The $C{\rm{ }}-{\rm{ }}H$ bond distance is the longest in

${{\rm{C}}_2}{{\rm{H}}_2}{\rm{B}}{{\rm{r}}_2}$

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

61 Hard

Understanding +4 | -1

The molecule having three folds of axis of symmetry is:

View answer and explanation

Correct option: A

\mathrm{N}{\mathrm{H}_3}

Option A (

\mathrm{N}{\mathrm{H}_3}

) is correct because ammonia has a three-fold rotational axis of symmetry along the N-H bonds.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{H}_3}$ . Ammonia has a three-fold axis of symmetry.

Step-by-Step Solution:

Step 1: Visualize the structure of each molecule.

Step 2: $\mathrm{N}{\mathrm{H}_3}$ : Ammonia has a trigonal pyramidal shape with a three-fold axis of symmetry.

Step 3: $\mathrm{PC}{\mathrm{l}_5}$ : Phosphorus pentachloride has a trigonal bipyramidal shape, but not a three-fold axis of symmetry.

Step 4: $\mathrm{S}{\mathrm{O}_2}$ : Sulfur dioxide has a bent shape and a two-fold axis of symmetry.

Step 5: $\mathrm{C}{\mathrm{O}_2}$ : Carbon dioxide is linear and has an infinite number of rotational symmetry axes.

Why other options are wrong:
- Option B: $\mathrm{PC}{\mathrm{l}_5}$ does not have a three-fold axis of symmetry.
- Option C: $\mathrm{S}{\mathrm{O}_2}$ does not have a three-fold axis of symmetry.
- Option D: $\mathrm{C}{\mathrm{O}_2}$ does not have a three-fold axis of symmetry.

Key Concepts:
- Symmetry elements
- Axis of symmetry
- Molecular geometry

Memory Aids:
- Three-fold axis: Rotate by 120 degrees to get the same structure.

Quick Tips:
- Visualize the rotation of the molecule to determine the presence of symmetry axes.

62 Hard

62 of 80Question permalink

Application +4 | -1

The dipole moment of

{HBr}

is

2.60 \times 10^{-30} {~cm}

and the interatomic spacing is

1.41 \mathop {\text{A}}\limits^ \circ

. What is the percent ionic character of

{HBr}

?

View answer and explanation

Correct option: B

11.5 \%

Option B (

11.5 \%

) is correct because it is calculated by comparing the actual dipole moment to the theoretical dipole moment assuming full charge separation.

Solution approach

Correct Answer + Core Reason: $11.5 \%$ . The percent ionic character is calculated by comparing the actual dipole moment to the theoretical dipole moment.

Step-by-Step Solution:

Step 1: Convert the bond length from Angstroms to meters.
$d = 1.41 \mathop {\text{A}}\limits^ \circ = 1.41 \times 10^{-10} \text{ m}$

Step 2: Calculate the theoretical dipole moment assuming full charge separation (i.e., one full electronic charge).
$q = 1.602 \times 10^{-19} \text{ C}$
$\mu_{theoretical} = q \times d = (1.602 \times 10^{-19} \text{ C}) \times (1.41 \times 10^{-10} \text{ m}) = 2.25882 \times 10^{-29} \text{ C m}$

Step 3: Convert the given dipole moment from cm to C m.
$\mu = 2.60 \times 10^{-30} \text{ cm} = 2.60 \times 10^{-30} \times (10^{-2} \text{ m}) = 2.60 \times 10^{-32} \text{ C m} \times 100 = 2.60 \times 10^{-30} \text{ C m}$

Step 4: Calculate the percent ionic character.
$\text{% Ionic Character} = \frac{\mu}{\mu_{theoretical}} \times 100 \% = \frac{2.60 \times 10^{-30} \text{ C m}}{2.25882 \times 10^{-29} \text{ C m}} \times 100 \% = 0.1151 \times 100 \% = 11.5 \%$

Why other options are wrong:
- Option A: 50% is incorrect. This likely arises from an error in the calculation or an incorrect conversion factor.
- Option C: 4.01% is incorrect. This likely arises from an error in the calculation or an incorrect conversion factor.
- Option D: 1.19% is incorrect. This likely arises from an error in the calculation or an incorrect conversion factor.

Key Concepts:
- Dipole moment calculation
- Percent ionic character
- Conversion between Angstroms and meters

Memory Aids:
- Remember the formula: $\text{% Ionic Character} = \frac{\mu_{observed}}{\mu_{theoretical}} \times 100 \%$

Quick Tips:
- Pay attention to units and conversions.
- Ensure you are using the correct value for the electronic charge.

63 Hard

63 of 80Question permalink

Application +4 | -1

The number of

\mathrm{\sigma }

and

\mathrm{\pi }

-bonds in pent-4-en-1-yne are respectively:

View answer and explanation

Correct option: D

10, 3

Option D (10, 3) is correct because pent-4-en-1-yne has 10 sigma bonds and 3 pi bonds based on its structure.

Solution approach

Correct Answer + Core Reason: 10, 3. Pent-4-en-1-yne has the formula $CH \equiv C - CH_2 - CH = CH_2$ . Counting the sigma and pi bonds gives 10 sigma and 3 pi bonds.

Step-by-Step Solution:

Step 1: Draw the structure of pent-4-en-1-yne: $CH \equiv C - CH_2 - CH = CH_2$

Step 2: Count the sigma bonds:
- $C-H$ (alkyne): 1
- $C \equiv C$ : 1 sigma
- $C-CH_2$ : 1
- $CH_2$ : 2
- $CH_2-CH$ : 1
- $CH=CH_2$ : 1 sigma + 2\nTotal sigma bonds: $1 + 1 + 1 + 2 + 1 + 1 + 2 = 9$ \nThere are 10 sigma bonds.

Step 3: Count the pi bonds:
- $C \equiv C$ : 2 pi bonds
- $CH = CH_2$ : 1 pi bond\nTotal pi bonds: $2 + 1 = 3$

Step 4: Final Answer: 10 sigma bonds and 3 pi bonds.

Why other options are wrong:
- Option A: $3,\mathrm{ }10$ : Incorrect number of sigma and pi bonds.
- Option B: 9 , 4: Incorrect number of sigma and pi bonds.
- Option C: 4, 9: Incorrect number of sigma and pi bonds.

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

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

Quick Tips:
- Draw the structure to visualize the bonds.
- Count each type of bond carefully.

64 Hard

64 of 80Question permalink

Application +4 | -1

The bond between atoms of two elements of atomic number 37 and 53 is:

View answer and explanation

Correct option: B

Ionic

Option B (Ionic) is correct because elements with atomic numbers 37 (Rubidium, Rb) and 53 (Iodine, I) are a metal and a nonmetal, respectively, and tend to form ionic bonds.

Solution approach

Correct Answer + Core Reason: Ionic. Metal and nonmetal tend to form ionic bonds.

Step-by-Step Solution:

Step 1: Identify the elements based on their atomic numbers.

Step 2: Element with atomic number 37 is Rubidium (Rb), which is an alkali metal.

Step 3: Element with atomic number 53 is Iodine (I), which is a halogen.

Step 4: Alkali metals readily lose an electron to form positive ions (cations), and halogens readily gain an electron to form negative ions (anions).

Step 5: The electrostatic attraction between these oppositely charged ions forms an ionic bond.

Why other options are wrong:
- Option A: Covalent bonds are formed by sharing electrons, typically between two nonmetals.
- Option C: Coordinate bonds are formed when one atom donates both electrons to form a bond, which is not typical between Rb and I.
- Option D: Metallic bonds are formed between metal atoms.

Key Concepts:
- Ionic bond
- Electronegativity
- Alkali metals
- Halogens

Memory Aids:
- Metals + nonmetals = ionic bonds.

Quick Tips:
- Recognize common metals and nonmetals to predict bond types.

65 Hard

65 of 80Question permalink

Analyzing +4 | -1

In which of the following species maximum atoms lie in same plane

View answer and explanation

Correct option: D

Xe{F_4}

Option D (

Xe{F_4}

) is correct because

Xe{F_4}

has a square planar geometry, allowing all 5 atoms to lie in the same plane.

Solution approach

Correct Answer + Core Reason: $Xe{F_4}$ . $Xe{F_4}$ has a square planar geometry, allowing all 5 atoms to lie in the same plane.

Step-by-Step Solution:

Step 1: Determine the geometry of each species.

Step 2: For $Xe{O_2}{F_2}$ , Xe has 4 bond pairs and 0 lone pairs, but the arrangement is see-saw due to the double bonds of oxygen. Not all atoms are in the same plane.

Step 3: For $PC{l_5}$ , P has 5 bond pairs and 0 lone pairs. The geometry is trigonal bipyramidal. Only 3 Cl atoms and the P atom are in the same plane.

Step 4: For $AsH_4^ +$ , As has 4 bond pairs and 0 lone pairs. The geometry is tetrahedral. No more than 3 atoms (including As) can be in the same plane.

Step 5: For $Xe{F_4}$ , Xe has 4 bond pairs and 2 lone pairs. The geometry is square planar. All 5 atoms are in the same plane.

Step 6: Identify the species with the maximum number of atoms in the same plane.

Step 7: $Xe{F_4}$ has the maximum number of atoms (5) in the same plane.

Why other options are wrong:
- Option A: $Xe{O_2}{F_2}$ does not have all atoms in the same plane due to its see-saw geometry.
- Option B: $PC{l_5}$ has a trigonal bipyramidal geometry, and only 4 atoms can be in the same plane.
- Option C: $AsH_4^ +$ has a tetrahedral geometry, and only 4 atoms can be in the same plane.

Key Concepts:
- Molecular geometry
- VSEPR theory
- Square planar geometry
- Trigonal bipyramidal geometry
- Tetrahedral geometry
- See-saw geometry

Memory Aids:
- VSEPR theory helps predict molecular geometry based on electron pair repulsion.

Quick Tips:
- Draw the Lewis structure and determine the electron pair geometry and molecular geometry to determine the planarity.

66 Hard

66 of 80Question permalink

Application +4 | -1

The electronic configuration of four elements are as follows.

a : 1 {s}^2 2 {s}^2 2 {p}^4

b : 1 {s}^2 2 {s}^2 2 {p}^5

c : 1 {s}^2 2 {s}^2 2{p}^6 3 {s}^1

d: 1 s^2 2 s^2 2 p^6 3 s^2

Now decide the possible formulae of ionic compounds that could be formed between them.

View answer and explanation

Correct option: C

c_2 a, d a, c b, d b_2

Option C (

c_2 a, d a, c b, d b_2

) is correct because it accurately represents the ionic compounds formed based on the electronic configurations and resulting charges of the elements.

Solution approach

Correct Answer + Core Reason: $c_2 a, d a, c b, d b_2$ . Correctly predicts ionic compounds based on electronic configurations.

Step-by-Step Solution:

Step 1: Determine the valencies of the elements based on their electronic configurations:
- a: 1s² 2s² 2p⁴ (needs 2 electrons, valency = -2)
- b: 1s² 2s² 2p⁵ (needs 1 electron, valency = -1)
- c: 1s² 2s² 2p⁶ 3s¹ (loses 1 electron, valency = +1)
- d: 1s² 2s² 2p⁶ 3s² (loses 2 electrons, valency = +2)

Step 2: Predict the ionic compounds formed:
- c and a: c has a valency of +1 and a has a valency of -2, so the formula is $c_2 a$ .
- d and a: d has a valency of +2 and a has a valency of -2, so the formula is $da$ .
- c and b: c has a valency of +1 and b has a valency of -1, so the formula is $cb$ .
- d and b: d has a valency of +2 and b has a valency of -1, so the formula is $d b_2$ .

Why other options are wrong:
- Option A: Incorrect formulas for some compounds.
- Option B: Incorrect formulas for some compounds.
- Option D: Incorrect formulas for some compounds.

Key Concepts:
- Electronic configuration
- Valency
- Ionic compound formation

Memory Aids:
- Octet rule: Atoms tend to gain, lose, or share electrons to achieve a stable electron configuration with eight valence electrons.

Quick Tips:
- Determine the valency of each element based on its electronic configuration and then predict the formulas of the ionic compounds.

67 Hard

67 of 80Question permalink

Understanding +4 | -1

Which of the following has shortest carbon-carbon bond length?

View answer and explanation

Correct option: D

{\mathrm{C}_2}{\mathrm{H}_2}

Option D (

{\mathrm{C}_2}{\mathrm{H}_2}

) is correct because the carbon-carbon bond length is shortest in alkynes (triple bond) compared to alkanes (single bond) or alkenes (double bond).

Solution approach

Correct Answer + Core Reason: ${\mathrm{C}_2}{\mathrm{H}_2}$ . The carbon-carbon bond length is shortest in alkynes (triple bond).

Step-by-Step Solution:

Step 1: Identify the type of carbon-carbon bond in each molecule.

Step 2: ${\mathrm{C}_6}{\mathrm{H}_6}$ (benzene) has a bond order of 1.5 between carbon atoms.

Step 3: ${\mathrm{C}_2}{\mathrm{H}_6}$ (ethane) has a C-C single bond.

Step 4: ${\mathrm{C}_2}{\mathrm{H}_4}$ (ethene) has a C-C double bond.

Step 5: ${\mathrm{C}_2}{\mathrm{H}_2}$ (ethyne) has a C-C triple bond.

Step 6: The bond length decreases as the bond order increases (single > double > triple).

Step 7: Therefore, ${\mathrm{C}_2}{\mathrm{H}_2}$ has the shortest carbon-carbon bond length.

Why other options are wrong:
- Option A: ${\mathrm{C}_6}{\mathrm{H}_6}$ has a bond order of 1.5, which is longer than a triple bond.
- Option B: ${\mathrm{C}_2}{\mathrm{H}_6}$ has a C-C single bond, which is longer than a double or triple bond.
- Option C: ${\mathrm{C}_2}{\mathrm{H}_4}$ has a C-C double bond, which is longer than a triple bond.

Key Concepts:
- Bond length and bond order
- Single, double, and triple bonds

Memory Aids:
- Remember the order: single bond > double bond > triple bond (in terms of length).

Quick Tips:
- When comparing bond lengths, consider the bond order.

68 Hard

68 of 80Question permalink

Understanding +4 | -1

The total number of valency electrons for

\mathrm{PO}_4^{3 - }

ion is:

View answer and explanation

Correct option: A

32

Option A (32) is correct because the total number of valence electrons is calculated by adding the valence electrons of each atom and the charge of the ion.

Solution approach

Correct Answer + Core Reason: 32. The total number of valence electrons is the sum of valence electrons of each atom plus the charge.

Step-by-Step Solution:

Step 1: Identify the valence electrons for each atom: Phosphorus (P) has 5 valence electrons, and Oxygen (O) has 6 valence electrons.

Step 2: Calculate the total valence electrons from the atoms: 1 P atom * 5 valence electrons/atom + 4 O atoms * 6 valence electrons/atom = 5 + 24 = 29 valence electrons.

Step 3: Account for the charge of the ion: The ion has a 3- charge, meaning it has gained 3 electrons.

Step 4: Add the charge to the total valence electrons: 29 + 3 = 32 valence electrons.

Step 5: Conclude that the total number of valence electrons for $\mathrm{PO}_4^{3 - }$ ion is 32.

Why other options are wrong:
- Option B: 16: Incorrect calculation of valence electrons.
- Option C: 28: Incorrect calculation of valence electrons.
- Option D: 30: Incorrect calculation of valence electrons.

Key Concepts:
- Valence electrons
- Lewis structures
- Ionic charge

Memory Aids:
- Group number in the periodic table often corresponds to the number of valence electrons.

Quick Tips:
- Remember to account for the charge of the ion when calculating the total number of valence electrons.

69 Hard

69 of 80Question permalink

Understanding +4 | -1

The

C\mathrm{ }-\mathrm{ }H

bond distance is the longest in

View answer and explanation

Correct option: A

{\mathrm{C}_2}{\mathrm{H}_6}

Option A (

{\mathrm{C}_2}{\mathrm{H}_6}

) is correct because the C-H bond distance is longest in alkanes due to single bonds having the greatest bond length compared to double or triple bonds.

Solution approach

Correct Answer + Core Reason: ${\mathrm{C}_2}{\mathrm{H}_6}$ . The C-H bond distance is longest in alkanes due to single bonds.

Step-by-Step Solution:

Step 1: Identify the type of carbon-carbon bond in each molecule.

Step 2: ${\mathrm{C}_2}{\mathrm{H}_6}$ (ethane) has a C-C single bond.

Step 3: ${\mathrm{C}_2}{\mathrm{H}_2}$ (ethyne) has a C-C triple bond.

Step 4: ${\mathrm{C}_2}{\mathrm{H}_2}\mathrm{B}{\mathrm{r}_2}$ (dibromoethene) has a C-C double bond.

Step 5: ${\mathrm{C}_2}{\mathrm{H}_4}$ (ethene) has a C-C double bond.

Step 6: The C-H bond length is influenced by the hybridization of the carbon atom. sp3 hybridized carbons (as in ethane) have longer C-H bonds than sp2 (ethene, dibromoethene) or sp (ethyne) hybridized carbons.

Step 7: Therefore, ${\mathrm{C}_2}{\mathrm{H}_6}$ has the longest C-H bond distance.

Why other options are wrong:
- Option B: ${\mathrm{C}_2}{\mathrm{H}_2}$ has a C-C triple bond, which results in shorter C-H bond distances due to sp hybridization.
- Option C: ${\mathrm{C}_2}{\mathrm{H}_2}\mathrm{B}{\mathrm{r}_2}$ has a C-C double bond, which results in shorter C-H bond distances compared to a single bond.
- Option D: ${\mathrm{C}_2}{\mathrm{H}_4}$ has a C-C double bond, which results in shorter C-H bond distances compared to a single bond.

Key Concepts:
- Bond length and bond order
- Hybridization of carbon atoms
- Single, double, and triple bonds

Memory Aids:
- Single bonds are longer than double bonds, which are longer than triple bonds.

Quick Tips:
- Consider the hybridization of the carbon atom when comparing C-H bond lengths.

70 Hard

70 of 80Question permalink

Analyzing +4 | -1

Among the following, the species with identical bond order are

View answer and explanation

Correct option: C

\mathrm{O}_2^{2-}

and

\mathrm{B}_2

Option C (

\mathrm{O}_2^{2-}

and

\mathrm{B}_2

) is correct because both species have a bond order of 1.

Solution approach

Correct Answer + Core Reason: $\mathrm{O}_2^{2-}$ and $\mathrm{B}_2$ . Both have a bond order of 1.

Step-by-Step Solution:

Step 1: Calculate the bond order for each species.

Step 2: $\mathrm{CO}$ : Bond order = (10 - 4)/2 = 3

Step 3: $\mathrm{O}_2^{2-}$ : Bond order = (10 - 8)/2 = 1

Step 4: $\mathrm{O}_2^{-}$ : Bond order = (10 - 7)/2 = 1.5

Step 5: $\mathrm{B}_2$ : Bond order = (6 - 4)/2 = 1

Step 6: $\mathrm{N}_2^{+}$ : Bond order = (9 - 4)/2 = 2.5

Step 7: Identify the pair with identical bond orders.

Why other options are wrong:
- Option A: $\mathrm{CO}$ has a bond order of 3, while $\mathrm{O}_2^{2-}$ has a bond order of 1.
- Option B: $\mathrm{O}_2^{-}$ has a bond order of 1.5, while $\mathrm{CO}$ has a bond order of 3.
- Option D: $\mathrm{CO}$ has a bond order of 3, while $\mathrm{N}_2^{+}$ has a bond order of 2.5.

Key Concepts:
- Bond Order Calculation

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

Quick Tips:
- Remember to consider the charge when calculating the number of electrons.

71 Hard

71 of 80Question permalink

Application +4 | -1

If

\mathrm{H} - X

bond length is

2.00 Å

and

\mathrm{H} - X

bond has dipole moment

5.12 \times {10^{ - 30}}\mathrm{C} - \mathrm{m}

, the percentage of ionic character in the molecule will be

View answer and explanation

Correct option: B

16\%

Option B is correct because the percentage of ionic character is calculated using the ratio of observed dipole moment to theoretical dipole moment.

Solution approach

Correct Answer + Core Reason: $16\%$ . Percentage ionic character is calculated by comparing the actual dipole moment with the theoretical dipole moment assuming 100% ionic character.

Step-by-Step Solution:

Step 1: Convert the bond length from Angstroms to meters: $2.00 \,\unicode{x212B} = 2.00 \times {10^{ - 10}}\,\mathrm{m}$

Step 2: Calculate the theoretical dipole moment assuming 100% ionic character: The theoretical dipole moment ( $\mu_{theoretical}$ ) is calculated as the product of the charge of an electron (e) and the bond length (d): $\mu_{theoretical} = e \times d$

Step 3: Substitute the values: $e = 1.602 \times {10^{ - 19}}\,\mathrm{C}$ and $d = 2.00 \times {10^{ - 10}}\,\mathrm{m}$
$\mu_{theoretical} = (1.602 \times {10^{ - 19}}\,\mathrm{C}) \times (2.00 \times {10^{ - 10}}\,\mathrm{m}) = 3.204 \times {10^{ - 29}}\,\mathrm{C} - \mathrm{m}$

Step 4: Calculate the percentage of ionic character: The percentage of ionic character is given by the formula:
$\text{{Percentage ionic character}} = \frac{{\mu_{observed}}}{{\mu_{theoretical}}} \times 100\%$

Step 5: Substitute the given observed dipole moment ( $\mu_{observed} = 5.12 \times {10^{ - 30}}\,\mathrm{C} - \mathrm{m}$ ) and the calculated theoretical dipole moment:
$\text{{Percentage ionic character}} = \frac{{5.12 \times {10^{ - 30}}\,\mathrm{C} - \mathrm{m}}}{{3.204 \times {10^{ - 29}}\,\mathrm{C} - \mathrm{m}}} \times 100\%$

Step 6: Calculate the percentage:
$\text{{Percentage ionic character}} = \frac{{5.12}}{{32.04}} \times 100\% \approx 16\%$

Why other options are wrong:
- Option A: $10\%$ is incorrect. It might result from an incorrect calculation or using wrong values.
- Option C: $18\%$ is incorrect. It might result from an incorrect calculation or using wrong values.
- Option D: $20\%$ is incorrect. It might result from an incorrect calculation or using wrong values.

Key Concepts:
- Dipole moment
- Percentage ionic character
- Theoretical dipole moment

Memory Aids:
- Percentage ionic character = (Observed dipole moment / Theoretical dipole moment) * 100

Quick Tips:
- Ensure correct unit conversions.
- Remember the formula for calculating percentage ionic character.

72 Hard

72 of 80Question permalink

Application +4 | -1

Total number of molecules/species from following which will be paramagnetic is ______.

\mathrm{O}_2,

\mathrm{O}_2^{+},

\mathrm{O}_2^{-},

\mathrm{NO},

\mathrm{NO}_2,

\mathrm{CO},

\mathrm{K}_2\left[\mathrm{NiCl}_4\right],

\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3,

\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]

View answer and explanation

Correct option: A

6

The correct answer is A (6) because the paramagnetic species are O2, O2+, O2-, NO, NO2, and [Co(NH3)6]Cl3.

Solution approach

Correct Answer + Core Reason: 6. Paramagnetic species have unpaired electrons.

Step-by-Step Solution:

Step 1: Identify the species and determine their electronic configurations or magnetic properties.

Step 2: $\mathrm{O}_2$ : Paramagnetic (2 unpaired electrons).

Step 3: $\mathrm{O}_2^+$ : Paramagnetic (1 unpaired electron).

Step 4: $\mathrm{O}_2^-$ : Paramagnetic (1 unpaired electron).

Step 5: $\mathrm{NO}$ : Paramagnetic (1 unpaired electron).

Step 6: $\mathrm{NO}_2$ : Paramagnetic (1 unpaired electron).

Step 7: $\mathrm{CO}$ : Diamagnetic (all electrons paired).

Step 8: $\mathrm{K}_2[\mathrm{NiCl}_4]$ : $\mathrm{Ni}^{2+}$ has a $d^8$ configuration. In a tetrahedral field, it is paramagnetic (2 unpaired electrons).

Step 9: $[\mathrm{Co}(\mathrm{NH}_3)_6]\mathrm{Cl}_3$ : $\mathrm{Co}^{3+}$ has a $d^6$ configuration. $\mathrm{NH}_3$ is a strong field ligand, causing pairing of electrons. However, since Co is +3, it can be either high spin or low spin. Since it is not specified, we assume it is paramagnetic (unpaired electrons).

Step 10: $\mathrm{K}_2[\mathrm{Ni}(\mathrm{CN})_4]$ : $\mathrm{Ni}^{2+}$ has a $d^8$ configuration. $\mathrm{CN}^-$ is a strong field ligand, causing pairing of electrons. Therefore, it is diamagnetic.

Step 11: Count the paramagnetic species: $\mathrm{O}_2, \mathrm{O}_2^+, \mathrm{O}_2^-, \mathrm{NO}, \mathrm{NO}_2, [\mathrm{Co}(\mathrm{NH}_3)_6]\mathrm{Cl}_3$ and $\mathrm{K}_2[\mathrm{NiCl}_4]$ . Total = 6.

Why other options are wrong:
- Option B: Incorrect count of paramagnetic species.
- Option C: Incorrect count of paramagnetic species.
- Option D: Incorrect count of paramagnetic species.

Key Concepts:
- Paramagnetism and Diamagnetism
- Molecular Orbital Theory
- Crystal Field Theory

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

Quick Tips:
- Determine the electronic configuration of each species.
- Identify the presence of unpaired electrons.

73 Hard

73 of 80Question permalink

Application +4 | -1

The number of unpaired electrons in a paramagnetic diatomic molecule of an element with atomic number 16 is:

View answer and explanation

Correct option: C

2

Option C (2) is correct because an element with atomic number 16 (Sulfur) forms

O_2

which has 2 unpaired electrons.

Solution approach

Correct Answer + Core Reason: 2. Diatomic molecule of element with atomic number 16 (Sulfur) is $S_2$ , which is paramagnetic with 2 unpaired electrons.

Step-by-Step Solution:

Step 1: Identify the element with atomic number 16.\nThe element is Sulfur (S).

Step 2: Consider the diatomic molecule formed by Sulfur, which is $S_2$ .

Step 3: Determine the electronic configuration of $S_2$ .
$S_2$ has 32 electrons. The electronic configuration is similar to $O_2$ but with more electrons. The relevant part is the filling of the $\pi^{\ast}$ orbitals, which will have 2 unpaired electrons.

Step 4: Determine the number of unpaired electrons.
$S_2$ has 2 unpaired electrons.

Why other options are wrong:
- Option A: 4 is incorrect. The diatomic molecule does not have 4 unpaired electrons.
- Option B: 1 is incorrect. The diatomic molecule does not have 1 unpaired electron.
- Option D: 3 is incorrect. The diatomic molecule does not have 3 unpaired electrons.

Key Concepts:
- Molecular Orbital Theory
- Paramagnetism and Unpaired Electrons

Memory Aids:
- Remember that $O_2$ and $S_2$ are paramagnetic with 2 unpaired electrons.

Quick Tips:
- Relate the electronic configuration to that of oxygen to determine unpaired electrons.

74 Hard

74 of 80Question permalink

Analyzing +4 | -1

The true statements from the following are
1.

\mathrm{P}{\mathrm{H}_5}

and

\mathrm{BiC}{\mathrm{l}_5}

do not exist
2.

p\pi - d\pi

bond is present in

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

3.Electrons travel at the speed of light
4.

\mathrm{Se}{\mathrm{F}_4}

and

\mathrm{C}{\mathrm{H}_4}

have same shape
5.

\mathrm{I}_3^ +

has bent geometry

View answer and explanation

Correct option: B

1,2,5

Option B (1,2,5) is correct because

PH_5

and

BiCl_5

do not exist due to the inert pair effect and the inability of P to expand its octet,

SO_2

has

p\pi - d\pi

bonding, and

I_3^+

has a bent geometry.

Solution approach

Correct Answer + Core Reason: 1,2,5. These statements are true based on chemical bonding principles.

Step-by-Step Solution:

Step 1: Analyze each statement.

Step 2: Statement 1: $PH_5$ and $BiCl_5$ do not exist. This is true. Phosphorus cannot expand its octet to form 5 bonds with hydrogen. Bismuth exhibits inert pair effect, making $BiCl_5$ unstable.

Step 3: Statement 2: $p\pi - d\pi$ bond is present in $SO_2$ . This is true. Sulfur can form $d\pi - p\pi$ bonds due to the presence of vacant d-orbitals.

Step 4: Statement 3: Electrons travel at the speed of light. This is false. Electrons move much slower than the speed of light.

Step 5: Statement 4: $SeF_4$ and $CH_4$ have same shape. This is false. $SeF_4$ has a see-saw shape, while $CH_4$ is tetrahedral.

Step 6: Statement 5: $I_3^+$ has bent geometry. This is true. $I_3^+$ has 2 bond pairs and 2 lone pairs, resulting in a bent geometry.

Step 7: Identify the true statements: 1, 2, and 5.

Why other options are wrong:
- Option A: Includes statement 3, which is false.
- Option C: Includes statement 3, which is false.
- Option D: Includes statement 4, which is false.

Key Concepts:
- Limitations of octet rule
- $p\pi - d\pi$ bonding
- Molecular geometry
- Inert pair effect

Memory Aids:
- Remember the shapes of common molecules.

Quick Tips:
- Consider the octet rule and the availability of d-orbitals.

75 Hard

75 of 80Question permalink

Understanding +4 | -1

Which has the minimum bond energy?

View answer and explanation

Correct option: C

I - I

Option C (

I - I

) is correct because the bond energy decreases as the size of the atom increases due to increased bond length and weaker attraction.

Solution approach

Correct Answer + Core Reason: $I - I$ . Bond energy decreases down the group due to increasing atomic size and bond length.

Step-by-Step Solution:

Step 1: Identify the trend of bond energy down the halogen group.

Step 2: Bond energy generally decreases as the size of the atom increases. This is because the bond length increases, leading to weaker attraction between the nuclei and the bonding electrons.

Step 3: Comparing the given options, $I-I$ has the largest atomic size and thus the weakest bond.

Step 4: $H-H$ has the smallest atoms and strongest bond. $H-Br$ and $H-I$ are intermediate, with $H-I$ being weaker than $H-Br$ .

Step 5: Therefore, $I-I$ has the minimum bond energy.

Why other options are wrong:
- Option A: $\mathrm{H} - \mathrm{Br}$ : This bond is stronger than $I-I$ due to the smaller size of Br.
- Option B: $\mathrm{H} - I$ : This bond is stronger than $I-I$ due to the smaller size of I compared to the $I-I$ bond.
- Option D: $\mathrm{H - H}$ : This bond is the strongest among the given options due to the small size of H and the short bond length.

Key Concepts:
- Bond energy
- Atomic size
- Bond length

Memory Aids:
- Down the group, bond energy decreases.

Quick Tips:
- Remember the trend of atomic size and bond length in the periodic table.

76 Hard

76 of 80Question permalink

Application +4 | -1

What is the formal charge on middle ‘

O

’ atom in ozone molecule?

View answer and explanation

Correct option: B

+1

Option B (+1) is correct because the formal charge on the central oxygen atom in ozone is +1.

Solution approach

Correct Answer + Core Reason: $+1$ . The formal charge is calculated based on the Lewis structure of ozone.

Step-by-Step Solution:

Step 1: Draw the Lewis structure of ozone ( $\mathrm{O}_3$ ). Ozone has two resonance structures: O=O-O <--> O-O=O. The central oxygen has one single bond and one double bond.

Step 2: Calculate the formal charge on the central oxygen atom. Formal charge = (Valence electrons) - (Non-bonding electrons) - (1/2 Bonding electrons) = 6 - 2 - (1/2 * 6) = 6 - 2 - 3 = +1.

Why other options are wrong:
- Option A: Zero is incorrect. The central oxygen has a formal charge of +1.
- Option C: -1 is incorrect. The central oxygen has a formal charge of +1.
- Option D: -2 is incorrect. The central oxygen has a formal charge of +1.

Key Concepts:
- Formal charge calculation
- Lewis structures
- Resonance structures

Memory Aids:
- Formal Charge = (Valence electrons) - (Non-bonding electrons) - (1/2 Bonding electrons)

Quick Tips:
- Draw the Lewis structure correctly before calculating formal charges.

77 Hard

77 of 80Question permalink

Application +4 | -1

The electronegativity of

{H}

and

{Cl}

are 2.1 and 3.0 respectively. The correct statement(s) about the nature of

{HCl}

is/are:

View answer and explanation

Correct option: A

{HCl}

is

17 \%

ionic

Option A (

{HCl}

is

17 \%

ionic) is correct because the percentage ionic character can be estimated using the electronegativity difference between H and Cl.

Solution approach

Correct Answer + Core Reason: ${HCl}$ is $17 \%$ ionic. The percentage ionic character can be estimated from the electronegativity difference using empirical relationships.

Step-by-Step Solution:

Step 1: Calculate the electronegativity difference: $|X_{Cl} - X_H| = |3.0 - 2.1| = 0.9$

Step 2: Use the formula to estimate the percentage ionic character:\nPercentage ionic character $= 16|X_A - X_B| + 3.5|X_A - X_B|^2$ \nPercentage ionic character $= 16(0.9) + 3.5(0.9)^2 = 14.4 + 3.5(0.81) = 14.4 + 2.835 = 17.235 \% \approx 17 \%$

Why other options are wrong:
- Option B: $83 \%$ ionic: This is incorrect; the percentage ionic character is closer to 17%.
- Option C: $50 \%$ : This is incorrect; the percentage ionic character is closer to 17%.
- Option D: $100 \%$ : This is incorrect; HCl is a polar covalent compound with partial ionic character.

Key Concepts:
- Electronegativity
- Electronegativity difference
- Percentage ionic character

Memory Aids:
- The larger the electronegativity difference, the more ionic the bond.

Quick Tips:
- Use the appropriate formula to estimate the percentage ionic character from the electronegativity difference.

78 Hard

78 of 80Question permalink

Understanding +4 | -1

Which of the following geometry is not possible when the central atom is having

s p^3 d

-hybridization?

View answer and explanation

Correct option: B

Trigonal planar

Option B (Trigonal planar) is correct because

sp^3d

hybridization typically leads to geometries such as trigonal bipyramidal, linear, T-shaped, or seesaw, but not trigonal planar.

Solution approach

Correct Answer + Core Reason: Trigonal planar. $sp^3d$ hybridization does not result in a trigonal planar geometry.

Step-by-Step Solution:

Step 1: Recall the geometries associated with $sp^3d$ hybridization.

Step 2: $sp^3d$ hybridization involves five hybrid orbitals, which can lead to the following geometries:
- Trigonal bipyramidal (TBP)
- Linear (with 3 lone pairs)
- T-shaped (with 2 lone pairs)

Step 3: Trigonal planar geometry is associated with $sp^2$ hybridization, which involves three hybrid orbitals.

Why other options are wrong:
- Option A: TBP is a possible geometry for $sp^3d$ hybridization.
- Option C: Linear is a possible geometry for $sp^3d$ hybridization (with 3 lone pairs).
- Option D: T-shaped is a possible geometry for $sp^3d$ hybridization (with 2 lone pairs).

Key Concepts:
- Hybridization
- Molecular Geometry

Memory Aids:
- Remember the geometries associated with each hybridization.

Quick Tips:
- Know the relationship between hybridization and geometry.

79 Hard

79 of 80Question permalink

Understanding +4 | -1

Choose the correct sequence of

{T}

and

{F}

for following statements. Here

{T}

stands for true statement and

{F}

stands for false statement.

(i) Formal charge in the Lewis structure helps in keeping track of the valence electrons in the molecule.

(ii) Formal charge indicates the actual charge separation within the molecule.

(iii) Formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures.

View answer and explanation

Correct option: B

{T} F {T}

Option B (T F T) is correct because formal charge helps track valence electrons and select the lowest energy structure, but it doesn't represent actual charge separation.

Solution approach

Correct Answer + Core Reason: ${T} F {T}$ . Formal charge helps track valence electrons and select the lowest energy structure, but it doesn't represent actual charge separation.

Step-by-Step Solution:

Step 1: Analyze statement (i): Formal charge in the Lewis structure helps in keeping track of the valence electrons in the molecule. This is true.

Step 2: Analyze statement (ii): Formal charge indicates the actual charge separation within the molecule. This is false. Formal charge is a theoretical concept and does not represent the actual charge.

Step 3: Analyze statement (iii): Formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures. This is true. The structure with the smallest formal charges is usually the most stable.

Step 4: Determine the correct sequence: True, False, True.

Why other options are wrong:
- Option A: ${T} T F$ : Statement (ii) is false.
- Option C: ${T} {T} {T}$ : Statement (ii) is false.
- Option D: $F T T$ : Statement (i) is true.

Key Concepts:
- Formal charge
- Lewis structures
- Valence electrons
- Resonance structures

Memory Aids:
- Formal Charge = (Valence electrons) - (Non-bonding electrons) - (1/2 Bonding electrons).

Quick Tips:
- Remember that formal charge is a theoretical concept and does not represent the actual charge on an atom.

80 Hard

80 of 80Question permalink

Application +4 | -1

In

\mathrm{PO}_4^{3 - }

, the formal charge on each on each oxygen atom and the

\mathrm{P} - \mathrm{O}

bond order respectively are

View answer and explanation

Correct option: C

- 0.75,1.25

Option C (-0.75, 1.25) is correct because the formal charge on each oxygen is -0.75 and the P-O bond order is 1.25.

Solution approach

Correct Answer + Core Reason: $- 0.75,1.25$ . The formal charge and bond order are calculated based on the resonance structures of $\mathrm{PO}_4^{3 - }$ .

Step-by-Step Solution:

Step 1: Draw the Lewis structure of $\mathrm{PO}_4^{3 - }$ .

Step 2: Calculate the formal charge on each oxygen atom. There are 4 oxygen atoms and a total charge of -3. So, the formal charge on each oxygen = -3/4 = -0.75.

Step 3: Determine the number of resonance structures. Phosphorus can form a double bond with one of the oxygen atoms, and this double bond can be in four different positions (with each of the four oxygen atoms). Therefore, there are four resonance structures.

Step 4: Calculate the bond order. In one resonance structure, there is one P=O double bond and three P-O single bonds. The total number of bonds is 5 (1 double + 3 single), and these bonds are distributed among 4 P-O linkages. Therefore, the bond order = 5/4 = 1.25.

Why other options are wrong:
- Option A: Incorrect formal charge and bond order.
- Option B: Incorrect formal charge and bond order.
- Option D: Incorrect formal charge. The formal charge is not -3 on each oxygen.

Key Concepts:
- Formal charge calculation
- Bond order calculation
- Resonance structures

Memory Aids:
- Formal Charge = (Valence electrons) - (Non-bonding electrons) - (1/2 Bonding electrons)
- Bond Order = (Total number of bonds) / (Number of bond locations)

Quick Tips:
- Remember to consider resonance when calculating bond order.

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