Covalent bonding Questions in English

(C) In $B_2H_6$ (diborane),the two bridging hydrogen atoms are involved in $3c-2e^-$ (three-center two-electron) bonds,also known as banana bonds.
$Al_2Cl_6$ and $Fe_2Cl_6$ contain $3c-4e^-$ bonds (due to coordinate bonds from chlorine to metal).
$Si_2H_6$ (disilane) contains normal $2c-2e^-$ covalent bonds.
Therefore,only $B_2H_6$ contains $3c-2e^-$ bonds.

(B) In $SO_3^{2-}$,the central sulfur atom has an expanded octet and utilizes its vacant $d$-orbitals to form $\pi$-bonds with oxygen atoms.
Specifically,the lone pair on the oxygen atom is donated into the vacant $d$-orbital of the sulfur atom,resulting in $p\pi-d\pi$ bonding.
In contrast,$NO_3^-$,$BO_3^{3-}$,and $CO_3^{2-}$ involve elements from the second period ($N$,$B$,$C$),which lack $d$-orbitals,and thus only exhibit $p\pi-p\pi$ bonding.
Therefore,the correct option is $B$.

(D) Lewis acid is defined as an electron-pair acceptor.
$PH_3$ and $NF_3$ have a lone pair on the central atom,making them Lewis bases.
$NaH$ is an ionic hydride containing the hydride ion $(H^-)$,which acts as a Lewis base.
In $B(CH_3)_3$,the central Boron atom has only $6$ valence electrons (an incomplete octet),as shown in the structure. Therefore,it can accept an electron pair to complete its octet,making it a Lewis acid.

(B) The $CaC_2$ compound consists of $Ca^{2+}$ and $C_2^{2-}$ ions.
In the $C_2^{2-}$ ion,the two carbon atoms are linked by a triple bond.
$A$ triple bond consists of one $\sigma$ bond and two $\pi$ bonds.
Therefore,the $C_2^{2-}$ ion contains one $\sigma$ bond and two $\pi$ bonds.

(C) Delocalization of $\pi-$ electrons requires a conjugated system where $p-$ orbitals overlap across adjacent atoms.
$1$. In the pyrrole cation,the nitrogen atom is part of a cyclic system with $4n+2$ $\pi-$ electrons,allowing for delocalization.
$2$. In the anilinium ion $(C_6H_5NH_3^+)$,the $\pi-$ electrons within the benzene ring remain delocalized.
$3$. In $H_2C=C=CH_2$ (allene),the central carbon is $sp$ hybridized. The two $\pi-$ bonds are perpendicular to each other,preventing continuous $p-$ orbital overlap across the entire molecule. Thus,the $\pi-$ electrons are localized in their respective $\pi-$ bonds.
$4$. In the borirene derivative,the boron atom has an empty $p-$ orbital,which allows for delocalization of the $\pi-$ electrons from the double bond into the boron $p-$ orbital.
Therefore,the molecule where $\pi-$ electrons are not delocalized across the entire system is allene $(H_2C=C=CH_2)$.

(C) The bond energy depends on the extent of orbital overlap and the bond length.
Oxygen has a smaller atomic size compared to sulfur,and carbon has a smaller size compared to phosphorus.
Due to the smaller size of the $O$ atom,the $2p-2p$ sideways overlap between $C$ and $O$ is more effective than the $2p-3p$ overlap in $C=S$ or other combinations.
Additionally,oxygen is highly electronegative,which contributes to a stronger bond.
Therefore,the $C = O$ bond has the highest bond energy among the given options.

(D) Due to the presence of vacant $d$-orbitals,excitation occurs in the $Cl$ atom.
It can expand its octet and have more than eight valence electrons when forming hypervalent compounds like $HClO_4$ or $PCl_5$.

(A) The set of compounds $BCl_3$,$SiCl_4$,and $PCl_3$ consists of molecules formed by covalent bonding between non-metals.
$(B)$ $NH_4Br$ is an ionic compound containing $NH_4^+$ and $Br^-$ ions.
$(C)$ $NaI$ is an ionic compound.
$(D)$ $Al_2O_3$ is an ionic compound with a giant lattice structure.

(B) The Lewis structure of $CS_2$ is $S=C=S$,where each sulfur atom has two lone pairs and the carbon atom is the central atom.
Each sulfur atom is bonded to the carbon atom by a double bond.
Therefore,all bonds in the molecule are double bonds.
Both sulfur atoms and the central carbon atom satisfy the octet rule.

(D) When the internuclear distance is decreased below the optimum distance,the electron clouds of the two atoms begin to overlap significantly,and the nuclei come closer to each other.
This leads to a sharp increase in the electrostatic repulsions between the electrons of the two atoms (interelectronic repulsion) and between the two positively charged nuclei (internuclear repulsion).
These repulsive forces dominate over the attractive forces,causing a steep increase in the potential energy of the system.

(C) In $BF_3$,the boron atom has an empty $2p$ orbital,which allows for $p\pi - p\pi$ back-bonding from the fluorine lone pairs to the boron atom.
This back-bonding imparts partial double bond character to the $B-F$ bond,thereby shortening its length.
In the other compounds ($BF_4^-$,$BF_3 \leftarrow NH_3$,and $BF_3 \leftarrow N(CH_3)_3$),the boron atom is $sp^3$ hybridized and does not have a vacant $p$-orbital,so no back-bonding occurs.
Therefore,the $B-F$ bond length is shortest in $BF_3$.

(D) The bond energy is inversely proportional to the extent of lone pair-lone pair repulsion between the bonded atoms.
As the number of lone pairs (non-bonded electrons) on the bonded atoms increases,the repulsion between them increases,which weakens the bond and decreases the bond energy.
Comparing the number of non-bonded electrons:
$C-C$ has $3+3 = 6$ non-bonded electrons (in the provided structure).
$N-N$ has $4+4 = 8$ non-bonded electrons (in the provided structure).
$H-H$ has $0$ non-bonded electrons.
$O-O$ has $6+6 = 12$ non-bonded electrons (in the provided structure).
Since the $O-O$ bond has the highest number of non-bonded electrons,the inter-electronic repulsion is maximum,making the $O-O$ bond the weakest with the lowest bond energy.

(A) $Mg_2C_3 + 4H_2O \longrightarrow 2Mg(OH)_2 + C_3H_4$ (Propyne)
The structure of the $C_3^{4-}$ ion is $[C \equiv C - C]^{4-}$.
In this structure,there is one triple bond (containing $1$ $\sigma$ and $2$ $\pi$ bonds) and one single bond (containing $1$ $\sigma$ bond).
Total bonds: $2$ $\sigma$ bonds and $2$ $\pi$ bonds.

(B) An electron-deficient molecule is one in which the central atom has fewer than $8$ electrons in its valence shell,or the total number of valence electrons is insufficient to form standard two-center two-electron $(2c-2e)$ bonds for all atoms.
In $B_2H_6$ (diborane),each boron atom is bonded to two terminal hydrogen atoms by normal $2c-2e$ covalent bonds. The two boron atoms are linked by two bridging hydrogen atoms,forming two $3c-2e$ (three-center two-electron) bonds. Because of these $3c-2e$ bonds,the molecule is electron-deficient.
$LiH$ is an ionic hydride,$LiBH_4$ contains the $BH_4^-$ ion (which is electron-precise),and $B_3N_3H_6$ (borazine) is an electron-precise molecule.
Therefore,option $B$ is correct.

(C) Bond strength is directly proportional to bond dissociation energy.
Since $Cl_2$ has the highest bond dissociation energy $(58 \ kcal/mol)$,it possesses the strongest bond.
Although $F_2$ is the smallest,the $F-F$ bond is weaker than the $Cl-Cl$ bond due to significant inter-electronic repulsion between the lone pairs of the small fluorine atoms.

(B) The $HCl$ molecule exhibits polar covalent bonding.
This is due to the difference in electronegativity between the $H$ atom and the $Cl$ atom,which causes an unequal sharing of the bonding electron pair.

(B) At $25\,^oC$,hydrogen chloride $(HCl)$ exists as a polar covalent gas.

(A) The bond strength depends on the effective overlapping of orbitals.
In the $F - F$ bond,there is an effective overlapping between the $2p - 2p$ orbitals of similar energy.
Although the $F - F$ bond is weakened by inter-electronic repulsion due to the small size of the fluorine atom,it is generally considered stronger than the other interhalogen bonds listed here due to the effective $2p - 2p$ overlap compared to $2p - 3p$ or $3p - 4p$ overlaps.

(B) $P\pi - P\pi$ bonding involves the overlap of $2p$ orbitals of oxygen with the $2p$ orbitals of another atom.
In $P_4O_{10}$,phosphorus uses $3d$ orbitals for $d\pi - p\pi$ bonding.
In $N_2O_5$,nitrogen uses $2p$ orbitals,but the structure involves $p\pi - p\pi$ bonding between $N$ and $O$.
However,specifically in the context of $P\pi - P\pi$ bonding,it refers to the overlap of $2p$ orbitals of both atoms.
In $N_2O_5$,the $N$ atom $(2p)$ and $O$ atom $(2p)$ form $p\pi - p\pi$ bonds.
Therefore,the correct option is $B$.

(B) Element $C$ has the electronic configuration $1s^2, 2s^2, 2p^6, 3s^2, 3p^5$,which corresponds to Chlorine $(Cl)$.
Chlorine is a halogen with $7$ valence electrons and requires $1$ electron to complete its octet.
Therefore,two Chlorine atoms share one electron each to form a stable diatomic molecule,$Cl_2$ (or $C_2$ in this context).
The molecular orbital configuration for $C_2$ (if $C$ were Carbon) would be $KK [\sigma_{2s}]^2 [\sigma_{2s}^*]^2 [\pi_{2px}]^2 [\pi_{2py}]^2$,but here $C$ represents a halogen,so it forms a stable diatomic molecule $C_2$.

(B) Element $B$ has the valence shell configuration $3s^2, 3p^3$,which means it has $5$ valence electrons. It needs $3$ electrons to complete its octet.
Element $C$ has the valence shell configuration $3s^2, 3p^5$,which means it has $7$ valence electrons. It needs $1$ electron to complete its octet.
Both $B$ and $C$ are non-metals. When two non-metals combine,they share electrons to complete their octets,resulting in the formation of a covalent bond.
Therefore,the bond between $B$ and $C$ will be a covalent bond.

(D) In compound $(I)$,the lone pair of nitrogen is involved in $p\pi-d\pi$ back-bonding with the empty $d$-orbitals of silicon.
This reduces the availability of the lone pair for $p\pi-p\pi$ back-bonding with boron.
In compound $(II)$,carbon has no empty $d$-orbitals,so the lone pair of nitrogen is fully available for $p\pi-p\pi$ back-bonding with boron.
Therefore,the $\pi$-bond strength between $B$ and $N$ is greater in $(II)$ than in $(I)$.

(D) In $SO_2Cl_2$,the central $S$ atom is $sp^3$ hybridized.
It forms two $S-Cl$ sigma bonds and two $S=O$ double bonds.
Each $S=O$ bond consists of one $\sigma$ bond (formed by $p-p$ overlap) and one $p\pi - d\pi$ back-bonding (formed by the overlap of the filled $2p$ orbital of $O$ and the empty $3d$ orbital of $S$).
Since there are no $p\pi - p\pi$ bonds (which would require $p-p$ overlap for $\pi$ bonding),the correct answer is $SO_2Cl_2$.

(B) The octet rule states that atoms tend to gain,lose,or share electrons to achieve a stable configuration of $8$ electrons in their valence shell.
In $BF_3$,the central Boron atom has $3$ valence electrons and forms $3$ covalent bonds with Fluorine atoms.
This results in a total of $3 \times 2 = 6$ electrons around the Boron atom.
Since $6 < 8$,$BF_3$ is an electron-deficient molecule and does not follow the octet rule.
In $CO_2$,$H_2O$,and $PCl_3$,the central atoms ($C$,$O$,and $P$ respectively) satisfy the octet rule.

(B) covalent bond is formed by the mutual sharing of electrons between two atoms to achieve a stable electronic configuration (usually an octet).
Therefore,in a covalent bond,electrons are shared between the participating atoms.

(C) The electronic configurations are:
$(I) \, 1s^2$: Helium $(He)$,a noble gas,inert.
$(II) \, 1s^2 2s^2 2p^2$: Carbon $(C)$,which has $4$ valence electrons. It primarily forms covalent bonds but can form carbides (ionic) with electropositive metals.
$(III) \, 1s^2 2s^2 2p^5$: Fluorine $(F)$,which has $7$ valence electrons. It forms ionic bonds with metals and covalent bonds with non-metals.
$(IV) \, 1s^2 2s^2 2p^6$: Neon $(Ne)$,a noble gas,inert.
Between $II$ and $III$,$III$ (Fluorine) is the most electronegative element and is well-known for forming strong ionic bonds with metals and covalent bonds with non-metals. However,in the context of typical chemistry curriculum questions regarding the ability to form both,$III$ is the standard answer.

(B) The chemical formula of carbon suboxide is $C_3O_2$.
To determine the Lewis structure,we calculate the total number of valence electrons: $3 \times 4 (C) + 2 \times 6 (O) = 12 + 12 = 24$ electrons.
The structure must satisfy the octet rule for all atoms.
The linear arrangement is $O=C=C=C=O$.
Counting the electrons in this structure: Each double bond represents $4$ electrons.
There are $4$ double bonds,so $4 \times 4 = 16$ electrons.
Each oxygen atom has $2$ lone pairs ($4$ electrons),so $2 \times 4 = 8$ electrons.
Total electrons = $16 + 8 = 24$ electrons.
This matches the total valence electrons,confirming the structure is $O=C=C=C=O$,which corresponds to option $B$.

(A) Covalent compounds are generally non-polar in nature. According to the principle of 'like dissolves like',non-polar substances dissolve in non-polar solvents (such as benzene,carbon tetrachloride,etc.). Therefore,covalent compounds are soluble in non-polar solvents.

(D) When two identical non-metallic atoms form a covalent bond,they have the same electronegativity.
Therefore,the shared pair of electrons is attracted equally by both nuclei.
This results in an equal sharing of electrons between the two atoms.

(C) In an $N_2$ molecule,each nitrogen atom has $5$ valence electrons. To complete its octet,each nitrogen atom needs $3$ more electrons. Therefore,each nitrogen atom shares $3$ electrons with the other to form a triple covalent bond $(N \equiv N)$. Thus,the number of electrons shared by each $N$ atom is $3$.

(C) In $H_2O_2$ (Hydrogen peroxide),the structure is $H-O-O-H$.
The $O-H$ bond is a polar covalent bond due to the electronegativity difference between $O$ and $H$.
The $O-O$ bond is a non-polar covalent bond because the electronegativity difference between two identical oxygen atoms is zero.
Therefore,$H_2O_2$ contains both polar and non-polar bonds.

(A) When two atomic orbitals,each containing one unpaired electron,overlap linearly (head-on),they form a $\sigma$-bond. This is the fundamental concept of Valence Bond Theory $(VBT)$ for the formation of covalent bonds.

(C) According to the Pauli Exclusion Principle and the valence bond theory,for a stable covalent bond to form,the electrons in the overlapping orbitals must have opposite spins. If the electrons have the same spin,the Pauli Exclusion Principle prevents them from occupying the same region of space with the same quantum numbers. Consequently,the orbitals do not overlap effectively,and no chemical bond is formed. Therefore,both statements $a$ and $b$ are correct.

(B) In $SO_3^{2-}$ (sulfite ion),the central sulfur atom has an oxidation state of $+4$ and its valence shell configuration is $3s^2 3p^4$.
Sulfur forms three $S-O$ sigma bonds using $sp^3$ hybridization.
One lone pair is present on sulfur.
The remaining $3d$ orbitals of sulfur overlap with the $2p$ orbitals of oxygen to form $p \pi - d \pi$ bonds.
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.

(C) bond is considered non-polar if there is no electronegativity difference between the bonded atoms.
In the bonds $C-H$,$N-H$,and $O-H$,there is a significant electronegativity difference between the two different atoms,making these bonds polar.
In the case of $F-F$,two identical atoms are bonded together. Therefore,the electronegativity difference is zero,making the $F-F$ bond non-polar.

(C) The covalent radii are calculated as $r_X = \frac{1.00}{2} = 0.50 \mathring{A}$ and $r_C = \frac{1.54}{2} = 0.77 \mathring{A}$.
Using the Schomaker-Stevenson equation for bond length: $d_{C-X} = r_C + r_X - 0.09 |\chi_C - \chi_X|$.
Substituting the values: $d_{C-X} = 0.77 + 0.50 - 0.09 |2.0 - 3.0|$.
$d_{C-X} = 1.27 - 0.09(1) = 1.18 \mathring{A}$.

(C) The bond energy depends on the strength of the bond between atoms. In the given molecules ($N-N$,$F-F$,$C-C$,$O-O$),all are connected by single bonds.
$1$. In $F-F$,$O-O$,and $N-N$,the atoms possess lone pairs of electrons. These lone pairs cause significant inter-electronic repulsion,which increases the bond length and decreases the bond dissociation energy.
$2$. Carbon atoms in $C-C$ do not have lone pairs,resulting in less repulsion and a stronger,more stable bond compared to the others.
Therefore,the molecule with the highest bond energy is $C-C$.

(D) In option $C$,the structure shown is for the dimer $Al_2Cl_6$. However,the label provided is 'Aluminium chloride' $(AlCl_3)$,which exists as a monomer in the gas phase at high temperatures but as a dimer in the solid state. The structure shown is specifically the dimer.
In option $D$,the structure for $BCl_3$ is shown with lone pairs on the Boron atom,which is incorrect. $BCl_3$ is a trigonal planar molecule with $sp^2$ hybridization on the Boron atom and no lone pair on the Boron atom. The Chlorine atoms have lone pairs,but the Boron atom is electron-deficient. Therefore,option $D$ is the most incorrectly matched structure.

(C) Carborundum is silicon carbide $(SiC)$.
In $SiC$,each silicon atom is bonded to four carbon atoms and each carbon atom is bonded to four silicon atoms through strong covalent bonds.
This forms a giant three-dimensional network structure similar to diamond.
Therefore,the type of bonding present is a covalent bond.

(B) Lewis acid is defined as an electron pair acceptor.
$AlCl_3$ is an electron-deficient compound because the central $Al$ atom has only $6$ electrons in its valence shell.
Therefore,it can accept a lone pair of electrons to complete its octet,making it a Lewis acid.
$NaCl$ is an ionic salt,while $NH_2^-$ and $F^-$ are Lewis bases because they have lone pairs of electrons to donate.

(A) $AlCl_3$ is an electron-deficient compound because the $Al$ atom has only $6$ electrons in its valence shell.
Due to this,it acts as a Lewis acid.
In its anhydrous form,$AlCl_3$ exists as a dimer $Al_2Cl_6$ with covalent bonds,where chlorine atoms act as bridges.
Therefore,$AlCl_3$ is anhydrous and covalent.

(D) An electron-deficient molecule is one in which the central atom has fewer than $8$ electrons in its valence shell,or the molecule has fewer electrons than required for conventional Lewis structures.
In $B_2H_6$ (diborane),each Boron atom is bonded to three terminal Hydrogen atoms and two bridging Hydrogen atoms.
Each Boron atom has only $6$ valence electrons in its bonding environment,making it an electron-deficient molecule (often referred to as a $3$-center-$2$-electron bond system).
$C_2H_6$,$SiH_4$,and $PH_3$ all follow the octet rule for their central atoms.

(D) In $BF_3$,boron is $sp^2$ hybridized and the molecule is planar.
Due to the vacant $2p$ orbital on the boron atom and the filled $2p$ orbitals on the fluorine atoms,there is a $p\pi - p\pi$ back-bonding (back-donation) from $F$ to $B$.
This gives the $B-F$ bond partial double bond character,making it shorter and stronger than a typical single bond.
Therefore,the statement that it is explained by $p\pi - d\pi$ interaction is incorrect,as boron does not have $d$-orbitals.

(B) In $BCl_3$,the boron atom has an incomplete octet. The lone pair of electrons on the chlorine atom is donated to the vacant $p$-orbital of the boron atom,resulting in $p\pi - p\pi$ back bonding. This stabilizes the molecule and satisfies the octet of boron,preventing dimerization. In $BH_3$,hydrogen does not have a lone pair to participate in back bonding,so $BH_3$ dimerizes to form $B_2H_6$ to complete the octet of boron.

(B) In diamond,each carbon atom is $sp^3$ hybridized and is bonded to four other carbon atoms through strong $C-C$ covalent bonds. This creates a rigid,three-dimensional network structure,which is responsible for the hardness of diamond. Therefore,the correct option is $B$.

(C) The Lewis structure of carbon monoxide $(CO)$ consists of a triple bond between the carbon and oxygen atoms.
This triple bond is composed of one $\sigma$ bond and two $\pi$ bonds.
Therefore,the atoms are bonded by both $\sigma$ and $\pi$ bonds.

(B) The bond dissociation enthalpy depends on the size of the atoms and the lone pair-lone pair repulsion.
For $O-O$,$N-N$,and $F-F$,the atoms are small and have lone pairs that cause significant inter-electronic repulsion,weakening the bond.
$C-C$ is a single covalent bond between carbon atoms which does not suffer from such lone pair repulsion.
Therefore,the $C-C$ bond has a higher bond dissociation enthalpy compared to the others listed.

(C) In $BF_3$,the Boron atom is electron-deficient and has an empty $2p$ orbital. The Fluorine atom has lone pairs in its $2p$ orbitals.
This allows for $p\pi - p\pi$ back-bonding (donation of electron density from $F$ to $B$),which imparts partial double bond character to the $B-F$ bond.
This additional bond character increases the bond strength and consequently the bond dissociation energy.
In $CF_4$,Carbon has a complete octet and no empty $p$ orbitals,so no such back-bonding occurs.
Therefore,the $B-F$ bond is stronger than the $C-F$ bond.

(D) The covalent character of a bond is related to the electronegativity difference and the formal charge on the atoms involved.
In $PH_4^+$,the phosphorus atom carries a positive charge.
$A$ positive charge on an atom increases its electronegativity,which decreases the electronegativity difference between $P$ and $H$.
However,the $P-H$ bond in $PH_4^+$ is more ionic in character compared to $PH_3$ due to the formal charge,making it the least covalent among the given options.

(B) The $NO_2$ molecule is an odd electron species with a paramagnetic character due to the presence of an unpaired electron on the nitrogen atom.
When two $NO_2$ molecules combine to form $N_2O_4$,the unpaired electrons on the nitrogen atoms of each $NO_2$ molecule pair up to form a $N-N$ single bond.
This $N-N$ bond is relatively weak due to the repulsion between the lone pairs on the oxygen atoms.
In the $N_2O_4$ structure,the four $N-O$ bonds are equivalent due to resonance,resulting in a planar structure.