04 रासायनिक बंधन और आणविक संरचना
अभ्यास
Question:
How is the bond strength related to the bond order?
Answer:
Step 1: Bond strength is a measure of the energy required to break a chemical bond.
Step 2: Bond order is the number of chemical bonds between two atoms in a molecule.
Step 3: The bond strength is directly proportional to the bond order, meaning that the higher the bond order, the stronger the bond. Therefore, the greater the number of bonds between two atoms, the greater the bond strength.
Question:
What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals :
Answer:
Answer: Hybridisation of atomic orbitals is a process in which two or more atomic orbitals combine to form a hybrid orbital, which is a combination of the original atomic orbitals.
The shapes of sp, sp2, sp3 hybrid orbitals are as follows:
sp Hybrid Orbital: The sp hybrid orbital is a combination of one s-orbital and one p-orbital, and has a linear shape.
sp2 Hybrid Orbital: The sp2 hybrid orbital is a combination of one s-orbital and two p-orbitals, and has a trigonal planar shape.
sp3 Hybrid Orbital: The sp3 hybrid orbital is a combination of one s-orbital and three p-orbitals, and has a tetrahedral shape.
Question:
Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C2H4 and C2H2 molecules.
Answer:
Double Bond in C2H4:
The formation of a double bond between two carbon atoms in a C2H4 molecule can be illustrated as follows:
Step 1: Two carbon atoms are connected by a single covalent bond.
Step 2: Each carbon atom shares two of its unpaired electrons with the other carbon atom, forming a double bond between them.
Step 3: The two hydrogen atoms are attached to the carbon atoms, completing the C2H4 molecule.
Triple Bond in C2H2:
The formation of a triple bond between two carbon atoms in a C2H2 molecule can be illustrated as follows:
Step 1: Two carbon atoms are connected by a single covalent bond.
Step 2: Each carbon atom shares three of its unpaired electrons with the other carbon atom, forming a triple bond between them.
Step 3: The two hydrogen atoms are attached to the carbon atoms, completing the C2H2 molecule.
Question:
Which hybrid orbitals are used by carbon atoms in the following molecules?
(a)CH3−CH3
(b)CH3−CH=CH2
(c)CH3−CH2−OH
(d)CH3−CHO
(e)CH3COOH
Answer:
(a) Carbon atoms in CH3−CH3 use sp3 hybrid orbitals.
(b) Carbon atoms in CH3−CH=CH2 use sp2 hybrid orbitals.
(c) Carbon atoms in CH3−CH2−OH use sp3 hybrid orbitals.
(d) Carbon atoms in CH3−CHO use sp2 hybrid orbitals.
(e) Carbon atoms in CH3COOH use sp3 hybrid orbitals.
Question:
What is meant by the term bond order ? Calculate the bond order of N2, O2 , O2+ and O2-.
Answer:
Answer:
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Bond order is the number of chemical bonds between two atoms in a molecule. It is calculated by subtracting the number of electrons in the antibonding molecular orbitals from the number of electrons in the bonding molecular orbitals.
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The bond order of N2 is 3, since it has six electrons in the bonding molecular orbitals and three electrons in the antibonding molecular orbitals.
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The bond order of O2 is 2, since it has four electrons in the bonding molecular orbitals and two electrons in the antibonding molecular orbitals.
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The bond order of O2+ is 1, since it has three electrons in the bonding molecular orbitals and two electrons in the antibonding molecular orbitals.
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The bond order of O2- is 2, since it has four electrons in the bonding molecular orbitals and two electrons in the antibonding molecular orbitals.
Question:
Compare the relative stability of the following species and indicate their magnetic properties : O2, O2+, O2- (superoxide) and O22- (peroxide).
Answer:
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O2 is a paramagnetic species, meaning it is attracted to a magnetic field.
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O2+ is a diamagnetic species, meaning it is repelled by a magnetic field.
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O2- (superoxide) is a paramagnetic species, meaning it is attracted to a magnetic field.
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O22- (peroxide) is a paramagnetic species, meaning it is attracted to a magnetic field.
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In terms of relative stability, O2 is the most stable, followed by O2- (superoxide), O2+ and then O22- (peroxide).
Question:
Use Lewis symbols to show electron transfer between the following atoms to form cations and anions: (a) K and S (b) Ca and O (c) Al and N.
Answer:
(a) K and S: Lewis symbol for K: [K] Lewis symbol for S: [S]
Electron transfer: [K] → [K]+ + e- [S] + e- → [S]2-
Resulting cation and anion: K+ and S2-
(b) Ca and O: Lewis symbol for Ca: [Ca] Lewis symbol for O: [O]
Electron transfer: [Ca] → [Ca]2+ + 2e- [O] + 2e- → [O]2-
Resulting cation and anion: Ca2+ and O2-
(c) Al and N: Lewis symbol for Al: [Al] Lewis symbol for N: [N]
Electron transfer: [Al] → [Al]3+ + 3e- [N] + 3e- → [N]3-
Resulting cation and anion: Al³⁺ and N³⁻
Question:
Distinguish between a sigma and a pi bond.
Answer:
Step 1: Understand what a sigma and pi bond are.
Step 2: A sigma bond is a type of covalent bond formed when two atoms share a single pair of electrons.
Step 3: A pi bond is a type of covalent bond formed when two atoms share two pairs of electrons.
Step 4: Compare the differences between the two types of bonds. A sigma bond is stronger than a pi bond and is the primary bond in a molecule. A pi bond is weaker than a sigma bond and is responsible for the shape of a molecule.
Question:
Define hydrogen bond. Is it weaker or stronger than the Van der Waals forces?
Answer:
Answer: A hydrogen bond is a type of attractive interaction between two molecules or atoms that occurs when a hydrogen atom is covalently bonded to an electronegative atom, such as nitrogen, oxygen, or fluorine. Hydrogen bonds are typically weaker than Van der Waals forces, which are attractive interactions between molecules or atoms that occur due to temporary dipole-dipole interactions.
Question:
Write Lewis dot symbols for atoms of the following elements : Mg, Na, B, O, N and Br.
Answer:
Mg: [Ne] 3s2 Na: [Ne] 3s1 B: [He] 2s2 2p1 O: [He] 22 2p4 N: [He] 2s2 2p3 Br: [Ar] 3d10 4s2 4p5
Question:
Define electronegativity? How does it differ from electron gain enthalpy ?
Answer:
Answer:
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Electronegativity is the measure of an atom’s ability to attract electrons when chemically combined with another atom.
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Electron gain enthalpy is the energy released when an electron is added to a neutral atom or molecule. It is the energy released when an electron is added to a neutral atom or molecule, while electronegativity is the measure of the ability of an atom to attract electrons.
Question:
What are the total number of σ and π bonds in the following molecules? (a) C2H2 (b) C2H4
Answer:
(a) C2H2 Number of σ bonds = 2 Number of π bonds = 0
Total number of σ and π bonds = 2
(b) C2H4 Number of σ bonds = 4 Number of π bonds = 0
Total number of σ and π bonds = 4
Question:
Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds?
Answer:
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In the case of PCl5, the hybridisation is sp3d.
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This means that the central phosphorus atom is hybridised with one s-orbital and three p-orbitals, forming four sp3 hybrid orbitals.
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These four hybrid orbitals are then used to form five sigma bonds with the five chlorine atoms.
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Out of the five sigma bonds, three are arranged in an equatorial plane and two are arranged in an axial plane.
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The axial bonds are longer than the equatorial bonds because the axial bonds are formed at a larger distance from the nucleus of the phosphorus atom as compared to the equatorial bonds. This is because the axial bonds are formed at a larger angle with respect to the nucleus of the phosphorus atom.
Question:
Define hydrogen bond. Is it weaker or stronger than the Van der Waals forces?
Answer:
Answer:
Definition of Hydrogen Bond: A hydrogen bond is a type of attractive force that occurs when a hydrogen atom, covalently bound to a strongly electronegative atom, such as nitrogen, oxygen, or fluorine, experiences the electrostatic attraction of another nearby strongly electronegative atom of the same type.
Comparison to Van der Waals Forces: Hydrogen bonds are typically stronger than Van der Waals forces, which are weaker attractions between molecules caused by temporary dipole-dipole interactions.
Question:
Explain why BeH2 molecule has a zero dipole moment although the Be−H bonds are polar.
Answer:
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A dipole moment is a measure of the separation of charge within a molecule.
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In the BeH2 molecule, the two hydrogen atoms are symmetrically arranged around the beryllium atom, creating a linear molecule.
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The two polar Be−H bonds cancel each other out, resulting in a zero dipole moment.
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This is because the two polar molecules are of equal magnitude, and therefore the positive and negative charges cancel each other out, resulting in a net zero dipole moment.
Question:
What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.
Answer:
Answer: Bond Pairs: Bond pairs are pairs of electrons that are shared between two atoms in a covalent bond. An example of a bond pair is the pair of electrons shared between two hydrogen atoms in a hydrogen molecule (H2).
Lone Pairs: Lone pairs are pairs of electrons that are not shared between two atoms, but rather are localized to a single atom. An example of a lone pair is the pair of electrons localized to the oxygen atom in a water molecule (H2O).
Question:
Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss :
Answer:
Step 1: NH3 and H2O molecules both have distorted tetrahedral geometries, with four atoms surrounding a central atom.
Step 2: The bond angle in water is less than that of ammonia. This is because the oxygen atom in water is more electronegative than the nitrogen atom in ammonia. This causes the electrons in the hydrogen-oxygen bond to be pulled closer to the oxygen atom, resulting in a slightly smaller bond angle.
Step 3: The difference in bond angles between NH3 and H2O is due to the difference in electronegativity between the two atoms. The higher the electronegativity of the central atom, the smaller the bond angle.
Question:
Define bond length.
Answer:
- Bond length is the distance between two atoms that are connected by a chemical bond.
- Bond length is usually measured in picometers (pm) or angstroms (Å).
- Bond length is an important factor in determining the strength of a chemical bond.
Question:
Write the significance of a plus and a minus sign shown in representing the orbitals.
Answer:
Answer:
Plus (+) sign: The plus sign indicates that the orbital has a positive spin, meaning that the electron located in the orbital is spinning in a clockwise direction.
Minus (-) sign: The minus sign indicates that the orbital has a negative spin, meaning that the electron located in the orbital is spinning in a counterclockwise direction.
Question:
Arrange the bonds in order of increasing ionic character in the molecules: LiF,K2O,N2,SO2 and ClF3
Answer:
- N2 (Non-polar covalent bond)
- LiF (Ionic bond)
- K2O (Ionic bond)
- SO2 (Polar covalent bond)
- ClF3 (Polar covalent bond)
Question:
Apart from tetrahedral geometry, another possible geometry of CH4 could be square planar with the four H atoms at the corners of the square and the C atom at its centre. Why CH4 is not square planar?
Answer:
Answer:
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CH4 has a tetrahedral geometry because the four C-H bonds are arranged at the four corners of a tetrahedron, with the C atom at the centre.
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The square planar geometry of CH4 is not possible because the four C-H bonds cannot be arranged in a plane.
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This is because the four C-H bonds are arranged at the four corners of a tetrahedron, which is a three-dimensional structure.
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Moreover, the C-H bonds are not equal in length, which means that the four C-H bonds cannot be arranged in a plane.
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Therefore, CH4 cannot adopt a square planar geometry and is limited to a tetrahedral geometry.
Question:
Which out of NH3 and NF3 has higher dipole moment and why?
Answer:
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Dipole moment is the measure of the separation between the positive and negative charges in a molecule.
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NH3 has a dipole moment of 1.47 D, while NF3 has a dipole moment of 0.6 D.
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Therefore, NH3 has a higher dipole moment than NF3.
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This is because NH3 has three hydrogen atoms, which are all slightly positively charged, and one nitrogen atom, which is slightly negatively charged. This creates a higher separation between the positive and negative charges than NF3, which has three fluorine atoms, which are all slightly negatively charged, and one nitrogen atom, which is slightly positively charged.
Question:
Describe the change in hybridisation (if any) of the Al atom in the following reaction: AlCl3+Cl-→AlCl-4.
Answer:
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The hybridization of the Al atom in AlCl3 is sp3.
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The hybridization of the Al atom in AlCl-4 is sp3d.
Question:
Considering x-axis as the internuclear axis, which out of the following will not form a sigma bond. A : 1s and 1s B : 1s and 2px C : 2py and 2py D : 1s and 2s
Answer:
A : 1s and 1s - This will not form a sigma bond because the orbitals must be of different types (e.g. 1s and 2s) in order to form a sigma bond.
Question:
Explain the formation of H2 molecule on the basis of valence bond theory.
Answer:
Step 1: Valence bond theory states that the formation of a chemical bond between two atoms is the result of the overlapping of two atomic orbitals.
Step 2: Hydrogen atoms have a single electron in their outermost shell. This electron is located in an s orbital, which is a spherical orbital.
Step 3: When two hydrogen atoms come close to each other, their s orbitals overlap. This creates a covalent bond between the two atoms, forming a hydrogen molecule (H2).
Step 4: The electron from each hydrogen atom is shared between the two atoms, forming a strong bond. This bond is known as a sigma bond, and it is the strongest type of covalent bond.
Question:
Write Lewis symbols for the following atoms and ions : (i) S and S2- (ii) Al and Al+3 (iii) H and H-
Answer:
(i) S: \overset{}{\circ}\overset{}{\bullet}\overset{}{\circ} S2-: \overset{-}{\circ}\overset{}{\bullet}\overset{}{\circ}\overset{-}{\circ}
(ii) Al: \overset{}{\circ}\overset{}{\bullet}\overset{}{\bullet}\overset{}{\circ}\overset{}{\bullet}\overset{}{\circ} Al+3: \overset{+3}{\circ}\overset{}{\bullet}\overset{}{\bullet}\overset{}{\circ}\overset{}{\bullet}\overset{}{\circ}
(iii) H: \overset{}{\circ}\overset{}{\bullet}\overset{}{\circ} H-: \overset{-}{\circ}\overset{}{\bullet}\overset{}{\circ}\overset{-}{\circ}
Question:
Draw the Lewis structures for the following molecules and ions : H2S, SiCl4, BeF2, CO32- and HCOOH
Answer:
H2S:
H-S-H
SiCl4:
Si-Cl-Cl-Cl-Cl
BeF2:
Be-F-F
CO32-:
C-O-O-O-
HCOOH:
H-C-O-O-H
Question:
Define octet rule and write its significance and limitations.
Answer:
Definition of Octet Rule: The octet rule states that atoms tend to combine in such a way that each atom has eight electrons in its outermost shell. This is also known as the Lewis structure of an atom.
Significance of Octet Rule: The octet rule is a useful tool for predicting the number of bonds that an atom will form with other atoms. This rule is used to explain the formation of covalent bonds between atoms, which are the most common type of chemical bonds found in molecules.
Limitations of Octet Rule: The octet rule does not always accurately predict the number of bonds formed between atoms. For example, atoms that have more than eight electrons in their outermost shell can form more than one bond with other atoms. Additionally, the octet rule does not take into account the effects of resonance, which can affect the number of bonds formed between atoms.
Question:
What is meant by the term bond order? Calculate the bond order of: N2,O2,O2+ and O2−?
Answer:
Answer:
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Bond order is a measure of the number of chemical bonds between two atoms in a molecule. It is calculated by subtracting the number of bonding electrons from the number of antibonding electrons.
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The bond order of N2 is 3, since there are 6 bonding electrons and no antibonding electrons.
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The bond order of O2 is 2, since there are 4 bonding electrons and 2 antibonding electrons.
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The bond order of O2+ is 1, since there are 2 bonding electrons and 1 antibonding electron.
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The bond order of O2- is 3, since there are 6 bonding electrons and no antibonding electrons.
Question:
Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals :
Answer:
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The wavefunctions of the atomic orbitals should be linearly combined in order to form the wavefunction of the molecular orbitals.
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The atomic orbitals should have the same symmetry as the molecular orbitals.
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The atomic orbitals should be of the same energy level.
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The coefficients of the linear combination should be determined by solving the Schrödinger equation.
Question:
Write the favourable factors for the formation of ionic bond :
Answer:
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Formation of ionic bonds requires an atom with a large difference in electronegativity between the two atoms.
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Ionic bonds are formed when one atom donates an electron to another atom.
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The atoms that form ionic bonds must have different charges, with one atom being positively charged and the other atom being negatively charged.
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The atoms must be held together by electrostatic attraction.
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Ionic bonds are usually formed between a metal and a nonmetal.
Question:
Determine the shape of the following molecules using the VSEPR model. BeCl2, BCl3, SiCl4, AsF5, H2S and PH3
Answer:
BeCl2: Linear BCl3: Trigonal Planar SiCl4: Tetrahedral AsF5: Trigonal Bipyramidal H2S: Bent PH3: Trigonal Pyramidal
Question:
Explain the important aspects of resonance with reference to the CO32- ion.
Answer:
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Resonance is a concept in chemistry that describes the delocalization of electrons within molecules or polyatomic ions.
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In the case of the CO32− ion, it is composed of two oxygen atoms and one carbon atom, which are bound together by covalent bonds.
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The electrons in the covalent bonds can move freely between the three atoms, creating a resonance structure.
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This means that the electrons are not localized to one particular atom, but instead spread out over the entire molecule.
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This delocalization of electrons gives the molecule greater stability and makes it more resistant to chemical reactions.
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Resonance also affects the bond length and bond strength of the molecule. The resonance structure of CO32− has a longer bond length and weaker bond strength than a single covalent bond.
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This is because the electrons are spread out over the entire molecule, making them less likely to form strong bonds with any one atom.
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Finally, resonance affects the molecular geometry of the molecule. The resonance structure of CO32− has a bent molecular geometry, which is different from the linear geometry of a single covalent bond.
Question:
Although both CO2 and H2O are triatomic molecules, the shape of H2O molecules is bent while that of CO2 is linear. Explain this on the basis of dipole moment of the significance/applications of dipole moment.
Answer:
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Dipole moment is a measure of the separation of charge within a molecule, which is determined by the electronegativity of the atoms in the molecule.
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The electronegativities of the atoms in CO2 are equal, thus the molecule has no net dipole moment and the shape of the molecule is linear.
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On the other hand, the electronegativities of the atoms in H2O are not equal, thus the molecule has a net dipole moment and the shape of the molecule is bent.
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The significance of dipole moment is that it determines the polarity of a molecule, which affects the types of interactions the molecule can form with other molecules. For example, molecules with a net dipole moment can form hydrogen bonds with other molecules, which can influence the physical properties of the molecules, such as boiling point and solubility.
Question:
Write the significance/application of dipole moment.
Answer:
Answer:
- Dipole moment is a measure of the separation of positive and negative charges within a molecule and the magnitude of the charge.
- It is an important parameter in determining the polarity of a molecule and its ability to interact with other molecules.
- Dipole moments are used to describe the structure and reactivity of molecules and to predict the properties of molecular solids.
- They can also be used to calculate the strength of intermolecular forces, such as hydrogen bonding, dipole-dipole interactions, and van der Waals forces.
- Dipole moments are important in the fields of chemistry, physics, and materials science, as they can be used to determine the structure and properties of molecules and materials.
Question:
Explain with the help of a suitable example polar covalent bond.
Answer:
Polar covalent bonds are chemical bonds that form between two atoms of different electronegativity. This means that the shared electrons in the bond are unequally distributed, creating an electric dipole moment. For example, when oxygen and hydrogen come together to form a water molecule, the oxygen atom has a higher electronegativity than the hydrogen atom, so the electrons are pulled closer to the oxygen atom. This creates a partial negative charge on the oxygen atom and a partial positive charge on the hydrogen atom, forming a polar covalent bond.
Question:
Is there any changes in the hybridisation of B and N atoms as a result of the following reaction.
Answer:
- B2H6 + N2O5 –> B(NO3)3 + H2O
Answer: Yes, there is a change in the hybridisation of B and N atoms as a result of the reaction. The B atom in B2H6 is sp3 hybridised, while the N atom in N2O5 is sp2 hybridised. After the reaction, the B atom in B(NO3)3 is sp2 hybridised and the N atom in B(NO3)3 is sp3 hybridised.
Question:
BF3+NH3→F3B−NH3?
Answer:
-
Balance the equation: BF3 + 3NH3 → F3B−NH3
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Determine the type of reaction: This is a double displacement reaction.
Question:
Give any two differences between σ and π bond.
Answer:
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A σ bond is a single covalent bond formed by the overlap of two atomic orbitals, while a π bond is a double covalent bond formed by the overlap of two parallel atomic orbitals.
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The electrons involved in a σ bond are located in the space between the nuclei of the two atoms, while the electrons involved in a π bond are located outside the internuclear axis, above and below the nuclei.
Question:
Use molecular orbital theory to explain why Be2 molecule does not exist.
Answer:
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Molecular orbital theory is based on the idea that electrons in a molecule are distributed among the atomic orbitals of the component atoms to form molecular orbitals.
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In Be2 molecule, the two Be atoms have the same number of valence electrons, so the molecular orbitals of the Be2 molecule would be symmetrical.
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This means that the electrons would be equally distributed among the two atoms, resulting in a zero net electric dipole moment.
-
Since the molecule has no net dipole moment, it cannot interact with other molecules and, therefore, does not exist.
जेईई अध्ययन सामग्री (रसायन विज्ञान)
01 रसायन विज्ञान की कुछ बुनियादी अवधारणाएँ
02 परमाणु की संरचना
03 तत्वों का वर्गीकरण और गुणों में आवधिकता
04 रासायनिक बंधन और आणविक संरचना
पदार्थ गैसों और तरल पदार्थों की 05 अवस्थाएँ
06 ऊष्मप्रवैगिकी
07 संतुलन
08 रेडॉक्स प्रतिक्रियाएं
09 हाइड्रोजन
10 एस ब्लॉक तत्व
11 कुछ पी ब्लॉक तत्व
12 कार्बनिक रसायन कुछ बुनियादी सिद्धांत और तकनीकें
13 हाइड्रोकार्बन
14 पर्यावरण