonds.
- Option D: Polar covalent bonds can be strong, but multiple bonds are generally stronger.

Key Concepts:
- Bond energy
- Sigma bond
- Multiple bond
- Coordinate bond

Memory Aids:
- Remember that multiple bonds are stronger than single bonds.

Quick Tips:
- Consider the number of shared electrons when comparing bond strengths.

13 Easy

13 of 220Question permalink

Remembering +4 | -1

s p^3

hybridization leads to which shape of the molecule

View answer and explanation

Correct option: A

Tetrahedron

Option A (Tetrahedron) is correct because

sp^3

hybridization results in four hybrid orbitals arranged in a tetrahedral geometry.

Solution approach

Correct Answer + Core Reason: Tetrahedron. $sp^3$ hybridization results in a tetrahedral shape due to the four hybrid orbitals.

Step-by-Step Solution:

Step 1: Identify the hybridization type: $sp^3$ .

Step 2: Recall the geometry associated with $sp^3$ hybridization.

Step 3: $sp^3$ hybridization involves one s orbital and three p orbitals, resulting in four hybrid orbitals.

Step 4: These four hybrid orbitals arrange themselves in space to minimize repulsion, leading to a tetrahedral geometry.

Step 5: Conclusion: $sp^3$ hybridization leads to a tetrahedral shape.

Why other options are wrong:
- Option B (Octahedron): Octahedral geometry is associated with $sp^3d^2$ hybridization.
- Option C (Linear): Linear geometry is associated with $sp$ hybridization.
- Option D (Plane triangle): Trigonal planar geometry is associated with $sp^2$ hybridization.

Key Concepts:
- Hybridization
- Molecular Geometry
- VSEPR Theory

Memory Aids:
- $sp$ - Linear
- $sp^2$ - Trigonal Planar
- $sp^3$ - Tetrahedral
- $sp^3d$ - Trigonal Bipyramidal
- $sp^3d^2$ - Octahedral

Quick Tips:
- Memorize the relationship between hybridization and molecular geometry.

14 Easy

14 of 220Question permalink

Application +4 | -1

Which species is paramagnetic?

View answer and explanation

Correct option: A

\mathrm{O}_2^ -

Option A (

\mathrm{O}_2^ -

) is correct because it has an odd number of electrons in its molecular orbitals, resulting in unpaired electrons and thus paramagnetism.

Solution approach

Correct Answer + Core Reason: $\mathrm{O}_2^ -$ . Paramagnetic species have unpaired electrons.

Step-by-Step Solution:

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

Step 2: The electronic configurations are:
$\mathrm{O}_2^ -$ : $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^2(\pi_{2p_y}^*)^1$
$N_2$ : $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2$
$CO$ : $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2$
$C_2$ : $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\pi_{2p_x})^2(\pi_{2p_y})^2$

Step 3: Identify the species with unpaired electrons.
$\mathrm{O}_2^ -$ has one unpaired electron in the $\pi_{2p}^*$ orbitals.
$N_2$ , $CO$ , and $C_2$ have all paired electrons.

Step 4: Paramagnetic species have unpaired electrons.

Why other options are wrong:
- Option B: $N_2$ is diamagnetic.
- Option C: $CO$ is diamagnetic.
- Option D: $C_2$ is diamagnetic.

Key Concepts:
- Molecular orbital theory
- Electronic configuration
- Paramagnetism and diamagnetism

Memory Aids:
- Paramagnetic: unpaired electrons; Diamagnetic: all paired electrons.

Quick Tips:
- Determine the electronic configuration and check for unpaired electrons.

15 Easy

15 of 220Question permalink

Understanding +4 | -1

Which statement is not correct?

View answer and explanation

Correct option: B

Sigma bond is weaker than

π

-bond.

Option B (Sigma bond is weaker than

π

-bond) is correct because sigma bonds are formed by head-on overlap, making them stronger than pi bonds formed by sideways overlap.

Solution approach

Correct Answer + Core Reason: Sigma bond is weaker than $\pi$ -bond. Sigma bonds are formed by head-on overlap, making them stronger than pi bonds formed by sideways overlap.

Step-by-Step Solution:

Step 1: A double bond consists of one sigma bond and one pi bond.

Step 2: Sigma bonds are formed by head-on overlap of atomic orbitals, resulting in a strong bond.

Step 3: Pi bonds are formed by sideways overlap of atomic orbitals, resulting in a weaker bond.

Step 4: Therefore, a sigma bond is stronger than a pi bond.

Step 5: Double bonds are shorter and stronger than single bonds.

Step 6: Covalent bonds are stronger than hydrogen bonds.

Why other options are wrong:
- Option A: Double bonds are indeed shorter than single bonds due to the presence of an additional bond.
- Option C: Double bonds are stronger than single (sigma) bonds because they contain both a sigma and a pi bond.
- Option D: Covalent bonds are much stronger than hydrogen bonds, which are intermolecular forces.

Key Concepts:
- Sigma bond
- Pi bond
- Bond strength
- Bond length

Memory Aids:
- Sigma = Stronger, Pi = Weaker

Quick Tips:
- Remember the difference between sigma and pi bonds and their relative strengths.

16 Easy

16 of 220Question permalink

Application +4 | -1

The pair from the following pairs having both compounds with net non-zero dipole moment is

View answer and explanation

Correct option: C

\mathrm{CH}_2 \mathrm{Cl}_2, \mathrm{CHCl}_3

Option C (

\mathrm{CH}_2 \mathrm{Cl}_2, \mathrm{CHCl}_3

) is correct because both dichloromethane and chloroform have non-zero dipole moments due to their asymmetrical structures and polar C-Cl bonds.

Solution approach

Correct Answer + Core Reason: $\mathrm{CH}_2 \mathrm{Cl}_2, \mathrm{CHCl}_3$ . Both molecules have asymmetrical structures and polar bonds.

Step-by-Step Solution:

Step 1: Analyze the molecular structure of each compound in the pairs.

Step 2: 1,4-Dichlorobenzene has a symmetrical structure where the dipole moments of the C-Cl bonds cancel each other out, resulting in a zero net dipole moment. 1,3-Dichlorobenzene has a non-zero dipole moment.

Step 3: cis-butene has a non-zero dipole moment, while trans-butene has a zero dipole moment due to symmetry.

Step 4: $\mathrm{CH}_2 \mathrm{Cl}_2$ (Dichloromethane) has a tetrahedral shape with polar C-Cl bonds. The bond dipoles do not cancel out, resulting in a net dipole moment. $\mathrm{CHCl}_3$ (Chloroform) also has a tetrahedral shape with polar C-Cl bonds and a C-H bond. The bond dipoles do not cancel out, resulting in a net dipole moment.

Step 5: Benzene has a symmetrical structure where the dipole moments of the C-H bonds cancel each other out, resulting in a zero net dipole moment. Anisidine has a non-zero dipole moment.

Why other options are wrong:
- Option A: 1,4-Dichlorobenzene has a zero dipole moment.
- Option B: trans-butene has a zero dipole moment.
- Option D: Benzene has a zero dipole moment.

Key Concepts:
- Dipole moment and molecular symmetry
- Electronegativity differences and bond polarity
- Molecular geometry and its effect on dipole moment

Memory Aids:
- Symmetrical molecules with identical bonds have zero dipole moment.

Quick Tips:
- Consider both bond polarity and molecular geometry when determining the net dipole moment.

17 Easy

17 of 220Question permalink

Remembering +4 | -1

Debye an unit of dipole moment is of the order of:

View answer and explanation

Correct option: B

{10^{ - 18}}

esu cm

Option B is correct because the Debye unit is defined as 1 × 10^-18 esu cm.

Solution approach

Correct Answer + Core Reason: ${10^{ - 18}}$ esu cm. The Debye unit is defined as the dipole moment resulting from two equal and opposite charges of magnitude $10^{-10}$ esu separated by a distance of 1 Å ( $10^{-8}$ cm).

Step-by-Step Solution:

Step 1: Recall the definition of Debye unit. 1 Debye (D) is defined as $10^{-18}$ esu cm.

Step 2: Conclusion: The Debye unit is of the order of $10^{-18}$ esu cm.

Why other options are wrong:
- Option A: ${10^{ - 10}}$ esu cm: This is incorrect because the Debye unit is $10^{-18}$ esu cm.
- Option C: ${10^{ - 6}}$ esu cm: This is incorrect because the Debye unit is $10^{-18}$ esu cm.
- Option D: ${10^{ - 12}}$ esu cm: This is incorrect because the Debye unit is $10^{-18}$ esu cm.

Key Concepts:
- Dipole Moment
- Debye Unit

Memory Aids:
- Remember that 1 Debye = $10^{-18}$ esu cm.

Quick Tips:
- The Debye unit is a common unit for expressing dipole moments.

18 Easy

18 of 220Question permalink

Remembering +4 | -1

Atoms or group of atoms which are electrically charged are known as:

View answer and explanation

Correct option: C

Ions

Option C (Ions) is correct because ions are atoms or groups of atoms that carry an electrical charge.

Solution approach

Correct Answer + Core Reason: Ions. Ions are atoms or groups of atoms that carry an electrical charge.

Step-by-Step Solution:

Step 1: Define ions.

Step 2: Ions are atoms or groups of atoms that have gained or lost electrons, resulting in an electrical charge. Cations are positively charged ions, and anions are negatively charged ions.

Why other options are wrong:
- Option A: Anions are negatively charged ions only.
- Option B: Cations are positively charged ions only.
- Option D: Atoms are neutral particles.

Key Concepts:
- Ions
- Cations
- Anions

Memory Aids:
- Cations are positive (t looks like a plus sign).
- Anions are negative.

Quick Tips:
- Remember the definitions of cations and anions.

19 Easy

19 of 220Question permalink

Understanding +4 | -1

In

\mathrm{PC}{\mathrm{l}_5}

molecule, P is:

View answer and explanation

Correct option: D

s{p^3}d -

hybridized

Option D (

s{p^3}d -

hybridized) is correct because in

\mathrm{PC}{\mathrm{l}_5}

, the phosphorus atom is

sp^3d

hybridized to accommodate five bonding pairs.

Solution approach

Correct Answer + Core Reason: $s{p^3}d -$ hybridized. P in PCl5 has 5 bond pairs and no lone pairs, thus $sp^3d$ hybridization.

Step-by-Step Solution:

Step 1: Determine the number of valence electrons of the central atom (P).

Step 2: Phosphorus has 5 valence electrons.

Step 3: Determine the number of atoms bonded to the central atom (Cl).

Step 4: Phosphorus is bonded to 5 chlorine atoms.

Step 5: Calculate the steric number: Steric number = number of bonded atoms + number of lone pairs = 5 + 0 = 5.

Step 6: Determine the hybridization based on the steric number.

Step 7: A steric number of 5 corresponds to $sp^3d$ hybridization.

Why other options are wrong:
- **Option A ( $s{p^3} -$ hybridized):** $sp^3$ hybridization corresponds to a steric number of 4, not 5.
- **Option B ( $ds{p^2} -$ hybridized): This hybridization is not typically used to describe the bonding in PCl5.
- Option C ( $d{s^3}p -$ hybridized):** This is an incorrect representation of the $sp^3d$ hybridization.

Key Concepts:
- Hybridization
- Steric Number
- VSEPR Theory

Memory Aids:
- Steric Number 5 = $sp^3d$

Quick Tips:
- Use the steric number to determine the hybridization of the central atom.

20 Easy

20 of 220Question permalink

Application +4 | -1

The hybridization of P in

\mathrm{PO}_4^{3 - }

is same as in:

View answer and explanation

Correct option: D

\mathrm{S\;in\;SO}_4^{2 - }

Option D is correct because both

\mathrm{PO}_4^{3 - }

and

\mathrm{SO}_4^{2 - }

have tetrahedral geometry and

sp^3

hybridization.

Solution approach

Correct Answer + Core Reason: $\mathrm{S in SO}_4^{2 - }$ . Both have $sp^3$ hybridization and tetrahedral geometry.

Step-by-Step Solution:

Step 1: Determine the hybridization of P in $\mathrm{PO}_4^{3 - }$ : The central atom P has 4 sigma bonds and no lone pairs. Therefore, the hybridization is $sp^3$ .

Step 2: Determine the hybridization of I in $\mathrm{ICl}_4^ -$ : The central atom I has 4 sigma bonds and 2 lone pairs. Therefore, the hybridization is $sp^3d^2$ .

Step 3: Determine the hybridization of S in $\mathrm{S}{\mathrm{O}_3}$ : The central atom S has 3 sigma bonds and no lone pairs. Therefore, the hybridization is $sp^2$ .

Step 4: Determine the hybridization of N in $\mathrm{NO}_3^ -$ : The central atom N has 3 sigma bonds and no lone pairs. Therefore, the hybridization is $sp^2$ .

Step 5: Determine the hybridization of S in $\mathrm{SO}_4^{2 - }$ : The central atom S has 4 sigma bonds and no lone pairs. Therefore, the hybridization is $sp^3$ .

Step 6: Compare the hybridizations: The hybridization of P in $\mathrm{PO}_4^{3 - }$ ( $sp^3$ ) is the same as the hybridization of S in $\mathrm{SO}_4^{2 - }$ ( $sp^3$ ).

Why other options are wrong:
- Option A: $\mathrm{I in ICl}_4^ -$ has $sp^3d^2$ hybridization, which is different from $sp^3$ .
- Option B: $\mathrm{S in S}{\mathrm{O}_3}$ has $sp^2$ hybridization, which is different from $sp^3$ .
- Option C: $\mathrm{N in NO}_3^ -$ has $sp^2$ hybridization, which is different from $sp^3$ .

Key Concepts:
- Hybridization
- Sigma bonds
- Lone pairs
- VSEPR theory

Memory Aids:
- Steric number = Number of sigma bonds + Number of lone pairs
- Hybridization can be determined from the steric number.

Quick Tips:
- Count the number of sigma bonds and lone pairs around the central atom to determine the hybridization.
- Remember the relationship between steric number and hybridization.

21 Easy

21 of 220Question permalink

Application +4 | -1

The phosphate of a metal has the formula

{MHPO}_4

. The formula of its chloride would be

View answer and explanation

Correct option: B

{MCl}_2

Option B (

{MCl}_2

) is correct because if MHPO4 has a +2 and -2 charge, then M has a +2 charge. Chloride has a -1 charge, so the formula will be MCl2.

Solution approach

Correct Answer + Core Reason: ${MCl}_2$ . If MHPO4 has a +2 and -2 charge, then M has a +2 charge. Chloride has a -1 charge, so the formula will be MCl2.

Step-by-Step Solution:

Step 1: Determine the charge of the metal M. Since the phosphate is ${MHPO}_4$ , and the $HPO_4^{2-}$ ion has a -2 charge, the metal M must have a +2 charge.

Step 2: Determine the charge of the chloride ion. The chloride ion is $Cl^-$ , so it has a -1 charge.

Step 3: Balance the charges to form a neutral compound. To balance the +2 charge of M and the -1 charge of chloride, we need 1 M ion and 2 chloride ions. This gives a formula of ${MCl}_2$ .

Why other options are wrong:
- **Option A ( ${MCl}$ ): This formula would result if M had a +1 charge and chloride had a -1 charge, which is incorrect.
- Option C ( ${MCl}_3$ ): This formula would result if M had a +3 charge and chloride had a -1 charge, which is incorrect.
- Option D ( ${M}_2 {Cl}_3$ ): This formula would result if M had a +3 charge and chloride had a -2 charge, which is incorrect.

Key Concepts:
- Ionic compound formation
- Balancing charges

Memory Aids:**
- Remember the charge of the $HPO_4^{2-}$ ion is -2 and the chloride ion is -1.

Quick Tips:
- Determine the charges of the ions involved and then criss-cross the charges to determine the subscripts in the formula.

22 Easy

22 of 220Question permalink

Analyzing +4 | -1

s{p^3}{d^2}

hybridization is not displayed by

View answer and explanation

Correct option: D

P{F_5}

Option D (

P{F_5}

) is correct because

PF_5

exhibits

sp^3d

hybridization, not

sp^3d^2

hybridization.

Solution approach

Correct Answer + Core Reason: $P{F_5}$ . $PF_5$ exhibits $sp^3d$ hybridization, not $sp^3d^2$ .

Step-by-Step Solution:

Step 1: Determine the hybridization of each molecule/ion.

Step 2: For $Br{F_5}$ , the central atom Br has 7+5(7) = 42 valence electrons. The Lewis structure has five Br-F bonds and one lone pair on Br. Steric number = 6, so it is $sp^3d^2$ hybridized.

Step 3: For $S{F_6}$ , the central atom S has 6+6(7) = 48 valence electrons. The Lewis structure has six S-F bonds and no lone pair on S. Steric number = 6, so it is $sp^3d^2$ hybridized.

Step 4: For ${\left[ {CrFr} \right]^{ - 3}}$ , it is likely a typo and should be ${\left[ {CrF_6} \right]^{ - 3}}$ . The central atom Cr has 6+6(7)+3 = 51 valence electrons. The Lewis structure has six Cr-F bonds and no lone pair on Cr. Steric number = 6, so it is $sp^3d^2$ hybridized.

Step 5: For $P{F_5}$ , the central atom P has 5+5(7) = 40 valence electrons. The Lewis structure has five P-F bonds and no lone pair on P. Steric number = 5, so it is $sp^3d$ hybridized.

Step 6: Identify the molecule/ion that does not have $sp^3d^2$ hybridization.

Why other options are wrong:
- Option A: $Br{F_5}$ has $sp^3d^2$ hybridization.
- Option B: $S{F_6}$ has $sp^3d^2$ hybridization.
- Option C: ${\left[ {CrFr} \right]^{ - 3}}$ (assuming it is ${\left[ {CrF_6} \right]^{ - 3}}$ ) has $sp^3d^2$ hybridization.

Key Concepts:
- Hybridization
- Steric number

Memory Aids:
- Steric Number = Number of Bond Pairs + Number of Lone Pairs

Quick Tips:
- Calculate the steric number to determine the hybridization.

23 Easy

23 of 220Question permalink

Understanding +4 | -1

s{p^3}

hybridisation is not found in

View answer and explanation

Correct option: C

\mathrm{BC}{\mathrm{l}_3}

Option C (

\mathrm{BC}{\mathrm{l}_3}

) is correct because Boron in

BCl_3

is

sp^2

hybridized, not

sp^3

hybridized.

Solution approach

Correct Answer + Core Reason: $\mathrm{BC}{\mathrm{l}_3}$ . Boron in $BCl_3$ is $sp^2$ hybridized.

Step-by-Step Solution:

Step 1: Determine the hybridization of the central atom in each molecule.

Step 2: For ${\mathrm{H}_2}\mathrm{O}$ : Oxygen has 2 bonding pairs and 2 lone pairs, so it is $sp^3$ hybridized.

Step 3: For $\mathrm{C}{\mathrm{H}_4}$ : Carbon has 4 bonding pairs and 0 lone pairs, so it is $sp^3$ hybridized.

Step 4: For $\mathrm{BC}{\mathrm{l}_3}$ : Boron has 3 bonding pairs and 0 lone pairs, so it is $sp^2$ hybridized.

Step 5: For $\mathrm{N}{\mathrm{H}_3}$ : Nitrogen has 3 bonding pairs and 1 lone pair, so it is $sp^3$ hybridized.

Why other options are wrong:
- **Option A ( ${\mathrm{H}_2}\mathrm{O}$ ):** Oxygen is $sp^3$ hybridized.
- **Option B ( $\mathrm{C}{\mathrm{H}_4}$ ):** Carbon is $sp^3$ hybridized.
- **Option D ( $\mathrm{N}{\mathrm{H}_3}$ ):** Nitrogen is $sp^3$ hybridized.

Key Concepts:
- Hybridization
- VSEPR theory

Memory Aids:
- Count bonding pairs and lone pairs to determine hybridization.

Quick Tips:
- Practice determining hybridization for common molecules.

24 Easy

24 of 220Question permalink

Application +4 | -1

Which of the following species has tetrahedral geometry.

View answer and explanation

Correct option: A

BH_4^ -

Option A (

BH_4^ -

) is correct because boron has four bond pairs and no lone pairs, resulting in a tetrahedral geometry.

Solution approach

Correct Answer + Core Reason: $BH_4^ -$ . Boron has four bond pairs and no lone pairs, resulting in a tetrahedral geometry.

Step-by-Step Solution:

Step 1: Determine the geometry of each species.

Step 2: $BH_4^ -$ : Boron has 4 bond pairs and no lone pairs, so it is tetrahedral.

Step 3: $NH_2^ -$ : Nitrogen has 2 bond pairs and 2 lone pairs, so it is bent.

Step 4: $CO_3^{2 - }$ : Carbon has 3 bond pairs and no lone pairs, so it is trigonal planar.

Step 5: ${H_3}{O^ + }$ : Oxygen has 3 bond pairs and 1 lone pair, so it is trigonal pyramidal.

Why other options are wrong:
- Option B: $NH_2^ -$ is bent.
- Option C: $CO_3^{2 - }$ is trigonal planar.
- Option D: ${H_3}{O^ + }$ is trigonal pyramidal.

Key Concepts:
- VSEPR theory
- Tetrahedral geometry

Memory Aids:
- Tetrahedral: 4 bond pairs, 0 lone pairs

Quick Tips:
- Count bond pairs and lone pairs to predict the shape.

25 Easy

25 of 220Question permalink

Application +4 | -1

Which has

s{p^2}

-hybridization?

View answer and explanation

Correct option: B

S{O_2}

Option B (

S{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: $S{O_2}$ . 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: $C{O_2}$ : Carbon has 2 bond pairs and no lone pairs, so it is sp hybridized.

Step 3: $S{O_2}$ : Sulfur has 2 bond pairs and 1 lone pair, so it is $sp^2$ hybridized.

Step 4: ${N_2}O$ : The central nitrogen has 2 bond pairs and no lone pairs, so it is sp hybridized.

Step 5: CO: Carbon has 1 triple bond, so it is sp hybridized.

Why other options are wrong:
- Option A: $C{O_2}$ is sp hybridized.
- Option C: ${N_2}O$ is sp hybridized.
- Option D: 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.

26 Easy

26 of 220Question permalink

Remembering +4 | -1

When two atomic orbitals combine, they form:

View answer and explanation

Correct option: B

Two molecular orbitals

Option B (Two molecular orbitals) is correct because when two atomic orbitals combine, they form two molecular orbitals: one bonding and one antibonding.

Solution approach

Correct Answer + Core Reason: Two molecular orbitals. The combination of two atomic orbitals results in the formation of two molecular orbitals.

Step-by-Step Solution:

Step 1: Understand the basic principle of molecular orbital theory.

Step 2: When atomic orbitals combine, they form molecular orbitals. The number of molecular orbitals formed is equal to the number of atomic orbitals that combine.

Step 3: In this case, two atomic orbitals combine, so two molecular orbitals are formed.

Why other options are wrong:
- Option A: One molecular orbital is incorrect because the number of molecular orbitals formed equals the number of atomic orbitals combined.
- Option C: Two bonding molecular orbitals is incorrect because it doesn't account for the formation of antibonding orbitals.
- Option D: Two antibonding molecular orbitals is incorrect because it doesn't account for the formation of bonding orbitals.

Key Concepts:
- Molecular orbital theory
- Linear combination of atomic orbitals (LCAO)
- Bonding and antibonding orbitals

Memory Aids:
- Number of atomic orbitals = Number of molecular orbitals.

Quick Tips:
- Remember that the combination of atomic orbitals always results in the same number of molecular orbitals.

27 Easy

27 of 220Question permalink

Application +4 | -1

Which of the following pair of compounds will have three

{sp}^2

hybrid orbitals?

View answer and explanation

Correct option: B

{SO}_3, {C}_2 {H}_4

Option B (

{SO}_3, {C}_2 {H}_4

) is correct because both

SO_3

and

C_2H_4

have central atoms with three

sp^2

hybrid orbitals.

Solution approach

Correct Answer + Core Reason: ${SO}_3, {C}_2 {H}_4$ . Both molecules have central atoms with three $sp^2$ hybrid orbitals.

Step-by-Step Solution:

Step 1: Determine the hybridization of the central atom in each compound.

Step 2: For ${SO}_2$ : Sulfur has 1 lone pair and 2 bond pairs (with oxygen atoms). Steric number = 3, so hybridization is $sp^2$ .\nFor ${CH}_4$ : Carbon has 4 bond pairs (with hydrogen atoms). Steric number = 4, so hybridization is $sp^3$ .\nFor ${SO}_3$ : Sulfur has 3 bond pairs (with oxygen atoms). Steric number = 3, so hybridization is $sp^2$ .\nFor ${C}_2 {H}_4$ : Each carbon has 3 sigma bonds (2 with H and 1 with C). Steric number = 3, so hybridization is $sp^2$ .\nFor ${BF}_3$ : Boron has 3 bond pairs (with fluorine atoms). Steric number = 3, so hybridization is $sp^2$ .\nFor ${SF}_4$ : Sulfur has 4 bond pairs and 1 lone pair. Steric number = 5, so hybridization is $sp^3d$ .\nFor ${I}_3^{-}$ : The central iodine has 2 bond pairs and 3 lone pairs. Steric number = 5, so hybridization is $sp^3d$ .

Step 3: Identify the pair with both compounds having $sp^2$ hybridization.
${SO}_3$ has $sp^2$ hybridization and ${C}_2 {H}_4$ has $sp^2$ hybridization.

Why other options are wrong:
- **Option A ( ${SO}_2, {CH}_4$ ):** ${SO}_2$ is $sp^2$ hybridized, but ${CH}_4$ is $sp^3$ hybridized.
- **Option C ( ${BF}_3, {SF}_4$ ):** ${BF}_3$ is $sp^2$ hybridized, but ${SF}_4$ is $sp^3d$ hybridized.
- **Option D ( ${I}_3^{-}, {SF}_4$ ):** Both ${I}_3^{-}$ and ${SF}_4$ are $sp^3d$ hybridized.

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

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.

28 Easy

28 of 220Question permalink

Application +4 | -1

Two elements

X

and

Y

have following electronic configurations

X=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6, 4 s^2

and

Y=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^5

. The expected compound formed by combination of

X

and

Y

is

View answer and explanation

Correct option: A

X Y_2

Option A (

X Y_2

) is correct because X has 2 valence electrons and Y needs 1 electron to complete its octet, so X forms a +2 ion and Y forms a -1 ion, leading to the formula XY2.

Solution approach

Correct Answer + Core Reason: $X Y_2$ . X has 2 valence electrons and Y needs 1 electron to complete its octet, so X forms a +2 ion and Y forms a -1 ion, leading to the formula XY2.

Step-by-Step Solution:

Step 1: Determine the valence electrons of X. The electronic configuration $X=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6, 4 s^2$ indicates that X has 2 valence electrons in the 4s orbital.

Step 2: Determine the charge of the X ion. Since X has 2 valence electrons, it will lose these 2 electrons to form a +2 ion ( $X^{2+}$ ).

Step 3: Determine the valence electrons of Y. The electronic configuration $Y=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^5$ indicates that Y has 7 valence electrons in the 3s and 3p orbitals.

Step 4: Determine the charge of the Y ion. Since Y has 7 valence electrons, it will gain one electron to form a -1 ion ( $Y^-$ ).

Step 5: Balance the charges to form a neutral compound. To balance the +2 charge of X and the -1 charge of Y, we need 1 X ion and 2 Y ions. This gives a formula of $X Y_2$ .

Why other options are wrong:
- **Option B ( ${X}_5 {Y}_2$ ): This formula would result if X had a +2/5 charge and Y had a -1 charge, which is incorrect.
- Option C ( $X_2 Y_5$ ): This formula would result if X had a +5 charge and Y had a -2 charge, which is incorrect.
- Option D ( ${XY}_5$ ): This formula would result if X had a +5 charge and Y had a -1 charge, which is incorrect.

Key Concepts:
- Ionic compound formation
- Balancing charges
- Electronic configuration and valence electrons

Memory Aids:
- Remember that group 2 elements form +2 ions and group 17 elements form -1 ions.

Quick Tips:**
- Determine the number of valence electrons and the charge of the ions involved and then criss-cross the charges to determine the subscripts in the formula.

29 Easy

29 of 220Question permalink

Application +4 | -1

Which is expected to show paramagnetism?

View answer and explanation

Correct option: A

\mathrm{Cl}{\mathrm{O}_2}

Option A (

\mathrm{Cl}{\mathrm{O}_2}

) is correct because it has an odd number of valence electrons, making it paramagnetic.

Solution approach

Correct Answer + Core Reason: $\mathrm{Cl}{\mathrm{O}_2}$ . It has an odd number of valence electrons.

Step-by-Step Solution:

Step 1: Define paramagnetism: Paramagnetism is the property of a substance being attracted to a magnetic field due to the presence of unpaired electrons.

Step 2: Determine the number of valence electrons for each molecule:
- $\mathrm{Cl}{\mathrm{O}_2}$ : Cl (7) + 2*O (6) = 7 + 12 = 19 (odd number of electrons)
- $\mathrm{S}{\mathrm{O}_2}$ : S (6) + 2*O (6) = 6 + 12 = 18 (even number of electrons)
- $\mathrm{C}{\mathrm{O}_2}$ : C (4) + 2*O (6) = 4 + 12 = 16 (even number of electrons)
- $\mathrm{Si}{\mathrm{O}_2}$ : Si (4) + 2*O (6) = 4 + 12 = 16 (even number of electrons)

Step 3: Identify the molecule with an odd number of electrons: $\mathrm{Cl}{\mathrm{O}_2}$ .

Step 4: Conclude that $\mathrm{Cl}{\mathrm{O}_2}$ is expected to show paramagnetism.

Why other options are wrong:
- Option B: $\mathrm{S}{\mathrm{O}_2}$ : Has an even number of electrons and is diamagnetic.
- Option C: $\mathrm{C}{\mathrm{O}_2}$ : Has an even number of electrons and is diamagnetic.
- Option D: $\mathrm{Si}{\mathrm{O}_2}$ : Has an even number of electrons and is diamagnetic.

Key Concepts:
- Paramagnetism
- Unpaired electrons
- Valence electrons

Memory Aids:
- Odd number of electrons usually indicates paramagnetism.

Quick Tips:
- Calculate the total number of valence electrons to determine if a molecule is likely to be paramagnetic.

30 Easy

30 of 220Question permalink

Understanding +4 | -1

In which of the following dative bond is not present

View answer and explanation

Correct option: D

{N_2}{H_4}

Option D (

{N_2}{H_4}

) is correct because in

{N_2}{H_4}

(hydrazine), the nitrogen atoms are connected by a single covalent bond and each nitrogen is also bonded to two hydrogen atoms through covalent bonds; there is no donation of electron pairs from one atom to another.

Solution approach

Correct Answer + Core Reason: ${N_2}{H_4}$ . In ${N_2}{H_4}$ , all bonds are covalent; there is no dative bond.

Step-by-Step Solution:

Step 1: Analyze the structure of each molecule:
- ${H_3}O^+$ : Oxygen donates a lone pair to $H^+$ , forming a dative bond.
- $N{H_4}^+$ : Nitrogen donates a lone pair to $H^+$ , forming a dative bond.
- $A{l_2}C{l_6}$ : Aluminum forms dative bonds with chlorine atoms from another $AlCl_3$ unit.
- ${N_2}{H_4}$ : Each nitrogen atom is bonded to two hydrogen atoms and the other nitrogen atom through covalent bonds. No dative bonds are present.

Why other options are wrong:
- **Option A ( ${H_3}O +$ ):** Oxygen donates a lone pair to $H^+$ , forming a dative bond.
- **Option B ( $N{H_4}^ +$ ):** Nitrogen donates a lone pair to $H^+$ , forming a dative bond.
- **Option C ( $A{l_2}C{l_6}$ ):** Aluminum forms dative bonds with chlorine atoms from another $AlCl_3$ unit.

Key Concepts:
- Dative Bond (Coordinate Covalent Bond)
- Covalent Bond

Memory Aids:
- Dative bonds involve the donation of a lone pair from one atom to another.

Quick Tips:
- Look for atoms donating lone pairs to form bonds.

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