Overlaping - s and p- bonds Questions in English

(C) 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 a fluorine molecule $(F_2)$,the half-filled $2p_z$ orbital of one fluorine atom overlaps with the half-filled $2p_z$ orbital of another fluorine atom.
This is an example of $p - p$ overlapping along the internuclear axis,resulting in the formation of a sigma bond.

(B) The correct answer is $B$.
$\pi$-bonds are formed by the lateral (sideways) overlapping of unhybridised $p-p$ orbitals.
Axial overlapping of orbitals results in the formation of a $\sigma$-bond.

(C) $\pi$ bond is formed by the lateral or sidewise overlapping of half-filled atomic orbitals (typically $p-$orbitals) that are parallel to each other and perpendicular to the internuclear axis.
This type of overlapping results in the formation of two electron clouds,one above and one below the plane of the nuclei.
Therefore,the correct statement is that it is formed by the sidewise overlapping of half-filled $p-$orbitals.

(B) In ethylene $(CH_2=CH_2)$,each carbon atom is $sp^2$ hybridized.
One $sp^2$ orbital of each carbon atom overlaps head-on to form a $C-C$ sigma $(\sigma)$ bond.
The remaining unhybridized $p$-orbital on each carbon atom overlaps laterally (sideways) to form a pi $(\pi)$ bond.
Therefore,a double bond consists of one sigma $(\sigma)$ bond and one pi $(\pi)$ bond.

(C) $\sigma$ bond is a covalent bond formed by the head-on (end-to-end) overlap of atomic orbitals along the internuclear axis.
This type of overlap allows for the maximum electron density to be concentrated between the two nuclei.

(A) $\sigma$ bond is formed by head-on overlapping of atomic orbitals, resulting in greater electron density between the nuclei, which makes it stronger than a $\pi$ bond formed by lateral overlapping.
Therefore, the statement that a $\sigma$ bond is weaker than a $\pi$ bond is incorrect.

(A) The strength of a covalent bond is directly proportional to the extent of overlapping of the atomic orbitals.
When atomic orbitals undergo maximum overlapping,a greater amount of energy is released,which leads to the formation of a more stable and stronger bond.

(D) The $p-p$ orbital overlapping occurs when the $p$-orbitals of two atoms overlap along the internuclear axis.
In the $Cl_2$ molecule,each chlorine atom has an unpaired electron in its $3p_z$ orbital.
The bond is formed by the head-on overlapping of the $3p_z$ orbital of one chlorine atom with the $3p_z$ orbital of another chlorine atom.
Therefore,$Cl_2$ exhibits $p-p$ orbital overlapping.

(A) The electronic configuration of nitrogen $(N)$ is $1s^2 2s^2 2p^3$.
To complete its octet,each nitrogen atom shares $3$ electrons with another nitrogen atom,forming a triple bond $(N \equiv N)$.
$A$ triple bond consists of $1$ $\sigma$ bond (formed by head-on overlap of $p_z$ orbitals) and $2$ $\pi$ bonds (formed by lateral overlap of $p_x$ and $p_y$ orbitals).
Therefore,the $N_2$ molecule contains $1$ $\sigma$ and $2$ $\pi$ bonds.

(C) The nitrogen molecule $(N_2)$ has a triple bond between the two nitrogen atoms.
According to valence bond theory,the triple bond consists of one sigma $(\sigma)$ bond and two pi $(\pi)$ bonds.
Therefore,the correct composition is $1\sigma$ and $2\pi$ bonds,which corresponds to $1s$ and $2p$ orbitals involved in the bonding.

(C) $H-C \equiv C-H$ has $2 \pi$-bonds.
$(b)$ $CH_2=CH-CHO$ has $2 \pi$-bonds (one in $C=C$ and one in $C=O$).
$(c)$ $CH_3CH=CH_2$ has $1 \pi$-bond.
$(d)$ $CH_3CH=CH-COOH$ has $2 \pi$-bonds (one in $C=C$ and one in $C=O$).

(C) In a carbon-carbon triple bond,the first bond formed is a $\sigma$ bond due to the head-on overlap of $sp$ hybrid orbitals.
The remaining two bonds are formed by the lateral overlap of unhybridized $p$-orbitals,resulting in two $\pi$ bonds.
Therefore,a triple bond consists of one $\sigma$ bond and $2 \pi$ bonds.

(B) In the structure of ethene $(CH_2=CH_2)$,the carbon-carbon double bond is formed by the overlap of atomic orbitals.
One bond is a $\sigma$ (sigma) bond formed by the head-on overlap of $sp^2$ hybrid orbitals.
The second bond is a $\pi$ (pi) bond formed by the lateral (sideways) overlap of unhybridized $p$-orbitals.
Therefore,a double bond consists of one $\sigma$ bond and one $\pi$ bond.

(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,totaling $5$ sigma $(\sigma)$ bonds.
There is $1$ $C-C$ pi $(\pi)$ bond present in the double bond.
Therefore,the total number of bonds is $5\sigma$ and $1\pi$.

(A) The $p$-orbitals $(p_x, p_y, p_z)$ are oriented along the $x, y,$ and $z$ axes respectively.
Since these axes are mutually perpendicular to each other,the angle between any two $p$-orbitals is $90^{\circ}$.

(C) $\pi$ bond is formed by the lateral or sideways overlapping of $P-P$ orbitals that are perpendicular to the internuclear axis.

(A) The chemical formula of benzene is $C_6H_6$.
In the structure of benzene,there are $6$ $C-C$ $\sigma$ bonds and $6$ $C-H$ $\sigma$ bonds,making a total of $12$ $\sigma$ bonds.
Additionally,there are $3$ $C=C$ double bonds,each containing one $\pi$ bond,resulting in a total of $3$ $\pi$ bonds.
Therefore,the structure contains $12$ $\sigma$ bonds and $3$ $\pi$ bonds.

(D) The internuclear axis is $Z$. The $P_x$ orbital has its electron density along the $X$-axis,and the $P_y$ orbital has its electron density along the $Y$-axis. Since these orbitals are perpendicular to each other and to the internuclear axis,they have different symmetries with respect to the internuclear axis. Therefore,they cannot overlap to form a bond. Thus,no bond will be formed.

(B) The statement that the sigma bond is weaker than the pi bond is incorrect. In reality,the sigma bond is stronger than the pi bond because of the greater extent of orbital overlap in a head-on collision compared to the lateral overlap in a pi bond. Furthermore,the bond energy values provided in option $C$ are also incorrect as the sigma bond energy is typically higher than the pi bond energy.

(A) The electronic configuration of $F$ $(Z=9)$ is $1s^2 2s^2 2p_x^2 2p_y^2 2p_z^1$.
In the $F_2$ molecule,the half-filled $2p_z$ orbital of one $F$ atom overlaps with the half-filled $2p_z$ orbital of another $F$ atom along the internuclear axis.
This is known as $p-p$ axial overlapping,which results in the formation of a $\sigma$ bond.

(B) In $PO_4^{3-}$,the central atom is Phosphorus $(P)$,which has vacant $d$-orbitals available for bonding.
Oxygen atoms have $p$-orbitals that can overlap with the $d$-orbitals of Phosphorus to form $p\pi - d\pi$ bonds.
In $NO_3^-$,$CO_3^{2-}$,and $NO_2^-$,the central atoms are Nitrogen $(N)$ and Carbon $(C)$,which belong to the second period and do not have $d$-orbitals.
Therefore,only $PO_4^{3-}$ exhibits $p\pi - d\pi$ overlap.

(D) The $HCl$ molecule is formed by the overlap of the $1s$ orbital of the hydrogen atom and the $3p_z$ orbital of the chlorine atom. Therefore,the overlap involved is $s - p$.

(D) The effectiveness of overlapping depends on the $s$-character in the hybrid orbitals. Higher $s$-character leads to more effective overlapping. The $s$-character in $sp$,$sp^2$,and $sp^3$ hybrid orbitals is $50\%$,$33.3\%$,and $25\%$ respectively. Therefore,$sp - sp$ overlapping is the most effective.

(B) The lateral (side-by-side) overlap of $p-p$ orbitals results in the formation of a pi $(\pi)$ bond.

(A) The linear overlap of two hybrid orbitals results in the formation of a $Sigma$ bond.

(A) The $\pi$-bond is formed by the lateral overlap of $p$-orbitals perpendicular to the molecular plane.
Since the electron density of the $\pi$-bond is zero in the molecular plane,the molecular plane itself acts as the nodal plane for the $\pi$-bond in ethene.

(B) The chemical formula of benzene is $C_6H_6$.
In the structure of benzene,there are $6$ $C-C$ $\sigma$ bonds and $6$ $C-H$ $\sigma$ bonds,making a total of $12$ $\sigma$ bonds.
There are $3$ $\pi$ bonds present in the ring due to the alternating double bonds.
The ratio of $\sigma$ bonds to $\pi$ bonds is $\frac{12}{3} = 4$.

(A) In benzene $(C_6H_6)$,each carbon atom is $sp^2$ hybridized.
These $sp^2$ hybridized orbitals overlap with each other to form the $C-C$ sigma bonds in the hexagonal ring.
Additionally,the unhybridized $p$-orbitals on each carbon atom overlap laterally to form the $\pi$-electron cloud above and below the plane of the ring.
However,the question asks for the overlap that forms the backbone of the ring structure,which is the $sp^2-sp^2$ overlap.

(A) Ethyne $(HC \equiv CH)$ contains a triple bond between the two carbon atoms.
In a triple bond,there is $1$ sigma $(\sigma)$ bond and $2$ pi $(\pi)$ bonds.
The sigma bond is formed by the head-on overlap of $sp$ hybridized orbitals,while the two pi bonds are formed by the lateral overlap of two unhybridized $p$-orbitals on each carbon atom.

(A) In the ethene molecule $(C_2H_4)$,the two carbon atoms are $sp^2$ hybridized.
The $C-C$ $\sigma$-bond and the four $C-H$ $\sigma$-bonds lie in the same plane,known as the molecular plane.
The $\pi$-bond is formed by the lateral overlap of the unhybridized $2p_z$ orbitals of the carbon atoms,which are perpendicular to the molecular plane.
$A$ nodal plane is a plane where the probability of finding an electron is zero.
For the $\pi$-bond in ethene,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 structure of the compound is $CH_2=CH-CH=CH-C(=O)OH$.
Counting the $\pi$ bonds:
$1$. There is one $\pi$ bond in the first $C=C$ double bond.
$2$. There is one $\pi$ bond in the second $C=C$ double bond.
$3$. There is one $\pi$ bond in the $C=O$ double bond of the carboxylic acid group.
Total number of $\pi$ bonds = $1 + 1 + 1 = 3$.

(D) The given structure is benzonitrile $(C_6H_5CN)$.
Structure analysis:
- The benzene ring has $3$ $\pi$ bonds.
- The $-CN$ group has a triple bond,which contains $2$ $\pi$ bonds.
- Total $\pi$ bonds = $3 + 2 = 5$.
Counting $\sigma$ bonds:
- Benzene ring $(C_6H_6)$: $6$ $C-C$ $\sigma$ bonds + $5$ $C-H$ $\sigma$ bonds = $11$ $\sigma$ bonds.
- Bond between ring and $-CN$ group: $1$ $\sigma$ bond.
- $C \equiv N$ group: $1$ $\sigma$ bond.
- Total $\sigma$ bonds = $11 + 1 + 1 = 13$.
Thus,the number of $\pi$ bonds is $5$ and $\sigma$ bonds is $13$.

(A) The structure of tetracyanoethylene is $(NC)_2C=C(CN)_2$.
$\sigma$ bonds:
- $1$ bond from the central $C=C$ double bond.
- $4$ bonds from the $C-C$ single bonds connecting the central carbons to the cyano groups.
- $4$ bonds from the $C-N$ triple bonds (one $\sigma$ bond per triple bond).
Total $\sigma$ bonds = $1 + 4 + 4 = 9$.
$\pi$ bonds:
- $1$ bond from the central $C=C$ double bond.
- $8$ bonds from the $4$ $C \equiv N$ triple bonds (two $\pi$ bonds per triple bond).
Total $\pi$ bonds = $1 + 8 = 9$.
Thus,there are $9$ $\sigma$ and $9$ $\pi$ bonds.

(B) The sidewise overlapping of $p$-orbitals leads to the formation of a $\pi$-bond.
In the given options,we look for a species containing a $p\pi-p\pi$ bond involving a chalcogen (Group $16$) atom.
In $CS_3^{2-}$ (trithiocarbonate ion),the structure involves a central carbon atom double-bonded to sulfur atoms.
The $C=S$ bond is formed by the sidewise overlapping of the $2p$-orbital of carbon and the $2p$-orbital of sulfur.
Therefore,$CS_3^{2-}$ contains a bond formed by the sidewise overlapping of $2p$-orbitals.

(A) For the $z$-axis as the internuclear axis:
$1$. $p_z + p_z$ orbitals overlap head-on to form a $\sigma$-bond.
$2$. $p_y + p_y$ orbitals overlap laterally to form a $\pi$-bond.
$3$. $d_{z^2} + d_{z^2}$ orbitals overlap head-on to form a $\sigma$-bond.
$4$. $d_{xy} + d_{xz}$ orbitals do not have the correct symmetry for effective overlap along the $z$-axis,hence no $\sigma$ or $\pi$ bond is formed.

(C) $\pi$-bond is formed by the lateral or side-by-side overlapping of atomic orbitals.
For $p_x$-orbitals,the electron density is concentrated along the $x$-axis.
If the orbitals approach along the $x$-axis,they undergo head-on overlapping,which results in the formation of a $\sigma$-bond.
If the $p_x$-orbitals approach each other along the $y$-axis or $z$-axis,they will overlap laterally (side-by-side),resulting in the formation of a $\pi$-bond.
Therefore,the correct axis for $\pi$-bond formation between $p_x$-orbitals is the $y$-axis or $z$-axis.

(A) If the internuclear axis is the $y-$ axis:
$1$. Orbitals overlapping along the $y-$ axis form $\sigma$ bonds (e.g.,$p_y - p_y$,$s - p_y$,$d_{y^2} - d_{y^2}$).
$2$. Orbitals overlapping perpendicular to the $y-$ axis form $\pi$ bonds.
$3$. For option $A$: $d_{xy}$ orbitals have lobes in the $xy$ plane. When two $d_{xy}$ orbitals approach along the $y-$ axis,they overlap sideways to form a $\pi$ bond.
$4$. For option $B$: $p_z$ orbitals are perpendicular to the $y-$ axis,so they form $\pi$ bonds,not $\sigma$ bonds.
$5$. For option $C$: $d_{x^2 - y^2}$ orbitals have lobes along the $x$ and $y$ axes. Overlap along the $y-$ axis results in a $\sigma$ bond due to the $y$ component.
$6$. For option $D$: $p_y$ (along $y$) and $d_{zx}$ (in $zx$ plane) have no effective overlap symmetry for a $\pi$ bond.

(C) The structure of acetic acid $(CH_3COOH)$ is $CH_3-C(=O)-OH$.
Counting the bonds:
- There are $3$ $C-H$ $\sigma$ bonds.
- There is $1$ $C-C$ $\sigma$ bond.
- There is $1$ $C=O$ bond,which consists of $1$ $\sigma$ and $1$ $\pi$ bond.
- There is $1$ $C-O$ $\sigma$ bond.
- There is $1$ $O-H$ $\sigma$ bond.
Total $\sigma$ bonds = $3 + 1 + 1 + 1 + 1 = 7$.
Total $\pi$ bonds = $1$.
Thus,there are $7$ $\sigma$ and $1$ $\pi$ bond.

(B) If the molecular axis is the $x-$axis,then the $\pi-$ bond is formed by the lateral overlap of orbitals perpendicular to this axis,such as $p_y$ or $p_z$ orbitals.
For the strongest $\pi-$ bond,the overlap should involve orbitals with the same principal quantum number and effective spatial orientation.
$2p_x + 2p_x$ results in a $\sigma-$ bond because the orbitals are oriented along the molecular axis.
$2p_y + 2p_y$ results in a $\pi-$ bond with effective overlap due to the same principal quantum number $(n=2)$.
$2p_y + 3d_{xy}$ and $2p_z + 4p_z$ involve orbitals with different principal quantum numbers,leading to weaker overlap compared to $2p_y + 2p_y$.

(B) For bonding to occur,the overlapping orbitals must have the same phase (sign) in the region of overlap.
In option $(A)$,the $s$-orbital $(+)$ overlaps with the negative lobe of the $p$-orbital,leading to anti-bonding.
In option $(B)$,the $s$-orbital $(+)$ overlaps with the positive lobe of the $p$-orbital,which results in constructive interference and thus leads to bonding.
In option $(C)$,the $p$-orbitals overlap with opposite signs,leading to anti-bonding.
In option $(D)$,the $p$-orbital and $d$-orbital overlap with opposite signs,leading to anti-bonding.
Therefore,option $(B)$ represents a bonding interaction.

(A) To determine the hybridization of carbon atoms in the given molecules,we look at the structure of $CH_2=CH-C\equiv CH$ (Butenyne).
In Butenyne,the structure is $CH_2=CH-C\equiv CH$.
The carbon atoms are numbered as $C_1=C_2-C_3\equiv C_4$.
$C_1$ is $sp^2$ hybridized,$C_2$ is $sp^2$ hybridized,$C_3$ is $sp$ hybridized,and $C_4$ is $sp$ hybridized.
The bond between $C_2$ and $C_3$ is formed by the overlap of an $sp^2$ orbital from $C_2$ and an $sp$ orbital from $C_3$.
Therefore,Butenyne contains an $sp^2-sp$ sigma bond.

(D) In the molecule $HCN$,the structure is $H-C \equiv N$.
There is one $\sigma$ bond between $H$ and $C$.
There is one $\sigma$ bond and two $\pi$ bonds between $C$ and $N$.
Total bonds in $HCN$: $2\,\sigma$ bonds and $2\,\pi$ bonds.
Therefore,the correct option is $D$.

(D) The $d_{xy}$ orbital has lobes lying in the $xy$ plane between the $x$ and $y$ axes.
For two $d_{xy}$ orbitals to form a $\pi$ bond,they must approach each other such that their lobes overlap laterally.
If they approach along the $z$ axis,the lobes of the two $d_{xy}$ orbitals are parallel to each other,allowing for effective lateral overlap to form a $\pi$ bond.
Therefore,the $d_{xy}$ orbital forms a $\pi$ bond with another $d_{xy}$ orbital along the $z$ axis.

(A) In ethene $(C_2H_4)$,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 with hydrogen atoms and the other carbon atom.
The unhybridized $p_z$ orbitals are perpendicular to this molecular plane and overlap laterally to form the $\pi$ bond.
$A$ nodal plane is a plane where the probability of finding an electron is zero.
For a $\pi$ bond formed by the lateral overlap of $p$ orbitals,the molecular plane itself acts as the nodal plane because the electron density of the $\pi$ bond is concentrated above and below this plane,not within it.

(B) The fluorine molecule $(F_2)$ is formed by the axial overlap of $2p_z$ orbitals of two fluorine atoms.
This axial overlap is known as end-to-end overlap,which results in the formation of a sigma $(\sigma)$ bond.
Therefore,the correct description is $p-p$ orbitals (end-to-end overlap).

(D) Given the internuclear axis is the $Z$-axis:
$(A)$ $2p_y + 2p_y$ overlap leads to $\pi$-bond formation (Correct).
$(B)$ $2p_x + 2p_x$ overlap leads to $\pi$-bond formation,not $\sigma$ (Incorrect).
$(C)$ $3d_{xy} + 3d_{xy}$ overlap leads to $\delta$-bond formation,not $\pi$ (Incorrect).
$(D)$ $2s + 2p_y$ overlap results in no bond formation because of symmetry mismatch (Incorrect).
$(E)$ $3d_{xy} + 3d_{xy}$ overlap leads to $\delta$-bond formation (Correct).
$(F)$ $2p_z + 2p_z$ overlap leads to $\sigma$-bond formation (Correct).
Therefore,the incorrect overlaps are $(B)$,$(C)$,and $(D)$.

(C) The strength of a covalent bond depends on the extent of overlapping of atomic orbitals.
Greater the extent of overlapping,stronger is the bond.
For orbitals with the same principal quantum number $(n=2)$,the directional nature of the orbitals determines the extent of overlap.
The $p-p$ overlap (coaxial) is more effective than $p-s$ overlap,which in turn is more effective than $s-s$ overlap due to the directional character of $p$ orbitals.
Therefore,the order of bond strength is $p-p > p-s > s-s$.

(B) sigma bond is stronger than a $\pi$-bond because the extent of overlap is greater in sigma bond formation. This statement is correct.
$(B)$ The bond energy of a sigma bond is typically higher (approx. $347 \ kJ/mol$) than that of a $\pi$-bond (approx. $264 \ kJ/mol$). The statement in the option reverses these values,making it incorrect.
$(C)$ Free rotation of atoms about a sigma bond is allowed due to cylindrical symmetry,whereas a $\pi$-bond restricts rotation. This statement is correct.
$(D)$ A sigma bond determines the molecular geometry and direction between carbon atoms,while a $\pi$-bond does not. This statement is correct.

(D) The $s$-orbital is spherically symmetric and non-directional,allowing it to overlap with any orbital to form a bond.
For a bond formed along the $z$-axis,the orbitals must have a component along the $z$-axis.
$I$. $s$ and $p_x$ overlap: $p_x$ is perpendicular to the $z$-axis,resulting in zero net overlap.
$II$. $s$ and $p_z$ overlap: $p_z$ is directed along the $z$-axis,resulting in effective head-on overlap (bonding).
$III$. $p_y$ and $p_z$ overlap: These orbitals are mutually perpendicular,resulting in zero net overlap.
$IV$. $s$ and $s$ overlap: Both are spherically symmetric,resulting in effective overlap (bonding).
Therefore,combinations $II$ and $IV$ result in bonding.

(A) $1$. For $A$ (Tetracyanomethane,$C(CN)_4$): Structure is $C(CN)_4$. It has $4$ $C-C$ $\sigma$ bonds,$4$ $C\equiv N$ $\sigma$ bonds,and $8$ $\pi$ bonds. Total $\sigma = 8$,$\pi = 8$. Ratio $\sigma/\pi = 8/8 = 1$.
$2$. For $B$ (Carbon dioxide,$CO_2$): Structure is $O=C=O$. It has $2$ $\sigma$ bonds and $2$ $\pi$ bonds. Ratio $\sigma/\pi = 2/2 = 1$.
$3$. For $C$ (Benzene,$C_6H_6$): It has $12$ $\sigma$ bonds and $3$ $\pi$ bonds. Ratio $\sigma/\pi = 12/3 = 4$.
$4$. For $D$ ($1, 3$-Butadiene,$CH_2=CH-CH=CH_2$): It has $9$ $\sigma$ bonds and $2$ $\pi$ bonds. Ratio $\sigma/\pi = 9/2 = 4.5$.
Comparing the ratios: $A(1) = B(1) < C(4) < D(4.5)$.
Thus,the correct order is $A = B < C < D$.