Chemical Bonding and Molecular Structure MCQs with Answers – Part 1 (NEET Chemistry)

Explanation: The Kossel–Lewis approach relates chemical bond formation to the rearrangement of valence electrons and the tendency to achieve stable outer electronic configurations. Atom X has one electron in its outermost shell, whereas atom Y has seven. Removal of one electron from X gives the electronic configuration \(2,8\), corresponding to a noble-gas configuration. Addition of one electron to Y also produces the configuration \(2,8\). Thus, electron transfer produces oppositely charged ions. The electronic changes are: \[ \mathrm{X\rightarrow X^++e^-} \] \[ \mathrm{Y+e^-\rightarrow Y^-} \] The resulting ions possess opposite charges and can attract each other electrostatically. The octet rule explains the electron-counting tendency, but it does not independently establish that the overall compound-formation process is energetically favourable. The stability of an actual ionic compound also depends on the energetics of ion formation and the stabilisation provided by the ionic lattice. Therefore, X loses one electron and Y gains one electron, making option B correct.

Explanation: \(\textbf{Count the available valence electrons:}\) Oxygen belongs to Group 16 and possesses six valence electrons. Each hydrogen atom contributes one valence electron. Therefore, the total number of available valence electrons is: \[ N_v=6+2(1)=8 \] These eight electrons correspond to four electron pairs. \(\textbf{Distribute the electron pairs:}\) The conventional Lewis structure contains two O–H single bonds. Each O–H bond consists of one shared electron pair. Consequently, two of the four available pairs are bonding pairs. The remaining four electrons are retained on oxygen as two lone pairs. \(\textbf{Check the stable configurations:}\) Each hydrogen shares one pair and thereby has two electrons in its valence shell, satisfying the duplet requirement. Oxygen has four electrons from its two bonding pairs and four from its two lone pairs around it. Thus, oxygen is surrounded by eight electrons and satisfies the octet rule. \(\textbf{Final answer:}\) \[ \boxed{2\ \text{bonding pairs},\ 2\ \text{lone pairs}} \]

Explanation: \(\textbf{Evaluate Statement I:}\) Statement I is correct because the octet rule provides an electronic basis for understanding many chemical bonds. Atoms may lose, gain or share valence electrons to achieve configurations resembling those of noble gases. In sodium chloride, electron transfer gives sodium and chloride ions with stable outer electronic configurations. In chlorine molecules, two chlorine atoms share an electron pair so that each completes its octet. These examples illustrate how the octet rule is useful for constructing simple bonding descriptions. \(\textbf{Evaluate Statement II:}\) Statement II is incorrect because the octet rule does not provide quantitative information about the energy of a molecule or ionic solid. Producing a gaseous cation requires energy, and obtaining noble-gas configurations does not automatically compensate for every energetic cost. The formation of an ionic lattice introduces an additional stabilising contribution that must be considered. Likewise, two covalent substances satisfying the octet rule need not possess equal bond strengths or equal thermodynamic stability. The rule also cannot independently predict molecular shape. Furthermore, stable compounds such as \(\mathrm{BF_3}\) and \(\mathrm{SF_6}\) are familiar exceptions to the ordinary octet requirement. Thus, the octet rule is a useful electron-counting principle rather than a complete theory of chemical stability. Statement I is therefore correct, whereas Statement II is incorrect.

Explanation: \(\textbf{Determine the available valence electrons:}\) Carbon contributes four valence electrons. Each oxygen contributes six valence electrons. The ionic charge of \(-2\) requires the addition of two electrons to the neutral-atom total. \[ N_v=V_{\mathrm C}+3V_{\mathrm O}+2 \] \[ N_v=4+3(6)+2 \] \[ N_v=24 \] \(\textbf{Distribute electrons in the Lewis structure:}\) Carbon forms one double bond and two single bonds with oxygen. The double bond contains four bonding electrons, while the two single bonds contain another four. Thus, eight electrons are used in bonding. The remaining sixteen electrons occur as lone pairs on the oxygen atoms. The doubly bonded oxygen has two lone pairs, and each singly bonded oxygen has three. Carbon has eight electrons around it and therefore satisfies the octet rule. \(\textbf{Calculate formal charges:}\) For any atom: \[ FC=V-N-\frac{B}{2} \] Here, \(V\) is the valence-electron count, \(N\) is the number of nonbonding electrons, and \(B\) is the number of bonding electrons. For each singly bonded oxygen: \[ FC_{\mathrm O}=6-6-\frac{2}{2} \] \[ FC_{\mathrm O}=-1 \] For the doubly bonded oxygen: \[ FC_{\mathrm O}=6-4-\frac{4}{2}=0 \] For carbon: \[ FC_{\mathrm C}=4-0-\frac{8}{2}=0 \] \(\textbf{Verify charge conservation:}\) \[ FC_{\mathrm{total}}=0+0+(-1)+(-1)=-2 \] The result agrees with the overall charge on the carbonate ion. The particular oxygen atoms carrying negative formal charge depend on the Lewis contributor chosen; this does not mean that the actual ion has permanently localised negative charges on two specific oxygen atoms. \(\textbf{Final answer:}\) \[ \boxed{24\ \text{valence electrons};\ 2\ \text{singly bonded O atoms with }FC=-1} \]

Explanation: \(\textbf{Count electrons with the ionic charge:}\) Nitrogen contributes five valence electrons. Four hydrogen atoms contribute four additional electrons. Because ammonium carries a positive charge of one unit, one electron must be subtracted. \[ N_v=5+4(1)-1 \] \[ N_v=8 \] \(\textbf{Construct the electron-pair arrangement:}\) Nitrogen forms four N–H single bonds. Each bond contains two electrons, so all eight available electrons are used in bonding. The nitrogen atom consequently has no lone pair in the completed Lewis structure. It is surrounded by four shared electron pairs and therefore has an octet. Each hydrogen participates in one shared pair and satisfies the duplet requirement. \(\textbf{Calculate the nitrogen formal charge:}\) \[ FC=V-N-\frac{B}{2} \] For nitrogen, \(V=5\), \(N=0\) and \(B=8\). \[ FC_{\mathrm N}=5-0-\frac{8}{2} \] \[ FC_{\mathrm N}=+1 \] Each hydrogen has a formal charge of zero because it participates in one single bond. The sum of all atomic formal charges is consequently \(+1\), matching the ammonium ion's charge. A positive ion must not be treated as having one additional valence electron; positive charge indicates electron deficiency relative to the corresponding neutral-atom total. \(\textbf{Final answer:}\) \[ \boxed{N_v=8;\ FC_{\mathrm N}=+1} \]

Explanation: \(\textbf{Establish the valence-electron total:}\) Carbon contributes four valence electrons and oxygen contributes six. \[ N_v=4+6=10 \] \(\textbf{Identify the electron distribution:}\) A carbon–oxygen triple bond contains three shared electron pairs, representing six bonding electrons. The remaining four electrons constitute one lone pair on carbon and one on oxygen. Each atom therefore has eight electrons around it: six associated with the triple bond and two in its own lone pair. The Lewis arrangement satisfies the octet rule for both second-period atoms. \(\textbf{Apply the formal-charge relation:}\) \[ FC=V-N-\frac{B}{2} \] For carbon: \[ FC_{\mathrm C}=4-2-\frac{6}{2} \] \[ FC_{\mathrm C}=-1 \] For oxygen: \[ FC_{\mathrm O}=6-2-\frac{6}{2} \] \[ FC_{\mathrm O}=+1 \] \(\textbf{Check the total charge:}\) \[ FC_{\mathrm C}+FC_{\mathrm O}=-1+1=0 \] The formal charges therefore agree with the overall electrical neutrality of CO. The negative formal charge on carbon may appear unexpected because oxygen is more electronegative. However, formal charge is a bookkeeping quantity obtained by assigning half the bonding electrons to each bonded atom. It is not identical to the experimentally determined distribution of electron density or to actual partial atomic charge. Therefore, electronegativity alone cannot replace formal-charge calculation. \(\textbf{Final answer:}\) \[ \boxed{FC_{\mathrm C}=-1,\quad FC_{\mathrm O}=+1} \]

Explanation: \(\textbf{Determine the total valence-electron count:}\) Nitrogen contributes five valence electrons, and each oxygen atom contributes six. The negative charge adds one electron. \[ N_v=5+2(6)+1=18 \] \(\textbf{Distribute the bonding and nonbonding electrons:}\) The N=O double bond contains four bonding electrons, while the N–O single bond contains two. Thus, six electrons are involved in bonding, leaving twelve nonbonding electrons. The singly bonded oxygen has three lone pairs, accounting for six nonbonding electrons. The doubly bonded oxygen has two lone pairs, accounting for four nonbonding electrons. The remaining two electrons form one lone pair on nitrogen. Therefore, all eighteen valence electrons are accounted for, and each atom satisfies the octet rule. \(\textbf{Calculate the formal charges:}\) The formal charge on an atom is given by: \[ FC=V-N-\frac{B}{2} \] Here, \(V\) represents the valence electrons of the isolated atom, \(N\) represents nonbonding electrons, and \(B\) represents bonding electrons. For nitrogen, which has one lone pair and participates in one double bond and one single bond: \[ FC_{\mathrm N}=5-2-\frac{6}{2}=0 \] For the singly bonded oxygen, which has three lone pairs: \[ FC_{\mathrm{O(single)}}=6-6-\frac{2}{2}=-1 \] For the doubly bonded oxygen, which has two lone pairs: \[ FC_{\mathrm{O(double)}}=6-4-\frac{4}{2}=0 \] \(\textbf{Verify the overall ionic charge:}\) \[ FC_{\mathrm{total}}=0+(-1)+0=-1 \] The sum of the formal charges agrees with the overall charge of \(\mathrm{NO_2^-}\). \(\textbf{Interpret the resonance structures:}\) The nitrite ion has two equivalent resonance contributors in which the positions of the N=O double bond and N–O single bond are interchanged. In either contributor, the singly bonded oxygen carries a formal charge of \(-1\), while nitrogen and the doubly bonded oxygen have zero formal charge. However, the actual ion is a resonance hybrid with equivalent N–O bonds and delocalized negative charge. Therefore, the specified Lewis contributor has formal charges \(0\), \(-1\) and \(0\) on nitrogen, singly bonded oxygen and doubly bonded oxygen, respectively.

Explanation: \(\textbf{Check the total valence electrons:}\) Nitrogen contributes five electrons and three oxygen atoms contribute eighteen. The negative ionic charge contributes one additional electron. \[ N_v=5+3(6)+1=24 \] The proposed arrangement contains two double bonds, accounting for eight bonding electrons, and one single bond, accounting for two. Thus, ten electrons are used in the nitrogen–oxygen bonds. The two doubly bonded oxygen atoms each retain two lone pairs, accounting for eight more electrons altogether. The singly bonded oxygen retains three lone pairs, accounting for six electrons. The total is therefore \(10+8+6=24\), so the overall electron count is not the defect. \(\textbf{Examine nitrogen's valence shell:}\) Each double bond places two shared electron pairs around nitrogen. The single bond places one additional shared pair around nitrogen. Consequently, nitrogen is surrounded by five shared electron pairs, or ten electrons. Nitrogen is a second-period element and cannot exceed an octet in an ordinary Lewis representation. The proposed structure therefore violates the octet restriction even though its total valence-electron count is correct. \(\textbf{Check the competing formal-charge argument:}\) The proposed arrangement gives nitrogen formal charge zero, two doubly bonded oxygens formal charge zero, and the singly bonded oxygen formal charge \(-1\). Its formal-charge sum is therefore \(-1\), not zero. Nevertheless, obtaining a reasonable formal-charge sum cannot make a structure acceptable if it violates the octet restriction on a second-period atom. A conventional acceptable Lewis contributor contains one N=O double bond and two N–O single bonds. In that contributor, nitrogen possesses eight surrounding electrons, has formal charge \(+1\), and the two singly bonded oxygens each carry formal charge \(-1\). Thus, octet validity must be checked before preferring a structure merely because it appears to minimise formal-charge separation.

Explanation: \(\textbf{Examine the electron-deficient species:}\) Boron possesses three valence electrons. In the conventional Lewis structure of \(\mathrm{BF_3}\), boron forms three B–F single bonds. Each bond contains one shared electron pair, giving boron six surrounding electrons. The boron atom is therefore electron-deficient relative to the octet rule. This does not make \(\mathrm{BF_3}\) an impossible molecule; it is a recognised incomplete-octet species. \(\textbf{Consider fluoride addition:}\) A fluoride ion possesses a filled valence shell and can donate an electron pair to electron-deficient boron. The combination can be represented as: \[ \mathrm{BF_3+F^-\longrightarrow BF_4^-} \] The resulting conventional Lewis structure contains four B–F single bonds. Boron is now surrounded by four shared electron pairs, corresponding to eight electrons. \(\textbf{Calculate the boron formal charge:}\) Boron has three valence electrons in the isolated atom, no lone pairs in \(\mathrm{BF_4^-}\), and eight bonding electrons. \[ FC_{\mathrm B}=V-N-\frac{B}{2} \] \[ FC_{\mathrm B}=3-0-\frac{8}{2} \] \[ FC_{\mathrm B}=-1 \] Each fluorine has three lone pairs and one single bond, giving it a formal charge of zero. The total formal charge is consequently \(-1\), consistent with the charge on \(\mathrm{BF_4^-}\). \(\textbf{Interpret the electron-pair donation:}\) The fourth B–F bond may be represented initially as a coordinate covalent bond to show that both electrons were donated by fluoride. After formation, the four B–F bonds in the tetrafluoroborate ion are equivalent in the usual bonding description. The formal charge of \(-1\) on boron is not a claim that the actual excess electron density is completely localised on boron. The important reasoning is that fluoride donation completes boron's octet while the formal-charge bookkeeping reflects its four shared pairs. \(\textbf{Final answer:}\) \[ \boxed{6\longrightarrow8\ \text{electrons};\ FC_{\mathrm B}=-1} \]

Explanation: \(\textbf{Determine the two electron totals:}\) Nitrogen contributes five valence electrons and oxygen contributes six. For neutral nitric oxide: \[ N_v(\mathrm{NO})=5+6=11 \] An odd total number of electrons means that one electron must remain unpaired in an ordinary Lewis electron arrangement. For the positively charged nitrosonium ion, one electron is removed. \[ N_v(\mathrm{NO^+})=5+6-1=10 \] \(\textbf{Apply the octet requirement:}\) For a diatomic species consisting of two second-period atoms, both complete octets would require an appropriate distribution of shared and nonbonding electron pairs. A conventional Lewis structure of \(\mathrm{NO^+}\) contains an N≡O triple bond and one lone pair on each atom. The triple bond uses six electrons, and the two lone pairs use four more. Both atoms then have eight electrons around them. The formal charge is zero on nitrogen and \(+1\) on oxygen, matching the ionic charge. \(\textbf{Identify the limitation for neutral NO:}\) Neutral NO has eleven valence electrons and cannot accommodate all its electrons as paired bonding and nonbonding electrons while satisfying ordinary octets on both atoms. It is an odd-electron species and therefore provides an exception to the octet rule. The change from NO to \(\mathrm{NO^+}\) is not simply a change in the written charge symbol; it changes the actual number of electrons available for the Lewis structure. This electron-count distinction explains why the nitrosonium ion admits an octet-satisfying representation whereas neutral NO does not.

Explanation: \(\textbf{Evaluate Statement I:}\) Statement I is correct. Phosphorus has five valence electrons and forms three P–Cl single bonds in \(\mathrm{PCl_3}\). One lone pair remains on phosphorus. Thus, phosphorus is surrounded by six bonding electrons and two nonbonding electrons, satisfying the octet rule. \(\textbf{Evaluate Statement II:}\) Statement II is correct. In the conventional Lewis representation of \(\mathrm{PCl_5}\), phosphorus forms five P–Cl single bonds. Five shared pairs correspond to ten electrons around the central phosphorus atom. This is an expanded-octet representation of a third-period element. It is a standard textbook description and should not be interpreted as requiring an advanced orbital model to establish the Lewis electron count. \(\textbf{Evaluate Statement III:}\) Statement III is incorrect. Nitrogen is a second-period element and cannot accommodate ten surrounding electrons in a conventional octet-based Lewis structure. Five N–Cl single bonds would place ten bonding electrons around nitrogen. Therefore, the proposed representation violates the octet restriction. The fact that nitrogen and phosphorus both belong to Group 15 does not mean they have identical permitted Lewis bonding patterns. \(\textbf{Evaluate Statement IV:}\) Statement IV is incorrect. Boron contributes three valence electrons and forms three B–F single bonds in \(\mathrm{BF_3}\). Only six electrons surround boron in this conventional structure. \(\mathrm{BF_3}\) is consequently an incomplete-octet example. \(\textbf{Overall conclusion:}\) The comparison distinguishes a normal octet in \(\mathrm{PCl_3}\), an expanded octet in \(\mathrm{PCl_5}\), an invalid second-period expansion in the proposed \(\mathrm{NCl_5}\) structure, and an incomplete octet in \(\mathrm{BF_3}\). Only Statements I and II are correct.

Explanation: \(\textbf{Determine the stable ionic charges:}\) Element P has two electrons in its outermost shell. Under the stated simple-ion assumption, it loses both electrons to attain the electronic configuration \(2,8\). The corresponding ion is \(\mathrm{P^{2+}}\). Element Q has seven valence electrons. It gains one electron to complete its outer octet and forms \(\mathrm{Q^-}\). \(\textbf{Apply electrical neutrality:}\) One \(\mathrm{P^{2+}}\) ion possesses a charge of \(+2\). Each \(\mathrm{Q^-}\) ion possesses a charge of \(-1\). Two Q ions are therefore required to balance the charge on one P ion. \[ (+2)+2(-1)=0 \] Hence, the simplest neutral ionic composition is: \[ \mathrm{PQ_2} \] \(\textbf{Count transferred electrons:}\) Formation of one \(\mathrm{P^{2+}}\) ion involves the loss of two electrons. Formation of each \(\mathrm{Q^-}\) ion involves the gain of one electron. The two Q atoms together accept the two electrons lost by P. \[ \mathrm{P\rightarrow P^{2+}+2e^-} \] \[ \mathrm{2Q+2e^-\rightarrow2Q^-} \] Therefore, two electrons are transferred per empirical formula unit. \(\textbf{Check the ionic formula:}\) The formula does not imply that the crystalline solid consists of independent three-atom molecules. It represents the simplest ratio of its constituent ions. Electrical neutrality and electron conservation are both satisfied. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{PQ_2};\ 2\ \text{electrons transferred}} \]

Explanation: The electronic description of sodium chloride begins with electron transfer from sodium to chlorine. Sodium forms \(\mathrm{Na^+}\), and chlorine forms \(\mathrm{Cl^-}\). These oppositely charged ions experience electrostatic attraction. However, the existence of electron transfer does not mean that the resulting crystalline material contains separate molecules consisting of one sodium ion and one chloride ion. In a crystal, each ion interacts with several surrounding ions rather than only one specially paired counterion. The attractions operate throughout an extended three-dimensional ionic arrangement. The forces also include repulsions between ions of the same charge, and the stable crystalline structure reflects their combined interactions. The overall solid remains electrically neutral because the total positive and negative charges balance. For sodium chloride, this condition requires equal numbers of \(\mathrm{Na^+}\) and \(\mathrm{Cl^-}\) ions. Consequently, the chemical formula \(\mathrm{NaCl}\) gives the simplest whole-number ratio of the ions. It is a formula unit rather than the molecular formula of a discrete covalent species in the crystal. The electrostatic attraction persists after electron transfer and is fundamental to the stability of the ionic solid. Thus, ionic bonding must be distinguished from the electron-transfer event that produces its constituent ions.

Explanation: \(\textbf{Establish the total electron count:}\) Carbon dioxide contains one carbon atom and two oxygen atoms. \[ N_v=4+2(6)=16 \] Both proposed structures can be assigned sixteen valence electrons while satisfying octets on all three atoms. Therefore, total electron count and octet completion alone do not distinguish them. \(\textbf{Examine Structure I:}\) In \(\mathrm{O=C=O}\), carbon forms two double bonds. Each oxygen has two lone pairs and participates in one double bond. For carbon: \[ FC_{\mathrm C}=4-0-\frac{8}{2}=0 \] For either oxygen: \[ FC_{\mathrm O}=6-4-\frac{4}{2}=0 \] Thus, Structure I has zero formal charge on every atom. \(\textbf{Examine Structure II:}\) In \(\mathrm{O\equiv C-O}\), the triply bonded oxygen has one lone pair. The singly bonded oxygen has three lone pairs. For the triply bonded oxygen: \[ FC_{\mathrm{O(triple)}}=6-2-\frac{6}{2} \] \[ FC_{\mathrm{O(triple)}}=+1 \] For the singly bonded oxygen: \[ FC_{\mathrm{O(single)}}=6-6-\frac{2}{2}=-1 \] Carbon still has four bonding pairs and zero formal charge. Hence, the total formal charge of Structure II is zero, as required for neutral \(\mathrm{CO_2}\). \(\textbf{Compare the two Lewis representations:}\) Structure I avoids formal-charge separation altogether, whereas Structure II places opposite formal charges on the two oxygen atoms. For two otherwise acceptable octet-satisfying Lewis structures, minimisation of formal-charge separation generally favours the structure with smaller charge magnitudes. The conventional dominant Lewis representation of carbon dioxide is therefore Structure I. Structure II is not rejected for an incorrect total charge or an incomplete octet; it is less favourable as a Lewis contributor because of its formal-charge separation. \(\textbf{Final answer:}\) Structure I is preferred, and the triply bonded oxygen in Structure II carries formal charge \(+1\).

Explanation: \(\textbf{Count the electrons available:}\) Each carbon contributes four valence electrons, and each hydrogen contributes one. \[ N_v=2(4)+4(1)=12 \] The four C–H single bonds account for eight electrons. An additional C–C single bond accounts for two electrons. Thus, the proposed five single bonds use ten electrons, leaving two electrons to be distributed. \(\textbf{Test the proposed arrangement:}\) If both remaining electrons form a lone pair on one carbon, that carbon has three shared pairs and one lone pair around it. It therefore achieves an octet. The other carbon, however, participates in only three single bonds and has six electrons around it. The two carbon atoms have different electron environments, and the proposed arrangement produces nonzero formal charges. \(\textbf{Restore the carbon octets:}\) Instead of placing the final two electrons in a lone pair on one carbon, they can form an additional shared pair between the carbon atoms. The C–C single bond becomes a C=C double bond. The resulting connectivity is: \[ \mathrm{H_2C=CH_2} \] The structure contains four C–H single bonds and one C=C double bond. These bonds account for all twelve valence electrons. Each carbon is surrounded by two C–H bonding pairs and two C–C bonding pairs, giving eight electrons. Each hydrogen shares one bonding pair and satisfies the duplet requirement. \(\textbf{Verify formal charges:}\) Each carbon forms four bonds in total when bond multiplicity is counted. \[ FC_{\mathrm C}=4-0-\frac{8}{2}=0 \] Each hydrogen also has formal charge zero. Thus, the completed structure satisfies electron conservation, carbon octets, hydrogen duplets and overall neutrality. The correction illustrates why simply obtaining the right total number of electrons does not guarantee that a proposed Lewis structure is satisfactory. The remaining electrons must also be distributed so that the bonding arrangement meets the relevant valence requirements.

Explanation: \(\textbf{Identify the initial electron-pair arrangement:}\) A neutral water molecule contains two O–H single bonds and two lone pairs on oxygen. The oxygen atom has a complete octet because four shared electrons and four nonbonding electrons surround it. A proton, \(\mathrm{H^+}\), has no electron available to contribute to a new shared pair. Consequently, the additional O–H bond can be represented as forming through donation of one existing oxygen lone pair. This is the donor–acceptor description of coordinate covalent bond formation. \[ \mathrm{H_2O+H^+\longrightarrow H_3O^+} \] \(\textbf{Follow the electron-pair rearrangement:}\) One of the two oxygen lone pairs becomes the shared pair in the third O–H bond. The other oxygen lone pair remains nonbonding. The hydronium ion therefore contains three O–H bonding pairs and one lone pair on oxygen. Three bonding pairs supply six surrounding electrons, and the remaining lone pair supplies two. Thus, oxygen still satisfies the octet rule. \(\textbf{Verify the total valence-electron count:}\) \[ N_v=6+3(1)-1 \] \[ N_v=8 \] The three O–H bonds use six electrons, and the remaining lone pair uses two. \(\textbf{Calculate the formal charge:}\) Oxygen possesses six valence electrons in the isolated atom. In the hydronium Lewis structure, it has two nonbonding electrons and six bonding electrons. \[ FC_{\mathrm O}=6-2-\frac{6}{2} \] \[ FC_{\mathrm O}=+1 \] Each hydrogen has formal charge zero. The sum of the formal charges therefore agrees with the overall positive ionic charge. \(\textbf{Interpret the coordinate-bond description:}\) Although the third bond is initially represented as arising through donation from oxygen, it is not permanently a separate physical category of O–H bond. All three O–H bonds in the hydronium ion are equivalent in the usual molecular description. The coordinate-bond notation identifies the origin of the shared pair during bond formation, not a lasting distinction between otherwise equivalent bonds. The correct structure therefore contains three bonding pairs, one lone pair and a formal charge of \(+1\) on oxygen.

Explanation: \(\textbf{Verify the available valence electrons:}\) Hydrogen contributes one electron, carbon contributes four and nitrogen contributes five. \[ N_v=1+4+5=10 \] Each proposed structure contains one single bond, one triple bond and one lone pair. Thus, both arrangements use the same ten valence electrons. Each hydrogen has a duplet, while carbon and nitrogen have complete octets. Consequently, electron counting and octet completion alone cannot distinguish the relative favourability of these Lewis structures. \(\textbf{Evaluate Structure I:}\) In \(\mathrm{H-C\equiv N}\), carbon participates in four shared electron pairs and possesses no lone pair. \[ FC_{\mathrm C}=4-0-\frac{8}{2}=0 \] Nitrogen has one lone pair and three bonding pairs. \[ FC_{\mathrm N}=5-2-\frac{6}{2}=0 \] Hydrogen also has formal charge zero. Therefore, Structure I contains no formal-charge separation. \(\textbf{Evaluate Structure II:}\) In \(\mathrm{H-N\equiv C}\), nitrogen participates in four bonding pairs and possesses no lone pair. \[ FC_{\mathrm N}=5-0-\frac{8}{2}=+1 \] Carbon has one lone pair and participates in the triple bond. \[ FC_{\mathrm C}=4-2-\frac{6}{2}=-1 \] The total formal charge remains zero, so Structure II is electrically neutral. \(\textbf{Compare the structures:}\) Structure II contains opposite formal charges, whereas Structure I does not. The conventional connectivity of hydrogen cyanide is \(\mathrm{H-C\equiv N}\), consistent with Structure I. The alternative connectivity corresponds to a distinct species rather than merely another resonance contributor, because the attachment of hydrogen changes. Formal-charge minimisation supports Structure I, but different atomic connectivities must not be treated as interchangeable resonance forms. Therefore, option A correctly describes the formal-charge distributions.

Explanation: \(\textbf{Examine the central-atom electron counts:}\) Beryllium possesses two valence electrons and forms two Be–Cl single bonds in the conventional Lewis structure of \(\mathrm{BeCl_2}\). Each single bond contributes two electrons to the number surrounding the central atom. Beryllium is therefore surrounded by four electrons. Boron possesses three valence electrons and forms three B–Cl single bonds in \(\mathrm{BCl_3}\). Consequently, six electrons surround the boron atom. Carbon forms four C–Cl single bonds in \(\mathrm{CCl_4}\), giving eight surrounding electrons. \(\textbf{Calculate the formal charges independently:}\) The formal-charge formula is: \[ FC=V-N-\frac{B}{2} \] For beryllium: \[ FC_{\mathrm{Be}}=2-0-\frac{4}{2}=0 \] For boron: \[ FC_{\mathrm B}=3-0-\frac{6}{2}=0 \] For carbon: \[ FC_{\mathrm C}=4-0-\frac{8}{2}=0 \] The central atoms therefore all possess zero formal charge in the specified structures. \(\textbf{Interpret the octet exceptions:}\) A formal charge of zero does not imply that an atom has a complete octet. Beryllium in \(\mathrm{BeCl_2}\) has only four surrounding electrons. Boron in \(\mathrm{BCl_3}\) has six. Carbon in \(\mathrm{CCl_4}\) has eight and satisfies the octet rule. Rows A and C confuse electron deficiency with formal charge, while Row B assigns an incorrect electron count to beryllium. Thus, only Row D simultaneously satisfies the Lewis electron counts and formal-charge calculations.

Explanation: An ionic crystal contains positively and negatively charged ions arranged in a regular three-dimensional structure. Its cohesion arises primarily from the collective electrostatic interactions among these ions. Oppositely charged ions attract, while similarly charged ions repel. In an undisturbed stable crystal, the arrangement provides favourable interactions that help maintain the lattice. The question specifies a displacement that brings several ions of the same charge into close proximity. Such a displacement increases unfavourable electrostatic repulsions across the shifted region. The repulsions oppose the displaced configuration and can promote separation along a fracture plane. This explains the characteristic brittleness of many ionic crystals under suitable mechanical stress. The process does not require ionic bonds to transform into covalent bonds. Neither does the displacement remove electrons from every neighbouring ion or make the ions electrically neutral. The ions retain their charges, and the electrostatic interactions respond to their changed relative positions. The tendency to fracture should also not be confused with the statement that every ionic crystal has identical mechanical strength. Actual fracture behaviour depends on crystal structure, defects and the direction of the applied force. The decisive feature in the stated arrangement is the appearance of unfavourable like-charge interactions after displacement. Therefore, option C correctly connects ionic charge, crystal arrangement and mechanical behaviour.

Explanation: Sodium chloride is composed of sodium cations and chloride anions. Electrical conduction requires charged particles capable of moving through the material. In dry, solid sodium chloride, the ions occupy positions in an extended crystalline lattice. Although the solid contains charged particles, those ions cannot move freely through the crystal under ordinary conditions. Therefore, solid sodium chloride is a poor conductor of electricity. When sodium chloride melts, the rigid crystalline arrangement is disrupted. The sodium and chloride ions remain charged, but they acquire sufficient mobility to carry electrical current. Thus, molten sodium chloride conducts electricity through ionic motion. When sodium chloride dissolves in water, the crystalline arrangement separates into hydrated sodium and chloride ions. These dissolved ions can also migrate under an applied electric field. Consequently, an aqueous sodium chloride solution conducts electricity. The conduction mechanism in the molten and dissolved states is therefore associated with ion movement. It is incorrect to assume that melting neutralises the ions or destroys their electrical charges. It is equally incorrect to infer good conductivity merely from the presence of ions if those ions remain essentially immobile. The comparison demonstrates that the presence of charged species and their mobility are distinct requirements for ionic electrical conduction.