Mix Examples-Chemical Bonding Questions in English

(N/A) Lewis dot symbols represent the valence electrons of an atom as dots around the chemical symbol.
$Mg$ (Group $2$): $2$ valence electrons $\rightarrow \cdot Mg \cdot$
$Na$ (Group $1$): $1$ valence electron $\rightarrow Na \cdot$
$B$ (Group $13$): $3$ valence electrons $\rightarrow \cdot \dot{B} \cdot$
$O$ (Group $16$): $6$ valence electrons $\rightarrow \cdot \ddot{O} \cdot \cdot$
$N$ (Group $15$): $5$ valence electrons $\rightarrow \cdot \ddot{N} \cdot \cdot$
$Br$ (Group $17$): $7$ valence electrons $\rightarrow \cdot \ddot{Br} \cdot \cdot \cdot$

(N/A) The electronic theory of chemical bonding,developed by Kossel and Lewis in $1916$,explains how atoms combine to form chemical bonds.
$1$. Octet Rule: Atoms tend to combine by transferring valence electrons (gaining or losing) or by sharing valence electrons to achieve a stable octet ($8$ electrons) in their outermost shell.
$2$. Chemical Combination: This theory proposes that atoms achieve stability by attaining the noble gas configuration through the formation of ionic or covalent bonds.
$3$. Limitations: The theory does not explain the stability of molecules with expanded octets (e.g.,$PF_5$,where $P$ has $10$ electrons) or incomplete octets (e.g.,$BF_3$).

(N/A) The Lewis structures represent the valence electrons of atoms in a molecule or ion. The structures are as follows:

Name / Formula	Lewis Structure
Hydrogen sulphide $(H_2S)$	$H:\ddot{S}:H$
Silicon tetrachloride $(SiCl_4)$	$\ddot{Cl}:\ddot{Si}:\ddot{Cl}$ (with $Cl$ atoms at four corners)
Beryllium fluoride $(BeF_2)$	$:\ddot{F}:Be:\ddot{F}:$
Carbonate ion $(CO_3^{2-})$	$[ :\ddot{O}:\ddot{C}:\ddot{O}: ]^{2-}$ (with one double bond to $O$)
Formic acid $(HCOOH)$	$H:\ddot{C}:\ddot{O}:H$ (with a double bonded $O$ on $C$)

When a molecule or ion is represented by a Lewis dot structure,the charge on an atom is called the formal charge. If an atom has more electrons than valence electrons,it has a negative formal charge,and if it has fewer electrons than valence electrons,it has a positive formal charge.
Calculation of formal charge is expressed by the following equation:
$F.C = (\text{Total number of valence electrons}) - (\text{Total number of non-bonding electrons}) - \frac{1}{2}(\text{Total number of bonding electrons})$
Note: Formal charge is not a real charge in a molecule,but it provides information about the position and arrangement of electrons.
Example: Formal charge of oxygen in an ozone molecule $(O_3)$.
The central $O$ atom marked $1$:
$F.C = (\text{Valence electron of neutral oxygen}) - (\text{Non-bonding electrons of } O_1) - \frac{1}{2}(\text{Bonding electrons of } O_1)$
$F.C = 6 - 2 - \frac{1}{2}(6) = +1$
The end $O$ atom marked $2$:
$F.C = (\text{Valence electron of } O) - (\text{Non-bonding electrons of } O_2) - \frac{1}{2}(\text{Bonding electrons of } O_2) = 6 - 4 - \frac{1}{2}(4) = 0$
The end $O$ atom marked $3$:
$F.C = (\text{Valence electron of } O) - (\text{Non-bonding electrons of } O_3) - \frac{1}{2}(\text{Bonding electrons of } O_3) = 6 - 6 - \frac{1}{2}(2) = -1$
Hence,the structure of $O_3$ with formal charges is indicated.

The formal charge on an atom in a Lewis structure is calculated as: $FC = [\text{Total number of valence electrons}] - [\text{Total number of non-bonding electrons}] - \frac{1}{2} [\text{Total number of bonding electrons}]$.
For $CO_3^{2-}$:
$1$. Carbon atom: $FC = 4 - 0 - \frac{1}{2}(8) = 0$.
$2$. Double-bonded Oxygen: $FC = 6 - 4 - \frac{1}{2}(4) = 0$.
$3$. Single-bonded Oxygen atoms (two): $FC = 6 - 6 - \frac{1}{2}(2) = -1$.
For $HNO_3$ (in the structure $H-O-N(=O)_2$):
$1$. Hydrogen: $FC = 1 - 0 - \frac{1}{2}(2) = 0$.
$2$. Hydroxyl Oxygen: $FC = 6 - 4 - \frac{1}{2}(4) = 0$.
$3$. Nitrogen: $FC = 5 - 0 - \frac{1}{2}(8) = 1$.
$4$. Double-bonded Oxygen: $FC = 6 - 4 - \frac{1}{2}(4) = 0$.
$5$. Single-bonded Oxygen (dative bond): $FC = 6 - 6 - \frac{1}{2}(2) = -1$.

(N/A) Definition: The octet rule states that atoms of elements tend to combine in such a way that their atoms have eight electrons in their valence shells,giving them the same electronic configuration as a noble gas.
Significance of the octet rule:
- It helps in predicting the structure of molecules and ions.
- It explains the stability of atoms that achieve a complete octet.
- It is useful in calculating the formal charge of atoms in a molecule or ion.
- It provides a basic understanding of the bonding in organic compounds and compounds of second-period elements.
Limitations of the octet rule:
$I$. Incomplete octet of the central atom:
In some compounds,the number of electrons surrounding the central atom is less than eight. This is especially common for elements of groups $1, 2,$ and $13$ (e.g.,$LiCl, BeH_2, BeCl_2, BF_3, AlCl_3$).
$II$. Odd-electron molecules:
In molecules with an odd number of electrons,like nitric oxide $(NO)$ and nitrogen dioxide $(NO_2)$,the octet rule cannot be satisfied for all atoms.
$III$. The expanded octet:
Elements in and beyond the third period of the periodic table have $d$-orbitals available for bonding. These elements can accommodate more than eight electrons around the central atom,forming an expanded octet (e.g.,$PF_5, SF_6, H_2SO_4$).

(N/A) The ionic character of a bond depends on the electronegativity difference between the bonded atoms. $A$ larger electronegativity difference corresponds to a higher ionic character.
Increasing order of ionic character: $N_2 < ClF_3 < SO_2 < K_2O < LiF$

$N_2$	$100 \%$ covalent (homonuclear),$0 \%$ ionic.
$SO_2$	Electronegativity difference: $|3.5 - 2.5| = 1.0$.
$ClF_3$	Electronegativity difference: $|4.0 - 3.0| = 1.0$.
$K_2O$	Electronegativity difference: $|3.5 - 0.8| = 2.7$ (highly ionic).
$LiF$	Electronegativity difference: $|4.0 - 1.0| = 3.0$ (maximum ionic character).

Comparing $SO_2$ and $ClF_3$ (both have $\Delta EN = 1.0$): Based on Fajan's rule,smaller cations and larger anions increase covalent character. $S^{2+}$ is smaller than $Cl^{+}$,and $O^{2-}$ is larger than $F^{-}$,making $ClF_3$ more covalent than $SO_2$. Thus,$SO_2$ has slightly more ionic character than $ClF_3$.

(D) The limitations of the framework model are as follows:
$i$. It does not explicitly indicate the atoms present in the structure.
$ii$. This model does not provide any information regarding the relative atomic size.
$iii$. This model primarily emphasizes the pattern of bonds within a molecule.

(A) The permanganate ion $(MnO_4^-)$ has a tetrahedral geometry where the central $Mn$ atom is in the $+7$ oxidation state.
The electronic configuration of $Mn$ is $[Ar] 3d^5 4s^2$,and for $Mn^{7+}$,it is $[Ar] 3d^0 4s^0$.
$(a)$ There are $4$ $Mn-O$ $\sigma$ bonds formed by the overlap of $Mn$ $sd^3$ hybrid orbitals with $O$ $2p$ orbitals. Thus,$(a)$ is True $(T)$.
$(b)$ There are $3$ $Mn-O$ $d\pi-p\pi$ bonds. Thus,$(b)$ is True $(T)$.
$(c)$ Since it contains both $\sigma$ and $\pi$ bonds,$(c)$ is True $(T)$.
$(d)$ Since $Mn^{7+}$ has a $d^0$ configuration,there are no unpaired electrons,making it diamagnetic. Thus,$(d)$ is False $(F)$.

(N/A) In the solid state,$BeCl_2$ has a polymeric chain-like structure. Each $Be$ atom is surrounded by four $Cl$ atoms,where two $Cl$ atoms are bonded through covalent bonds and the other two are bonded through coordinate bonds. This is shown in structure $A$.
In the gaseous state,beryllium chloride exists as a dimer $(Be_2Cl_4)$ at lower temperatures,which has a bridged structure. At high temperatures (about $1200 \ K$),it dissociates into a monomer $(BeCl_2)$,which is linear in structure. This is shown in structure $B$.

(N/A) . $A$ covalent bond is formed by the overlapping of atomic orbitals. The direction of overlapping gives the direction of the bond. In an ionic bond,the electrostatic field of an ion is non-directional.
Each positive ion is surrounded by a number of anions in any direction depending upon its size and vice-versa. That is why covalent bonds are directional and ionic bonds are non-directional.
$B$. In $H_{2}O$,the oxygen atom is $sp^{3}$ hybridised with two lone pairs. The four $sp^{3}$ hybridised orbitals possess a tetrahedral geometry with two corners occupied by hydrogen atoms and the other two by the lone pairs.
The bond angle is reduced to $104.5^{\circ}$ from $109.5^{\circ}$ because of greater repulsive forces between $lp-lp$ (lone pair-lone pair) and the molecule gains a $V$-shape or bent structure (angular structure).
In the $CO_{2}$ molecule,the carbon atom is $sp$-hybridised. The two $sp$ hybrid orbitals are oriented in opposite directions forming an angle of $180^{\circ}$.
$O=C=O$
So,the $H_{2}O$ molecule has a bent structure and the $CO_{2}$ molecule is linear.
$C$. In the ethyne molecule,both carbon atoms are $sp$ hybridised,having two unhybridised orbitals,i.e.,$2p_{x}$ and $2p_{y}$. The two $sp$ hybrid orbitals of both carbon atoms are oriented in opposite directions forming an angle of $180^{\circ}$.
So,the ethyne molecule is linear.

(N/A) An ionic bond is the electrostatic force of attraction between oppositely charged ions formed by the complete transfer of one or more electrons from one atom to another.
Differences:

Property	Ionic Bond	Covalent Bond
Formation	Complete transfer of electrons.	Mutual sharing of electrons.
Examples	$NaCl, MgCl_2$	$Cl_2, CH_4$
Melting Point	High	Generally low

(N/A) In $1916$,$Kossel$ and $Lewis$ independently succeeded in providing a satisfactory explanation for the formation of chemical bonds in terms of electrons.

(N/A) 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.
$(1)$ $PCl_5$: Phosphorus has $10$ electrons in its valence shell (expanded octet). Rule not obeyed.
$(2)$ $SF_6$: Sulfur has $12$ electrons in its valence shell (expanded octet). Rule not obeyed.
$(3)$ $PF_5$: Phosphorus has $10$ electrons in its valence shell (expanded octet). Rule not obeyed.
$(4)$ $NH_4^+$: Nitrogen has $8$ electrons in its valence shell. Rule obeyed.
$(5)$ $BeCl_2$: Beryllium has $4$ electrons in its valence shell (incomplete octet). Rule not obeyed.
$(6)$ $NCl_3$: Nitrogen has $8$ electrons in its valence shell. Rule obeyed.

(B) To identify odd-electron molecules,we calculate the total number of valence electrons in each molecule:
$1$. $NO$: Valence electrons = $5 (N) + 6 (O) = 11$. Since $11$ is an odd number,$NO$ is an odd-electron molecule.
$2$. $NO_2$: Valence electrons = $5 (N) + 2 \times 6 (O) = 17$. Since $17$ is an odd number,$NO_2$ is an odd-electron molecule.
$3$. $CO$: Valence electrons = $4 (C) + 6 (O) = 10$. Since $10$ is an even number,$CO$ is not an odd-electron molecule.
$4$. $CO_2$: Valence electrons = $4 (C) + 2 \times 6 (O) = 16$. Since $16$ is an even number,$CO_2$ is not an odd-electron molecule.
Therefore,$NO$ and $NO_2$ are odd-electron molecules.

(N/A) Similarity: In both structures,the central sulfur atom is surrounded by four oxygen atoms,and the total number of valence electrons involved in bonding and lone pairs is consistent with the octet rule or expanded octet requirements for sulfur.
Difference: $H_2SO_4$ is a neutral molecule where two hydrogen atoms are covalently bonded to two oxygen atoms. $SO_4^{2-}$ is an ion with a $-2$ charge,where the two hydrogen atoms have been removed as $H^+$,leaving their electrons behind on the oxygen atoms,resulting in two oxygen atoms carrying a formal negative charge.

(A) . The bond length depends on the atomic radii and bond order.
In $C=O$ (as in carbon monoxide,$CO$),the bond length is approximately $1.128 \ \mathring{A}$.
In $N=O$ (as in nitric oxide,$NO$),the bond length is approximately $1.150 \ \mathring{A}$.
Therefore,$C=O$ is shorter than $N=O$. This is primarily due to the higher bond order and the specific electronic environment in $CO$ compared to the $NO$ radical.

(A) The decreasing order of bond length is: $HI > HBr > HCl > HF$.
Reason: As the size of the halogen atom increases from $F$ to $I$,the bond length increases. Therefore,$HI$ has the longest bond length and $HF$ has the shortest bond length.

(A) The bond enthalpy depends on the bond order and the size of the atoms involved. The bond enthalpies for the given molecules are as follows:

$Molecule$	$Bond \text{ } Enthalpy \text{ } (\Delta H) \text{ } (kJ \text{ } mol^{-1})$
$N_2$	$946$
$H_2$	$435.8$
$O_2$	$408$

Comparing these values,the decreasing order of bond enthalpy is $N_2 > H_2 > O_2$.

(N/A) The Lewis approach is successful in explaining the sharing of electrons between atoms to form covalent bonds and the formation of simple molecules.
However,it fails to explain the following:
$1$. It does not explain the shape of polyatomic molecules.
$2$. It does not provide information about the energy of the bond (bond enthalpy) or the bond length.
$3$. It fails to explain the stability of molecules based on the electronic configuration.

(N/A) Structure of ethene $(C_2H_4)$:
$C-C$ bond length is $134 \text{ pm}$.
Bond angles: $\angle H-C-H = 117.6^{\circ}$ and $\angle H-C-C = 121^{\circ}$.
Structure of $PCl_5$:
$PCl_5$ has a trigonal bipyramidal geometry.
Three $P-Cl$ bonds lie in the same equatorial plane with a bond angle of $120^{\circ}$ $(\angle Cl-P-Cl)$.
Two $P-Cl$ bonds are axial, perpendicular to the equatorial plane, with bond angles of $90^{\circ}$ with the equatorial bonds.

(N/A) $(i)$ Chemical bond
$(ii)$ Kossel and Lewis
$(iii)$ Valence electrons
$(iv)$ Lewis symbol

(N/A) $(i)$ $ns^{2} np^{6}$
$(ii)$ Electrovalent bond (or Ionic bond)
$(iii)$ Lewis octet rule
$(iv)$ Bond length

(N/A) $(i)$ Lattice enthalpy
$(ii)$ Bond angle
$(iii)$ Bond enthalpy
$(iv)$ Sidgwick and Powell

(N/A) $(i)$ Valence Bond Theory $(VBT)$.
$(ii)$ $T$-shaped.
$(iii)$ $4$.
$(iv)$ $0.79 \ D$.

(N/A) $(i)$ Valence Bond Theory and Molecular Orbital Theory.
$(ii)$ $[He] 2s^{1} 2p_{x}^{1} 2p_{y}^{1} 2p_{z}^{1}$.
$(iii)$ Covalent bonds.
$(iv)$ Tetrahedral.

(D) The correct matches are:
$(1)$ Kossel and Lewis proposed the $(D)$ Octet Rule.
$(2)$ Langmuir introduced the concept of $(C)$ Covalent Bond.
$(3)$ Sidgwick and Powell proposed the $(A)$ $VSEPR$ theory.
$(4)$ Heitler and London proposed the $(E)$ Valence Bond Theory.
$(5)$ $F$. Hund and Mulliken proposed the $(B)$ Molecular Orbital Theory.
Thus,the correct sequence is $(1-D, 2-C, 3-A, 4-E, 5-B)$.

(D) The correct matches are:
$(1)$ Hydrogen bond $\rightarrow (D) \text{ } {\rm{HF}}$
$(2)$ Resonance $\rightarrow (E) \text{ } {\rm{O}}_3$
$(3)$ Ionic solid $\rightarrow (B) \text{ } {\rm{LiF}}$
$(4)$ Covalent solid $\rightarrow (A) \text{ } {\rm{C}}$
Therefore,the correct sequence is $(1-D, 2-E, 3-B, 4-A)$.

(N/A) 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.
$(1) BeF_4^{2-}$: Obeyed. The central $Be$ atom shares $4$ pairs of electrons,resulting in $8$ electrons in its valence shell.
$(2) BF_3$: Not obeyed. The central $B$ atom has only $6$ electrons in its valence shell (electron-deficient molecule).
$(3) CO_2$: Obeyed. Both $C$ and $O$ atoms complete their octets through double bonds.
$(4) NO$: Not obeyed. This is an odd-electron molecule with $11$ valence electrons,leaving the $N$ atom with $7$ electrons.
$(5) NH_3$: Obeyed. The central $N$ atom has $8$ electrons in its valence shell ($3$ shared pairs and $1$ lone pair).

(N/A) Bond strength refers to the amount of energy required to break a chemical bond,also known as bond dissociation enthalpy. Higher bond strength indicates a more stable bond.
Reactivity refers to the ease with which a bond can be broken or participate in a chemical reaction. Generally,a weaker bond is more reactive because it requires less energy to break.

(A) Lewis base is defined as a species that can donate a lone pair of electrons.
$BF_{3}$ is an electron-deficient molecule with an incomplete octet on the central boron atom,making it a Lewis acid,not a Lewis base.
$NH_{3}$,$H_{2}O$,and $OH^{-}$ all possess lone pairs of electrons that they can donate,thus acting as Lewis bases.

(A) Lewis base is defined as a substance that can donate a lone pair of electrons.
$NH_3$,$H_2O$,and $OH^-$ all possess at least one lone pair of electrons on the central atom,allowing them to act as Lewis bases.
$BF_3$ is an electron-deficient molecule with an incomplete octet (only $6$ electrons around the central $B$ atom).
Therefore,$BF_3$ acts as a Lewis acid because it accepts an electron pair,not a Lewis base.

(A) Lewis base is a substance that can donate a lone pair of electrons.
In $PF_{3}$,the phosphorus atom has one lone pair of electrons.
Therefore,$PF_{3}$ can act as a Lewis base.
In contrast,$SF_{4}$,$SiF_{4}$,and $BF_{3}$ are Lewis acids because they have vacant orbitals or are electron-deficient and can accept electron pairs.

(C) The chemical formula of perchloric acid is $HClO_4$.
In the structure of perchloric acid,the central chlorine atom is bonded to one hydroxyl group $(-OH)$ and three oxygen atoms via double bonds $(Cl=O)$.
Therefore,there are $3$ $Cl=O$ bonds in the molecule.
The correct option is $C$.

(A) The $S-S$ bond length depends on the oxidation state of sulfur and the repulsion between lone pairs on the sulfur atoms.
In $S_2O_4^{2-}$ (dithionite),each sulfur atom has one lone pair,leading to significant $L.P.-L.P.$ repulsion between the two sulfur atoms,which increases the $S-S$ bond length.
In $S_2O_5^{2-}$ (disulfite),one sulfur has a lone pair and the other does not,resulting in less repulsion compared to $S_2O_4^{2-}$.
In $S_2O_6^{2-}$ (dithionate),both sulfur atoms are in a higher oxidation state and have no lone pairs involved in $S-S$ repulsion,resulting in the shortest $S-S$ bond length.
Therefore,the correct order of $S-S$ bond length is $I > II > III$.

(B) Lewis base is a chemical species that has the capability to donate an electron pair.
In $NF_{3}$,the central atom $N$ has one lone pair.
In $SF_{4}$,the central atom $S$ has one lone pair.
In $ClF_{3}$,the central atom $Cl$ has two lone pairs.
Therefore,$NF_{3}$,$SF_{4}$,and $ClF_{3}$ can act as Lewis bases.
In $PCl_{5}$,the central atom $P$ is in its maximum oxidation state $(+5)$ and has no lone pairs available for donation. Thus,it cannot act as a Lewis base.

(B) Isostructural species have the same geometry and hybridization.
$A.$ $SO_{4}^{2-}$ and $CrO_{4}^{2-}$ are both tetrahedral ($sp^3$ hybridization).
$B.$ $SiCl_{4}$ and $TiCl_{4}$ are both tetrahedral ($sp^3$ hybridization).
$C.$ $NH_{3}$ is trigonal pyramidal $(sp^3)$,while $NO_{3}^{-}$ is trigonal planar $(sp^2)$.
$D.$ $BCl_{3}$ is trigonal planar $(sp^2)$,while $BrCl_{3}$ is $T$-shaped $(sp^3d)$.
Therefore,the isostructural pairs are $A$ and $B$.

(D) The reaction is $Na_{(g)} + Br_{(g)} \rightarrow NaBr_{(s)}$.
According to the Born-Haber cycle,the enthalpy change for this reaction is given by the sum of the ionization enthalpy of $Na_{(g)}$,the electron gain enthalpy of $Br_{(g)}$,and the lattice enthalpy of $NaBr_{(s)}$.
$\Delta H = IE_{1} + \Delta H_{eg} + LE$
$\Delta H = 495.8 \ kJ \ mol^{-1} + (-325.0 \ kJ \ mol^{-1}) + (-728.4 \ kJ \ mol^{-1})$
$\Delta H = 495.8 - 325.0 - 728.4 = -557.6 \ kJ \ mol^{-1}$
To express this in the form $x \times 10^{-1} \ kJ \ mol^{-1}$,we have:
$-557.6 \ kJ \ mol^{-1} = -5576 \times 10^{-1} \ kJ \ mol^{-1}$
Thus,the value is $5576$.

(D) $1$. The bond angle order $CH_{4} (109.5^{\circ}) > NH_{3} (107^{\circ}) > H_{2}O (104.5^{\circ}) > H_{2}S (92^{\circ})$ is correct due to the presence of lone pairs.
$2$. The bond order for $O_{2}$ species is $O_{2}^{+} (2.5) > O_{2} (2.0) > O_{2}^{-} (1.5) > O_{2}^{2-} (1.0)$,which is correct.
$3$. The strength of $H$-bond depends on electronegativity. $HF > H_{2}O > NH_{3}$ is correct,but $HCl$ does not form significant $H$-bonds,so the order is correct.
$4$. In $CO_{2}$,the structure $O=C=O$ is more stable than $\overline{O}-C \equiv \stackrel{+}{O}$ because $O=C=O$ has no formal charges and satisfies the octet rule for all atoms. Therefore,option $D$ is incorrect.

(D) Electron-deficient species are those in which the central atom has an incomplete octet,meaning it has fewer than $8$ electrons in its valence shell.
$1$. $PH_{3}$: $P$ has $8$ electrons (octet complete).
$2$. $B_{2}H_{6}$: Boron has $6$ electrons (electron-deficient).
$3$. $CCl_{4}$: $C$ has $8$ electrons (octet complete).
$4$. $NH_{3}$: $N$ has $8$ electrons (octet complete).
$5$. $LiH$: $Li$ has $2$ electrons (electron-deficient).
$6$. $BCl_{3}$: $B$ has $6$ electrons (electron-deficient).
Therefore,the electron-deficient molecules are $B_{2}H_{6}$,$LiH$,and $BCl_{3}$.
The total number of electron-deficient molecules is $3$.

(C) To determine the paramagnetism,we check for the presence of unpaired electrons:
$Na_2O$: Diamagnetic (all electrons paired).
$KO_2$: Paramagnetic (contains $O_2^-$ ion with one unpaired electron).
$NO_2$: Paramagnetic (odd number of valence electrons,$17$ valence electrons).
$N_2O$: Diamagnetic (all electrons paired).
$ClO_2$: Paramagnetic (odd number of valence electrons,$19$ valence electrons).
$NO$: Paramagnetic (odd number of valence electrons,$11$ valence electrons).
$SO_2$: Diamagnetic (all electrons paired).
$Cl_2O$: Diamagnetic (all electrons paired).
The paramagnetic oxides are $KO_2$,$NO_2$,$ClO_2$,and $NO$.
Therefore,the total number of paramagnetic oxides is $4$.

(B) An odd-electron molecule is one that has an odd number of valence electrons,such as $NO$ (Nitric oxide,$5+6=11$ valence electrons).
An expanded octet molecule is one where the central atom has more than $8$ electrons in its valence shell,such as $H_{2}SO_{4}$ (Sulfuric acid).
In $H_{2}SO_{4}$,the central sulfur atom is bonded to four oxygen atoms (two double bonds and two single bonds),resulting in $12$ electrons in its valence shell.
Thus,the pair containing an odd-electron molecule $(NO)$ and an expanded octet molecule $(H_{2}SO_{4})$ is $NO$ and $H_{2}SO_{4}$.

(C) Isoelectronic species are those that have the same number of electrons.
Let us calculate the total number of electrons for each pair:
$(A) \ CO, N_2$: $CO = 6 + 8 = 14$; $N_2 = 7 + 7 = 14$. (Isoelectronic)
$(B) \ O_2, NO$: $O_2 = 8 + 8 = 16$; $NO = 7 + 8 = 15$. (Not isoelectronic)
$(D) \ F_2, HCl$: $F_2 = 9 + 9 = 18$; $HCl = 1 + 17 = 18$. (Isoelectronic)
Since both $(A)$ and $(D)$ represent isoelectronic pairs,the correct choice is $(C)$.

(C) The molecular formula of hydrazine is $NH_2-NH_2$.
In the structure of hydrazine,each nitrogen atom has one lone pair,so there are $2$ lone pairs in total.
The bond pairs consist of $4$ $N-H$ bonds and $1$ $N-N$ bond,totaling $5$ bond pairs.
Therefore,the number of lone pairs is $2$ and the number of bond pairs is $5$.

(D) The oxygen-oxygen bond lengths are determined by the bond order and the electronic environment.
$1$. $O_2$: Bond order is $2.0$,bond length $\approx 121 \text{ pm}$.
$2$. $KO_2$: Contains superoxide ion $O_2^-$,bond order is $1.5$,bond length $\approx 128 \text{ pm}$.
$3$. $H_2O_2$: Contains peroxide ion $O_2^{2-}$,bond order is $1.0$,bond length $\approx 148 \text{ pm}$.
$4$. $F_2O_2$: Bond order is $1.0$,but due to the high electronegativity of $F$,the $O-O$ bond length is $\approx 122 \text{ pm}$.
$5$. $BaO_2$: Contains peroxide ion $O_2^{2-}$,bond order is $1.0$,bond length $\approx 149 \text{ pm}$.
Comparing the values,$KO_2$ $(128 \text{ pm})$ and $F_2O_2$ $(122 \text{ pm})$ have the most similar bond lengths among the given choices.

(A) Peroxodisulphuric acid $(H_2S_2O_8)$ has the structure $HO-SO_2-O-O-SO_2-OH$. Each sulfur atom is double-bonded to two oxygen atoms,resulting in $4$ $\pi$-bonds.
Pyrosulphuric acid $(H_2S_2O_7)$ has the structure $HO-SO_2-O-SO_2-OH$. Each sulfur atom is double-bonded to two oxygen atoms,resulting in $4$ $\pi$-bonds.
Total number of $\pi$-bonds $= 4 + 4 = 8$.

(B) The Lewis structure of ozone $(O_3)$ consists of a central oxygen atom bonded to two other oxygen atoms.
One oxygen atom has a double bond with the central oxygen atom and possesses $2$ lone pairs.
The central oxygen atom has a positive charge and possesses $1$ lone pair.
The third oxygen atom has a single bond with the central oxygen atom and possesses $3$ lone pairs.
Total number of lone pairs $= 2 + 1 + 3 = 6$.

(A) To determine if a molecule or ion has an odd number of electrons,we calculate the total valence electrons:
$A. NO_2$: $5 + 2(6) = 17$ (Odd)
$B. ICl_4^{-}$: $7 + 4(7) + 1 = 36$ (Even)
$C. BrF_3$: $7 + 3(7) = 28$ (Even)
$D. ClO_2$: $7 + 2(6) = 19$ (Odd)
$E. NO_2^{+}$: $5 + 2(6) - 1 = 16$ (Even)
$F. NO$: $5 + 6 = 11$ (Odd)
The species with an even number of electrons are $ICl_4^{-}$,$BrF_3$,and $NO_2^{+}$.
Therefore,the total count is $3$.

(D) The $O-O$ bond length in $H_2O_2$ is $1.48 \ \mathring{A}$,while in $F_2O_2$ it is $1.22 \ \mathring{A}$. Thus,the $O-O$ bond in $H_2O_2$ is longer $(X = \text{longer})$.
The $O-H$ bond length in $H_2O_2$ is approximately $0.95 \ \mathring{A}$,whereas the $O-F$ bond length in $F_2O_2$ is approximately $1.41 \ \mathring{A}$. Thus,the $O-H$ bond is shorter $(Y = \text{shorter})$.
Therefore,the correct option is $D$.

(C) Let us determine the number of lone pairs for each molecule:
$1$. $H_2O$: Oxygen has $2$ lone pairs.
$2$. $N_2$: Each Nitrogen atom has $1$ lone pair,total $2$ lone pairs.
$3$. $CO$: Carbon has $1$ lone pair and Oxygen has $1$ lone pair,total $2$ lone pairs.
$4$. $XeF_4$: Xenon has $2$ lone pairs.
$5$. $NH_3$: Nitrogen has $1$ lone pair.
$6$. $NO$: Nitrogen has $1$ lone pair and Oxygen has $2$ lone pairs (total $3$ lone pairs).
$7$. $CO_2$: Carbon has $0$ lone pairs,each Oxygen has $2$ lone pairs (total $4$ lone pairs).
$8$. $F_2$: Each Fluorine has $3$ lone pairs (total $6$ lone pairs).
Thus,the molecules with exactly $2$ lone pairs are $H_2O, N_2, CO,$ and $XeF_4$.
The total count is $4$.