Q.Which of the following attain the linear structure: (Note: more than one of the given options may be correct.)

We need to find which pair of molecules both have an $\mathrm{sp^2}$-hybridized central atom and a bent (V-shaped) molecular geometry. The trick is that "bent" can arise from two different hybridization patterns: $\mathrm{sp^3}$ (like water, with two lone pairs) or $\mathrm{sp^2}$ (like ozone, with one lone pair). So we must check each molecule individually.

Concept & Intuition

Hybridization is determined by the steric number:

$\text{Steric number} = (\text{number of atoms bonded to central atom}) + (\text{number of lone pairs on central atom})$.

• Steric number 2 → $\mathrm{sp}$ → linear
• Steric number 3 → $\mathrm{sp^2}$ → trigonal planar (if 0 lone pairs) or bent (if 1 lone pair)
• Steric number 4 → $\mathrm{sp^3}$ → tetrahedral (0 lone pairs), trigonal pyramidal (1 lone pair), or bent (2 lone pairs)

So for a molecule to be $\mathrm{sp^2}$ and bent, the central atom must have exactly 3 regions of electron density (steric number 3), with one of them being a lone pair.

Step-by-step analysis

1. $\mathbf{H_2O}$

   Central atom: oxygen.

   Bonds: 2 H atoms.

   Lone pairs on O: 2.

   Steric number = 2 + 2 = 4 → $\mathrm{sp^3}$ hybridization.

   Geometry: bent (due to two lone pairs), but not $\mathrm{sp^2}$.

   → Eliminate any option containing $\mathrm{H_2O}$ if we need $\mathrm{sp^2}$ bent. That rules out (A), (C), and (D) because they all include $\mathrm{H_2O}$.

2. $\mathbf{SO_2}$

   Central atom: sulfur.

   Bonds: 2 O atoms (one double bond, one coordinate bond — but both count as 1 bonding region each).

   Lone pairs on S: 1.

   Steric number = 2 + 1 = 3 → $\mathrm{sp^2}$ hybridization.

   Geometry: bent (trigonal planar with one vertex occupied by a lone pair).

   → $\mathrm{SO_2}$ fits.

3. $\mathbf{O_3}$ (ozone)

   Central atom: the middle oxygen.

   Bonds: 2 O atoms (one double bond, one single bond — again, two bonding regions).

   Lone pairs on central O: 1. …

Concept & Intuition

Bond order tells us the net number of bonding pairs between two atoms. For diatomic molecules like O₂ and its ions, we use molecular orbital (MO) theory. The key insight: adding electrons to O₂ fills antibonding orbitals (which decrease bond order), while removing electrons removes antibonding electrons (which increase bond order). So O₂²⁺ (missing two electrons) has a higher bond order than neutral O₂, while O₂⁻ and O₂²⁻ (added electrons) have lower bond orders. The problem gives us a relationship between these values.

Step-by-step reasoning

1. Recall the MO electron configuration for O₂ and its ions For O₂ (16 electrons total), the valence MO order is:

Bond order = $\frac{\text{(bonding electrons)} - \text{(antibonding electrons)}}{2}$.

For neutral O₂: bonding = 10 (σ₂s², σ₂p_z², π₂p⁴), antibonding = 6 (σ₂s², π₂p²) → bond order = (10-6)/2 = 2.

2. Find bond order of O₂⁻ O₂⁻ has 17 electrons. The extra electron goes into a π* orbital:

Antibonding count becomes 7. Bond order = (10 - 7)/2 = 1.5.

3. Find bond order of O₂²⁻

   O₂²⁻ has 18 electrons. Both π* orbitals are filled: π*⁴.

   Antibonding count = 8. Bond order = (10 - 8)/2 = 1.0.

4. Sum of bond orders for O₂⁻ and O₂²⁻

5. Find bond order of O₂²⁺ …

Concept & Intuition

This question tests your knowledge of the trends in the oxygen family (Group 16: O, S, Se, Te, Po). As you go down the group, the oxides become more metallic, less volatile, and their acidic/basic character shifts. For example, SO₂ is a gas, SeO₂ is a solid, and TeO₂ is also a solid. Additionally, the oxidizing power of the +4 oxidation state increases down the group: Te(IV) is more easily reduced than S(IV). So we need to check each statement against these trends.

Step-by-step reasoning

1. Statement (A): TeO₂ is an oxidizing agent

   • Tellurium in TeO₂ is in the +4 oxidation state. Down the group, the stability of the +4 state decreases relative to the +6 state, so Te(IV) tends to get reduced to Te(0) or Te(-II), making it an oxidizing agent. For example, TeO₂ can oxidize SO₂ to SO₄²⁻. This statement is correct.

2. Statement (B): SeO₃ is acidic in nature

   • SeO₃ is the anhydride of selenic acid (H₂SeO₄). It dissolves in water to give a strongly acidic solution. All higher oxides of Group 16 (SO₃, SeO₃, TeO₃) are acidic. This statement is correct.

3. Statement (C): SeO₂ is a gas …

The concept first

This is a structure-recognition question from Chemistry in Everyday Life. The trick is to identify each drawn molecule by its functional-group pattern and then recall its therapeutic class.

Aspirin is made by acetylating salicylic acid (2-hydroxybenzoic acid) with acetic anhydride:

So its fingerprint is: carboxylic acid + ester, ortho to each other on a benzene ring.

Step-by-step through the options

(A) Aspirin — $-\mathrm{O\text{-}CO\text{-}CH_3}$ and $-\mathrm{CO_2H}$ ortho. ✓ Aspirin is an analgesic and antipyretic, but its most famous modern use is as an antiplatelet agent: it irreversibly acetylates the enzyme cyclo-oxygenase (COX) in platelets, shutting down synthesis of thromboxane $\mathrm{A_2}$, the prostaglandin that makes platelets aggregate. Fewer platelet clumps means fewer clots in the coronary arteries — hence a daily low dose is prescribed to prevent heart attacks. This is the compound the question wants.

(B) Serotonin — indole ring with a phenolic $-\mathrm{OH}$ and a $-\mathrm{CH_2CH_2NH_2}$ side chain. ✗ This is a neurotransmitter; low levels are associated with depression (antidepressants act on serotonin pathways). Nothing to do with preventing myocardial infarction. …

The key to this question is understanding how the properties of oxides change as you go down Group 16 (the chalcogens: O, S, Se, Te, Po). The stability of oxidation states, the nature of the oxides (acidic/basic), and their physical states all follow predictable trends.

Let's examine each statement one by one.

1. Statement (A): TeO$_2$ is an oxidizing agent.

   Tellurium is below sulfur in the group. Down the group, the +4 oxidation state becomes more stable relative to the +6 state. This means Te(IV) in TeO$_2$ has a tendency to get reduced to a lower state (like Te metal) or to accept electrons — it acts as an oxidizing agent. This is correct. In fact, TeO$_2$ is a mild oxidizing agent.

2. Statement (B): SO$_2$ is a reducing agent.

   Sulfur in SO$_2$ is in the +4 oxidation state. Sulfur can go to +6 (in SO$_3$ or H$_2$SO$_4$) by losing electrons, so SO$_2$ commonly acts as a reducing agent. For example, it reduces Fe$^{3+}$ to Fe$^{2+}$ or decolourizes KMnO$_4$. This statement is correct.

3. Statement (C): SeO$_3$ is acidic in nature. …

The key here is to recall the physical properties of each substance mentioned. The question tests factual knowledge from inorganic chemistry, especially about noble gas compounds, common gases, and periodic trends. Let’s examine each option carefully.

1. Option (A): XeO₃ is a colourless explosive gas

   Xenon trioxide (XeO₃) is indeed a powerful explosive and colourless, but it is not a gas at room temperature. It is a white crystalline solid that detonates easily when dry. The phrase “explosive gas” is therefore incorrect.

   Watch out

   A common mistake is to assume all xenon compounds are gases because xenon itself is a gas. In fact, XeO₃ and XeF₂ are solids.

2. Option (B): SO₂ is highly soluble in water

   Sulfur dioxide (SO₂) dissolves readily in water to form sulfurous acid (H₂SO₃). Its solubility is about 40 g per 100 mL at 20°C, which qualifies as “highly soluble.” This statement is correct.

3. Option (C): Noble gases have very low boiling points …

Concept & Intuition

Hybridization is a model that explains the geometry of molecules by mixing atomic orbitals to form new, equivalent orbitals. The key is the steric number (SN): the number of atoms bonded to the central atom plus the number of lone pairs on it.

• SN = 2 → sp hybridization (linear)
• SN = 3 → sp² hybridization (trigonal planar)
• SN = 4 → sp³ hybridization (tetrahedral)
• SN = 5 → sp³d hybridization (trigonal bipyramidal)
• SN = 6 → sp³d² hybridization (octahedral)

We apply this to each molecule.

1. BF₃ (boron trifluoride)

   • Boron is the central atom. It forms three single bonds with three fluorine atoms.
   • Boron has no lone pairs (it has only 3 valence electrons, all used in bonding).
   • Steric number = 3 (three bonded atoms) → hybridization = sp².
   • Geometry: trigonal planar.

2. BeF₂ (beryllium difluoride)

   • Beryllium is the central atom. It forms two single bonds with two fluorine atoms.
   • Beryllium has no lone pairs (its two valence electrons are both used in bonding).
   • Steric number = 2 → hybridization = sp.
   • Geometry: linear.

3. BrF₃ (bromine trifluoride)

   • Bromine is the central atom. It forms three single bonds with three fluorine atoms.
   • Bromine has 7 valence electrons; three are used in bonds, leaving 4 electrons as two lone pairs. …

The key here is understanding how nitrates of different metals behave when heated. For most metal nitrates, the decomposition products depend on the position of the metal in the reactivity series. Magnesium is a fairly reactive metal — it lies above copper in the series but below sodium and potassium. Its nitrate does not decompose to the metal or the metal peroxide; instead, it gives the metal oxide, along with nitrogen dioxide and oxygen.

Let’s walk through the reasoning step by step.

1. Recall the general decomposition pattern for nitrates of metals like magnesium. For metals that are moderately reactive (from magnesium to copper in the reactivity series), heating the nitrate yields the metal oxide, nitrogen dioxide ($\mathrm{NO_2}$), and oxygen ($\mathrm{O_2}$). The general equation is:

where M is a divalent metal.

2. Apply this to magnesium nitrate. Magnesium nitrate is $\mathrm{Mg(NO_3)_2}$. Following the pattern above:

So the products are magnesium oxide ($\mathrm{MgO}$), nitrogen dioxide ($\mathrm{NO_2}$), and oxygen ($\mathrm{O_2}$).

3. Check the options against this result.
   • (A) $\mathrm{MgO_2, NO_2, O_2}$ — magnesium peroxide is not formed here; that would require different conditions.
   • (B) $\mathrm{MgO_2, NO, O_2}$ — again, wrong oxide and wrong nitrogen oxide. …

The acidic or basic nature of an oxide depends on the metallic or non-metallic character of the element forming the oxide.

• Basic oxides are typically formed by metals, especially alkali and alkaline earth metals. They react with acids to form salt and water.
• Acidic oxides are typically formed by non-metals. They react with bases to form salt and water.
• Amphoteric oxides are formed by certain metals or metalloids. They can react with both acids and bases.

Periodic Trends in Acidic Nature of Oxides:

1. Across a Period (Left to Right): As we move from left to right across a period, the non-metallic character of elements increases, and electronegativity increases. Consequently, the acidic nature of their oxides generally increases. The trend typically goes from strongly basic to amphoteric to weakly acidic to strongly acidic.
2. Down a Group: As we move down a group, the metallic character of elements increases. Therefore, the basic nature of their oxides generally increases, or the acidic nature decreases.

Let's apply these concepts to the given oxides.

1. Identify the elements and their positions:

   The oxides given are $\mathrm{Li_2O}$, $\mathrm{BeO}$, $\mathrm{CO_2}$, $\mathrm{B_2O_3}$, $\mathrm{N_2O_3}$. The elements involved are Lithium (Li), Beryllium (Be), Boron (B), Carbon (C), and Nitrogen (N). All these elements belong to the second period of the periodic table:

   • Li (Group 1, metal)
   • Be (Group 2, metal)
   • B (Group 13, metalloid)
   • C (Group 14, non-metal)
   • N (Group 15, non-metal)

2. Determine the nature of each oxide based on its element's position:

   Following the trend across Period 2:

   • $\mathrm{Li_2O}$: Lithium is an alkali metal. Its oxide is strongly basic.
   • $\mathrm{BeO}$: Beryllium is an alkaline earth metal, but due to its small size and high charge density, $\mathrm{BeO}$ is amphoteric.
   • $\mathrm{B_2O_3}$: Boron is a metalloid. Its oxide is weakly acidic.
   • $\mathrm{CO_2}$: Carbon is a non-metal. Its oxide is acidic.
   • $\mathrm{N_2O_3}$: Nitrogen is a non-metal. Its oxide is acidic.

3. Compare the acidic nature of $\mathrm{CO_2}$ and $\mathrm{N_2O_3}$: …

Hybridization is a concept used to explain the bonding in molecules where atomic orbitals mix to form new hybrid orbitals. These hybrid orbitals are more suitable for forming bonds and determining molecular geometry. The type of hybridization of a central atom is determined by its steric number, which is the sum of the number of sigma ($\sigma$) bonds and the number of lone pairs of electrons around that central atom.

Here's how the steric number relates to hybridization:

• Steric number 2: $sp$ hybridization
• Steric number 3: $sp^2$ hybridization
• Steric number 4: $sp^3$ hybridization
• Steric number 5: $sp^3d$ hybridization
• Steric number 6: $sp^3d^2$ hybridization

To find the hybridization of the central atom in each compound, we will:

1. Identify the central atom.
2. Draw the Lewis structure to determine the number of sigma bonds and lone pairs on the central atom.
3. Calculate the steric number.
4. Determine the hybridization based on the steric number.

Let's analyze each compound:

1. $\mathrm{H_2CO_3}$ (Carbonic acid)

• Central atom: Carbon (C).
• Lewis structure: Carbon is bonded to three oxygen atoms. One oxygen is double-bonded to carbon, and the other two oxygens are single-bonded to carbon and also to hydrogen atoms (forming -OH groups).
  • The carbon atom forms one $\mathrm{C=O}$ double bond and two $\mathrm{C-OH}$ single bonds.
• Sigma bonds on C: A double bond consists of one sigma bond and one pi bond. A single bond is one sigma bond. So, C has $1 (\text{from C=O}) + 2 (\text{from C-OH}) = 3$ sigma bonds.
• Lone pairs on C: Carbon has no lone pairs in this structure.
• Steric number: $3 (\text{sigma bonds}) + 0 (\text{lone pairs}) = 3$.
• Hybridization: A steric number of 3 corresponds to $sp^2$ hybridization.

2. $\mathrm{SiF_4}$ (Silicon tetrafluoride)

• Central atom: Silicon (Si).
• Lewis structure: Silicon is bonded to four fluorine atoms, each with a single bond.
• Sigma bonds on Si: Silicon forms 4 single bonds with fluorine atoms, so it has 4 sigma bonds.
• Lone pairs on Si: Silicon has no lone pairs in this structure.
• Steric number: $4 (\text{sigma bonds}) + 0 (\text{lone pairs}) = 4$.
• Hybridization: A steric number of 4 corresponds to $sp^3$ hybridization.

3. $\mathrm{BF_3}$ (Boron trifluoride)

• Central atom: Boron (B).
• Lewis structure: Boron is bonded to three fluorine atoms, each with a single bond. Boron is an exception to the octet rule and typically forms three bonds.
• Sigma bonds on B: Boron forms 3 single bonds with fluorine atoms, so it has 3 sigma bonds.
• Lone pairs on B: Boron has no lone pairs in this structure.
• Steric number: $3 (\text{sigma bonds}) + 0 (\text{lone pairs}) = 3$. …

The key to this question lies in understanding the relationship between bond order and bond length: higher bond order means more electron density between nuclei, pulling them closer together and shortening the bond. We need to examine the structure of each oxide to determine which has the strongest N–N bond.

Let me work through the structure of each molecule:

1. $\mathrm{N_2O}$ (nitrous oxide)

   The Lewis structure is $\mathrm{N≡N–O}$ with resonance forms, but the dominant picture shows a nitrogen–nitrogen triple bond. The bond order is approximately 3 (or slightly less due to resonance, but still very high). This is the familiar "laughing gas" with a linear geometry.

2. $\mathrm{N_2O_3}$ (dinitrogen trioxide)

   This molecule has the structure $\mathrm{O=N–N=O}$ (with one oxygen bonded to each nitrogen). The central N–N bond is a single bond. The molecule exists in equilibrium with $\mathrm{NO}$ and $\mathrm{NO_2}$ because this single bond is relatively weak. Bond order ≈ 1.

3. $\mathrm{N_2O_4}$ (dinitrogen tetroxide)

   The structure is $\mathrm{O_2N–NO_2}$, essentially two $\mathrm{NO_2}$ groups joined by a central N–N bond. This central bond is a single bond and is actually quite long and weak (which is why $\mathrm{N_2O_4}$ readily dissociates into two $\mathrm{NO_2}$ radicals). Bond order = 1.

4. $\mathrm{N_2O_5}$ (dinitrogen pentoxide) …

The key to this question is VSEPR (Valence Shell Electron Pair Repulsion) theory. Molecular shape is determined not just by the number of atoms bonded to the central atom, but by the total number of electron domains (bonding pairs and lone pairs) around it. A see-saw shape arises specifically from 5 electron domains with one lone pair; a linear shape arises from 2 electron domains (or from a molecule with a triple bond, like acetylene).

Let’s check each option systematically.

1. Option (A): CH₄ and SO₃

   • CH₄: Carbon has 4 bonding pairs, zero lone pairs — 4 electron domains. That gives a tetrahedral shape, not see-saw.
   • SO₃: Sulfur has 3 bonding pairs (double bonds count as one domain each) and zero lone pairs — 3 domains. That gives a trigonal planar shape, not linear. So this pair is wrong.

2. Option (B): XeF₄ and CS₂

   • XeF₄: Xenon has 4 bonding pairs and 2 lone pairs — 6 electron domains. The lone pairs occupy opposite positions (axial), giving a square planar shape, not see-saw.
   • CS₂: Carbon has 2 double bonds (2 domains), zero lone pairs — that’s linear. But the first molecule fails. So this pair is wrong.

3. Option (C): SF₄ and C₂H₂

   • SF₄: Sulfur has 4 bonding pairs and 1 lone pair — 5 electron domains. The lone pair occupies an equatorial position, distorting the trigonal bipyramid into a see-saw shape.
   • C₂H₂ (acetylene): Each carbon is sp-hybridized with 2 sigma bonds and 2 pi bonds — effectively 2 electron domains. The molecule is linear (H–C≡C–H). Both match perfectly. This is the correct pair. …