Test: Bond Parameters (May 23) - JEE MCQ

Concept:
Ionization energy - The minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom is called the ionization energy.

• It is endothermic in nature.
• It is measured in unit KJmol^-1.

First ionization energy - Energy required by an atom to form an M+ ion is called first ionization energy
M_(g) → M⁺_(g)  + e^- ; ΔH_1st 
Explanation:
The factor affecting the value of ionization energy are-

1. Nuclear charge
2. Atomic size
3. Penetration effect
4. Screening effect
5. Arrangement of electrons in an orbital (stable configuration or not)

→ Depending upon these factors, the value of ionization energy decreases down a group and is increasing with some exceptions in a period.
In the given configurations, [Ne]3s²3p³ has the highest value of ionization energy because-

• In a period, the ionization energy value increases.
• [Ne]3s²3p³ is a half-filled configuration which is a stable configuration and it becomes difficult to remove electrons from a stable configuration. 

Conclusion:
Thus, [Ne]3s²3p³ is having the highest value of ionization energy among the others.
Hence, the correct answer is option 2.
Additional Information
Electronic Configuration - 
The distribution of the electrons in the orbitals of an element is described by the electronic configuration.

• It is a notation in which the arrangement of electrons of an element is shown.
• It is useful for predicting the properties of an element or group of elements.
• It is used as the identity of an element.

• The shape of a molecule is determined by the electron pairs in the valence shell of the central atom—both those involved in bonding and those not involved in bonding (i.e., lone pairs).
• Valence Shell Electron Pair Repulsion (VSEPR) Theory explains the shape of a molecule. According to the VSEPR theory, electron pairs in the valence shell of any atom repel each other and try to stay as far apart as possible.
• The VSEPR theory classifies molecules based on the number of bonding groups and lone pairs. For example, if there are 3 bonding groups and no lone pair on the central atom, it will be trigonal planar (examples include BF₃, SO₃, etc.)

The shape of a molecule is not determined by the other options given:

• The adjacent atom: While the adjacent atoms do participate in bonds, the overall shape of a molecule is more due to how the valence electrons distribute themselves to minimize repulsion, rather than which atoms are present.
• The surroundings: The immediate surroundings or the external environment doesn't impact the shape of a molecule. The molecular shape is a result of the intrinsic characteristics of the atoms in the molecule.
• The atmosphere: The atmospheric conditions may affect other properties of a molecule, but they do not influence the inherent shape of a molecule.
   

However, polatomic ions or molecules having multiple centres do not always follow the VSEPR theory as the total charge distribution and symmetry play a role in determining shape.
In summary, the shape of a molecule depends fundamentally on the electron configuration of the central atom, specifically the distribution of valence electrons in the 3D space to minimize the electron-electron repulsion. This is the key idea of the VSEPR theory, which helps us predict the shape of a molecule.
Conclusion:-
So, The shape of the molecule depends on the valence electrons.

Concept:-

• Isoelectronic Definition: Understanding that isoelectronic molecules have the same number of electrons, which is crucial in identifying the correct structure.
• Aromaticity: Recognizing the aromatic nature of naphthalene, which has two fused benzene rings, and applying this concept to select a molecule with a similar structure.
• Heterocyclic Rings: Identifying the presence of heterocyclic rings (imidazole) in benzimidazole and understanding their role in maintaining isoelectronicity.

Explanation:-

• Naphthalene is a polycyclic aromatic hydrocarbon with two fused benzene rings. A molecule is isoelectronic with another when it has the same number of electrons. For naphthalene (C₁₀H₈), a molecule with the same number of electrons would be isoelectronic.
• 
• Among the given options, Benzimidazole is isoelectronic with naphthalene. Benzimidazole has a fused benzene ring and an imidazole ring, providing the same number of electrons as naphthalene. 
• 
• Both naphthalene and benzimidazole have a fused benzene ring, contributing to their isoelectronic nature. Benzimidazole has additional heteroatoms (nitrogen) in the imidazole ring, maintaining the overall electron count.

Conclusion:-
So, Benzimidazole is isoelectronic with naphthalene, sharing a similar structure with a fused benzene ring and additional heterocyclic rings.

Concept:
 According to Fajan's rule:

• The smaller the cation, the bigger the anion, and the more the covalent character.
• Bigger cations and smaller anions will favor ionic character.

Explanation:
In the given options all have the same anion (Cl^-), the character of the compounds decided by cation only.
Na, K, Li, and Cs belong to I group and their order of atomic size is:
Li < Na < K < Cs.
According to Fajan's rule: Bigger cations and smaller anions will favor ionic character.
So, CsCl has maximum ionic character.

Isoelectronic species :

• Those species having the same number of electrons but differ in nuclear charge are called isoelectronic species.
• Example: Mg2+, Na+, F–, and O2–  have 10 electrons each so they are isoelectronic.
• But, their radius is different due to differences in nuclear charges. 
• Ionic radius decreases along a period for Isoelectronic ions.

​Isostructural species:

• The species which have different atoms or elements in them but have the same hybridization and structure are called isostructural species.
• Examples are NF3 and NH3, they have both sp3 hybridization and pyramidal shape.

Explanation:

• The species and their structures and number of electrons are given below:
  
• From the table above we see that CH₃^-, NH₃, H₃O⁺ all have 10 electrons in them.
• Hence, they are isoelectronic species.

Thus the correct option is II, III, IV.

The geometry of a molecule:

• The geometry of a molecule depends on the arrangement of bonds about its center in space.
• The arrangement further depends on the type of hybridization the center atom is undergoing.
• The orientation of the hybrid orbitals is different in different cases.
• As bonds are formed via overlap of these orbitals, the bonds have directional nature.
• Therefore, hybridization is directly linked to the geometry of the molecule.

Hybridization and bond angles:

• According to VSEPR theory, the electron groups arrange themselves around each other so as to minimize repulsion.
• The electron group includes the bond pairs as well as lone pairs of electrons.
• If repulsion is more, the energy of the system is raised and the molecule becomes unstable.
• So, the arrangement in which there are minimum repulsion and maximum attraction is the most stable structure.
• The arrangement in space gives some angles between the central atom and the bonded atoms which are known as the bond angles.
• There exist anomalies in predicted bond angles when there is the presence of lone pair electrons.

Explanation:
H₂O:

• In water, the center atom oxygen is sp³ hybridized.
• Out of the four bond pairs, two are bond pairs and the other two are lone pairs.
• Due to the presence of lone electrons, there is repulsion between bond pairs and lone pairs.
• The repulsion between l.p is > b.p and thus pushes the two bonds to come closer decreasing the bond angle from ideal 109.8°.

H2S:

• In water, the center atom oxygen is sp³ hybridized.
• Out of the three bond pairs, two are bond pairs and the other two are lone pairs.
• Due to the presence of lone electrons, there is repulsion between bond pairs and lone pairs.
• The repulsion between l.p is > b.p and thus pushes the two bonds to come closer decreasing the bond angle from ideal 109.8°.
• As sulfur is a larger molecule than oxygen, its electron density is more dispersed. It is more easily distorted by the repulsion from the lone pair of electrons and the bond pairs come closer to each other.
• This decreases the bond angle more than in water.
• The bond angle in H₂S is 92.1⁰.

NH3:

• In ammonia, the center atom oxygen is sp³ hybridized.
• Out of the three bond pairs, three are bond pairs and the other is lone pair.
• Due to the presence of lone electrons, there is repulsion between bond pairs and lone pairs.
• The repulsion between l.p > b.p and thus pushes the two bonds to come closer decreasing the bond angle from ideal 109.8° to 107^o.

CH4

• CH₄ molecule has sp³ hybridization.
• Each of these 4 sp³ hybrid orbitals overlaps with one 1s atomic orbital of hydrogen.
• The arrangement of the atoms gives tetrahedral geometry.
• There is no lone pair and the bond angle is ideal 109.5 °.

Hence, the minimum bond angle is present in H₂S.

Important Points

• Distortion in shape occurs when there is presence of lone pair of electrons as the table shows:

Concept:​

Fajan's Rule:

• Coulombic attraction between the cations and anions in certain cases leads to the deformation of the ions.
• This deformation caused by one molecule to the other is called polarization.
• The extent to which the molecule is able to polarise the other is called its polarisation power.
• The extent to which a molecule can get polarised is called its polarisability.
• The rise in deformity of ions may give rise to increased electron density between the ions and this leads to a considerable amount of covalent bonding.

Explanation:
(A) For a compound to have an ionic bond, a low positive charge on the cation and a large cation radius are preferred. This results in the cation having a low polarizing power, and the anion being less polarizable, preventing covalent character and resulting in an ion bond. Thus, A is correct.
(B) Conversely, for a compound to have a covalent bond, a small, highly positively charged cation is preferred as it has a large polarizing power and can polarize the electron cloud of the anion, resulting in some sharing of electrons that gives the bond a covalent character. Thus, B is also correct.
(C) This statement is incorrect. It is not the smaller cation with high charge that has less polarizing power, but rather the larger cation or the cation with a lower positive charge that would have less polarizing power.
So Option 1) A and B is the correct answer according to Fajan’s rule.
The concepts involved here include ionic bonding, covalent bonding, and the theory of polarization involving polarizing power and polarizability of ions. Fajan's rules help make the distinction between ionic and covalent character in compounds.

Dipole moment:

• It is a vector quantity that gives a measure of the polarity of the bond.
• It arises due to differences in the electronegativities of the constituent atom of a chemical bond.
• Since it is a vector quantity it has both magnitudes as well as direction.
• The dipole moment in a molecule can be zero as it is a vector quantity, two opposite dipoles can cancel each other.
• The direction of the dipole moment is shown with the help of an arrow.
• Dipole moment of a molecule having a number of atoms is the vector sum of all the dipoles present in the molecule.

Explanation:
The dipole moment of a molecule is the vector sum of all the dipoles present in the molecule.

→ BF₃

• BF₃ has sp² hybridisation and trigonal planar geometry.
• In its structure, it has three σ bonds (B-F), separated by 120° each.
• As the dipole moment in a molecule is the vector sum of all dipoles present in the molecule, the BF₃ molecule is having zero dipole moment.

→ Dipole moment of NH₃ and NF₃ -

• Both NH₃ and NF₃ have sp³ hybridisation and trigonal pyramidal shape.
• Both are having one lone pair each.
• But there is a difference in electronegativities of Hydrogen and Fluorine.
• In the case of NH₃, the dipole moment due to the N-H bond and due to the lone pair are in the same direction.
• But in the case of NF₃, the dipole moment due to N-F bonds are opposite to the direction of the dipole moment due to the lone pair. Thus, it is having a lesser dipole moment than NH₃.
• The dipole moment of NH₃ is greater than NF₃.

→ Dipole moment of H₂O -

• Oxygen is more electronegative than Hydrogen.
• It has two lone pairs.
• All dipole moments are in the same direction.
• It has a dipole moment greater than NH₃ due to having more number of lone pairs and also the electronegativity of oxygen is greater than nitrogen.
• Hence, H₂O has more dipole moment than NH₃.

So, the order of dipole moment in the above molecules is -

∴ The correct answer is option (4).