Mix Examples-Chemical Bonding Questions in English

(A) The compound $GaAlCl_4$ is an ionic salt consisting of $Ga^+$ and $[AlCl_4]^-$ ions.
In this structure,$Ga$ exists as a $Ga^+$ cation,meaning its oxidation state is $+1$,not $+3$.
The $[AlCl_4]^-$ anion consists of an $Al$ atom coordinated to four $Cl$ atoms.
Since $Ga$ is the cationic part of the salt,it is not coordinated to $Cl$ in the same way $Al$ is within the anion.
Therefore,$Ga$ is present as a cationic part of the salt $GaAlCl_4$.

(D) . $NF_3$ has a trigonal pyramidal shape due to the presence of one lone pair on the $N$ atom.
$B$. The bond order of $N_2$ is $3$ and $O_2$ is $2$. Higher bond order implies shorter bond length,so $N_2$ has a shorter bond length than $O_2$. Statement $B$ is correct.
$C$. Isoelectronic species (e.g.,$N_2$ and $CO$) have identical bond orders. Statement $C$ is correct.
$D$. The dipole moment of $H_2O$ $(1.85 \ D)$ is higher than that of $H_2S$ $(0.95 \ D)$ due to the higher electronegativity of oxygen compared to sulfur. Statement $D$ is incorrect.
Therefore,statements $B$ and $C$ are correct.

(D) Pyridine $(C_5H_5N)$ has a hexagonal ring structure similar to benzene where one $CH$ group is replaced by a nitrogen atom.
$1$. $\sigma$ bonds: There are $5$ $C-H$ bonds,$5$ $C-C/C-N$ single bonds,and $1$ $C-N$ double bond (which contains $1$ $\sigma$ bond). Total $\sigma$ bonds $= 5 + 5 + 1 = 11$.
$2$. $\pi$ bonds: There are $3$ double bonds in the ring,each containing $1$ $\pi$ bond. Total $\pi$ bonds $= 3$.
$3$. Lone pair of electrons: The nitrogen atom in pyridine has $1$ lone pair of electrons.
Thus,the number of $\sigma$ bonds,$\pi$ bonds,and lone pairs are $11, 3, 1$ respectively.

(B) To determine the number of species that do not follow the octet rule (i.e.,do not have $8$ electrons around the central atom),we analyze the valence electrons of each molecule:
$1$. $NH_3$: Nitrogen has $5$ valence electrons and forms $3$ bonds with Hydrogen,plus $1$ lone pair. Total electrons = $3 \times 2 + 2 = 8$ electrons.
$2$. $AlCl_3$: Aluminum has $3$ valence electrons and forms $3$ bonds with Chlorine. Total electrons = $3 \times 2 = 6$ electrons (Incomplete octet).
$3$. $BeCl_2$: Beryllium has $2$ valence electrons and forms $2$ bonds with Chlorine. Total electrons = $2 \times 2 = 4$ electrons (Incomplete octet).
$4$. $CCl_4$: Carbon has $4$ valence electrons and forms $4$ bonds with Chlorine. Total electrons = $4 \times 2 = 8$ electrons.
$5$. $PCl_5$: Phosphorus has $5$ valence electrons and forms $5$ bonds with Chlorine. Total electrons = $5 \times 2 = 10$ electrons (Expanded octet).
Species with $NOT$ $8$ electrons are $AlCl_3$,$BeCl_2$,and $PCl_5$.
Therefore,the total number of such species is $3$.

(C) The ionic character of a bond depends on the difference in electronegativity between the bonded atoms. Larger electronegativity difference leads to higher ionic character.
$1$. $N_2$ (non-polar covalent,$\Delta EN = 0$)
$2$. $SO_2$ (polar covalent,$\Delta EN \approx 0.9$)
$3$. $ClF_3$ (polar covalent,$\Delta EN \approx 1.0$)
$4$. $K_2O$ (ionic,$\Delta EN \approx 2.6$)
$5$. $LiF$ (ionic,$\Delta EN \approx 3.0$)
Therefore,the increasing order of ionic character is: $N_2 < SO_2 < ClF_3 < K_2O < LiF$.

(A) The nitrogen atom in the $NO_2^-$ ion is bonded to two oxygen atoms,one via a double bond and one via a single bond.
It also possesses one lone pair of electrons.
Total electrons around $N = (2 \times 2 \text{ from double bond}) + (1 \times 2 \text{ from single bond}) + (1 \times 2 \text{ from lone pair}) = 4 + 2 + 2 = 8$ electrons.

(B) $1$. $SO_2Cl_2$: The central $S$ atom has $4$ bonding pairs (two $S=O$ and two $S-Cl$ bonds) and $0$ lone pairs. Hybridization is $sp^3$,resulting in a Tetrahedral shape $(III)$.
$2$. $NO$: Total valence electrons = $5 + 6 = 11$. It has an odd number of electrons,making it Paramagnetic $(I)$.
$3$. $NO_2^-$: The central $N$ atom has $2$ bonding pairs and $1$ lone pair. It is Diamagnetic $(II)$.
$4$. $I_3^-$: The central $I$ atom has $2$ bonding pairs and $3$ lone pairs. Hybridization is $sp^3d$,resulting in a Linear shape $(IV)$.
Matching: $A-III, B-I, C-II, D-IV$.

(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.
Let us analyze each molecule:
$1$. $CO_2$: Carbon has $8$ electrons (complete octet).
$2$. $NO_2$: Nitrogen has an odd number of electrons ($7$ electrons,exception).
$3$. $H_2SO_4$: Sulfur has an expanded octet ($12$ electrons,exception).
$4$. $BF_3$: Boron has an incomplete octet ($6$ electrons,exception).
$5$. $CH_4$: Carbon has $8$ electrons (complete octet).
$6$. $SiF_4$: Silicon has $8$ electrons (complete octet).
$7$. $ClO_2$: Chlorine has an odd number of electrons ($7$ electrons,exception).
$8$. $PCl_5$: Phosphorus has an expanded octet ($10$ electrons,exception).
$9$. $BeF_2$: Beryllium has an incomplete octet ($4$ electrons,exception).
$10$. $C_2H_6$: Carbon has $8$ electrons (complete octet).
$11$. $CHCl_3$: Carbon has $8$ electrons (complete octet).
$12$. $CBr_4$: Carbon has $8$ electrons (complete octet).
The molecules that are exceptions to the octet rule are: $NO_2, H_2SO_4, BF_3, ClO_2, PCl_5, BeF_2$.
Total number of exceptions = $6$.

(B) . $Al_2(CH_3)_6$ (dimer of $Al(CH_3)_3$) contains $Al-C-Al$ bridges which are $3c-2e$ bonds.
$B$. $B_2H_6$ (dimer of $BH_3$) contains $B-H-B$ bridges which are $3c-2e$ bonds.
$C$. $Al_2Cl_6$ (dimer of $AlCl_3$) contains $Al-Cl-Al$ bridges,but these are $3c-4e$ bonds (due to lone pairs on $Cl$),not $3c-2e$ bonds.
$D$. $BCl_3$ is a stronger Lewis acid than $AlCl_3$ because the back-bonding in $BCl_3$ is less effective than in $AlCl_3$ due to the size mismatch between $B$ $(2p)$ and $Cl$ $(3p)$ orbitals compared to $Al$ $(3p)$ and $Cl$ $(3p)$.
Thus,statements $A, B,$ and $D$ are correct.

(C) $Q$ is $SnCl_2$.
$1. SnCl_2 + Cl^{-} \rightarrow [SnCl_3]^{-} (X)$. The $Sn$ atom in $[SnCl_3]^{-}$ has $3$ bond pairs and $1$ lone pair,so it is $sp^3$ hybridized with pyramidal geometry. Statement $(1)$ is correct.
$2. SnCl_2 + Me_3N \rightarrow SnCl_2 \cdot NMe_3 (Y)$. This is an adduct where $N$ donates a lone pair to $Sn$,forming a coordinate bond. Statement $(4)$ is correct.
$3. SnCl_2 + 2CuCl_2 \rightarrow SnCl_4 (Z) + 2CuCl$. The oxidation state of $Sn$ in $SnCl_4$ is $+4$. Statement $(2)$ is incorrect.
$4. SnCl_4$ has $4$ bond pairs and $0$ lone pairs. Statement $(3)$ is incorrect.
Therefore,statements $(1)$ and $(4)$ are correct.

(A) Melamine is a heterocyclic organic compound with the formula $C_3H_6N_6$.
It consists of a $1,3,5$-triazine ring substituted with three amino groups at the $2, 4,$ and $6$ positions.
In the structure of melamine:
$1$. Each of the $3$ nitrogen atoms in the triazine ring has $1$ lone pair of electrons.
$2$. Each of the $3$ nitrogen atoms in the amino $(-NH_2)$ groups has $1$ lone pair of electrons.
Therefore,the total number of lone pairs of electrons in melamine is $3 + 3 = 6$.

(D) To determine the total number of lone pairs in $N_2O_3$,we look at its Lewis structure.
In the structure of $N_2O_3$ $(O=N-O-N=O)$,the distribution of lone pairs is as follows:
$1$. Each terminal oxygen atom $(=O)$ has $2$ lone pairs.
$2$. The central oxygen atom $(-O-)$ has $2$ lone pairs.
$3$. Each nitrogen atom has $1$ lone pair.
Total lone pairs = $(2 \times 2) + 2 + (2 \times 1) = 4 + 2 + 2 = 8$.
Thus,the total number of lone pairs is $8$.

(C) To determine which molecules follow the Octet Rule,we check the valence electrons around the central atom:
$1$. $CO_2$: $C$ has $8$ electrons (follows octet).
$2$. $C_2H_4$: $C$ has $8$ electrons (follows octet).
$3$. $HCl$: $Cl$ has $8$ electrons (follows octet).
$4$. $NO$: $N$ has $7$ electrons (odd electron molecule,violates octet).
$5$. $NO_2$: $N$ has $7$ electrons (odd electron molecule,violates octet).
$6$. $O_3$: Central $O$ has $8$ electrons (follows octet).
$7$. $H_2SO_4$: $S$ has $12$ electrons (expanded octet,violates octet).
$8$. $BCl_3$: $B$ has $6$ electrons (incomplete octet,violates octet).
Evaluating options:
$(A)$ $CO_2, C_2H_4, HCl$ follow octet ($3$ molecules). $NO$ does not.
$(B)$ $O_3, HCl$ follow octet. $NO_2, H_2SO_4$ do not.
$(C)$ $NO, NO_2, BCl_3, H_2SO_4$ do not follow octet.
$(D)$ $CO_2, O_3, C_2H_4$ follow octet ($3$ molecules). $BCl_3$ does not.
Thus,options $(A)$ and $(D)$ contain at least three molecules that follow the Octet Rule.

(A) $A \rightarrow IV$: $CCl_4$ and $CO_2$ follow the octet rule as all atoms have $8$ electrons in their valence shell.
$B \rightarrow II$: $BCl_3$ and $AlCl_3$ have incomplete octets with only $6$ electrons around the central atom.
$C \rightarrow I$: $NO$ and $NO_2$ are odd-electron molecules containing an unpaired electron.
$D \rightarrow III$: $H_2SO_4$ and $PCl_5$ have expanded octets where the central atom has more than $8$ electrons.
Therefore,the correct matching is $A-IV, B-II, C-I, D-III$.

(D) $1$. All central atoms ($O$,$N$,$C$) in $H_2O$,$NH_3$,and $CH_4$ have $sp^3$ hybridization. Thus,statement $A$ is correct.
$2$. The bond angles are $104.5^{\circ}$ $(H_2O)$,$107.5^{\circ}$ $(NH_3)$,and $109.5^{\circ}$ $(CH_4)$. Thus,statement $B$ is correct.
$3$. Dipole moments are $CH_4$ $(0 \ D)$,$NH_3$ $(1.47 \ D)$,and $H_2O$ $(1.85 \ D)$. The increasing order is $CH_4 < NH_3 < H_2O$. Thus,statement $C$ is correct.
$4$. $H_2O$ and $NH_3$ have lone pairs,so they act as Lewis bases,not acids. $CH_4$ does not act as a Lewis base. Thus,statement $D$ is incorrect.
$5$. In $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$,$NH_3$ accepts a proton (base) and $H_2O$ donates a proton (acid). Thus,statement $E$ is correct.
$6$. Statements $A$,$B$,$C$,and $E$ are correct.

(A) The total number of valence electrons in $NO_2^{-}$ is calculated as follows:
Valence electrons of $N = 5$
Valence electrons of $O = 6 \times 2 = 12$
Negative charge = $1$
Total valence electrons = $5 + 12 + 1 = 18$.
In the $NO_2^{-}$ structure,the nitrogen atom is bonded to two oxygen atoms with one single bond and one double bond $(O=N-O^{-})$.
Each double bond involves $4$ electrons (bonding) and each single bond involves $2$ electrons (bonding).
Total bonding electrons = $4 + 2 = 6$.
Total non-bonded (lone pair) electrons = Total valence electrons - Total bonding electrons = $18 - 6 = 18 - 6 = 12$.
Thus,the total number of non-bonded electrons is $12$.

(A) is correct: The dipole moment order is $H_2O$ $(1.85 \ D)$ $> NH_3$ $(1.47 \ D)$ $> CHCl_3$ $(1.01 \ D)$.
$(B)$ is incorrect: The number of lone pairs on the central $Xe$ atom are: $XeF_4$ $(2)$,$XeO_3$ $(1)$,and $XeF_2$ $(3)$. Thus,the order is $XeF_2 > XeF_4 > XeO_3$.
$(C)$ is incorrect: The bond length order is $N-O > C-H > O-H$.
$(D)$ is correct: Bond enthalpy depends on bond order. The bond orders are $N_2$ $(3)$,$O_2$ $(2)$,and $H_2$ $(1)$. Thus,the order is $N_2 > O_2 > H_2$.

(D) Bond length is inversely proportional to the bond order and directly proportional to the atomic size of the bonded atoms.
In option $D$,the bond lengths are $C-N$ $(1.47 \ \mathring{A})$,$C-O$ $(1.43 \ \mathring{A})$,and $C-H$ $(1.09 \ \mathring{A})$.
The correct order of bond length is $C-H < C-O < C-N$.
Therefore,the order $C-N < C-O < C-H$ is incorrect.

(D) $1$. In $F-F$ vs $Cl-Cl$: Due to the small size of $F$ atoms,there is significant inter-electronic repulsion between the lone pairs,making the $F-F$ bond longer and weaker than expected,but compared to $Cl-Cl$,$F-F$ has a shorter bond length. This option does not satisfy the condition.
$2$. In $C-C$ vs $Si-Si$: $C-C$ has a smaller atomic radius than $Si-Si$,so $C-C$ has a shorter bond length and higher bond energy. This does not satisfy the condition.
$3$. In $P-P$ vs $N-N$: $N-N$ has a very small bond length due to the small size of $N$ atoms,but it has a very low bond energy due to strong lone pair-lone pair repulsions. $P-P$ has a larger bond length and higher bond energy than $N-N$. Thus,the $I^{st}$ species $(P-P)$ has a larger bond length and higher bond energy than the $II^{nd}$ species $(N-N)$.
$4$. Upon re-evaluating the question criteria: The question asks for $I < II$ for both bond length and bond energy. None of the provided options satisfy this condition correctly based on standard chemical trends.

(A) True: Hybrid orbitals have more directional character and overlap more effectively,forming stronger bonds than pure atomic orbitals.
$(b)$ False: Canonical structures (resonance structures) differ only in the arrangement of electrons,not in the arrangement of atoms.
$(c)$ False: $A$ symmetrical molecule like $PCl_5$ (trigonal bipyramidal) is symmetrical but contains different bond angles $(180^{\circ}, 120^{\circ}, 90^{\circ})$.
$(d)$ True: $VSEPR$ theory accounts for the lone pairs on $Xe$ in $XeF_4$,predicting its square planar geometry.

(B) . $BF_3$ is non-polar $(\mu = 0)$ while $PF_3$ has a dipole moment,so $BF_3 < PF_3$ is correct.
$B$. $N_2$ has a triple bond with high bond energy,while $P_2$ is unstable and exists as $P_4$,so $N_2 > P_2$ in bond strength. Thus,$N_2 < P_2$ is incorrect.
$C$. $NH_3$ forms hydrogen bonds with water,making it highly soluble,while $PH_3$ does not,so $PH_3 < NH_3$ in solubility. Thus,$PH_3 > NH_3$ is incorrect.
$D$. $Ne $ has a higher effective nuclear charge than $Ne$,so $I.P. (Ne) < I.P. (Ne^ )$ is correct.
Note: Both $B$ and $C$ are technically incorrect statements based on standard chemical properties.

(A) $I$. The $C-C$ sigma bond in ethene $(CH_2=CH_2)$ is formed by the head-on overlap of $sp^2$ hybrid orbitals of two carbon atoms. This is correct.
$II$. The $C-H$ sigma bond is formed by the overlap of the $sp^2$ hybrid orbital of carbon and the $1s$ orbital of hydrogen. This is correct.
$III$. The $\pi$ bond is formed by the lateral overlap of unhybridized $2p_z$ orbitals,which is a $p\pi-p\pi$ bond. This is correct.
$IV$. In ethene,the $H-C-H$ bond angle is approximately $117.6^\circ$ and the $H-C-C$ bond angle is approximately $121.2^\circ$. Thus,the $H-C-H$ bond angle is smaller than the $H-C-C$ bond angle. This statement is incorrect.
$V$. The $\pi$ bond has a nodal plane that is perpendicular to the molecular plane. This statement is incorrect.

(A) To be isoelectronic,species must have the same number of electrons.
To be isostructural,species must have the same hybridization and geometry.
$1$. $NO_3^{-}$: Electrons = $7 + (3 \times 8) + 1 = 32$. Hybridization = $sp^2$,Geometry = Trigonal planar.
$2$. $CO_3^{2-}$: Electrons = $6 + (3 \times 8) + 2 = 32$. Hybridization = $sp^2$,Geometry = Trigonal planar.
$3$. $ClO_3^{-}$: Electrons = $17 + (3 \times 8) + 1 = 42$. Hybridization = $sp^3$,Geometry = Trigonal pyramidal.
$4$. $SO_3$: Electrons = $16 + (3 \times 8) = 40$. Hybridization = $sp^2$,Geometry = Trigonal planar.
$5$. $NO_2^{-}$: Electrons = $7 + (2 \times 8) + 1 = 24$. Hybridization = $sp^2$,Geometry = Bent.
Comparing these,$NO_3^{-}$ and $CO_3^{2-}$ both have $32$ electrons and are trigonal planar.

(C) $1$. $IF_7$: Pentagonal bipyramidal geometry,non-planar and non-polar $(P \rightarrow d)$.
$2$. $SO_2$: Bent shape,planar and polar $(Q \rightarrow a)$.
$3$. $SF_4$: See-saw shape,non-planar and polar $(R \rightarrow b)$.
$4$. $CS_2$: Linear shape,planar and non-polar $(S \rightarrow c)$.
Therefore,the correct matching is $P$ $\rightarrow d, Q$ $\rightarrow a, R$ $\rightarrow b, S$ $\rightarrow c$.

(C) $1$. The sulfur trioxide molecule $(SO_3)$ has a central sulfur atom bonded to three oxygen atoms.
$2$. The sulfur atom is $sp^2$ hybridised,resulting in a trigonal planar geometry with bond angles of $120^{\circ}$.
$3$. In $SO_3$,all electrons are paired,making the molecule diamagnetic,not paramagnetic.
$4$. Therefore,the statement that $SO_3$ is paramagnetic is incorrect.

(A) $1$. $ICl$ is an interhalogen compound,which is covalent in nature,not ionic. Thus,the statement '$ICl$ is an ionic compound' is incorrect.
$2$. The $Cl-Cl$ bond is stronger than the $Br-Br$ bond due to smaller atomic size and less inter-electronic repulsion in $Cl_2$.
$3$. $ICl$ is polar due to the electronegativity difference between $I$ and $Cl$,whereas $Br_2$ is a homonuclear diatomic molecule and is non-polar.
$4$. Ionisation energy decreases down the group,so $I$ has lower ionisation energy than $Br$.

(B) Let us analyze the hybridization and geometry of each species:
$NH_3$: $sp^3$ (Pyramidal)
$PCl_4^{\oplus}$: $sp^3$ (Tetrahedral)
$NH_2^{-}$: $sp^3$ (Bent)
$PCl_3$: $sp^3$ (Pyramidal)
$SO_2$: $sp^2$ (Bent)
$SO_3$: $sp^2$ (Planar)
$BF_3$: $sp^2$ (Planar)
$IF_7$: $sp^3d^3$ (Pentagonal bipyramidal)
$1$. Hybridization: $NH_3, PCl_4^{\oplus}, NH_2^{-}, PCl_3$ are all $sp^3$ hybridized. So,option $A$ is correct.
$2$. Geometry: $NH_3, PCl_4^{\oplus}, NH_2^{-}, PCl_3, IF_7$ are non-planar. Thus,the statement that all species are planar is incorrect. So,option $B$ is incorrect.
$3$. $p \pi-p \pi$ and $p \pi-d \pi$ bonds: $SO_2$ and $SO_3$ contain both types of back bonding. So,option $C$ is correct.
$4$. Angles: $IF_7$ has $72^{\circ}$ angles in its pentagonal plane. So,option $D$ is correct.

(A) $H_2O$ has a bent structure ($sp^3$ hybridization) and possesses a plane of symmetry (the molecular plane). It is a polar molecule due to the net dipole moment.
$XeF_4$ has a square planar structure ($sp^3d^2$ hybridization) and possesses multiple planes of symmetry (including the molecular plane). It is non-polar because the individual bond dipoles cancel each other out.
Both statements are true,but the polarity of the molecules is not the reason for the existence of a plane of symmetry. Therefore,the Reason is not the correct explanation of the Assertion.

(D) In $CH_4$,the central atom is $C$ (Carbon). The ground state electronic configuration of $C$ is $[He] 2s^2 2p^2$. To form $4$ bonds,$C$ undergoes excitation to the $2s^1 2p^3$ state,which is the $I^{st}$ excited state. However,$CH_4$ is formed by $sp^3$ hybridization involving this state.
In $BF_3$,$B$ (Boron) is in the $I^{st}$ excited state $(2s^1 2p^2)$ to form $3$ bonds.
In $ICl_3$,$I$ (Iodine) is in the $I^{st}$ excited state to form $3$ bonds.
In $SF_6$,$S$ (Sulfur) is in the $II^{nd}$ excited state to form $6$ bonds.
Therefore,$SF_6$ is the molecule that does not form in the $I^{st}$ excited state,as it requires the $II^{nd}$ excited state.

(A) $1$. $PH_5, NCl_5$ and $OF_6$ do not exist because $P$ cannot expand its octet to $5$ bonds due to steric hindrance,$N$ lacks $d$-orbitals to form $5$ bonds,and $O$ cannot form $6$ bonds due to its small size and lack of $d$-orbitals. Thus,this statement is true.
$2$. $I_3^{-}$ has a linear geometry ($sp^3d$ hybridization with $3$ lone pairs on the central $I$ atom). Thus,this statement is false.
$3$. $XeF_4$ has a square planar geometry,which is symmetric,making it non-polar. Thus,this statement is false.
$4$. The bond order of $O_2$ is $2.0$ and the bond order of $O_2^{2-}$ (peroxide ion) is $1.0$. Thus,this statement is false.

(C) 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 $CO_2$,$C$ has $8$ electrons and $O$ has $8$ electrons.
In $CHCl_3$,$C$ has $8$ electrons,$H$ has $2$ electrons (duet rule),and $Cl$ has $8$ electrons.
In $NH_3$,$N$ has $8$ electrons and $H$ has $2$ electrons.
In $ClF_3$,the central atom $Cl$ has $10$ electrons in its valence shell ($3$ bonding pairs and $2$ lone pairs),which is an expanded octet.
Therefore,$ClF_3$ does not obey the octet rule.

(C) $NO$ (nitric oxide) contains a total of $15$ valence electrons ($5$ from $N$ and $10$ from $O$).
Since the total number of valence electrons is odd,it is an odd electron molecule and does not obey the octet rule.
The structure can be represented as $\dot{N}=\ddot{O}$.

(A) is the correct answer.
$1$. In $CH_{4}$ (Methane),the central carbon atom forms $4$ covalent bonds,sharing $8$ electrons in its valence shell,thus completing its octet.
$2$. In $SF_{6}$ (Sulphur hexafluoride),the central sulphur atom has $12$ electrons,which is an expanded octet.
$3$. In $AlCl_{3}$ (Aluminium chloride) and $BF_{3}$ (Boron trifluoride),the central atoms ($Al$ and $B$) have only $6$ electrons in their valence shells,representing an incomplete octet.

(C) Ammonium chloride $(NH_{4}Cl)$ contains both covalent and ionic bonds.
In the ammonium ion $(NH_{4}^{+})$,the nitrogen atom is bonded to four hydrogen atoms by covalent bonds.
The ammonium ion $(NH_{4}^{+})$ and the chloride ion $(Cl^{-})$ are held together by an ionic bond.

(A) An odd electron molecule is a molecule that contains an unpaired electron in its valence shell.
$1$. $CO$: Total valence electrons = $4 + 6 = 10$ (even). It is not an odd electron molecule.
$2$. $NO$: Total valence electrons = $5 + 6 = 11$ (odd). It is an odd electron molecule.
$3$. $NO_2$: Total valence electrons = $5 + 6 + 6 = 17$ (odd). It is an odd electron molecule.
$4$. $ClO_2$: Total valence electrons = $7 + 6 + 6 = 19$ (odd). It is an odd electron molecule.
Therefore,$CO$ is the molecule that is $NOT$ an odd electron molecule.

(B) The formula for formal charge $(F.C.)$ is:
$F.C. = V.E. - N.E. - \frac{1}{2} B.E.$
Where:
$V.E.$ = Valence electrons of the atom ($1$ for $H$)
$N.E.$ = Non-bonding electrons ($0$ for $H$)
$B.E.$ = Bonding electrons ($2$ for $H$)
Substituting the values for the hydrogen atom in $H_2O$:
$F.C. = 1 - 0 - \frac{1}{2}(2) = 1 - 1 = 0$
Therefore,the formal charge on the hydrogen atom is $0$.

$(A)$ To determine the bond length, we look at the bond order and the atomic radii of the atoms involved.
$1$. The bond orders are: $N_2$ $(3)$, $O_2$ $(2)$, and $Cl_2$ $(1)$.
$2$. Bond length is inversely proportional to bond order for similar atoms, but here we must also consider atomic size.
$3$. The atomic radius of $Cl$ $(99 \text{ pm})$ is significantly larger than that of $N$ $(75 \text{ pm})$ and $O$ $(73 \text{ pm})$.
$4$. Due to the large atomic size of $Cl$, the $Cl-Cl$ bond length $(199 \text{ pm})$ is much greater than the $O=O$ bond length $(121 \text{ pm})$ and the $N \equiv N$ bond length $(110 \text{ pm})$.
$5$. Comparing $O_2$ and $N_2$, $N_2$ has a higher bond order $(3)$ than $O_2$ $(2)$, so $N_2$ has a shorter bond length.
$6$. Therefore, the decreasing order of bond length is $Cl_2 > O_2 > N_2$.

(C) $Li$ in $LiCl$ has less than eight electrons in its valence shell. It has only two electrons shared in the bond. Hence,it has an incomplete octet.
$SF_6$,$PCl_5$,and $H_2SO_4$ are examples of molecules with an expanded octet,where the central atom has more than eight electrons.

(B) The bond length of the $C-F$ bond in a $CH_3F$ (fluoromethane) molecule is experimentally determined to be approximately $139 \ pm$.

(B) The Lewis structure of carbon monoxide is $:C \equiv O:$.
Formal charge $= [\text{Total no. of valence } e^- \text{ in free state}] - [\text{Total no. of lone pair } e^-] - \frac{1}{2}[\text{Total no. of bonding } e^-]$.
For oxygen atom:
Valence electrons $= 6$.
Lone pair electrons $= 2$.
Bonding electrons $= 6$.
Formal charge $= 6 - 2 - \frac{1}{2} \times 6 = 6 - 2 - 3 = +1$.

(A) The formula for calculating the formal charge is:
$\text{Formal charge} = [\text{Total number of valence electrons in free state}] - [\text{Total number of lone pair electrons}] - \frac{1}{2} [\text{Number of bonding electrons on that particular atom}]$
For the $N$ atom in $NH_4^{+}$:
- The number of valence electrons in a free $N$ atom is $5$.
- The number of lone pair electrons on the $N$ atom is $0$.
- The number of bonding electrons around the $N$ atom is $8$ (since there are $4$ covalent bonds).
$\text{Formal charge on } N = 5 - 0 - \frac{1}{2} \times 8 = 5 - 4 = +1$.

(D) The formula for formal charge is: $\text{Formal charge} = [ \text{Total no. of valence } e^- \text{ in free state} ] - [ \text{Total no. of lone pair electrons} ] - \frac{1}{2} [ \text{Total no. of bonding electrons} ]$.
For the carbon atom $(C)$ in the structure $S \equiv C - \ddot{N} :$,the number of valence electrons is $4$.
The carbon atom has $0$ lone pair electrons and forms $4$ bonds (a triple bond with $S$ and a single bond with $N$),which corresponds to $8$ bonding electrons.
Substituting these values: $\text{Formal charge} = 4 - 0 - \frac{1}{2} \times 8 = 4 - 4 = 0$.

(B) The formal charge $(FC)$ of an atom in a molecule or ion is calculated using the formula:
$FC = V - L - \frac{B}{2}$
Where:
$V$ = Total number of valence electrons in the free atom.
$L$ = Total number of non-bonding (lone pair) electrons.
$B$ = Total number of bonding (shared) electrons.
For the Carbon $(C)$ atom in the given structure of $SCN^-$ $(S \equiv C - N:)$:
$V = 4$ (Carbon is in group $14$)
$L = 0$ (Carbon has no lone pairs)
$B = 8$ (Carbon forms $3$ bonds with $S$ and $1$ bond with $N$,total $4$ bonds = $8$ electrons)
$FC = 4 - 0 - \frac{8}{2} = 4 - 4 = 0$.
Thus,the formal charge on the $C$ atom is $0$.

(A) The formal charge $(FC)$ is calculated using the formula: $FC = V - L - \frac{B}{2}$.
For the nitrogen $(N)$ atom in the $[S-C \equiv N:]^-$ ion:
Valence electrons $(V)$ of $N = 5$.
Lone pair electrons $(L)$ on $N = 2$.
Bonding electrons $(B)$ for $N = 6$ (due to the triple bond).
$FC = 5 - 2 - \frac{6}{2} = 5 - 2 - 3 = 0$.

(B) The structure of benzoic acid $(C_6H_5COOH)$ consists of a benzene ring attached to a carboxylic acid group.
In the benzene ring,there are $3$ $\pi-$bonds due to the alternating double bonds.
In the carboxylic acid group $(-COOH)$,there is one $C=O$ double bond,which contains $1$ $\pi-$bond.
Therefore,the total number of $\pi-$bonds in a benzoic acid molecule is $3 + 1 = 4$.

(C) The chemical formula for thiosulfuric acid is $H_2S_2O_3$.
In its structure,the central sulfur atom is bonded to another sulfur atom via a double bond $(S=S)$.
The central sulfur atom is also bonded to two hydroxyl groups $(-OH)$ via single bonds and one oxygen atom via a double bond $(S=O)$.
Thus,the bonds formed by the central sulfur with oxygen atoms are two $S=O$ double bonds and two $S-OH$ single bonds.
However,considering the standard representation of $H_2S_2O_3$ as $O=S(OH)_2=S$,the central sulfur atom forms two double bonds with oxygen/sulfur and two single bonds with oxygen atoms.
Therefore,the correct description is two double bonds and two single bonds.

(D) In the $H_2SO_4$ molecule,the sulfur atom is bonded to four oxygen atoms.
There are two single bonds with two $-OH$ groups and two double bonds with two oxygen atoms.
Each single bond contributes $2$ electrons and each double bond contributes $4$ electrons to the valence shell of sulfur.
Total electrons around sulfur = $(2 \times 2) + (2 \times 4) = 4 + 8 = 12$ electrons.

(C) The compound $NaNC$ (sodium isocyanide) contains the following bonds:
$1$. Ionic bond: Between $Na^+$ and $[NC]^-$ ions.
$2$. Covalent bond: Between $N$ and $C$ atoms (triple bond).
$3$. Coordinate bond: The lone pair of electrons on $N$ is donated to $C$ to form the triple bond structure in the isocyanide ion $[N \equiv C]^-$.
Thus,$NaNC$ contains all three types of bonds.

(B) The compound $NH_4Cl$ (ammonium chloride) consists of an ammonium ion $(NH_4^+)$ and a chloride ion $(Cl^-)$.
$1$. The bond between $NH_4^+$ and $Cl^-$ is an ionic bond.
$2$. Within the $NH_4^+$ ion,there are three $N-H$ covalent bonds.
$3$. The fourth $N-H$ bond is formed by the donation of a lone pair of electrons from the nitrogen atom to the $H^+$ ion,which is a coordinate (dative) covalent bond.
Therefore,$NH_4Cl$ contains all three types of linkages.

(D) During the formation of a chemical bond,the system moves to a state of lower potential energy to achieve stability. Therefore,the energy of the system decreases as the atoms approach each other and form a bond.