Covalent bonding Questions in English

(A) In the $H_2S$ molecule,the sulfur atom $(S)$ shares its valence electrons with two hydrogen atoms $(H)$ to complete its octet.
Since the bond is formed by the sharing of electrons between two non-metals,it is a covalent bond.
Therefore,the correct option is $A$.

(B) In $SO_3^{2-}$ (sulfite ion),the central sulfur atom has an oxidation state of $+4$. The electronic configuration of sulfur is $[Ne] 3s^2 3p^4$.
In the formation of $SO_3^{2-}$,sulfur undergoes $sp^3$ hybridization.
One of the $p$-orbitals of oxygen overlaps with the empty $d$-orbital of sulfur to form a $p\pi - d\pi$ bond.
In $NO_3^-$,$BO_3^{3-}$,and $CO_3^{2-}$,the central atoms $(N, B, C)$ belong to the second period and do not have $d$-orbitals available for bonding,so they only exhibit $p\pi - p\pi$ bonding.

(B) Calcium carbide $(CaC_2)$ is an ionic compound consisting of $Ca^{2+}$ ions and acetylide ions $(C_2^{2-})$.
The structure of the acetylide ion is $[C \equiv C]^{2-}$.
In the $C \equiv C$ triple bond,there is one sigma $(\sigma)$ bond and two pi $(\pi)$ bonds.
Therefore,the correct option is $B$.

(C) The bond length of carbon-halogen bonds increases as the size of the halogen atom increases.
The order of bond lengths is: $C-F < C-Cl < C-Br < C-I$.
Since bond energy is inversely proportional to bond length,the bond energy decreases as the bond length increases.
Therefore,the order of bond energies in descending order is: $C-F > C-Cl > C-Br > C-I$.

(N/A) Step $1$. Count the total number of valence electrons of carbon and oxygen atoms. The outer (valence) shell configurations of carbon and oxygen atoms are: $2s^{2} 2p^{2}$ and $2s^{2} 2p^{4}$ respectively. The valence electrons available are $4+6=10$.
Step $2$. The skeletal structure of $CO$ is written as: $C-O$.
Step $3$. Draw a triple bond between $C$ and $O$ atoms to satisfy the octet rule for both. Each atom contributes electrons to form three shared pairs,and each atom retains one lone pair.
The Lewis dot structure is: $:C \equiv O:$

(N/A) The Lewis structures are drawn by representing valence electrons as dots and bonds as lines:
$1. H_2S$: Sulfur is the central atom with two lone pairs and two $S-H$ single bonds.
$2. SiCl_4$: Silicon is the central atom bonded to four chlorine atoms by single bonds,with each chlorine atom having three lone pairs.
$3. BeF_2$: Beryllium is the central atom bonded to two fluorine atoms by single bonds,with each fluorine atom having three lone pairs.
$4. CO_3^{2-}$: Carbon is the central atom bonded to three oxygen atoms. One oxygen is double-bonded to carbon,and two are single-bonded with a negative charge,with all oxygens having lone pairs.
$5. HCOOH$: Carbon is the central atom bonded to one hydrogen,one oxygen (double bond),and one hydroxyl group $(-OH)$.

(N/A) The octet rule,or the electronic theory of chemical bonding,was developed by Kossel and Lewis. According to this rule,atoms combine either by the transfer of valence electrons from one atom to another or by sharing their valence electrons to attain the nearest noble gas configuration by having an octet in their valence shell.
Significance: The octet rule successfully explained the formation of chemical bonds depending upon the nature of the element.
Limitations of the octet theory:
$(a)$ The rule failed to predict the shape and relative stability of molecules.
$(b)$ It is based upon the inert nature of noble gases. However,some noble gases like xenon and krypton form compounds such as $XeF_{2}$,$KrF_{2}$,etc.
$(c)$ The octet rule cannot be applied to elements in and beyond the third period of the periodic table. The elements present in these periods have more than eight valence electrons around the central atom. For example: $PF_{5}$,$SF_{6}$,etc.
$(d)$ The octet rule is not satisfied for all atoms in a molecule having an odd number of electrons. For example,$NO$ and $NO_{2}$ do not satisfy the octet rule.
$(e)$ This rule cannot be applied to those compounds in which the number of electrons surrounding the central atom is less than eight. For example,$LiCl$,$BeH_{2}$,$AlCl_{3}$,etc.,do not obey the octet rule.

(N/A) Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.
Bond lengths are expressed in terms of $\mathring{A}$ $(10^{-10} \ m)$ or picometer $(10^{-12} \ m)$ and are measured by spectroscopic $X$-ray diffraction and electron-diffraction techniques.
In an ionic compound,the bond length is the sum of the ionic radii of the constituting atoms $(d = r_+ + r_-)$. In a covalent compound,it is the sum of their covalent radii $(d = r_A + r_B)$.

(N/A) When two dissimilar atoms having different electronegativities combine to form a covalent bond,the bond pair of electrons is not shared equally.
The bond pair shifts towards the nucleus of the atom having greater electronegativity. As a result,the electron distribution becomes distorted and the electron cloud is displaced towards the more electronegative atom.
Consequently,the electronegative atom acquires a partial negative charge $(\delta^-)$,while the other atom acquires a partial positive charge $(\delta^+)$. Thus,opposite poles are developed in the molecule,and this type of bond is called a polar covalent bond.
Example: $HCl$ contains a polar covalent bond. The chlorine atom is more electronegative than the hydrogen atom. Hence,the bond pair lies closer to the chlorine atom,which acquires a partial negative charge,while the hydrogen atom acquires a partial positive charge: $H^{\delta+} - Cl^{\delta-}$.

(N/A) In the given structure,the carbon atom is shown to have five bonds,which is incorrect as carbon is tetravalent. The correct Lewis structure for acetic acid $(CH_{3}COOH)$ involves a double bond between the carbonyl carbon and the oxygen atom,and a single bond between the carbonyl carbon and the hydroxyl oxygen. The correct structure is:
$H_{3}C-C(=O)OH$

(N/A) Let us assume that two hydrogen atoms ($A$ and $B$) with nuclei ($N_{A}$ and $N_{B}$) and electrons ($e_{A}$ and $e_{B}$) are taken to undergo a reaction to form a hydrogen molecule.
When $A$ and $B$ are at a large distance,there is no interaction between them. As they begin to approach each other,the attractive and repulsive forces start operating.
Attractive force arises between:
$(a)$ Nucleus of one atom and its own electron i.e.,$N_{A}-e_{A}$ and $N_{B}-e_{B}$
$(b)$ Nucleus of one atom and electron of another atom i.e.,$N_{A}-e_{B}$ and $N_{B}-e_{A}$
Repulsive force arises between:
$(a)$ Electrons of two atoms i.e.,$e_{A}-e_{B}$
$(b)$ Nuclei of two atoms i.e.,$N_{A}-N_{B}$
The force of attraction brings the two atoms together,whereas the force of repulsion tends to push them apart.
The magnitude of the attractive forces is more than that of the repulsive forces. Hence,the two atoms approach each other. As a result,the potential energy decreases. Finally,a state is reached when the attractive forces balance the repulsive forces and the system acquires minimum energy. This leads to the formation of a dihydrogen molecule.

(N/A) Langmuir $(1919)$ refined the Lewis postulations by abandoning the idea of the stationary cubical arrangement of the octet and by introducing the term covalent bond.
According to the Lewis-Langmuir theory,atoms can combine by sharing electrons between them. This type of bond is called a covalent bond,and each atom contributes an equal number of electrons for sharing in order to acquire the configuration of the nearest noble gas.
Example: Covalent bond in a dichlorine $(Cl_{2})$ molecule.
The $Cl$ atom with electronic configuration: $[Ne] \ 3s^{2} 3p^{5}$,is one electron short of the argon noble gas configuration.
The formation of the $Cl_{2}$ molecule can be understood in terms of the sharing of $1$ pair of electrons between the two chlorine atoms,each chlorine atom contributing one electron to the shared pair.
The electron pair takes part in the octet of both $Cl$ atoms.
So,both chlorine atoms get an octet like $Ar$: $[Ne] \ 3s^{2} 3p^{6}$ or $[Ar]$.
Note: '$\circ$' and '$x$' symbols are used to represent electrons of both chlorine atoms. '-' is used to represent the covalent bond between two $Cl$ atoms.

(N/A) In water $(H_{2}O)$,there are two $O-H$ covalent bonds. In carbon tetrachloride $(CCl_{4})$,there are four $C-Cl$ covalent bonds.
$H_{2}O$ Structure:
- Each $H$ atom attains a duplet (two electrons).
- The $O$ atom attains an octet (eight electrons).
$CCl_{4}$ Structure:
- Each $C-Cl$ bond consists of one electron from $C$ and one electron from $Cl$,satisfying the octet for all atoms.

(N/A) multiple bond is a covalent bond formed by the sharing of more than one pair of electrons between two atoms. When two atoms share two or more electron pairs,it is called a multiple bond (e.g.,double or triple bonds).
| Feature | Single Bond | Double Bond | Triple Bond |
| :--- | :--- | :--- | :--- |
| Electron pairs shared | $1$ pair ($2$ electrons) | $2$ pairs ($4$ electrons) | $3$ pairs ($6$ electrons) |
| Representation | $-$ | $=$ | $\equiv$ |
| Example | $Cl_2$ $(Cl-Cl)$ | $O_2$ $(O=O)$ | $N_2$ $(N\equiv N)$ |

(N/A) Carbon dioxide $(CO_{2})$: In the $CO_{2}$ molecule,two double bonds are present. Each $C=O$ bond involves two electron pairs (four electrons) shared between $C$ and $O$ atoms.
$(b)$ Ethene $(C_{2}H_{4})$: In this molecule,one double bond $(C=C)$ is present between two carbon atoms,and four single $C-H$ bonds are present. Between the two carbon atoms,two electron pairs (four electrons) are shared,forming a double bond.
$(c)$ Dinitrogen $(N_{2})$: In the $N_{2}$ molecule,a triple bond $(N \equiv N)$ is present between the two nitrogen atoms,involving three electron pairs (six electrons).
$(d)$ Ethyne $(C_{2}H_{2})$: In this molecule,one triple bond $(C \equiv C)$ is present between two carbon atoms,and two single $C-H$ bonds are present. Between the two carbon atoms,three electron pairs (six electrons) are shared,forming a triple bond.

(N/A) $1$. Each covalent bond is formed as a result of the sharing of an electron pair between the atoms.
$2$. Each combining atom contributes at least one electron to the shared pair.
$3$. The combining atoms attain the outer shell noble gas configuration (octet or duplet) as a result of the sharing of electrons.

(N/A) Number of total electrons in a neutral molecule: The total number of electrons required for writing the structures is obtained by adding the valence electrons of the combining atoms. For example,in the $CH_{4}$ molecule,there are eight valence electrons available for bonding ($4$ from carbon and $4$ from the four hydrogen atoms).
Number of electrons in a negative ion: Each negative charge means the addition of one electron to the total number of valence electrons. e.g.,For the $CO_{3}^{2-}$ ion,in Lewis representation,there are a total of $24$ electrons,in which $4$ $e^{-}$ are from one carbon,$18$ $e^{-}$ are from three oxygen atoms,and $2$ additional $e^{-}$ are from the two negative charges.
$(\text{Number of electrons in negative ion}) = (\text{Total valence electrons of all atoms}) + (\text{Number of negative charges})$
Number of electrons in a positive ion:
$(\text{Number of electrons in positive ion}) = (\text{Total valence electrons of all atoms}) - (\text{Number of positive charges})$
e.g.,$(\text{Total electrons in } NH_{4}^{+}) = (\text{Valence electrons of } N \text{ and } 4 H) - 1$
$= (5 + 4) - 1 = 8$
Distribution of electrons: Knowing the chemical symbols of the combining atoms and having knowledge of the skeletal structure of the compound (known or guessed intelligently),it is easy to distribute the total number of electrons as bonding shared pairs between the atoms in proportion to the total bonds.
Position of electrons: After accounting for the shared pairs of electrons for single bonds,the remaining electron pairs are either utilized for multiple bonding or remain as lone pairs. The basic requirement is that each bonded atom gets an octet of electrons.

(N/A) $1$. Total number of electrons in a neutral molecule: The total number of electrons required for writing the structures is obtained by adding the valence electrons of the combining atoms. For example,in the $CH_{4}$ molecule,there are eight valence electrons available for bonding ($4$ from carbon and $4$ from the four hydrogen atoms).
$2$. Number of electrons in a negative ion: Each negative charge means the addition of one electron to the total number of valence electrons. For example,for the $CO_{3}^{2-}$ ion,in the Lewis representation,there are a total of $24$ electrons: $4 \ e^{-}$ from one carbon,$18 \ e^{-}$ from three oxygen atoms,and $2$ additional $e^{-}$ due to the two negative charges.
$\left(\begin{array}{c} \text{Number of} \\ \text{electrons in} \\ \text{negative ion} \end{array}\right) = \left(\begin{array}{c} \text{Total valence} \\ \text{electrons} \\ \text{of all atoms} \end{array}\right) + \left(\begin{array}{c} \text{Number of} \\ \text{negative} \\ \text{charges} \end{array}\right)$
$3$. Number of electrons in a positive ion: Each positive charge means the subtraction of one electron from the total number of valence electrons.
$\left(\begin{array}{c} \text{Number of} \\ \text{electrons in} \\ \text{positive ion} \end{array}\right) = \left(\begin{array}{c} \text{Total valence} \\ \text{electrons} \\ \text{of all atoms} \end{array}\right) - \left(\begin{array}{c} \text{Number of} \\ \text{positive} \\ \text{charges} \end{array}\right)$
For example,$\left(\begin{array}{c} \text{Total} \\ \text{electrons} \\ \text{in} \ NH_{4}^{+} \end{array}\right) = \left(\begin{array}{c} \text{Valence} \\ \text{electrons of} \\ N \ \text{and} \ 4 \ H \end{array}\right) - 1 = (5+4)-1 = 8$.
$4$. Distribution of electrons: Knowing the chemical symbols of the combining atoms and the skeletal structure of the compound,it is easy to distribute the total number of electrons as bonding shared pairs between the atoms.
$5$. Position of electrons: After accounting for the shared pairs of electrons for single bonds,the remaining electron pairs are either utilized for multiple bonding or remain as lone pairs. The basic requirement is that each bonded atom attains an octet of electrons.

(N/A) The Lewis structures represent the valence electrons of atoms in a molecule or ion. The structures are as follows:
$(i) H_2: H:H$
$(ii) O_2: :\ddot{O}::\ddot{O}:$
$(iii) O_3: :\ddot{O}-\ddot{O}=\ddot{O}:$
$(iv) NF_3: \text{The central } N \text{ atom is bonded to three } F \text{ atoms with one lone pair on } N.$
$(v) CO_3^{2-}: \text{The central } C \text{ atom is double bonded to one } O \text{ and single bonded to two } O^- \text{ atoms.}$
$(vi) HNO_3: \text{The central } N \text{ atom is bonded to one } =O, \text{ one } -OH, \text{ and one } -O^- \text{ (dative bond).}$

(N/A) The correct Lewis structure for acetic acid $(CH_3COOH)$ requires a double bond between the carbonyl carbon and one oxygen atom $(C=O)$,while the other oxygen is single-bonded to both the carbon and a hydrogen atom $(C-OH)$. The methyl group $(CH_3)$ consists of three $C-H$ single bonds. The structure is represented as follows:
$H_3C-C(=O)OH$

(N/A) $(i)$ $C$ has electronic configuration $[He] 2s^2 2p^2$,so valence electrons $= 4$.
$O$ has electronic configuration $[He] 2s^2 2p^4$,so valence electrons $= 6$.
Therefore,total valence electrons in $CO = 4 + 6 = 10$.
$(ii)$ The skeletal structure of $CO$ is $C-O$.
$(iii)$ If we place a single bond between $C$ and $O$,the octet is not satisfied for $C$. By forming a triple bond between $C$ and $O$,we share $6$ electrons.
$(iv)$ The final Lewis structure is $:C \equiv O:$,where $C$ and $O$ both complete their octets through sharing $6$ electrons and having one lone pair each.

(N/A) Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.
Bond lengths are measured using experimental techniques such as spectroscopic methods,$X$-ray diffraction,and electron diffraction.

(N/A) According to the Lewis theory of covalent bonding,the bond order is defined as the number of chemical bonds between two atoms in a molecule.

Molecule and Structure	Bond Order
$H_2$ (dihydrogen): $H-H$	$1$
$O_2$ (dioxygen): $O=O$	$2$
$N_2$ (dinitrogen): $N \equiv N$	$3$
$CO$ (carbon monoxide): $C \equiv O$	$3$
$NO^{+}$: $[N \equiv O]^+$	$3$

(N/A) The existence of a $100\%$ ionic or covalent bond represents an ideal situation. In reality,no bond or compound is either completely covalent or ionic. Even in the case of a covalent bond between two hydrogen atoms,there is some ionic character.
Non-polar covalent bond: When a covalent bond is formed between two similar atoms,the shared pair of electrons is equally attracted by the two atoms. As a result,the electron pair is situated exactly between the two identical nuclei. The bond so formed is called a non-polar covalent bond.
Example: Non-polar covalent bonds in $H_2, N_2, O_2, F_2, Cl_2$.

Molecule	$H_2, N_2, O_2, F_2, Cl_2$
Bond structure	$H-H, N \equiv N, O=O, F-F, Cl-Cl$

Polar covalent bond: When atoms of two different elements (heteronuclear) combine by a covalent bond,the electron pair in the bond is attracted more by the more electronegative atom. Consequently,it acquires a partial negative charge $(-\delta)$ and the other atom acquires a partial positive charge $(+\delta)$. Such a covalent bond is called a polar covalent bond.
In an $AB$ molecule,if $A$ is less electronegative and $B$ is more electronegative,the $A-B$ bond will be $A^{+\delta}-B^{-\delta}$ (polar bond).

Molecule	$HF$	$HCl$	$CO$	$NO$
Real bond	$H^{+\delta}-F^{-\delta}$	$H^{+\delta}-Cl^{-\delta}$	$C^{+\delta} \equiv O^{-\delta}$	$N^{+\delta}=O^{-\delta}$

(N/A) Bond pairs are the shared electrons between two atoms in a covalent bond. For example,in the $H-H$ molecule,the shared pair of electrons between the two $H$ atoms constitutes a bond pair.
Lone pairs are valence shell electron pairs that are not involved in bonding and remain on a single atom. For example,in the $H_2O$ molecule,the oxygen atom has two lone pairs of electrons that do not participate in bonding.

(N/A) Consider two hydrogen atoms $A$ and $B$ approaching each other, having nuclei $N_{A}$ and $N_{B}$, and electrons represented by $e_{A}$ and $e_{B}$.
When the two atoms are at a large distance from each other, there is no interaction between them. At this stage, the energy of the system is equal to the sum of the energies of both isolated $H$ atoms. In each atom, the nucleus and electron have an attractive interaction ($N_{A}-e_{A}$ and $N_{B}-e_{B}$).
When both atoms $H_{A}$ and $H_{B}$ approach each other, both attractive and repulsive forces are generated:
$(i)$ Attractive forces: These occur between the nucleus of one atom and its own electron ($N_{A}-e_{A}$, $N_{B}-e_{B}$) and between the nucleus of one atom and the electron of the other atom ($N_{A}-e_{B}$ and $N_{B}-e_{A}$).
$(ii)$ Repulsive forces: These occur between the electrons of the two atoms $(e_{A}-e_{B})$ and between the nuclei of the two atoms $(N_{A}-N_{B})$.
- Attractive forces pull the atoms toward each other, while repulsive forces push them apart.
- Experimental evidence shows that the magnitude of attractive forces is greater than that of repulsive forces. Consequently, the atoms move closer, and the potential energy of the system decreases.
$H_{2}$ formation: $(i)$ The atoms approach until the attractive and repulsive forces are balanced. $(ii)$ The system attains minimum potential energy. $(iii)$ At this stage, the two hydrogen atoms combine. $(iv)$ A stable hydrogen molecule is formed. $(v)$ The equilibrium bond length is $74 \ pm$.

(N/A) The $VBT$ was introduced by Heitler and London $(1927)$ and further developed by Pauling and others. It addresses the limitations of the Lewis approach and $VSEPR$ theory.
Key postulates of $VBT$:
$1$. Formation of a covalent bond: It occurs due to the overlapping of half-filled atomic orbitals of the participating atoms.
$2$. Energy minimization: As two atoms approach each other,attractive and repulsive forces act between them. At a specific distance,the system reaches a state of minimum potential energy,resulting in a stable covalent bond. For example,in $H_2$,the energy decreases by $435.8 \ kJ \ mol^{-1}$.
$3$. Extent of overlapping: The strength of a covalent bond is directly proportional to the extent of overlapping of the atomic orbitals.
$4$. Directional properties: The geometry of a molecule is determined by the directional nature of the overlapping atomic orbitals (e.g.,tetrahedral geometry in $CH_4$).
$5$. Types of bonds: Overlapping along the internuclear axis results in a $\sigma$ bond,while lateral (perpendicular) overlapping results in a $\pi$ bond.
$6$. Hybridization: Atomic orbitals undergo hybridization to form equivalent hybrid orbitals,which then overlap to form bonds.

(N/A) Electron-deficient compounds are those in which the central atom has an incomplete octet,meaning it has fewer than $8$ electrons in its valence shell.
$(i)$ $BCl_3$: In $BCl_3$,the Boron atom has $3$ valence electrons. After forming $3$ covalent bonds with chlorine atoms,it has $6$ electrons in its valence shell. Since $6 < 8$,$BCl_3$ is an electron-deficient species.
$(ii)$ $SiCl_4$: Silicon has $4$ valence electrons. After forming $4$ covalent bonds with chlorine atoms,it completes its octet with $8$ electrons. Therefore,$SiCl_4$ is not an electron-deficient species.

(B) In the ball and stick model,atoms are represented by balls and bonds are represented by sticks. For double bonds $(C=C)$ and triple bonds $(C \equiv C)$,springs are used instead of rigid sticks to represent the curvature and flexibility of the multiple bonds.

(D) The maximum covalency of the first-period elements (specifically $Li$ to $Be$ and $B$) is limited to $4$.
This is because they possess only four valence orbitals ($one$ $2s$ and $three$ $2p$ orbitals) available for bonding.
For example,in the case of boron,it can form the $[BF_4]^-$ ion.

(A) The elements of the second period ($Li$ to $Ne$) have only $2s$ and $2p$ orbitals in their valence shell.
Since there are only four orbitals available ($one$ $2s$ and $three$ $2p$),they can accommodate a maximum of $8$ electrons,limiting their maximum covalency to $4$.

(N/A) The Lewis structures and formal charges on each atom are calculated using the formula: $\text{Formal charge} = [\text{Total valence electrons}] - [\text{Total non-bonding electrons}] - \frac{1}{2} [\text{Total shared electrons}]$.
$(i)$ $HNO_3$:
Formal charge on $H = 1 - 0 - \frac{1}{2}(2) = 0$
Formal charge on $N = 5 - 0 - \frac{1}{2}(8) = +1$
Formal charge on $O(1) = 6 - 4 - \frac{1}{2}(4) = 0$
Formal charge on $O(2) = 6 - 4 - \frac{1}{2}(4) = 0$
Formal charge on $O(3) = 6 - 6 - \frac{1}{2}(2) = -1$
$(ii)$ $NO_2$:
Formal charge on $O(1) = 6 - 4 - \frac{1}{2}(4) = 0$
Formal charge on $N = 5 - 1 - \frac{1}{2}(6) = +1$
Formal charge on $O(2) = 6 - 6 - \frac{1}{2}(2) = -1$
$(iii)$ $H_2SO_4$:
In $H_2SO_4$,the central $S$ atom is bonded to two $OH$ groups and two $O$ atoms via double bonds. All atoms follow the octet rule (except $H$). The formal charge on $S = 6 - 0 - \frac{1}{2}(12) = 0$. The formal charge on $OH$ oxygen atoms $= 6 - 4 - \frac{1}{2}(4) = 0$. The formal charge on double-bonded oxygen atoms $= 6 - 4 - \frac{1}{2}(4) = 0$.

(N/A) Valence bond theory $(VBT)$ was introduced by Heitler and London $(1927)$ and developed further by Pauling and others. $VBT$ is based on the knowledge of atomic orbitals, electronic configurations of elements, the overlap criteria of atomic orbitals, the hybridization of atomic orbitals, and the principles of variation and superposition.
Consider two hydrogen atoms $A$ and $B$ approaching each other, having nuclei $N_{A}$ and $N_{B}$, and electrons represented by $e_{A}$ and $e_{B}$. When the two atoms are at a large distance from each other, there is no interaction between them.
As these two atoms approach each other, new attractive and repulsive forces begin to operate.
- Attractive forces arise between:
$(i)$ Nucleus of one atom and its own electron: i.e., $N_{A}-e_{A}$ and $N_{B}-e_{B}$
$(ii)$ Nucleus of one atom and the electron of the other atom: i.e., $N_{A}-e_{B}$ and $N_{B}-e_{A}$
- Similarly, repulsive forces arise between:
$(i)$ Electrons of the two atoms: $e_{A}-e_{B}$
$(ii)$ Nuclei of the two atoms: $N_{A}-N_{B}$
Experimentally, it is found that the magnitude of the new attractive forces is greater than the new repulsive forces. As a result, the two atoms approach each other and the potential energy decreases. A stage is reached where the net force of attraction balances the force of repulsion, and the system attains minimum energy. At this stage, the two $H$ atoms are bonded together to form a stable molecule having a bond length of $74 \text{ pm}$.
Energy is released when the bond is formed between two hydrogen atoms, making the hydrogen molecule more stable than isolated hydrogen atoms.

(N/A) The Lewis covalent bond theory,while useful for understanding the formation of covalent bonds,has several limitations:
$1$. It does not explain the shape and geometry of molecules.
$2$. It fails to explain the stability of molecules with an odd number of electrons (e.g.,$NO$,$NO_2$).
$3$. It cannot explain the stability of molecules where the central atom has an incomplete octet (e.g.,$BeCl_2$,$BF_3$) or an expanded octet (e.g.,$PF_5$,$SF_6$).
$4$. It does not provide any information regarding the energy of the molecule or the relative stability of different structures.
$5$. It does not account for the magnetic properties of molecules,such as the paramagnetic nature of $O_2$.

(B) The term 'covalent bond' was coined by $Irving \ Langmuir$ in $1919$.
The electronic theory of covalent bonding was proposed by $G.N. \ Lewis$ in $1916$.

(N/A) covalent bond is formed by the sharing of unpaired electrons between two atoms. This sharing allows each atom to attain a stable noble gas electronic configuration.

(N/A) When a covalent bond is formed,both atoms involved in the bond achieve the stable electronic configuration of the nearest noble gas by sharing electrons to complete their octets (or duplets in the case of hydrogen).

(N/A) In a covalent bond,each participating atom contributes one electron to the shared pair. After the bond is formed,these shared electrons are counted towards the octet of both bonded atoms.

(B) multiple bond is said to exist between two atoms when more than one pair of electrons is shared between them. For example,a double bond involves the sharing of $2$ electron pairs,and a triple bond involves the sharing of $3$ electron pairs.

(B) Multiple bonds are double and triple bonds.
They are called multiple bonds because they involve the sharing of more than one electron pair between two atoms.
$A$ double bond involves the sharing of $2$ electron pairs,while a triple bond involves the sharing of $3$ electron pairs.

(N/A) The number of covalent bonds in a molecule is determined by the number of shared electron pairs between atoms to achieve a stable octet or duplet configuration.
$1$. In $H_2O$: The central oxygen atom shares two electrons with two hydrogen atoms,forming $2$ $O-H$ covalent bonds.
$2$. In $NH_3$: The central nitrogen atom shares three electrons with three hydrogen atoms,forming $3$ $N-H$ covalent bonds.
$3$. In $CCl_4$: The central carbon atom shares four electrons with four chlorine atoms,forming $4$ $C-Cl$ covalent bonds.
The number of bonds is determined by the valency of the central atom and the requirement to complete the valence shell of each atom involved in the bonding.

(N/A) In $NH_{3}$,there are three $N-H$ single bonds. $A$ triple bond requires three bonds between the same two atoms,whereas in $NH_{3}$,the nitrogen atom is bonded to three separate hydrogen atoms.

(B) The nitrogen atom has an electronic configuration of $1s^2 2s^2 2p^3$.
To complete its octet,each nitrogen atom requires $3$ more electrons.
Therefore,two nitrogen atoms share $3$ pairs of electrons to form a triple bond $(N \equiv N)$.

(N/A) In $CO_2$ (carbon dioxide),the carbon atom is at the center and is bonded to two oxygen atoms.
The structure is $O=C=O$.
There are two double bonds present in the molecule.
Each $C=O$ bond consists of one sigma $(\sigma)$ bond and one pi $(\pi)$ bond.

(A) Yes,$CO_2$ contains multiple bonds.
In the $CO_2$ molecule,the carbon atom is central and is bonded to each oxygen atom by a double bond.
Therefore,the structure is $O=C=O$,which contains two $C=O$ double bonds.

(N/A) Ethyne $(C_2H_2)$ contains a triple bond between the two carbon atoms.
This is because there are three shared electron pairs between the two carbon atoms.
The structure is represented as $H-C \equiv C-H$.

(N/A) In $BCl_3$,the $B$ atom has an incomplete octet (sextet),while each $Cl$ atom has a complete octet.
In $BH_3$,the $B$ atom has an incomplete octet (sextet),and the $H$ atoms have a duplet configuration (not an octet).
In $BeCl_2$,the $Be$ atom has an incomplete octet (quartet),while each $Cl$ atom has a complete octet.
In $BeH_2$,the $Be$ atom has an incomplete octet (quartet),and the $H$ atoms have a duplet configuration.
Thus,in both pairs,the central atom ($B$ or $Be$) has an incomplete octet,but the terminal atoms differ in their valence shell completion ($Cl$ completes its octet,whereas $H$ completes its duplet).

(A) The bond length depends on the bond order. As the bond order increases,the bond length decreases.
$C-C$ (single bond) has a bond order of $1$.
$C=C$ (double bond) has a bond order of $2$.
$C\equiv C$ (triple bond) has a bond order of $3$.
Therefore,the increasing order of bond length is: $C\equiv C < C=C < C-C$.

(C) During the formation of an $H_{2}$ molecule from two hydrogen atoms ($H_{A}$ and $H_{B}$),two types of forces operate:
$1$. Attractive forces: Between the nucleus of one atom and its own electron,and between the nucleus of one atom and the electron of the other atom $(N_{A}-e_{A}, N_{B}-e_{B}, N_{A}-e_{B}, N_{B}-e_{A})$.
$2$. Repulsive forces: Between the nuclei of the two atoms $(N_{A}-N_{B})$ and between the electrons of the two atoms $(e_{A}-e_{B})$.
When the attractive forces exceed the repulsive forces,the potential energy of the system decreases,leading to the formation of a stable $H_{2}$ molecule.

$(A)$ As two hydrogen atoms approach each other, the potential energy of the system decreases due to the attractive forces between the nucleus of one atom and the electron of the other.
At a specific internuclear distance (the bond length, $74 \text{ pm}$), the potential energy reaches a minimum, representing the most stable state of the $H_{2}$ molecule.
If the atoms get closer than this distance, the repulsive forces between the nuclei and between the electrons dominate, causing the potential energy to increase sharply.