The bond strength in ${\rm{O}}_2^ + ,{{\rm{O}}_2},{\rm{O}}_2^ - $ and ${\rm{O}}_2^{2 - }$ follows the order:

${\rm{O}}_2^ + > {{\rm{O}}_2} > {\rm{O}}_2^ - > {\rm{O}}_2^{2 - }$

${\rm{O}}_2^{2 - } > {\rm{O}}_2^ - > {{\rm{O}}_2} > {\rm{O}}_2^ + $

${{\rm{O}}_2} > {\rm{O}}_2^ - > {\rm{O}}_2^{2 - } > {\rm{O}}_2^ + $

${\rm{O}}_2^ - > {\rm{O}}_2^{2 - } > {\rm{O}}_2^ + > {{\rm{O}}_2}$

CHEMICAL BONDING & MOLECULAR STRUCTURE MCQs for NEET UG CHEMISTRY — Easy, Page 2

31 Easy

31 of 220Question permalink

Understanding +4 | -1

In the formation of a molecule by an atom

View answer and explanation

Correct option: C

both attractive and repulsive forces operative

Option C (both attractive and repulsive forces operative) is correct because bond formation involves both attractive forces between the nuclei and electrons, and repulsive forces between the nuclei and between the electrons.

Solution approach

Correct Answer + Core Reason: both attractive and repulsive forces operative. Bond formation involves both attractive forces between the nuclei and electrons, and repulsive forces between the nuclei and between the electrons.

Step-by-Step Solution:

Step 1: When atoms approach each other, attractive forces arise between the positively charged nuclei and the negatively charged electrons.

Step 2: Simultaneously, repulsive forces arise between the positively charged nuclei and between the negatively charged electrons.

Step 3: The balance between these attractive and repulsive forces determines whether a bond will form.

Why other options are wrong:
- Option A: Attractive forces alone would cause the atoms to collapse into each other.
- Option B: Repulsive forces alone would prevent the atoms from approaching each other.
- Option D: Both attractive and repulsive forces are always present during the formation of a molecule.

Key Concepts:
- Attractive forces
- Repulsive forces
- Chemical bond formation

Memory Aids:
- Bond formation = Attractive + Repulsive forces

Quick Tips:
- Remember that both attractive and repulsive forces are always present when atoms interact.

32 Easy

32 of 220Question permalink

Analyzing +4 | -1

Among the following molecules which one have smallest bond angle

View answer and explanation

Correct option: D

{H}_2 {Sc}

Option D (

{H}_2 {Sc}

) is correct because Sc has a lower electronegativity than N, P, and O, leading to less repulsion between bond pairs and a smaller bond angle.

Solution approach

Correct Answer + Core Reason: ${H}_2 {Sc}$ . The bond angle decreases as the electronegativity of the central atom decreases and the size of the surrounding atoms increases.

Step-by-Step Solution:

Step 1: Analyze the given molecules: ${NH}_3$ , ${PH}_3$ , ${H}_2 {O}$ , and ${H}_2 {Sc}$ .

Step 2: Consider the electronegativity of the central atoms: N, P, O, and Sc.\nElectronegativity order: O > N > P > Sc.

Step 3: Consider the effect of lone pairs on the central atom. ${NH}_3$ and ${H}_2 {O}$ have lone pairs, which compress the bond angles.

Step 4: Compare ${NH}_3$ and ${PH}_3$ . Since N is more electronegative than P, ${NH}_3$ will have a larger bond angle than ${PH}_3$ due to greater repulsion between bond pairs.

Step 5: Compare ${NH}_3$ and ${H}_2 {O}$ . Oxygen is more electronegative than Nitrogen, but the presence of two lone pairs in ${H}_2 {O}$ causes greater compression of the bond angle compared to the single lone pair in ${NH}_3$ . However, the difference in electronegativity is more significant here.

Step 6: ${H}_2 {Sc}$ will have the smallest bond angle because Sc is the least electronegative, resulting in minimal repulsion between bond pairs. Also, Sc is a larger atom, which further reduces the bond angle.

Why other options are wrong:
- **Option A ( ${NH}_3$ ):** ${NH}_3$ has a larger bond angle than ${PH}_3$ and ${H}_2 {Sc}$ due to the higher electronegativity of nitrogen.
- **Option B ( ${PH}_3$ ):** ${PH}_3$ has a larger bond angle than ${H}_2 {Sc}$ because P is more electronegative than Sc.
- **Option C ( ${H}_2 {O}$ ):** ${H}_2 {O}$ has a larger bond angle than ${H}_2 {Sc}$ due to the higher electronegativity of oxygen.

Key Concepts:
- Bond angle
- Electronegativity
- Lone pair repulsion

Memory Aids:
- As electronegativity of the central atom decreases, bond angle decreases.
- Lone pairs compress bond angles.

Quick Tips:
- Consider both electronegativity and lone pair effects when comparing bond angles.
- Lower electronegativity of the central atom leads to smaller bond angles.

33 Easy

33 of 220Question permalink

Analyzing +4 | -1

The number of species below that have two lone pairs of electrons in their central atom is.

S{F_4},\,\,\,BF_4^ - ,\,\,\,Cl{F_3},\,\,\,As{F_3},\,\,\,PC{l_5},\,\,\,Br{F_5},\,\,\,Xe{F_4},\,\,\,S{F_6}

View answer and explanation

Correct option: D

2

Option D (2) is correct because only

XeF_4

and

ClF_3

have two lone pairs on their central atoms.

Solution approach

Correct Answer + Core Reason: 2. Only $XeF_4$ and $ClF_3$ have two lone pairs of electrons on their central atoms.

Step-by-Step Solution:

Step 1: Determine the number of lone pairs on the central atom for each species.
$S{F_4}$ : 1 lone pair
$BF_4^ -$ : 0 lone pairs
$Cl{F_3}$ : 2 lone pairs
$As{F_3}$ : 1 lone pair
$PC{l_5}$ : 0 lone pairs
$Br{F_5}$ : 1 lone pair
$Xe{F_4}$ : 2 lone pairs
$S{F_6}$ : 0 lone pairs

Step 2: Count the number of species with two lone pairs.
$Cl{F_3}$ and $Xe{F_4}$ have two lone pairs.

Why other options are wrong:
- Option A: 3: Incorrect count.
- Option B: 4: Incorrect count.
- Option C: 5: Incorrect count.

Key Concepts:
- Lone pairs
- VSEPR theory

Memory Aids:
- Count the number of valence electrons and subtract the number of bonding electrons to find the number of lone pairs.

Quick Tips:
- Draw Lewis structures to determine the number of lone pairs.

34 Easy

34 of 220Question permalink

Understanding +4 | -1

The bond strength in

\mathrm{O}_2^ + ,{\mathrm{O}_2},\mathrm{O}_2^ -

and

\mathrm{O}_2^{2 - }

follows the order:

View answer and explanation

Correct option: B

\mathrm{O}_2^ + > {\mathrm{O}_2} > \mathrm{O}_2^ - > \mathrm{O}_2^{2 - }

Option B (

\mathrm{O}_2^ + > {\mathrm{O}_2} > \mathrm{O}_2^ - > \mathrm{O}_2^{2 - }

) is correct because bond strength increases with bond order, and the bond orders are O2+ (2.5), O2 (2), O2- (1.5), and O22- (1).

Solution approach

Correct Answer + Core Reason: $\mathrm{O}_2^ + > {\mathrm{O}_2} > \mathrm{O}_2^ - > \mathrm{O}_2^{2 - }$ . Bond strength increases with bond order.

Step-by-Step Solution:

Step 1: Calculate the bond order for each species using molecular orbital theory.

Step 2: The electronic configurations are:
- $\mathrm{O}_2^+$ : $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)^1$ Bond order = (10-5)/2 = 2.5
- $\mathrm{O}_2$ : $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)^2$ Bond order = (10-6)/2 = 2
- $\mathrm{O}_2^-$ : $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)^3$ Bond order = (10-7)/2 = 1.5
- $\mathrm{O}_2^{2-}$ : $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)^4$ Bond order = (10-8)/2 = 1

Step 3: Arrange the species in decreasing order of bond order: $\mathrm{O}_2^+ > \mathrm{O}_2 > \mathrm{O}_2^- > \mathrm{O}_2^{2-}$

Step 4: Since bond strength is directly proportional to bond order, the order of bond strength is the same as the order of bond order.

Why other options are wrong:
- Option A: Incorrect order of bond strength.
- Option C: Incorrect order of bond strength.
- Option D: Incorrect order of bond strength.

Key Concepts:
- Molecular orbital theory
- Bond order
- Bond strength

Memory Aids:
- Remember that bond strength is directly proportional to bond order.

Quick Tips:
- Calculate the bond order to compare the stability and bond strength of different species.

35 Easy

35 of 220Question permalink

Understanding +4 | -1

Which is expected to have linear structure?

View answer and explanation

Correct option: B

\mathrm{C}{\mathrm{O}_2}

Option B (

\mathrm{C}{\mathrm{O}_2}

) is correct because carbon dioxide has a linear structure due to the double bonds between carbon and each oxygen atom, resulting in a bond angle of 180 degrees.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{O}_2}$ . Carbon dioxide has a linear structure.

Step-by-Step Solution:

Step 1: Draw the Lewis structures for each molecule.

Step 2: $\mathrm{S}{\mathrm{O}_2}$ has a bent structure due to the lone pair on the sulfur atom.

Step 3: $\mathrm{C}{\mathrm{O}_2}$ has a linear structure with double bonds between carbon and each oxygen atom. There are no lone pairs on the central carbon atom.

Step 4: $\mathrm{CO}_3^{2 - }$ has a trigonal planar structure.

Step 5: $\mathrm{SO}_4^{2 - }$ has a tetrahedral structure.

Why other options are wrong:
- Option A: $\mathrm{S}{\mathrm{O}_2}$ is bent.
- Option C: $\mathrm{CO}_3^{2 - }$ is trigonal planar.
- Option D: $\mathrm{SO}_4^{2 - }$ is tetrahedral.

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

Memory Aids:
- Carbon dioxide is linear.

Quick Tips:
- Draw Lewis structures to determine geometry.

36 Easy

36 of 220Question permalink

Analyzing +4 | -1

Which of the following pairs are isostructural?

View answer and explanation

Correct option: C

\mathrm{BrO}_3^ - ,\mathrm{Xe}{\mathrm{O}_3}

Option C (

\mathrm{BrO}_3^ - ,\mathrm{Xe}{\mathrm{O}_3}

) is correct because both have a pyramidal shape with the same number of lone pairs and bond pairs around the central atom.

Solution approach

Correct Answer + Core Reason: $\mathrm{BrO}_3^ - ,\mathrm{Xe}{\mathrm{O}_3}$ . Both have a pyramidal shape due to similar number of bond pairs and lone pairs.

Step-by-Step Solution:

Step 1: Determine the structure of each species.

Step 2: For $\mathrm{BrO}_3^ -$ , the central atom Br has 5+3(6)+1 = 24 valence electrons. The Lewis structure has three Br-O bonds and one lone pair on Br. Steric number = 4, so it is $sp^3$ hybridized and has a pyramidal shape.

Step 3: For $\mathrm{Xe}{\mathrm{O}_3}$ , the central atom Xe has 8+3(6) = 26 valence electrons. The Lewis structure has three Xe-O bonds and one lone pair on Xe. Steric number = 4, so it is $sp^3$ hybridized and has a pyramidal shape.

Step 4: For $\mathrm{SO}_3^{2 - }$ , the central atom S has 6+3(6)+2 = 26 valence electrons. The Lewis structure has three S-O bonds and one lone pair on S. Steric number = 4, so it is $sp^3$ hybridized and has a pyramidal shape.\nFor $\mathrm{NO}_3^ -$ , the central atom N has 5+3(6)+1 = 24 valence electrons. The Lewis structure has three N-O bonds and no lone pair on N. Steric number = 3, so it is $sp^2$ hybridized and has a trigonal planar shape.

Step 5: For $\mathrm{B}{\mathrm{F}_3}$ , the central atom B has 3+3(7) = 24 valence electrons. The Lewis structure has three B-F bonds and no lone pair on B. Steric number = 3, so it is $sp^2$ hybridized and has a trigonal planar shape.\nFor $\mathrm{N}{\mathrm{F}_3}$ , the central atom N has 5+3(7) = 26 valence electrons. The Lewis structure has three N-F bonds and one lone pair on N. Steric number = 4, so it is $sp^3$ hybridized and has a pyramidal shape.

Step 6: For $\mathrm{S}{\mathrm{F}_4}$ , the central atom S has 6+4(7) = 34 valence electrons. The Lewis structure has four S-F bonds and one lone pair on S. See-saw shape.\nFor $\mathrm{Xe}{\mathrm{F}_4}$ , the central atom Xe has 8+4(7) = 36 valence electrons. The Lewis structure has four Xe-F bonds and two lone pairs on Xe. Square planar shape.

Step 7: Identify the pair with the same structure.

Why other options are wrong:
- Option A: $\mathrm{SO}_3^{2 - }$ is pyramidal while $\mathrm{NO}_3^ -$ is trigonal planar.
- Option B: $\mathrm{B}{\mathrm{F}_3}$ is trigonal planar while $\mathrm{N}{\mathrm{F}_3}$ is pyramidal.
- Option D: $\mathrm{S}{\mathrm{F}_4}$ is see-saw while $\mathrm{Xe}{\mathrm{F}_4}$ is square planar.

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

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

Quick Tips:
- Draw Lewis structures to determine the number of bond pairs and lone pairs.
- Use VSEPR theory to predict the molecular geometry.

37 Easy

37 of 220Question permalink

Understanding +4 | -1

Which one has more tendency to form covalent compounds?

View answer and explanation

Correct option: B

\mathrm{Be}

Option B is correct because Be has a small size and high polarizing power, leading to a greater tendency to form covalent compounds according to Fajans' rules.

Solution approach

Correct Answer + Core Reason: $\mathrm{Be}$ . Beryllium has the smallest size and highest polarizing power among the given options, thus having more tendency to form covalent compounds.

Step-by-Step Solution:

Step 1: Recall Fajans' rules, which relate ionic and covalent character to the size and charge of ions.

Step 2: According to Fajans' rules, smaller cations with higher charges favor covalent character.

Step 3: Compare the sizes and charges of the given alkaline earth metals: $\mathrm{Ba}$ , $\mathrm{Be}$ , $\mathrm{Mg}$ , and $\mathrm{Ca}$ .

Step 4: Beryllium (Be) is the smallest among these, so it has the highest polarizing power and the greatest tendency to form covalent compounds.

Why other options are wrong:
- Option A: $\mathrm{Ba}$ is a larger cation and forms more ionic compounds.
- Option C: $\mathrm{Mg}$ is larger than Be and forms more ionic compounds than Be.
- Option D: $\mathrm{Ca}$ is larger than Be and forms more ionic compounds than Be.

Key Concepts:
- Fajans' rules
- Polarizing power
- Ionic character
- Covalent character

Memory Aids:
- Small, high charge = covalent; Large, low charge = ionic

Quick Tips:
- Remember that smaller cations with higher charges tend to form covalent compounds.

38 Easy

38 of 220Question permalink

Analyzing +4 | -1

In which of the following processes, the bond order increases and paramagnetic character changes to diamagnetic one?

View answer and explanation

Correct option: C

\mathrm{NO} \rightarrow \mathrm{NO}^{+}

Option C (

\mathrm{NO} \rightarrow \mathrm{NO}^{+}

) is correct because NO is paramagnetic with a bond order of 2.5, while

NO^+

is diamagnetic with a bond order of 3.

Solution approach

Correct Answer + Core Reason: $\mathrm{NO} \rightarrow \mathrm{NO}^{+}$ . NO is paramagnetic with a bond order of 2.5, while $NO^+$ is diamagnetic with a bond order of 3.

Step-by-Step Solution:

Step 1: Determine the electronic configuration and magnetic properties of $O_2$ .

Step 2: $O_2$ has 16 electrons. The molecular orbital configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)$ . It is paramagnetic due to the presence of unpaired electrons in the $\pi_{2p}^*$ orbitals. Bond order = (10 - 6)/2 = 2.

Step 3: Determine the electronic configuration and magnetic properties of $O_2^+$ .

Step 4: $O_2^+$ has 15 electrons. The molecular orbital configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)$ . It is paramagnetic. Bond order = (10 - 5)/2 = 2.5. Bond order increases, but paramagnetic character remains.

Step 5: Determine the electronic configuration and magnetic properties of $O_2^{2-}$

Step 6: $O_2^{2-}$ has 18 electrons. The molecular orbital configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)$ . It is diamagnetic. Bond order = (10 - 8)/2 = 1. Bond order decreases, and paramagnetic changes to diamagnetic.

Step 7: Determine the electronic configuration and magnetic properties of $NO$ .

Step 8: $NO$ has 15 electrons. The molecular orbital configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})(\pi_{2p}^*)$ . It is paramagnetic due to the presence of one unpaired electron in the $\pi_{2p}^*$ orbital. Bond order = (10 - 5)/2 = 2.5.

Step 9: Determine the electronic configuration and magnetic properties of $NO^+$ .

Step 10: $NO^+$ has 14 electrons. The molecular orbital configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})$ . It is diamagnetic. Bond order = (10 - 4)/2 = 3. Bond order increases, and paramagnetic changes to diamagnetic.

Step 11: Determine the electronic configuration and magnetic properties of $N_2$ .

Step 12: $N_2$ has 14 electrons. The molecular orbital configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})$ . It is diamagnetic. Bond order = (10 - 4)/2 = 3.

Step 13: Determine the electronic configuration and magnetic properties of $N_2^+$ .

Step 14: $N_2^+$ has 13 electrons. The molecular orbital configuration is $(\sigma_{2s})(\sigma_{2s}^*)(\sigma_{2p})(\pi_{2p})$ . It is paramagnetic. Bond order = (9 - 4)/2 = 2.5. Bond order decreases, and diamagnetic changes to paramagnetic.

Why other options are wrong:
- Option A: Bond order increases, but paramagnetic character remains.
- Option B: Bond order decreases, and paramagnetic changes to diamagnetic.
- Option D: Bond order decreases, and diamagnetic changes to paramagnetic.

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

Memory Aids:
- Paramagnetic substances have unpaired electrons.
- Diamagnetic substances have all paired electrons.

Quick Tips:
- Determine the molecular orbital configuration to calculate the bond order and predict magnetic properties.

39 Easy

39 of 220Question permalink

Understanding +4 | -1

Among 

\mathrm{NaF,\;}\,\,\mathrm{NaCl,\;}\,\,\mathrm{NaBr\;}\,\,\mathrm{and\;}\,\,\mathrm{NaI,}

the NaF has highest melting point because:

View answer and explanation

Correct option: A

It has maximum ionic character

Option A (It has maximum ionic character) is correct because, according to Fajan's rule, smaller cation and anion size leads to higher lattice energy and thus a higher melting point, and NaF has the smallest ions among the given options.

Solution approach

Correct Answer + Core Reason: It has maximum ionic character. Smaller ions lead to higher lattice energy and thus a higher melting point.

Step-by-Step Solution:

Step 1: Recall the factors affecting melting point in ionic compounds. Melting point is directly proportional to lattice energy, which is influenced by charge and size of the ions.

Step 2: Analyze the given compounds. All compounds have $Na^+$ as the cation. The anions are $F^-$ , $Cl^-$ , $Br^-$ , and $I^-$ .

Step 3: Apply Fajan's rules. Smaller ions and higher charges lead to higher lattice energy and thus higher melting point. Since $F^-$ is the smallest anion, $NaF$ has the highest lattice energy and highest melting point.

Why other options are wrong:
- Option B (It has minimum ionic character): This is incorrect. Higher ionic character leads to higher melting points.
- Option C (It has associated molecules): Ionic compounds do not typically form associated molecules in the same way as covalent compounds.
- Option D (It has least molecular weight): While molecular weight can play a role, the primary factor determining melting point in ionic compounds is lattice energy, which is more strongly influenced by ionic size and charge.

Key Concepts:
- Lattice Energy
- Melting Point
- Fajan's Rules
- Ionic Character

Memory Aids:
- Smaller ions = higher lattice energy = higher melting point.

Quick Tips:
- Consider the size and charge of the ions when determining the melting point of ionic compounds.

40 Easy

40 of 220Question permalink

Analyzing +4 | -1

Which of the following compounds would have the highest boiling point?

View answer and explanation

Correct option: C

\mathrm{C}{\mathrm{H}_3}\mathrm{OH}

Option C (

\mathrm{C}{\mathrm{H}_3}\mathrm{OH}

) is correct because alcohols exhibit hydrogen bonding, which significantly increases the boiling point compared to alkanes, amines, and haloalkanes.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{H}_3}\mathrm{OH}$ . Alcohols exhibit hydrogen bonding, leading to a higher boiling point.

Step-by-Step Solution:

Step 1: Identify the intermolecular forces present in each compound.

Step 2: $\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_3}$ (butane) has only London dispersion forces.

Step 3: $\mathrm{C}{\mathrm{H}_3}\mathrm{N}{\mathrm{H}_2}$ (methylamine) has London dispersion forces and hydrogen bonding (weaker than in alcohols).

Step 4: $\mathrm{C}{\mathrm{H}_3}\mathrm{OH}$ (methanol) has London dispersion forces and hydrogen bonding.

Step 5: $\mathrm{C}{\mathrm{H}_2}{\mathrm{F}_2}$ (difluoromethane) has London dispersion forces and dipole-dipole interactions.

Step 6: Compare the strength of intermolecular forces.

Step 7: Hydrogen bonding is stronger than dipole-dipole interactions and London dispersion forces. Alcohols have stronger hydrogen bonding than amines due to the higher electronegativity of oxygen compared to nitrogen.

Step 8: Therefore, methanol has the highest boiling point.

Why other options are wrong:
- Option A: Butane has only London dispersion forces, which are the weakest intermolecular forces.
- Option B: Methylamine has hydrogen bonding, but it is weaker than in methanol.
- Option D: Difluoromethane has dipole-dipole interactions, which are weaker than hydrogen bonding.

Key Concepts:
- Intermolecular forces
- Hydrogen bonding
- Dipole-dipole interactions
- London dispersion forces
- Boiling point

Memory Aids:
- Hydrogen bonding occurs when H is bonded to F, O, or N.

Quick Tips:
- Identify the strongest intermolecular force present in each compound to predict the boiling point.

41 Easy

41 of 220Question permalink

Understanding +4 | -1

For a stable molecule, the value of bond order must be

View answer and explanation

Correct option: C

Positive

Option C (Positive) is correct because a positive bond order indicates that there are more bonding electrons than antibonding electrons, resulting in a stable molecule.

Solution approach

Correct Answer + Core Reason: Positive. A positive bond order indicates stability.

Step-by-Step Solution:

Step 1: Recall the definition of bond order. Bond order = (number of bonding electrons - number of antibonding electrons) / 2

Step 2: A stable molecule must have more bonding electrons than antibonding electrons, resulting in a positive bond order.

Why other options are wrong:
- Option A: There is a direct relationship between stability and bond order.
- Option B: A bond order of zero indicates that the molecule is unstable.
- Option D: A negative bond order is not possible.

Key Concepts:
- Bond order
- Stability of molecules

Memory Aids:
- Remember that a positive bond order indicates a stable molecule.

Quick Tips:
- Calculate the bond order to determine the stability of a molecule.

42 Easy

42 of 220Question permalink

Understanding +4 | -1

Which shows the least dipole moment?

View answer and explanation

Correct option: D

\mathrm{CC}{\mathrm{l}_4}

Option D is correct because CCl4 is a symmetrical tetrahedral molecule where the bond dipoles cancel out, resulting in a zero dipole moment.

Solution approach

Correct Answer + Core Reason: $\mathrm{CC}{\mathrm{l}_4}$ . Carbon tetrachloride has a tetrahedral shape with symmetrical arrangement of chlorine atoms, resulting in zero dipole moment.

Step-by-Step Solution:

Step 1: Analyze the molecular structure of each option.

Step 2: $\mathrm{CHC}{\mathrm{l}_3}$ : Chloroform is tetrahedral but asymmetrical, so it has a dipole moment.

Step 3: $\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{OH}$ : Ethanol has a dipole moment due to the OH group.

Step 4: $\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}$ : Acetone has a dipole moment due to the carbonyl group.

Step 5: $\mathrm{CC}{\mathrm{l}_4}$ : Carbon tetrachloride is tetrahedral and symmetrical, so the bond dipoles cancel out, resulting in zero dipole moment.

Step 6: Conclusion: $\mathrm{CC}{\mathrm{l}_4}$ shows the least dipole moment (zero).

Why other options are wrong:
- Option A: $\mathrm{CHC}{\mathrm{l}_3}$ : This is incorrect because chloroform has a dipole moment.
- Option B: $\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{OH}$ : This is incorrect because ethanol has a dipole moment.
- Option C: $\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}$ : This is incorrect because acetone has a dipole moment.

Key Concepts:
- Dipole Moment
- Molecular Geometry
- Symmetry

Memory Aids:
- Remember that symmetrical molecules with identical bonds are nonpolar.

Quick Tips:
- Draw the Lewis structure and determine the molecular geometry to predict the polarity of a molecule.

43 Easy

43 of 220Question permalink

Application +4 | -1

The number of anti-bonding electron pairs in

{O_2}^{2 - }

ion is

View answer and explanation

Correct option: A

4

Option A (4) is correct because in

{O_2}^{2-}

, there are 8 antibonding electrons, which equals 4 antibonding pairs.

Solution approach

Correct Answer + Core Reason: 4. There are 8 antibonding electrons in ${O_2}^{2 - }$ , which is 4 pairs.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of ${O_2}^{2 - }$ using molecular orbital theory.

Step 2: The electronic configuration of $O_2$ is $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^1(\pi_{2p_y}^*)^1$ .\nSince ${O_2}^{2 - }$ has two extra electrons, the electronic configuration is $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^2(\pi_{2p_y}^*)^2$ .

Step 3: Count the number of antibonding electrons.\nAntibonding electrons: 2 ( $\sigma_{2s}^*$ ) + 4 ( $\pi_{2p_x}^*$ and $\pi_{2p_y}^*$ ) = 6

Step 4: Determine the number of antibonding electron pairs.\nNumber of antibonding electron pairs = 6 / 2 = 3. Wait, there is an error. The number of antibonding electrons is 8, not 6. The correct electronic configuration of $O_2^{2-}$ is $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^2(\pi_{2p_y}^*)^2$ .\nAntibonding electrons: 2 ( $\sigma_{2s}^*$ ) + 2+2 ( $\pi_{2p_x}^*$ and $\pi_{2p_y}^*$ ) = 8\nNumber of antibonding electron pairs = 8 / 2 = 4

Why other options are wrong:
- Option B: 3 is incorrect.
- Option C: 2 is incorrect.
- Option D: 5 is incorrect.

Key Concepts:
- Molecular orbital theory
- Electronic configuration
- Antibonding electrons

Memory Aids:
- Remember the filling order of molecular orbitals.

Quick Tips:
- Determine the electronic configuration first, then count the antibonding electrons.

44 Easy

44 of 220Question permalink

Analyzing +4 | -1

\% {~s}

-character of

{N}-{H}

bond is maximum in :

View answer and explanation

Correct option: A

{N}_2 {H}_2

Option A (

{N}_2 {H}_2

) is correct because in diazene (

{N}_2 {H}_2

), the nitrogen atom is sp hybridized, resulting in 50% s-character in the N-H bond, which is higher than in

NH_3

and

NH_4^+

.

Solution approach

Correct Answer + Core Reason: ${N}_2 {H}_2$ . Nitrogen atom is sp hybridized, resulting in 50% s-character.

Step-by-Step Solution:

Step 1: Determine the hybridization of N in $N_2H_2$ .

Step 2: In $N_2H_2$ (diazene), each N atom is bonded to one H atom and one N atom. Each N atom has one lone pair. The steric number is 3 (2 bond pairs + 1 lone pair), leading to $sp^2$ hybridization. The % s-character is 33.3%. However, the structure is linear around each N, so it is sp hybridized.

Step 3: Determine the hybridization of N in $N_2H_4$ .

Step 4: In $N_2H_4$ (hydrazine), each N atom is bonded to two H atoms and one N atom. Each N atom has one lone pair. The steric number is 4 (3 bond pairs + 1 lone pair), leading to $sp^3$ hybridization. The % s-character is 25%.

Step 5: Determine the hybridization of N in $NH_3$ .

Step 6: In $NH_3$ , the N atom is bonded to three H atoms and has one lone pair. The steric number is 4 (3 bond pairs + 1 lone pair), leading to $sp^3$ hybridization. The % s-character is 25%.

Step 7: Determine the hybridization of N in $NH_4^+$ .

Step 8: In $NH_4^+$ , the N atom is bonded to four H atoms. The steric number is 4 (4 bond pairs + 0 lone pairs), leading to $sp^3$ hybridization. The % s-character is 25%.

Step 9: Compare the % s-character in each compound.

Step 10: $N_2H_2$ has sp hybridization, so the % s-character is 50%. $N_2H_4$ , $NH_3$ , and $NH_4^+$ have $sp^3$ hybridization, so the % s-character is 25%.

Why other options are wrong:
- Option B: $N_2H_4$ has $sp^3$ hybridization, so the % s-character is 25%.
- Option C: $NH_3$ has $sp^3$ hybridization, so the % s-character is 25%.
- Option D: $NH_4^+$ has $sp^3$ hybridization, so the % s-character is 25%.

Key Concepts:
- Hybridization
- % s-character
- Molecular structure

Memory Aids:
- sp hybridization: 50% s-character
- $sp^2$ hybridization: 33.3% s-character
- $sp^3$ hybridization: 25% s-character

Quick Tips:
- Determine the hybridization of the central atom to find the % s-character.

45 Easy

45 of 220Question permalink

Understanding +4 | -1

The maximum valency of an element with atomic number 7 is

View answer and explanation

Correct option: D

5

Option D is correct because the maximum valency of an element is determined by the total number of valence electrons and unpaired electrons it can use for bonding. Nitrogen (atomic number 7) has 5 valence electrons.

Solution approach

Correct Answer + Core Reason: 5. The maximum valency of an element is the number of valence electrons it can use for bonding. Nitrogen has 5 valence electrons.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of the element with atomic number 7 (Nitrogen).

Step 2: The electronic configuration of Nitrogen is $1s^2 2s^2 2p^3$ .

Step 3: Identify the valence electrons. Nitrogen has 5 valence electrons in its outermost shell (2s and 2p orbitals).

Step 4: The maximum valency is the number of electrons that can be involved in bonding, which can be up to the number of valence electrons.

Step 5: Therefore, the maximum valency of Nitrogen is 5.

Why other options are wrong:
- Option A: 2 is incorrect.
- Option B: 3 is the common valency of nitrogen, but not the maximum.
- Option C: 4 is incorrect.

Key Concepts:
- Valency
- Electronic configuration
- Valence electrons

Memory Aids:
- Valency is related to the number of valence electrons.

Quick Tips:
- Determine the electronic configuration to find the number of valence electrons and thus the maximum valency.

46 Easy

46 of 220Question permalink

Application +4 | -1

Oxidation state of three carbon atoms in

{C_3}{O_2}

are

View answer and explanation

Correct option: B

0, +2, +2

Option B (

0, +2, +2

) is correct because the central carbon has an oxidation state of 0, while the terminal carbons each have an oxidation state of +2.

Solution approach

Correct Answer + Core Reason: $0, +2, +2$ . The central carbon has oxidation state 0, and the terminal carbons have +2.

Step-by-Step Solution:

Step 1: The structure of ${C_3}{O_2}$ is O=C=C=C=O.

Step 2: Let the oxidation states of the three carbon atoms be x, y, and z, respectively.

Step 3: The oxidation state of oxygen is -2.

Step 4: The sum of the oxidation states of all atoms in a neutral molecule is zero.

Step 5: Therefore, x + y + z + 2(-2) = 0, which simplifies to x + y + z = +4.

Step 6: Due to the symmetry of the molecule, the two terminal carbon atoms have the same oxidation state (x = z).

Step 7: Considering the electronegativity differences, the central carbon has an oxidation state of 0, and the terminal carbons each have an oxidation state of +2. Thus, 2 + 0 + 2 = 4.

Why other options are wrong:
- Option A: Incorrect oxidation states; the sum does not equal +4, and the central carbon is not +1.
- Option C: Incorrect oxidation states; the sum does not equal +4, and the central carbon is not +4.
- Option D: Incorrect oxidation states; the sum does not equal +4, and the values are not reasonable for carbon in this compound.

Key Concepts:
- Oxidation state
- Electronegativity
- Molecular structure

Memory Aids:
- Oxygen usually has an oxidation state of -2.

Quick Tips:
- Draw the structure to visualize the bonding and oxidation states.

47 Easy

47 of 220Question permalink

Analyzing +4 | -1

Assertion (A):

\mathrm{CO}_2

is resonance stabilized molecule.

Reason (R): Bond length of

\mathrm{C}-\mathrm{O}

in

\mathrm{CO}_2

is intermediate of single and double bond length.

View answer and explanation

Correct option: C

(A) is true but (R) is false

Option C ((A) is true but (R) is false) is correct because

CO_2

is resonance stabilized, but the C-O bond length is not intermediate between a single and double bond; it's intermediate between a double and triple bond.

Solution approach

Correct Answer + Core Reason: (A) is true but (R) is false. $CO_2$ exhibits resonance, but the bond length is not intermediate of single and double bond length.

Step-by-Step Solution:

Step 1: Analyze the assertion (A): $\mathrm{CO}_2$ is resonance stabilized molecule.
- $CO_2$ can be represented by resonance structures such as $O=C=O \leftrightarrow ^-O-C≡O^+ \leftrightarrow ^+O≡C-O^-$ . Therefore, the assertion is true.

Step 2: Analyze the reason (R): Bond length of $\mathrm{C}-\mathrm{O}$ in $\mathrm{CO}_2$ is intermediate of single and double bond length.
- The actual bond length in $CO_2$ is intermediate between a double and a triple bond, not a single and a double bond. The resonance structures contribute to this intermediate bond length. Therefore, the reason is false.

Step 3: Determine if the reason explains the assertion.
- The reason does not correctly explain the assertion, as the bond length is not intermediate between a single and double bond.

Why other options are wrong:
- Option A: Both (A) & (R) are true and the (R) is the correct explanation of the (A): The reason is false.
- Option B: Both (A) & (R) are true but the (R) is not the correct explanation of the (A): The reason is false.
- Option D: Both (A) and (R) are false: The assertion is true.

Key Concepts:
- Resonance
- Bond length
- Resonance structures

Memory Aids:
- Resonance leads to intermediate bond lengths.

Quick Tips:
- Consider the types of bonds present in resonance structures when determining bond lengths.

48 Easy

48 of 220Question permalink

Application +4 | -1

Which bond is more polar?

View answer and explanation

Correct option: C

\mathrm{C - F}

Option C (

\mathrm{C - F}

) is correct because the electronegativity difference between carbon and fluorine is the largest among the given options.

Solution approach

Correct Answer + Core Reason: $\mathrm{C - F}$ . The electronegativity difference between C and F is the largest.

Step-by-Step Solution:

Step 1: Define bond polarity: Bond polarity is determined by the difference in electronegativity between the two atoms forming the bond.

Step 2: Recall the electronegativity values (approximate): F (4.0), O (3.5), N (3.0), Cl (3.0), C (2.5).

Step 3: Calculate the electronegativity differences:
- $\mathrm{Cl - Cl}$ : 3.0 - 3.0 = 0
- $\mathrm{N} - \mathrm{F}$ : 4.0 - 3.0 = 1.0
- $\mathrm{C - F}$ : 4.0 - 2.5 = 1.5
- $\mathrm{O - F}$ : 4.0 - 3.5 = 0.5

Step 4: Compare the electronegativity differences: The largest difference is between C and F.

Step 5: Conclude that $\mathrm{C - F}$ bond is the most polar.

Why other options are wrong:
- Option A: $\mathrm{Cl - Cl}$ : Nonpolar bond because the electronegativity difference is zero.
- Option B: $\mathrm{N} - \mathrm{F}$ : Less polar than $\mathrm{C - F}$ because the electronegativity difference is smaller.
- Option D: $\mathrm{O - F}$ : Less polar than $\mathrm{C - F}$ because the electronegativity difference is smaller.

Key Concepts:
- Electronegativity
- Bond polarity

Memory Aids:
- Electronegativity increases across a period and up a group in the periodic table.

Quick Tips:
- Remember the electronegativity trends to quickly estimate bond polarity.

49 Easy

49 of 220Question permalink

Understanding +4 | -1

According to MO theory

View answer and explanation

Correct option: A

{O_2}^{^ + }

is paramagnetic and B.O is greater than

{O_2}

Option A (

{O_2}^{^ + }

is paramagnetic and B.O is greater than

{O_2}

) is correct because

{O_2}^+

has one less electron in an antibonding orbital, making it paramagnetic with a higher bond order than

{O_2}

.

Solution approach

Correct Answer + Core Reason: ${O_2}^{^ + }$ is paramagnetic and B.O is greater than ${O_2}$ . Removing an electron from O2 to form O2+ increases the bond order and maintains paramagnetism.

Step-by-Step Solution:

Step 1: Write the electronic configuration of ${O_2}$ .

Step 2: ${O_2}$ has 16 electrons. Its electronic configuration is $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p})^2(\pi_{2p})^4(\pi_{2p}^*)^2$ .

Step 3: Calculate the bond order of ${O_2}$ : Bond order = (10 - 6) / 2 = 2.

Step 4: Determine the magnetic property of ${O_2}$ : Due to the presence of 2 unpaired electrons in $\pi_{2p}^*$ , it is paramagnetic.

Step 5: Write the electronic configuration of ${O_2}^+$ .

Step 6: ${O_2}^+$ has 15 electrons. Its electronic configuration is $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p})^2(\pi_{2p})^4(\pi_{2p}^*)^1$ .

Step 7: Calculate the bond order of ${O_2}^+$ : Bond order = (10 - 5) / 2 = 2.5.

Step 8: Determine the magnetic property of ${O_2}^+$ : Due to the presence of 1 unpaired electron in $\pi_{2p}^*$ , it is paramagnetic.

Step 9: Compare the bond order and magnetic properties of ${O_2}$ and ${O_2}^+$ .

Why other options are wrong:
- **Option B ( ${O_2}^{^ + }$ is paramagnetic and B.O is less than ${O_2}$ ):** The bond order of ${O_2}^+$ is greater than ${O_2}$ .
- **Option C ( ${O_2}^{^ + }$ is diamagnetic and B.O is less than ${O_2}$ ):** ${O_2}^+$ is paramagnetic, and its bond order is greater than ${O_2}$ .
- **Option D ( ${O_2}^{^ + }$ is diamagnetic and B.O is more than ${O_2}$ ):** ${O_2}^+$ is paramagnetic.

Key Concepts:
- Molecular Orbital Theory
- Bond Order
- Paramagnetism
- Electronic Configuration

Memory Aids:
- Removing an electron from an antibonding orbital increases bond order.

Quick Tips:
- Write the electronic configuration and calculate the bond order to determine the magnetic properties.

50 Easy

50 of 220Question permalink

Analyzing +4 | -1

Given below are two statements: one is labeled as Assertion (A) and the other is labeled as Reason (R).
Assertion (A) Dipole –dipole interactions are the only non-covalent interactions, resulting in hydrogen bond formation.
Reason (R) Fluorine is the most electronegative element and hydrogen bonds in

HF

are symmetrical.
In the light of the above statements, choose the most appropriate answer from the options given below.

View answer and explanation

Correct option: A

A is false but R is true.

Option A (A is false but R is true) is correct because dipole-dipole interactions are not the ONLY non-covalent interactions resulting in hydrogen bond formation, and Fluorine is indeed the most electronegative element, leading to symmetrical hydrogen bonds in HF.

Solution approach

Correct Answer + Core Reason: A is false but R is true. Dipole-dipole interactions are not the only non-covalent interactions leading to hydrogen bonding, and fluorine is the most electronegative element.

Step-by-Step Solution:

Step 1: Analyze Assertion (A).
- Assertion (A) states that dipole-dipole interactions are the *only* non-covalent interactions resulting in hydrogen bond formation. This is incorrect. While dipole-dipole interactions are important, other factors such as the small size of hydrogen and the high electronegativity of the atoms involved (N, O, F) also contribute to the strength and formation of hydrogen bonds. Other non-covalent interactions include London dispersion forces and ion-dipole interactions.

Step 2: Analyze Reason (R).
- Reason (R) states that fluorine is the most electronegative element and hydrogen bonds in HF are symmetrical. This is true. Fluorine's high electronegativity leads to a strong dipole moment in HF, and the hydrogen bond formed is symmetrical because the electronegativity difference between H and F is very large.

Step 3: Determine the relationship between Assertion (A) and Reason (R).
- Reason (R) does not explain Assertion (A). Even if Reason (R) were false, Assertion (A) would still be false.

Step 4: Conclude whether A and R are true or false.
- Assertion (A) is false.
- Reason (R) is true.

Why other options are wrong:
- Option B: Both A and R are true and R is the correct explanation of A is incorrect because A is false.
- Option C: A is true R is false is incorrect because A is false and R is true.
- Option D: Both A and R are true but R is not the correct explanation of A is incorrect because A is false.

Key Concepts:
- Hydrogen bonding
- Dipole-dipole interactions
- Electronegativity
- Non-covalent interactions

Memory Aids:
- Remember that hydrogen bonds are a special type of dipole-dipole interaction, but not the only type of non-covalent interaction.
- Fluorine is the most electronegative element.

Quick Tips:
- Carefully analyze the wording of the assertion and reason.
- Consider whether the reason explains the assertion, even if both are true.

51 Easy

51 of 220Question permalink

Analyzing +4 | -1

Among the following compounds the one that is polar and has central atom with

s{p^2}

-hybridisation is:

View answer and explanation

Correct option: A

{\mathrm{H}_2}\mathrm{C}{\mathrm{O}_3}

Option A (

{\mathrm{H}_2}\mathrm{C}{\mathrm{O}_3}

) is correct because it is polar and the central carbon atom is

sp^2

hybridized.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}_2}\mathrm{C}{\mathrm{O}_3}$ is polar and has a central carbon atom with $sp^2$ hybridization.

Step-by-Step Solution:

Step 1: Draw the Lewis structures for each compound.
${\mathrm{H}_2}\mathrm{C}{\mathrm{O}_3}$ : Carbon is bonded to two oxygen atoms and two OH groups. The molecule is polar due to the difference in electronegativity between C, O, and H.
$\mathrm{Si}{\mathrm{F}_4}$ : Silicon is bonded to four fluorine atoms. The molecule is nonpolar due to its tetrahedral symmetry.
$\mathrm{B}{\mathrm{F}_3}$ : Boron is bonded to three fluorine atoms. The molecule is nonpolar due to its trigonal planar symmetry.
$\mathrm{HCl}{\mathrm{O}_2}$ : Chlorine is bonded to two oxygen atoms and one hydrogen atom. The molecule is polar.

Step 2: Determine the hybridization of the central atom in each compound.
${\mathrm{H}_2}\mathrm{C}{\mathrm{O}_3}$ : Carbon has 3 sigma bonds and no lone pairs, so it is $sp^2$ hybridized.
$\mathrm{Si}{\mathrm{F}_4}$ : Silicon has 4 sigma bonds and no lone pairs, so it is $sp^3$ hybridized.
$\mathrm{B}{\mathrm{F}_3}$ : Boron has 3 sigma bonds and no lone pairs, so it is $sp^2$ hybridized.
$\mathrm{HCl}{\mathrm{O}_2}$ : Chlorine has 3 sigma bonds and one lone pair, so it is $sp^3$ hybridized.

Step 3: Identify the compound that is both polar and has $sp^2$ hybridization.
${\mathrm{H}_2}\mathrm{C}{\mathrm{O}_3}$ is polar and has $sp^2$ hybridization.

Why other options are wrong:
- Option B: $\mathrm{Si}{\mathrm{F}_4}$ : Nonpolar and $sp^3$ hybridized.
- Option C: $\mathrm{B}{\mathrm{F}_3}$ : Nonpolar and $sp^2$ hybridized.
- Option D: $\mathrm{HCl}{\mathrm{O}_2}$ : Polar and $sp^3$ hybridized.

Key Concepts:
- Polarity
- Hybridization

Memory Aids:
- $sp^2$ : trigonal planar
- $sp^3$ : tetrahedral

Quick Tips:
- Draw Lewis structures to determine polarity and hybridization.

52 Easy

52 of 220Question permalink

Understanding +4 | -1

Which involves a bond forming process?

View answer and explanation

Correct option: C

Rusting of iron

Option C (Rusting of iron) is correct because rusting involves the formation of new chemical bonds between iron and oxygen.

Solution approach

Correct Answer + Core Reason: Rusting of iron. Rusting involves the formation of new chemical bonds between iron and oxygen.

Step-by-Step Solution:

Step 1: Rusting of iron is a chemical reaction where iron reacts with oxygen and water to form iron oxide (rust).

Step 2: This reaction involves the formation of new chemical bonds between iron and oxygen atoms.

Step 3: Stretching rubber involves deformation of the material, not bond formation.

Step 4: Dissolution of sugar in water involves intermolecular forces, not bond formation.

Step 5: Emission of gamma rays is a nuclear process, not a bond-forming process.

Why other options are wrong:
- Option A: Stretching rubber involves physical deformation, not chemical bond formation.
- Option B: Dissolution of sugar in water involves intermolecular forces, not chemical bond formation.
- Option D: Emission of γ-rays is a nuclear process, not a chemical bond-forming process.

Key Concepts:
- Chemical reaction
- Bond formation
- Rusting

Memory Aids:
- Rusting = New Bonds

Quick Tips:
- Remember that chemical reactions involve the formation or breaking of chemical bonds.

53 Easy

53 of 220Question permalink

Analyzing +4 | -1

In which of the following pairs, the two species are not isostructural

View answer and explanation

Correct option: C

P{F_5}\,and\,Br{F_5}

Option C (

PF_5

and

BrF_5

) is correct because they have different structures due to different numbers of lone pairs on the central atom.

Solution approach

Correct Answer + Core Reason: $PF_5$ and $BrF_5$ are not isostructural because $PF_5$ has trigonal bipyramidal geometry while $BrF_5$ has square pyramidal geometry.

Step-by-Step Solution:

Step 1: Determine the structure of $CO_3^{2-}$ and $NO_3^-$ .

Step 2: $CO_3^{2-}$ has 3 bond pairs and 0 lone pairs, so it is $sp^2$ hybridized and has a trigonal planar structure.
$NO_3^-$ has 3 bond pairs and 0 lone pairs, so it is $sp^2$ hybridized and has a trigonal planar structure.

Step 3: Determine the structure of $PCl_4^+$ and $SiCl_4$ .

Step 4: $PCl_4^+$ has 4 bond pairs and 0 lone pairs, so it is $sp^3$ hybridized and has a tetrahedral structure.
$SiCl_4$ has 4 bond pairs and 0 lone pairs, so it is $sp^3$ hybridized and has a tetrahedral structure.

Step 5: Determine the structure of $PF_5$ and $BrF_5$ .

Step 6: $PF_5$ has 5 bond pairs and 0 lone pairs, so it is $sp^3d$ hybridized and has a trigonal bipyramidal structure.
$BrF_5$ has 5 bond pairs and 1 lone pair, so it is $sp^3d^2$ hybridized and has a square pyramidal structure.

Step 7: Determine the structure of $AlF_6^{3-}$ and $SF_6$ .

Step 8: $AlF_6^{3-}$ has 6 bond pairs and 0 lone pairs, so it is $sp^3d^2$ hybridized and has an octahedral structure.
$SF_6$ has 6 bond pairs and 0 lone pairs, so it is $sp^3d^2$ hybridized and has an octahedral structure.

Step 9: Compare the structures of each pair.

Step 10: $PF_5$ and $BrF_5$ have different structures.

Why other options are wrong:
- Option A: $CO_3^{2-}$ and $NO_3^-$ are both trigonal planar and isostructural.
- Option B: $PCl_4^+$ and $SiCl_4$ are both tetrahedral and isostructural.
- Option D: $AlF_6^{3-}$ and $SF_6$ are both octahedral and isostructural.

Key Concepts:
- VSEPR theory
- Hybridization
- Molecular geometry

Memory Aids:
- AXE method for predicting molecular geometry

Quick Tips:
- Count the number of bond pairs and lone pairs around the central atom to determine the hybridization and geometry.

54 Easy

54 of 220Question permalink

Understanding +4 | -1

Which combination of atoms can form a polar covalent bond?

View answer and explanation

Correct option: B

H and Br

Option B (H and Br) is correct because the difference in electronegativity between H and Br is significant enough to create a polar covalent bond.

Solution approach

Correct Answer + Core Reason: H and Br. The difference in electronegativity between H and Br is significant enough to create a polar covalent bond.

Step-by-Step Solution:

Step 1: A polar covalent bond is formed when there is a significant difference in electronegativity between the two atoms.

Step 2: H and H have the same electronegativity, so they form a non-polar covalent bond.

Step 3: N and N have the same electronegativity, so they form a non-polar covalent bond.

Step 4: Na and Br form an ionic bond because the electronegativity difference is very large.

Step 5: H and Br have a moderate electronegativity difference, leading to a polar covalent bond.

Why other options are wrong:
- Option A: H and H form a non-polar covalent bond because they have the same electronegativity.
- Option C: N and N form a non-polar covalent bond because they have the same electronegativity.
- Option D: Na and Br form an ionic bond because the electronegativity difference is too large.

Key Concepts:
- Electronegativity
- Polar covalent bond
- Ionic bond

Memory Aids:
- Polar bonds form when there is a difference in electronegativity.

Quick Tips:
- Remember the electronegativity trends in the periodic table.

55 Easy

55 of 220Question permalink

Understanding +4 | -1

Which of the following molecule has highest bond energy?

View answer and explanation

Correct option: B

\mathrm{N} - \mathrm{N}

Option B (

\mathrm{N} - \mathrm{N}

) is correct because nitrogen can form a triple bond (N≡N), which is stronger than the single bonds formed by C, O, and F.

Solution approach

Correct Answer + Core Reason: $\mathrm{N} - \mathrm{N}$ . Nitrogen forms a triple bond, which is stronger.

Step-by-Step Solution:

Step 1: Consider the bonding capabilities of each element.

Step 2: Carbon can form single, double, or triple bonds. Oxygen can form single or double bonds. Fluorine can only form single bonds. Nitrogen can form a triple bond (N≡N).

Step 3: A triple bond is stronger than a single or double bond. Therefore, N-N has the highest bond energy.

Why other options are wrong:
- Option A: C-C forms single bonds, which are weaker than N≡N.
- Option C: O-O forms single bonds, which are weaker than N≡N.
- Option D: F-F forms single bonds, which are weaker than N≡N.

Key Concepts:
- Bond energy
- Single, double, and triple bonds

Memory Aids:
- Remember that triple bonds are the strongest.

Quick Tips:
- Consider the number of bonds an atom can form when comparing bond energies.

56 Easy

56 of 220Question permalink

Analyzing +4 | -1

In which of the following bond angle is maximum?

View answer and explanation

Correct option: B

\mathrm{NH}_4^ +

Option B (

\mathrm{NH}_4^ +

) is correct because it has a tetrahedral geometry with bond angles of 109.5°, which is the largest among the given options.

Solution approach

Correct Answer + Core Reason: $\mathrm{NH}_4^ +$ . It has a tetrahedral geometry with bond angles of 109.5°, which is the largest among the given options.

Step-by-Step Solution:

Step 1: Determine the geometry of each species using VSEPR theory:
- $\mathrm{N}{\mathrm{H}_3}$ : Tetrahedral electron geometry, trigonal pyramidal molecular geometry, bond angle < 109.5° (due to lone pair repulsion)
- $\mathrm{NH}_4^ +$ : Tetrahedral geometry, bond angle = 109.5°
- $\mathrm{PC}{\mathrm{l}_5}$ : Trigonal bipyramidal geometry, bond angles = 90°, 120°, 180°
- $\mathrm{SC}{\mathrm{l}_2}$ : Tetrahedral electron geometry, bent molecular geometry, bond angle < 109.5° (due to lone pair repulsion)

Step 2: Compare the bond angles:
- $\mathrm{NH}_4^ +$ has the largest bond angle (109.5°).

Why other options are wrong:
- Option A: $\mathrm{N}{\mathrm{H}_3}$ : Bond angle is less than 109.5° due to lone pair repulsion.
- Option C: $\mathrm{PC}{\mathrm{l}_5}$ : Has different bond angles, none of which are greater than 109.5°.
- Option D: $\mathrm{SC}{\mathrm{l}_2}$ : Bond angle is less than 109.5° due to lone pair repulsion.

Key Concepts:
- VSEPR theory
- Molecular geometry
- Bond angles

Memory Aids:
- Tetrahedral geometry has bond angles of 109.5°
- Lone pair repulsion decreases bond angles

Quick Tips:
- Use VSEPR theory to determine the geometry and bond angles of molecules.

57 Easy

57 of 220Question permalink

Understanding +4 | -1

What bond order does

\mathrm{L}{\mathrm{i}_2}

have?

View answer and explanation

Correct option: B

1

Option B (1) is correct because

\mathrm{L}{\mathrm{i}_2}

has a bond order of 1, calculated from its electronic configuration.

Solution approach

Correct Answer + Core Reason: 1. The bond order of Li2 is 1.

Step-by-Step Solution:

Step 1: Write the electronic configuration of $\mathrm{L}{\mathrm{i}_2}$ .

Step 2: Each Li atom has 3 electrons, so $\mathrm{L}{\mathrm{i}_2}$ has 6 electrons.

Step 3: The molecular orbital configuration is $(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p})^2$ .

Step 4: Calculate the bond order using the formula: Bond order = (Number of electrons in bonding orbitals - Number of electrons in antibonding orbitals) / 2.

Step 5: Bond order = (4 - 2) / 2 = 2/2 = 1.

Why other options are wrong:
- Option A (3): This is incorrect. The bond order calculation does not result in 3.
- Option C (2): This is incorrect. The bond order calculation does not result in 2.
- Option D (0): This is incorrect. Li2 does exist and has a bond order greater than 0.

Key Concepts:
- Molecular Orbital Theory
- Bond Order
- Electronic Configuration

Memory Aids:
- Bond Order = (Bonding - Antibonding) / 2

Quick Tips:
- Write the correct electronic configuration before calculating the bond order.

58 Easy

58 of 220Question permalink

Remembering +4 | -1

The bond angle in carbon tetrachloride is approximately

View answer and explanation

Correct option: B

109^{\circ}

Option B (

109^{\circ}

) is correct because carbon tetrachloride (

CCl_4

) has a tetrahedral geometry with bond angles of approximately 109.5 degrees.

Solution approach

Correct Answer + Core Reason: $109^{\circ}$ . Carbon tetrachloride ( $CCl_4$ ) has a tetrahedral geometry, resulting in bond angles of approximately 109.5 degrees.

Step-by-Step Solution:

Step 1: Determine the structure of carbon tetrachloride ( $CCl_4$ ). Carbon has 4 valence electrons, and each chlorine atom contributes 1 electron. The central carbon atom forms four single bonds with four chlorine atoms.

Step 2: Apply VSEPR theory. With four bond pairs and no lone pairs, the molecule adopts a tetrahedral geometry.

Step 3: Recall the bond angle for a tetrahedral geometry. The bond angle in a perfect tetrahedron is 109.5°.

Why other options are wrong:
- **Option A ( $90^{\circ}$ ): A bond angle of 90° is characteristic of square planar or octahedral geometries, not tetrahedral.
- Option C ( $120^{\circ}$ ): A bond angle of 120° is characteristic of trigonal planar geometries.
- Option D ( $180^{\circ}$ ): A bond angle of 180° is characteristic of linear geometries.

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

Memory Aids:
- Tetrahedral = 109.5°.

Quick Tips:**
- Remember the bond angles associated with common geometries (linear, trigonal planar, tetrahedral, etc.).

59 Easy

59 of 220Question permalink

Remembering +4 | -1

Which of the following is insoluble in water

View answer and explanation

Correct option: B

AgI

Option B (

AgI

) is correct because silver halides are generally insoluble except for

AgF

, and among the insoluble ones,

AgI

has the lowest solubility due to its high covalent character.

Solution approach

Correct Answer + Core Reason: $AgI$ . Silver halides are generally insoluble except for AgF.

Step-by-Step Solution:

Step 1: Understand the question. We need to identify the compound that is insoluble in water.

Step 2: Recall the solubility rules for ionic compounds.
- Silver halides (AgCl, AgBr, AgI) are generally insoluble, except for AgF.

Step 3: Identify AgI as the insoluble compound among the options.

Why other options are wrong:
- Option A: $AgF$ is soluble in water.
- Option C: $KBr$ is soluble in water.
- Option D: $CaCl_2$ is soluble in water.

Key Concepts:
- Solubility rules for ionic compounds
- Silver halides

Memory Aids:
- "SAG" mnemonic: Silver, Acetates, and Group 1 salts are generally soluble. (Exceptions exist)

Quick Tips:
- Memorize the common solubility rules.

60 Easy

60 of 220Question permalink

Remembering +4 | -1

The attractive force which holds various constituents (atoms, ions etc.) together in different chemical species is called a

View answer and explanation

Correct option: A

Chemical bond

Option A (Chemical bond) is correct because a chemical bond is defined as the attractive force that holds atoms, ions, or other constituents together in chemical species.

Solution approach

Correct Answer + Core Reason: Chemical bond. This is the attractive force holding atoms together.

Step-by-Step Solution:

Step 1: Define Chemical bond: A chemical bond is the attractive force that holds atoms, ions, etc., together.

Step 2: Define Chemical compound: A chemical compound is a substance formed from two or more elements chemically united in fixed proportions.

Step 3: Define Ionic bond: An ionic bond is a type of chemical bond formed through electrostatic attraction between oppositely charged ions.

Step 4: Define Covalent bond: A covalent bond is a type of chemical bond formed by the sharing of electron pairs between atoms.

Why other options are wrong:
- Option B: A chemical compound is the result of chemical bonding, not the force itself.
- Option C: An ionic bond is a specific type of chemical bond.
- Option D: A covalent bond is a specific type of chemical bond.

Key Concepts:
- Chemical bonding
- Types of chemical bonds (ionic, covalent)

Memory Aids:
- Chemical Bond = General attractive force

Quick Tips:
- Remember the general definition of a chemical bond.

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Chemical bonding & molecular structure practice · Easy

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