Overlaping - s and p- bonds Questions in English

(C) The molecule $CH_2=C=C=CH_2$ is an allene derivative (specifically,buta$-1,2,3-$triene).
In this molecule,the terminal $CH_2$ groups are perpendicular to each other.
The two $\pi$-bonds are formed by the overlap of $p$-orbitals.
One $\pi$-bond is formed by the overlap of $p_y$ orbitals,and the other $\pi$-bond is formed by the overlap of $p_z$ orbitals.
The nodal plane of a $\pi$-bond is the plane containing the internuclear axis and perpendicular to the plane of the $p$-orbitals.
For the two $\pi$-bonds in this system,one nodal plane lies in the molecular plane (containing the $C-C$ $\sigma$-bonds),and the other nodal plane is perpendicular to the molecular plane.

(A) The structure of sulfuric acid $(H_2SO_4)$ is:
$HO-S(=O)_2-OH$.
In this structure,there are $6$ single bonds (each being a $\sigma$ bond) and $2$ double bonds (each containing one $\sigma$ and one $\pi$ bond).
Total $\sigma$ bonds = $4$ (from $S-O$ and $S-OH$ bonds) + $2$ (from $O-H$ bonds) = $6\sigma$ bonds.
Total $\pi$ bonds = $2$ (from the two $S=O$ double bonds).
Therefore,the number of $\sigma$ and $\pi$ bonds are $6$ and $2$ respectively.

(A, C, D) In $NO_3^-$,$CO_3^{2-}$,and $BO_3^{3-}$,the central atoms ($N$,$C$,$B$) belong to the second period and form $p\pi - p\pi$ bonds with oxygen atoms ($2p$ orbitals).
However,in $SO_3^{2-}$,the sulfur atom belongs to the third period and forms $d\pi - p\pi$ bonds with oxygen atoms.
Therefore,all options except $B$ contain $p\pi - p\pi$ bonding.
Given the standard nature of this question,it is often asked to identify which species does $NOT$ contain $p\pi - p\pi$ bonding,or the question implies identifying the species with $d\pi - p\pi$ bonding.
Assuming the question asks for the species with $d\pi - p\pi$ bonding,the answer is $B$.

(C) In an ethylene $(C_2H_4)$ molecule,the carbon atoms are $sp^2$ hybridized.
The $p_z$ orbitals of both carbon atoms overlap sideways to form a $\pi$ bond.
The nodal plane of this $\pi$ bond is the molecular plane itself,which contains the two carbon nuclei and the four hydrogen nuclei.

(C) In the $PO_4^{3-}$ ion,the central phosphorus atom has vacant $d$-orbitals available for bonding.
The $\pi$-bond in the $P=O$ bond is formed by the sidewise overlapping of the $d$-orbital of $P$ and the $p$-orbital of oxygen,which is a $p\pi - d\pi$ overlap.
In $NO_2^-$,$NO_3^-$,and $CO_3^{2-}$,the central atoms ($N$ and $C$) belong to the second period and do not have vacant $d$-orbitals,so they cannot form $p\pi - d\pi$ bonds.

(D) The strength of a covalent bond formed by head-on overlapping depends on the extent of overlapping.
Greater the extent of overlapping,stronger is the bond.
For $s$ and $p$ orbitals,the $1s$ orbital is the smallest and closest to the nucleus,leading to the maximum extent of overlapping.
Therefore,the $1s-1s$ overlapping is the strongest among the given options.

(A) In ethene $(CH_2=CH_2)$,the carbon atoms are $sp^2$ hybridized.
The three $sp^2$ hybrid orbitals of each carbon atom lie in the same plane (the molecular plane) and form $\sigma$-bonds.
The unhybridized $2p_z$ orbitals are perpendicular to this molecular plane.
The $\pi$-bond is formed by the lateral overlap of these $2p_z$ orbitals.
$A$ nodal plane is a plane where the probability of finding an electron is zero.
For a $\pi$-bond,the molecular plane itself acts as the nodal plane because the electron density of the $\pi$-bond is concentrated above and below this plane.

(C) The correct statement regarding bond energies is that the $\sigma$ bond is stronger than the $\pi$ bond.
Typically,the bond energy of a $\sigma$ bond is higher than that of a $\pi$ bond.
Option $C$ states that the bond energy of a $\sigma$ bond is $264 \, kJ/mol$ and a $\pi$ bond is $347 \, kJ/mol$,which is factually incorrect as it reverses the values and their relative strengths.
Therefore,statement $C$ is not correct.

(D) In a diatomic molecule,if the internuclear axis is the $z$-axis,then the $p_z$ orbitals undergo head-on overlap to form a $\sigma$ bond.
The $p_x$ and $p_y$ orbitals are oriented perpendicular to the $z$-axis.
Since $p_x$ and $p_y$ orbitals are orthogonal to each other and to the $z$-axis,they cannot overlap effectively to form a bond.
Therefore,the overlap of $p_x$ and $p_y$ orbitals results in no bond formation.

(A) The atomic number of fluorine $(F)$ is $9$. Its electronic configuration is $1s^2 2s^2 2p_x^2 2p_y^2 2p_z^1$.
In the formation of the $F_2$ molecule,the half-filled $2p_z$ orbital of one fluorine atom overlaps axially with the half-filled $2p_z$ orbital of another fluorine atom.
This axial overlap results in the formation of a sigma $(\sigma)$ bond.
Therefore,the $F_2$ molecule is formed by the axial overlap of $p-p$ orbitals.

(A) The structure is $CH_3-CH=CH-C\equiv CH$.
Counting the bonds:
$C-H$ bonds: $3$ (in $CH_3$) $+ 1$ (in $CH$) $+ 1$ (in $CH$) $+ 1$ (in $CH$) $= 6$ $\sigma$ bonds.
$C-C$ bonds: $1$ $(C-C)$ $+ 1$ $(C=C)$ $+ 1$ $(C-C)$ $+ 1$ $(C\equiv C)$ $= 4$ $\sigma$ bonds.
Total $\sigma$ bonds $= 6 + 4 = 10$.
$\pi$ bonds: $1$ (in $C=C$) $+ 2$ (in $C\equiv C$) $= 3$ $\pi$ bonds.
Therefore,the total is $10\sigma$ and $3\pi$ bonds.

(C) The extent of overlapping depends on the size of the orbitals and the type of overlap.
Smaller orbitals overlap more effectively than larger orbitals because the electron density is closer to the nucleus.
Thus,the order of size is $1s < 2s < 2p$.
For the same shell,the order of overlapping strength is $p-p \text{ (axial)} > s-p > s-s$.
Comparing $1s-1s$ with others,$1s$ is the smallest orbital,leading to the strongest overlap.
Therefore,the correct order is $1s-1s > 2p-2p \text{ (axial)} > 2s-2p > 2s-2s$.

(B) The strength of a covalent bond depends on the extent of overlapping of atomic orbitals.
Axial overlapping is always stronger than sidewise overlapping.
Among axial overlaps,the $p-p$ (axial) overlap is stronger than $s-p$ and $s-s$ overlaps because $p-$orbitals are more directional in nature,allowing for a greater extent of overlap compared to the spherical $s-$orbitals.

(A) The structure of sulfuric acid $(H_2SO_4)$ consists of a central sulfur atom bonded to two hydroxyl groups $(-OH)$ and two oxygen atoms via double bonds $(S=O)$.
Specifically,there are $2$ $S-OH$ bonds and $2$ $S=O$ bonds.
Each single bond is a $\sigma$ bond,and each double bond contains one $\sigma$ bond and one $\pi$ bond.
Counting the $\sigma$ bonds: $2$ $(S-OH)$ + $2$ ($S-O$ single part of $S=O$) + $2$ $(O-H)$ = $6$ $\sigma$ bonds.
Counting the $\pi$ bonds: $2$ $\pi$ bonds (one from each $S=O$ bond).
Thus,the number of $\sigma$ and $\pi$ bonds are $6$ and $2$ respectively.

(C) The Assertion is correct because a sigma $(\sigma)$ bond is formed by the axial overlap of atomic orbitals,which is more extensive than the lateral overlap that forms a pi $(\pi)$ bond. Therefore,the sigma $(\sigma)$ bond is stronger than the pi $(\pi)$ bond.
The Reason is incorrect because atoms cannot rotate freely around a pi $(\pi)$ bond. The presence of a pi $(\pi)$ bond restricts rotation,as rotation would require breaking the lateral overlap of the $p$-orbitals.

(N/A) single bond is a result of the axial overlap of bonding orbitals. Hence,it contributes a sigma $(\sigma)$ bond.
$A$ multiple bond (double or triple bond) is always formed as a result of the sidewise overlap of orbitals,which contributes a pi $(\pi)$ bond.
$1.$ Structure of $C_{2}H_{2}$ (Ethyne): $H-C \equiv C-H$
In $C_{2}H_{2}$,there are $3$ sigma $(\sigma)$ bonds ($2$ $C-H$ and $1$ $C-C$) and $2$ pi $(\pi)$ bonds.
$2.$ Structure of $C_{2}H_{4}$ (Ethene): $CH_{2}=CH_{2}$
In $C_{2}H_{4}$,there are $5$ sigma $(\sigma)$ bonds ($4$ $C-H$ and $1$ $C-C$) and $1$ pi $(\pi)$ bond.

(C) $2p_{y}$ and $2p_{y}$ orbitals will not form a sigma $(\sigma)$ bond.
Since the $x$-axis is the internuclear axis,the $2p_{y}$ orbitals are oriented perpendicular to the internuclear axis.
Therefore,$2p_{y}$ and $2p_{y}$ orbitals will undergo lateral (sideways) overlapping,resulting in the formation of a pi $(\pi)$ bond instead of a sigma $(\sigma)$ bond.

The following are the differences between sigma and pi-bonds:

Sigma $(\sigma)$ Bond	Pi $(\pi)$ Bond
$(a)$ It is formed by the end-to-end overlap of orbitals.	$(a)$ It is formed by the lateral overlap of orbitals.
$(b)$ The orbitals involved in the overlapping are $s-s, s-p,$ or $p-p$.	$(b)$ These bonds are formed by the overlap of $p-p$ orbitals only.
$(c)$ It is a strong bond.	$(c)$ It is a weak bond.
$(d)$ The electron cloud is symmetrical about the line joining the two nuclei.	$(d)$ The electron cloud is not symmetrical about the internuclear axis.
$(e)$ It consists of one electron cloud,which is symmetrical about the internuclear axis.	$(e)$ There are two electron clouds lying above and below the plane of the atomic nuclei.
$(f)$ Free rotation about $\sigma$ bonds is possible.	$(f)$ Rotation is restricted in case of $\pi$ bonds.

(N/A) For $(a)$ $HC \equiv C-CH=CH-CH_3$:
Total $\sigma$ bonds = $4$ ($C$-$C$) + $6$ ($C$-$H$) = $10$ $\sigma$ bonds.
Total $\pi$ bonds = $2$ ($\pi$ in $C \equiv C$) + $1$ ($\pi$ in $C=C$) = $3$ $\pi$ bonds.
For $(b)$ $CH_2=C=CH-CH_3$:
Total $\sigma$ bonds = $3$ ($C$-$C$) + $6$ ($C$-$H$) = $9$ $\sigma$ bonds.
Total $\pi$ bonds = $2$ ($\pi$ in $C=C=C$) = $2$ $\pi$ bonds.

(A) $(i)$ $C_{6}H_{6}$: There are $6$ $C-C$ $\sigma$ bonds,$6$ $C-H$ $\sigma$ bonds,and $3$ $\pi$ bonds.
$(ii)$ $C_{6}H_{12}$: There are $6$ $C-C$ $\sigma$ bonds and $12$ $C-H$ $\sigma$ bonds.
$(iii)$ $CH_{2}Cl_{2}$: There are $2$ $C-H$ $\sigma$ bonds,$2$ $C-Cl$ $\sigma$ bonds,and $2$ $C-H$ $\sigma$ bonds (total $4$ $\sigma$ bonds).
$(iv)$ $CH_{2}=C=CH_{2}$: There are $4$ $C-H$ $\sigma$ bonds,$2$ $C-C$ $\sigma$ bonds,and $2$ $\pi$ bonds.
$(v)$ $CH_{3}NO_{2}$: There are $3$ $C-H$ $\sigma$ bonds,$1$ $C-N$ $\sigma$ bond,$1$ $N-O$ $\sigma$ bond,and $1$ $N=O$ $\pi$ bond.
$(vi)$ $HCONHCH_{3}$: There are $4$ $C-H$ $\sigma$ bonds,$1$ $N-H$ $\sigma$ bond,$1$ $C-N$ $\sigma$ bond (in $C-N$),$1$ $C-N$ $\sigma$ bond (in $C-NH$),$1$ $C-O$ $\sigma$ bond,and $1$ $C=O$ $\pi$ bond.

(N/A) In a $\pi$ bond formation,parallel orientation of the two $p$-orbitals on adjacent atoms is necessary for a proper sideways overlap.
Example: For the formation of a $\pi$ bond in $H_2C=CH_2$:
- All the atoms must be in the same plane.
- The $p$-orbitals are mutually parallel.
- Both $p$-orbitals are perpendicular to the plane of the molecule.
Effect of rotation on $C-C$ bond: In $CH_2=CH_2$,the rotation of one $CH_2$ fragment with respect to the other interferes with the maximum overlap of $p$-orbitals; therefore,such rotation about the carbon-carbon double bond $(C=C)$ is restricted.
Availability of electrons by $\pi$ bond and reactive centers: The electron charge cloud of the $\pi$ bond is located above and below the plane of the bonding atoms. This results in the electrons being easily available to attacking reagents. In general,$\pi$ bonds provide the most reactive centers in molecules containing multiple bonds.

(N/A) The overlapping of atomic orbitals is the partial interpenetration of atomic orbitals of two atoms when they come close to each other to achieve a state of minimum energy.
$A$ covalent bond is formed between two atoms by the pairing of electrons present in the valence shells that have opposite spins.
The extent of this overlap determines the strength of the covalent bond; greater the overlap,the stronger is the bond.

(N/A) Atomic orbital overlapping is classified based on the sign of the wave function ($+$ or $-$) and the orientation of the orbitals:
$1$. Positive Overlapping (Bonding): Occurs when orbitals with the same sign ($+$ with $+$ or $-$ with $-$) overlap. This leads to a lower energy state and bond formation.
$2$. Negative Overlapping (Antibonding): Occurs when orbitals with opposite signs ($+$ with $-$) overlap. This leads to a higher energy state and no bond formation.
$3$. Zero Overlapping: Occurs when the orientation of orbitals is such that there is no net overlap (e.g.,$p_x$ and $s$ or $p_x$ and $p_y$ in certain orientations). The net overlap is zero.

(N/A)

$\sigma$ bond	$\pi$ bond
This type of covalent bond is formed by the head-on (end-to-end) overlapping of atomic orbitals along the internuclear axis.	This type of covalent bond is formed by the lateral (sideways) overlapping of atomic orbitals,where the axes of the orbitals remain parallel to each other and perpendicular to the internuclear axis.
The electron cloud is cylindrically symmetrical around the internuclear axis.	The electron cloud is distributed above and below the plane of the internuclear axis.
$\sigma$ molecular orbitals are symmetrical around the internuclear axis.	$\pi$ molecular orbitals are not symmetrical around the internuclear axis.
$\sigma$ bonds are formed by $s-s$,$s-p$,and $p-p$ (axial) overlapping.	$\pi$ bonds are formed by $p-p$,$p-d$,and $d-d$ (lateral) overlapping.
Such a bond is not broken by rotation around the bond axis.	Such a bond breaks upon rotation around the bond axis.
$\sigma$ bonds are stronger due to greater extent of overlapping.	$\pi$ bonds are weaker due to less effective lateral overlapping.
$A$ single bond between two atoms is always a $\sigma$ bond.	In multiple bonds,the additional bonds are $\pi$ bonds ($1$ in double,$2$ in triple).

(N/A) The covalent bond formed by the overlapping of atomic orbitals can be classified into two types:
$\sigma$-bond: This type of covalent bond is formed by the end-to-end (head-on) overlapping of two atomic orbitals along the internuclear axis. This is also known as axial overlapping. It occurs in the following ways:
- $s-s$ overlapping: Overlapping of two half-filled $s$-orbitals.
- $s-p$ overlapping: Overlapping between a half-filled $s$-orbital and a half-filled $p$-orbital.
- $p-p$ overlapping: Overlapping of two half-filled $p$-orbitals along the internuclear axis.
$\pi$-bond: This type of covalent bond is formed by the sidewise (lateral) overlapping of atomic orbitals. In this case,the axes of the overlapping atomic orbitals remain parallel to each other and perpendicular to the internuclear axis.

(N/A) The covalent bond formed by the overlapping of atomic orbitals can be classified into two types:
$1$. $\sigma$-bond: This type of covalent bond is formed by the end-to-end (head-on) overlapping of two atomic orbitals along the internuclear axis. It is also known as axial overlapping.
$a$. $s-s$ overlapping: This involves the overlapping of two half-filled $s$-orbitals along the internuclear axis.
$b$. $s-p$ overlapping: This involves the overlapping between a half-filled $s$-orbital of one atom and a half-filled $p$-orbital of another atom along the internuclear axis.
$c$. $p-p$ overlapping: This involves the head-on overlapping of two half-filled $p$-orbitals along the internuclear axis.
$2$. $\pi$-bond: In this type of covalent bond formation,the axes of the overlapping atomic orbitals remain parallel to each other and perpendicular to the internuclear axis,resulting in lateral (side-wise) overlapping.

(A) The $\sigma$ bond is stronger than the $\pi$ bond.
The strength of a covalent bond depends on the extent of overlapping of atomic orbitals.
In the formation of a $\sigma$ bond,the atomic orbitals overlap along the internuclear axis (head-on overlap),which results in a greater extent of overlapping.
In the formation of a $\pi$ bond,the atomic orbitals overlap sideways (lateral overlap),which results in a smaller extent of overlapping.
Since the strength of a bond is directly proportional to the magnitude of overlapping,the $\sigma$ bond is stronger than the $\pi$ bond.

(N/A) In $C_2H_4$ (Ethene),the carbon atoms undergo $sp^2$ hybridization. One $C-C$ $\sigma$ bond is formed by the overlap of $sp^2$ orbitals,and one $\pi$ bond is formed by the lateral overlap of unhybridized $2p_x$ orbitals.
In $C_2H_2$ (Ethyne),the carbon atoms undergo $sp$ hybridization. One $C-C$ $\sigma$ bond is formed by the overlap of $sp$ orbitals,and two $\pi$ bonds are formed by the lateral overlap of unhybridized $2p_x$ and $2p_y$ orbitals.

(N/A) The structure of $C_2H_2$ is $H-C \equiv C-H$.
In the triple bond between the two carbon atoms,there is $1$ $\sigma$ bond and $2$ $\pi$ bonds.
There are also $2$ $C-H$ $\sigma$ bonds.
Thus,$C_2H_2$ has a total of $3$ $\sigma$ bonds and $2$ $\pi$ bonds.
$(b)$ The structure of $C_2H_4$ is $H_2C=CH_2$.
There are $4$ $C-H$ $\sigma$ bonds and in the $C=C$ double bond,there is $1$ $\sigma$ bond and $1$ $\pi$ bond.
Thus,$C_2H_4$ has a total of $5$ $\sigma$ bonds and $1$ $\pi$ bond.

(C) sigma $(\sigma)$ bond is formed by the head-on (axial) overlap of atomic orbitals along the internuclear axis.
If the $X$-axis is the internuclear axis,$2p_y$ orbitals (which are oriented along the $Y$-axis) are perpendicular to the internuclear axis.
Their lateral (sideways) overlap results in the formation of a pi $(\pi)$ bond,not a sigma $(\sigma)$ bond.
On the other hand,$s-s$ and $s-p_x$ overlaps along the $X$-axis are axial and form $\sigma$ bonds.

(N/A) In figure $(i)$,the $s$-orbital overlaps with both lobes of the $p_x$-orbital. The overlap between the $s$-orbital and the positive lobe of the $p_x$-orbital is positive,while the overlap with the negative lobe is negative. Since these two overlaps are equal in magnitude and opposite in sign,the net overlap is zero,preventing bond formation.
In figure $(ii)$,the $p_x$-orbital and $p_y$-orbital have different symmetries with respect to the internuclear axis ($z$-axis). Due to this mismatch in symmetry,they cannot overlap effectively to form a bond.

(N/A) According to the $VB$ theory,a bond is not formed if the two atomic orbitals are not oriented along the common internuclear axis,as this leads to zero overlap.
Examples:
$1$. The overlap between an $s$-orbital and a $p_x$-orbital oriented perpendicular to the internuclear axis ($z$-axis) results in zero overlap.
$2$. The overlap between a $p_x$-orbital and a $p_y$-orbital oriented along the same internuclear axis ($z$-axis) also results in zero overlap.

(A) The number of $\sigma$ and $\pi$ bonds are calculated as follows:
$(i)$ Methane $(CH_4)$: It has $4$ $C-H$ single bonds. Total: $4\sigma, 0\pi$.
$(ii)$ Ethane $(C_2H_6)$: It has $1$ $C-C$ single bond and $6$ $C-H$ single bonds. Total: $7\sigma, 0\pi$.
$(iii)$ Ethene $(C_2H_4)$: It has $1$ $C=C$ double bond $(1\sigma, 1\pi)$ and $4$ $C-H$ single bonds. Total: $5\sigma, 1\pi$.
$(iv)$ Ethyne $(C_2H_2)$: It has $1$ $C\equiv C$ triple bond $(1\sigma, 2\pi)$ and $2$ $C-H$ single bonds. Total: $3\sigma, 2\pi$.

(A) $\sigma$-bond is formed by the head-on (axial) overlap of atomic orbitals.
$1$. $1s-1s$ overlap: Forms a $\sigma$-bond.
$2$. $2s-2s$ overlap: Forms a $\sigma$-bond.
$3$. $1s-2s$ overlap: Forms a $\sigma$-bond.
$4$. $2p_z-2p_z$ overlap (assuming $z$-axis as the internuclear axis): Forms a $\sigma$-bond.
$2p_x-2p_x$ and $2p_y-2p_y$ overlaps result in $\pi$-bonds because they overlap laterally (sideways).

(N/A) $\pi$-bond is formed by the lateral or side-wise overlapping of $2p_x-2p_x$ or $2p_y-2p_y$ orbitals.
This overlapping occurs only when these participating orbitals are parallel to each other and perpendicular to the internuclear axis.

(NONE) No bond is formed between $2p_x$ and $2p_y$ orbitals.
Because $2p_x$ and $2p_y$ orbitals are mutually perpendicular to each other,they cannot overlap effectively to form a chemical bond.

(N/A) Resemblance: Both $\sigma$ and $\pi$ bonds are covalent bonds formed by the overlapping of atomic orbitals. Both involve the sharing of electron pairs between two atoms.
Differences:
$1$. $\sigma$ bond is formed by the head-on (axial) overlapping of atomic orbitals, whereas $\pi$ bond is formed by the lateral (sideways) overlapping of atomic orbitals.
$2$. $\sigma$ bond allows free rotation of atoms around the bond axis, while $\pi$ bond restricts rotation.
$3$. $\sigma$ bond is stronger due to greater extent of overlapping, whereas $\pi$ bond is weaker due to lesser extent of overlapping.
$4$. $\sigma$ bond can exist independently, whereas $\pi$ bond is always formed in addition to a $\sigma$ bond.

(B) By convention,the internuclear axis is taken as the $z$-axis.
$(i) \ 2p_y - 2p_y$ overlap: Since the $p_y$ orbitals are perpendicular to the $z$-axis,they undergo lateral (sideways) overlap,forming a $\pi$ bond,not a $\sigma$ bond.
$(ii) \ 2p_z - 2p_z$ overlap: These orbitals are oriented along the internuclear axis,resulting in head-on (axial) overlap,which forms a $\sigma$ bond.
$(iii) \ 2s - 2p_x$ overlap: The $2s$ orbital is spherical and the $2p_x$ orbital is perpendicular to the $z$-axis. Their overlap is zero (non-bonding) because of symmetry.
$(iv) \ 2p_x - 2p_x$ overlap: Similar to $p_y$,these orbitals are perpendicular to the $z$-axis and undergo lateral overlap,forming a $\pi$ bond,not a $\sigma$ bond.
Therefore,combinations $(i)$,$(iii)$,and $(iv)$ do not form a $\sigma$ bond. Given the options,$(i)$ and $(iv)$ are the most appropriate choices for lateral overlap.

(N/A) The number of $\sigma$ and $\pi$ bonds are calculated as follows:
$(i)$ $O_2$ $(O=O)$: $1 \sigma, 1 \pi$.
$(ii)$ $N_2$ $(N\equiv N)$: $1 \sigma, 2 \pi$.
$(iii)$ $C_6H_6$ (Benzene): It has $12 \sigma$ bonds ($6$ $C-C$ and $6$ $C-H$) and $3 \pi$ bonds.
$(iv)$ $O_3$ (Ozone): The resonance structure shows $2 \sigma$ bonds and $1 \pi$ bond (delocalized over the structure).

(N/A) In an alkene,each carbon atom undergoes $sp^{2}$ hybridization.
The $\sigma$-bond between the two carbon atoms is formed by the head-on (axial) overlapping of one $sp^{2}$ hybrid orbital from each carbon atom.
The $\pi$-bond is formed by the lateral (sideways) overlapping of the unhybridized $2p$ orbitals of the two carbon atoms,which are parallel to each other and perpendicular to the molecular plane.

(D) Zero overlap occurs when the orbitals do not have the proper symmetry or orientation to overlap effectively.
Assertion $A$ is false because zero overlap is not necessarily an out-of-phase overlap (which refers to destructive interference of wave functions). Zero overlap often occurs due to improper orientation or symmetry,such as the overlap between $s$ and $p_y$ orbitals along the $x$-axis.
Reason $R$ is true because zero overlap is indeed caused by the different orientation or direction of approach of the orbitals relative to each other,as shown in the provided figure where $p_x$ and $p_y$ orbitals do not overlap effectively.
Therefore,$A$ is false but $R$ is true.

(A) When orbitals of two atoms come close to form a bond,their overlap may be positive,negative,or zero depending upon the sign and the direction of orientation of the amplitude of the orbital wave function in space.
In the case of non-bonding overlap,there is no effective overlap because the positive and negative regions cancel each other out,resulting in a net overlap of zero.
This occurs when the symmetry of the orbitals is such that they cannot interact effectively,such as the overlap between a $p_x$ orbital and a $p_z$ orbital,where the positive lobe of one overlaps with both the positive and negative lobes of the other equally,leading to zero net overlap as shown in option $(A)$.

(D) The chemical formula of ethylene is $CH_2=CH_2$.
In this molecule,there are $4$ $C-H$ $\sigma$ bonds and $1$ $C-C$ $\sigma$ bond,making a total of $5$ $\sigma$ bonds.
There is also $1$ $C-C$ $\pi$ bond.
Therefore,the number of $\sigma$ and $\pi$ bonds is $5$ and $1$ respectively.

(D) Statement $(I)$ is incorrect because the oxidation state is defined based on electronegativity,not electron gain enthalpy.
Statement $(II)$ is correct because elements of the second period (like $C, N, O$) have small atomic sizes,which allow for effective side-on overlap of $2p$ orbitals,leading to strong $p\pi-p\pi$ bonding.
Therefore,Statement $(I)$ is incorrect but Statement $(II)$ is correct.

(C) Based on the orbital overlap diagrams:
$P$: Two $d$-orbitals overlapping axially in the same phase result in $d-d$ $\sigma$ bonding $(P-2)$.
$Q$: $A$ $p$-orbital and a $d$-orbital overlapping laterally in the same phase result in $p-d$ $\pi$ bonding $(Q-3)$.
$R$: $A$ $p$-orbital and a $d$-orbital overlapping laterally in opposite phases result in $p-d$ $\pi$ antibonding $(R-1)$.
$S$: Two $d$-orbitals overlapping axially in opposite phases result in $d-d$ $\sigma$ antibonding $(S-4)$.
Thus,the correct matching is $P-2, Q-3, R-1, S-4$.

(D) $\sigma$ bond is formed by the head-on (axial) overlapping of atomic orbitals along the internuclear axis.
Given the internuclear axis is $z$:
$1$. $p_{z} \& d_{z^2}$ overlap head-on along the $z$-axis to form a $\sigma$ bond.
$2$. $p_{z} \& p_{z}$ overlap head-on along the $z$-axis to form a $\sigma$ bond.
$3$. $s \& p_{z}$ overlap head-on along the $z$-axis to form a $\sigma$ bond.
Since all the given combinations result in head-on overlapping along the $z$-axis,all of them form a $\sigma$ bond.

(A) Zero overlap occurs when the symmetry of the atomic orbitals is such that the positive and negative regions of the overlapping orbitals cancel each other out,or when the orbitals are oriented in a way that they cannot effectively interact to form a bond.
In option $A$,the $P_x$ orbital lies in the $xy$-plane and the $P_y$ orbital also lies in the $xy$-plane. When placed along the $Z$-axis,their symmetry is such that they cannot overlap to form a bond,resulting in zero overlap.
In option $C$,two $P_x$ orbitals are placed side-by-side along the $Z$-axis. Since they are oriented perpendicular to the internuclear axis ($Z$-axis),they can undergo lateral overlap to form a $\pi$-bond,so this is not a zero-overlap condition.
Therefore,the correct representation of zero overlap is given by the combination of $P_x$ and $P_y$ orbitals.

(D) The chemical formula of $2-$formylbenzoic acid is $C_8H_6O_3$.
Its structure consists of a benzene ring substituted with a carboxylic acid group $(-COOH)$ and an aldehyde group $(-CHO)$ at the ortho position.
Counting the bonds:
- There are $12$ $\sigma$ bonds in the benzene ring and its substituents (including $C-H$,$C-C$,$C=O$,$C-O$,and $O-H$ bonds).
- Specifically,the structure contains $17$ $\sigma$ bonds and $5$ $\pi$ bonds (three from the benzene ring,one from the $C=O$ of the aldehyde,and one from the $C=O$ of the carboxylic acid).

(D) Aspirin is acetylsalicylic acid with the chemical formula $C_9H_8O_4$.
Its structure consists of a benzene ring,a carboxylic acid group $(-COOH)$,and an ester group $(-OCOCH_3)$.
Counting the bonds:
$1$. The benzene ring has $6$ $C-C$ $\sigma$ bonds,$3$ $C-C$ $\pi$ bonds,and $4$ $C-H$ $\sigma$ bonds.
$2$. The carboxylic acid group has $1$ $C-C$ $\sigma$ bond,$1$ $C=O$ $\sigma$ bond,$1$ $C=O$ $\pi$ bond,$1$ $C-O$ $\sigma$ bond,and $1$ $O-H$ $\sigma$ bond.
$3$. The ester group has $1$ $C-O$ $\sigma$ bond,$1$ $C=O$ $\sigma$ bond,$1$ $C=O$ $\pi$ bond,$1$ $C-C$ $\sigma$ bond,and $3$ $C-H$ $\sigma$ bonds.
Total $\sigma$ bonds = $6 (ring) + 4 (C-H) + 1 (C-C) + 1 (C=O) + 1 (C-O) + 1 (O-H) + 1 (C-O) + 1 (C=O) + 1 (C-C) + 3 (C-H) = 21$.
Total $\pi$ bonds = $3 (ring) + 1 (C=O) + 1 (C=O) = 5$.
Therefore,the number of $\sigma$ and $\pi$ bonds are $21$ and $5$ respectively.

(B) The chemical formula of acetylene is $C_2H_2$,and its structural formula is $H-C \equiv C-H$.
In the triple bond between the two carbon atoms,there is one $\sigma$ bond (formed by the head-on overlap of $sp$ hybrid orbitals) and two $\pi$ bonds (formed by the lateral overlap of unhybridized $2p$ orbitals).