JEE Chemical Bonding & Molecular Structure Questions

Molecular orbital (MO) theory gives a direct relation between electronic configuration and measurable properties such as bond order, bond energy and bond length.
Bond order $$= \dfrac{n_b - n_a}{2}$$ where $$n_b$$ and $$n_a$$ are the numbers of electrons in bonding and antibonding MOs respectively.

Option A Ne$$_2$$

Each Ne atom has eight valence electrons (2s$$^2$$ 2p$$^6$$); Ne$$_2$$ therefore has 16 valence electrons.

For elements with atomic number $$\gt 7$$ (O, F, Ne) the MO order is
$$\sigma_{2s}\lt\sigma^{*}_{2s}\lt\sigma_{2p_z}\lt\pi_{2p_x}=\pi_{2p_y}\lt \pi^{*}_{2p_x}=\pi^{*}_{2p_y}\lt\sigma^{*}_{2p_z}$$

Filling 16 electrons gives
bonding electrons $$n_b = 8$$ (σ$$_{2s}$$, σ$$_{2p_z}$$, π$$_{2p_x}$$, π$$_{2p_y}$$)
antibonding electrons $$n_a = 8$$ (σ$$_{2s}^{*}$$, π$$_{2p_x}^{*}$$, π$$_{2p_y}^{*}$$, σ$$_{2p_z}^{*}$$)

Hence $$\text{Bond order} = \dfrac{8-8}{2}=0$$; Ne$$_2$$ is not a stable molecule. Statement A is therefore correct (so it is NOT an incorrect statement).

Option B F$$_2$$

Each F atom contributes seven valence electrons; F$$_2$$ has 14 valence electrons. Using the same MO order as above:

σ$$_{2s}$$ (2) → σ$$_{2s}^{*}$$ (2) → σ$$_{2p_z}$$ (2) → π$$_{2p_x}$$ (2) & π$$_{2p_y}$$ (2) → π$$_{2p_x}^{*}$$ (2) & π$$_{2p_y}^{*}$$ (2)

The highest-energy filled level (HOMO) is the degenerate pair $$\pi^{*}_{2p_x},\;\pi^{*}_{2p_y}$$—both are $$\pi$$-type antibonding orbitals, not σ-type. Hence statement B is incorrect.

Option C Comparison of O$$_2$$ and O$$_2^{+}$$

O$$_2$$ has 12 valence electrons. Configuration: σ$$_{2s}^{2}$$ σ$$_{2s}^{*2}$$ σ$$_{2p_z}^{2}$$ π$$_{2p_x}^{2}$$ π$$_{2p_y}^{2}$$ π$$_{2p_x}^{*1}$$ π$$_{2p_y}^{*1}$$

For O$$_2$$ $$n_b=8,\; n_a=4 \Longrightarrow \text{BO}=\dfrac{8-4}{2}=2$$

O$$_2^{+}$$ has one electron fewer; the electron is removed from a π$$_{2p}^{*}$$ antibonding orbital:

O$$_2^{+}$$ π$$_{2p_x}^{*1}$$, π$$_{2p_y}^{*0}$$

Now $$n_b=8,\; n_a=3 \Longrightarrow \text{BO}=\dfrac{8-3}{2}=2.5$$

Greater bond order implies stronger bond (larger bond energy) and shorter bond length. Therefore O$$_2^{+}$$ has a higher (not smaller) bond energy than O$$_2$$. Statement C is incorrect.

Option D Li$$_2$$ vs. B$$_2$$

Li$$_2$$ (two valence electrons) σ$$_{2s}^{2}$$ ⇒ $$n_b=2,\; n_a=0$$, BO = 1

B$$_2$$ (six valence electrons) σ$$_{2s}^{2}$$ σ$$_{2s}^{*2}$$ π$$_{2p_x}^{1}$$ π$$_{2p_y}^{1}$$ ⇒ $$n_b=4,\; n_a=2$$, BO = 1

Though both have the same bond order, the π(2p)-π(2p) overlap in B$$_2$$ is stronger than the σ(2s)-σ(2s) overlap in Li$$_2$$, producing a shorter bond length for B$$_2$$. Thus Li$$_2$$ indeed has the larger bond length. Statement D is correct (so it is NOT an incorrect statement).

Hence the INCORRECT statements are:
Option B (HOMO of F$$_2$$ is σ-type), Option C (Bond energy of O$$_2^{+}$$ is smaller than that of O$$_2$$).

Option B, Option C

We need to evaluate two statements about bond lengths in ozone ($$O_3$$).

Analysis of Statement I: "The experimentally determined oxygen-oxygen bond lengths in $$O_3$$ are found to be the same, and the bond length is greater than that of $$O=O$$ (double bond) but less than that of a single $$O-O$$ bond."

Ozone has two resonance structures where a double bond alternates between the two terminal oxygen atoms. Due to resonance, both O-O bonds in ozone are equivalent, with a bond order of 1.5 (intermediate between single and double bonds). The bond length in $$O_3$$ is approximately 1.278 angstroms, which is indeed:

- Greater than the $$O=O$$ double bond length (1.21 angstroms in $$O_2$$)

- Less than the $$O-O$$ single bond length (1.48 angstroms in $$H_2O_2$$)

Statement I is TRUE.

Analysis of Statement II: "The strong lone pair-lone pair repulsion between oxygen atoms is solely responsible for the bond length in ozone being smaller than a double bond but more than a single bond."

This statement is incorrect for two reasons. First, the intermediate bond length in ozone is primarily due to resonance, not lone pair-lone pair repulsion. The delocalization of electrons across the two O-O bonds gives each bond a bond order of 1.5. Second, the statement says "solely responsible," which is factually wrong since resonance is the dominant factor.

Statement II is FALSE.

The correct answer is Option 3: Statement I is true but Statement II is false.

For a molecule to be paramagnetic, it must contain at least one unpaired electron. The Molecular Orbital (MO) approach is the most reliable way to count unpaired electrons in homonuclear diatomic molecules.

Rule: Fill MOs in the order of increasing energy while obeying Hund’s rule (maximum multiplicity) and the Pauli exclusion principle. In period-2 diatomics the ordering changes after $$N_2$$:
• For $$B_2,\,C_2,\,N_2$$ the sequence is $$\sigma_{2p_z}$$ > $$\pi_{2p_x} = \pi_{2p_y}$$.
• For $$O_2,\,F_2$$ and beyond (including $$S_2,\,Cl_2$$) the sequence is $$\pi_{2p_x} = \pi_{2p_y}$$ > $$\sigma_{2p_z}$$.

Case A: $$O_2$$
Total valence electrons per atom = 6, so molecule has $$12$$.
Filling the energy-correct MO sequence (after $$N_2$$) gives:
$$\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\sigma_{2p_z}^2\,\pi_{2p_x}^2\,\pi_{2p_y}^2\,\pi_{2p_x}^{*1}\,\pi_{2p_y}^{*1}$$
Two electrons remain unpaired in the degenerate $$\pi_{2p}^{*}$$ orbitals, so $$O_2$$ is paramagnetic.

Case B: $$N_2$$
Total valence electrons = $$10$$.
Using the first energy sequence (before $$O_2$$):
$$\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\pi_{2p_x}^2\,\pi_{2p_y}^2\,\sigma_{2p_z}^2$$
All electrons are paired ⇒ $$N_2$$ is diamagnetic.

Case C: $$F_2$$
Total valence electrons = $$14$$.
Filling after $$O_2$$ sequence:
$$\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\sigma_{2p_z}^2\,\pi_{2p_x}^2\,\pi_{2p_y}^2\,\pi_{2p_x}^{*2}\,\pi_{2p_y}^{*2}$$
All electrons are paired ⇒ $$F_2$$ is diamagnetic.

Case D: $$S_2$$
Sulphur is one period below oxygen but forms a similar $$S_2$$ molecule with valence 6 e⁻ per atom ⇒ $$12$$ electrons, and the same MO ordering used for $$O_2$$ applies (the 3p orbitals follow the same pattern).
Hence configuration of $$S_2$$ ends with $$\pi_{3p_x}^{*1}\,\pi_{3p_y}^{*1}$$.
Two unpaired electrons are present, so $$S_2$$ is paramagnetic.

Case E: $$Cl_2$$
Chlorine contributes 7 valence electrons; $$Cl_2$$ has $$14$$. Filling the 3p MO diagram analogous to $$F_2$$ gives all electrons paired ⇒ $$Cl_2$$ is diamagnetic.

Summary: The molecules with unpaired electrons are $$O_2$$ and $$S_2$$ only.

Therefore, the correct option is Option D: A & D only.

We have to identify the molecule that simultaneously satisfies three conditions:
(a) central atom undergoes $$sp^3d$$ hybridization,
(b) the molecule possesses more than one type of bond length, and
(c) the central atom contains at least one lone pair of electrons.

• Central atom $$P$$: valence electrons = $$5$$, all used for $$5$$ $$P-F$$ bonds.
• Hybridization: $$sp^3d$$ (trigonal-bipyramidal).
• Lone pairs on $$P$$: $$0$$, so condition (c) is not met.
Hence $$PF_5$$ is rejected.

• Central atom $$Xe$$: valence electrons = $$8$$, uses $$4$$ for bonds and retains $$2$$ lone pairs.
• Hybridization: $$sp^3d^2$$ (octahedral arrangement, square-planar shape).
• Geometry is symmetric; all $$Xe-F$$ bonds are equivalent in length.
Condition (a) fails (hybridization not $$sp^3d$$), so $$XeF_4$$ is rejected.

• Central atom $$S$$: valence electrons = $$6$$.
  - $$4$$ electrons form $$4$$ $$S-F$$ sigma bonds.
  - $$2$$ electrons remain as one lone pair.
• Steric number $$= 5$$ ⇒ hybridization $$sp^3d$$ leading to a trigonal-bipyramidal electron geometry.
• Molecular shape: seesaw (because one equatorial position is occupied by the lone pair).
• In a trigonal-bipyramidal framework, axial bonds (aligned vertically) are longer than equatorial bonds (lying in the trigonal plane) due to greater repulsion along the axial-equatorial angle $$90^{\circ}$$.
Therefore $$SF_4$$ possesses two distinct $$S-F$$ bond lengths.
Conditions (a), (b) and (c) are all satisfied.

• Central atom $$Xe$$: valence electrons = $$8$$.
  - $$2$$ electrons pairs form $$2$$ $$Xe-F$$ bonds.
  - $$3$$ lone pairs remain.
• Steric number $$= 5$$ ⇒ hybridization $$sp^3d$$.
• All three lone pairs occupy equatorial positions, giving a linear molecule F-Xe-F.
• Both $$Xe-F$$ bonds occupy identical axial positions; thus the two bond lengths are equal.
Condition (b) fails, so $$XeF_2$$ is rejected.

Only $$SF_4$$ fulfils all three requirements.

Hence the correct option is Option C.

We need to identify molecules with square pyramidal geometry.

$$BrF_5$$: Br has 7 valence electrons. With 5 F atoms: 5 bond pairs + 1 lone pair = 6 electron pairs. Arrangement: octahedral. With one lone pair, the shape is square pyramidal.

$$PCl_5$$: P has 5 valence electrons. With 5 Cl atoms: 5 bond pairs + 0 lone pairs. Shape is trigonal bipyramidal, not square pyramidal.

$$SbF_5$$: Sb has 5 valence electrons. With 5 F atoms: 5 bond pairs + 0 lone pairs. Shape is trigonal bipyramidal, not square pyramidal.

$$XeOF_4$$: Xe has 8 valence electrons. With 1 O (double bond) + 4 F atoms: 5 bond pairs + 1 lone pair = 6 electron pairs. Arrangement: octahedral. With one lone pair, the shape is square pyramidal.

$$BrF_5$$ and $$XeOF_4$$ both have square pyramidal geometry.

The correct answer is Option D) $$BrF_5$$ and $$XeOF_4$$.

We must locate, from the given list, the molecule that satisfies two conditions:
  • its dipole moment $$\mu$$ is non-zero,
  • its central atom possesses the greatest possible number of lone pairs.

List of species with the data required (use VSEPR rules):

Case 1: $$SO_2$$ - Structure is angular (bent).
  Hybridization of S: $$sp^2$$ (AX2E).
  Lone pairs on S: 1.
  Dipole moment: non-zero (bent shape).

Case 2: $$NF_3$$ - Pyramidal (AX3E).
  Hybridization of N: $$sp^3$$.
  Lone pairs on N: 1.
  Dipole moment: non-zero.

Case 3: $$NH_3$$ - Also pyramidal (AX3E).
  Hybridization of N: $$sp^3$$.
  Lone pairs on N: 1.
  Dipole moment: non-zero.

Case 4: $$XeF_2$$ - Linear (AX2E3).
  Hybridization of Xe: $$sp^3d$$.
  Lone pairs on Xe: 3 (highest so far).
  Dipole moment: $$\mu = 0$$, because the axial $$\mathrm{Xe-F}$$ bonds are colinear and cancel.

Case 5: $$ClF_3$$ - T-shaped (AX3E2).
  Hybridization of Cl: $$sp^3d$$.
  Lone pairs on Cl: 2.
  Dipole moment: non-zero (T-shape is not symmetrical enough for cancellation).

Case 6: $$SF_4$$ - See-saw (AX4E).
  Hybridization of S: $$sp^3d$$.
  Lone pairs on S: 1.
  Dipole moment: non-zero.

Among the systems having non-zero dipole moment, the largest number of lone pairs is 2 (present in $$ClF_3$$). All other dipolar molecules posses only 1 lone pair, while $$XeF_2$$—though it has 3 lone pairs—has $$\mu = 0$$ and therefore does not meet the first criterion.

Therefore the required molecule is $$ClF_3$$.

Hybridization of the central atom in $$ClF_3$$ is $$sp^3d$$.

Hence, the correct option is Option D, $$sp^3d$$.

Cellulose, the main constituent of cotton fibres, is a polysaccharide chain carrying abundant $$-OH$$ groups on every glucose unit.
Each $$-OH$$ group can act both as a hydrogen-bond donor and acceptor, so a single cellulose chain can form a large number of intermolecular hydrogen bonds with surrounding water molecules.

Nylon is a polyamide. Although each repeat unit contains one $$-CONH$$ group capable of hydrogen bonding, the rest of the chain is a long, non-polar hydrocarbon segment. The total number of hydrogen-bonding sites per unit mass of nylon is therefore much smaller than in cellulose, and the hydrophobic backbone further reduces water affinity.

Because of the extensive hydrogen bonding, cotton absorbs and retains far more water than nylon. Extra heat (or time) is needed to break these numerous hydrogen bonds during drying, so wet cotton garments take a longer time to dry than wet nylon garments.

Thus:
Statement I: Cotton clothes indeed take longer to dry — TRUE.
Statement II: Hydrogen bonding with water is actually stronger and more numerous in cotton, not in nylon — FALSE.

The correct choice is Option B: Statement I is true but Statement II is false.

Hex-1-en-4-yne has the structure: $$CH_2=CH-CH_2-C \equiv C-CH_3$$

Let us count the bonds:

Sigma bonds:

- C1=C2: 1 σ bond

- C2-C3: 1 σ bond

- C3-C4: 1 σ bond

- C4≡C5: 1 σ bond

- C5-C6: 1 σ bond

- C1-H bonds: 2 σ bonds

- C2-H bond: 1 σ bond

- C3-H bonds: 2 σ bonds

- C6-H bonds: 3 σ bonds

Total σ bonds = 5 + 8 = 13

Pi bonds:

- C1=C2: 1 π bond

- C4≡C5: 2 π bonds

Total π bonds = 3

The correct answer is Option 3: 13 and 3.

The hybridization of a given atom can be predicted from its steric number.
Steric number $$=\,$$ (number of $$\sigma$$-bonds around the atom) $$+\,$$ (number of lone pairs on that atom).

• Central atom: $$S$$.
• Lewis structure: $$O = S - O$$ with one lone pair on $$S$$.
• $$\sigma$$-bonds on $$S = 2$$ (one to each $$O$$).
• Lone pairs on $$S = 1$$.
Therefore, steric number $$= 2 + 1 = 3$$.

Steric number 3 corresponds to $$sp^2$$ hybridization, giving a bent (V-shaped) molecule.

• Central atom: $$N$$.
• Lewis structure: $$O = N - O^-$$ (resonance forms) with one lone pair on $$N$$.
• $$\sigma$$-bonds on $$N = 2$$.
• Lone pairs on $$N = 1$$.
Steric number $$= 2 + 1 = 3$$.

Steric number 3 again implies $$sp^2$$ hybridization, giving a bent ion.

• Central atom: the middle $$N$$ of $$N - N - N$$.
• Resonance structures involve $$N \equiv N^+ - N^-$$ and $$N^- - N^+ \equiv N$$, but the central $$N$$ always has:
  $$\sigma$$-bonds = 2$$\,$$(to the terminal nitrogens).
  Lone pairs = 0 (all its electrons are in bonds or as formal charges on terminals).
Steric number $$= 2 + 0 = 2$$.

Steric number 2 corresponds to $$sp$$ hybridization, giving a linear ion.

Thus, the hybridizations are $$sp^2$$ in $$SO_2$$, $$sp^2$$ in $$NO_2^-$$ and $$sp$$ in $$N_3^-$$.

Matching with the options, we select Option A.

Answer: Option A

We analyze each statement about $$H_2O$$, $$NH_3$$, and $$CH_4$$:

A. Central atoms of all molecules are $$sp^3$$ hybridized.

$$H_2O$$: O has 2 bond pairs + 2 lone pairs = $$sp^3$$. $$NH_3$$: N has 3 bond pairs + 1 lone pair = $$sp^3$$. $$CH_4$$: C has 4 bond pairs = $$sp^3$$. Correct.

B. Bond angles are $$104.5°$$, $$107.5°$$, and $$109.5°$$ respectively.

H-O-H = $$104.5°$$, H-N-H = $$107°$$ (approximately $$107.5°$$), H-C-H = $$109.5°$$. Correct.

C. Increasing order of dipole moment: $$CH_4 < NH_3 < H_2O$$.

$$CH_4$$ has zero dipole moment (symmetric). $$NH_3$$ has $$\mu = 1.47$$ D. $$H_2O$$ has $$\mu = 1.85$$ D. So $$CH_4 < NH_3 < H_2O$$. Correct.

D. Both $$H_2O$$ and $$NH_3$$ are Lewis acids and $$CH_4$$ is a Lewis base.

$$H_2O$$ and $$NH_3$$ have lone pairs and act as Lewis BASES (not acids). $$CH_4$$ is neither a good Lewis acid nor base. Incorrect.

E. A solution of $$NH_3$$ in $$H_2O$$ is basic. $$NH_3$$ and $$H_2O$$ act as Lowry-Bronsted acid and base respectively.

In the reaction $$NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$$, $$NH_3$$ acts as a Bronsted BASE (accepts $$H^+$$) and $$H_2O$$ acts as a Bronsted ACID (donates $$H^+$$). The statement says $$NH_3$$ is the acid and $$H_2O$$ is the base, which is reversed. Incorrect.

Correct statements: A, B, and C only.

The correct answer is Option A: A, B and C only.

The hybridisation of a central atom is obtained from its steric number (total number of $$\sigma$$-bonds + lone pairs).
For transition-metal complexes we also look at the observed geometry.

Case A: $$PF_5$$
Phosphorus forms five $$\sigma$$-bonds with fluorine and has no lone pair.
Hence steric number $$=5$$  $$\Rightarrow$$  hybridisation $$= sp^3d$$ (trigonal bipyramidal).
Therefore $$PF_5$$ matches List-II entry II.

Case B: $$SF_6$$
Sulphur forms six $$\sigma$$-bonds with fluorine and has no lone pair.
Steric number $$=6$$  $$\Rightarrow$$  hybridisation $$= sp^3d^2$$ (octahedral).
Therefore $$SF_6$$ matches List-II entry III.

Case C: $$Ni(CO)_4$$
Nickel in $$Ni(CO)_4$$ is in the zero oxidation state and the complex is tetrahedral.
A tetrahedral arrangement uses four equivalent hybrid orbitals: $$sp^3$$.
Therefore $$Ni(CO)_4$$ matches List-II entry IV.

Case D: $$[PtCl_4]^{2-}$$
The ion is square-planar; Pt(II) uses one $$d$$, one $$s$$ and two $$p$$ orbitals forming $$dsp^2$$ hybrids.
Therefore $$[PtCl_4]^{2-}$$ matches List-II entry I.

Collecting all results:
A → II,  B → III,  C → IV,  D → I.

The correct option is Option A.

The central atom $$Cl$$ in $$ClF_3$$ has $$7$$ valence electrons. Three of them are used to form $$\sigma$$-bonds with three $$F$$ atoms and the remaining $$4$$ electrons constitute two lone pairs. Hence the steric number is $$5$$ ( $$= 3\; \text{bond pairs} + 2\; \text{lone pairs}$$ ).

For steric number $$5$$, the electron-pair geometry is $$\text{trigonal bipyramidal}$$. The five positions are

• two axial (directly opposite, $$180^{\circ}$$ apart)
• three equatorial (coplanar, $$120^{\circ}$$ apart)

Placing two lone pairs and three bond pairs in these five positions can, in principle, give the following three arrangements:

Case I: both lone pairs axial  $$\rightarrow$$  all three $$F$$ atoms equatorial.

Case II: one lone pair axial, one lone pair equatorial  $$\rightarrow$$  two $$F$$ atoms equatorial, one axial.

Case III: both lone pairs equatorial  $$\rightarrow$$  one $$F$$ atom equatorial, two $$F$$ atoms axial (T-shaped).

Thus, as statement (I) says, all three structures can indeed be drawn; so Statement I is correct.

According to VSEPR theory the order of repulsions is $$lp\!-\!lp \; \gt \; lp\!-\!bp \; \gt \; bp\!-\!bp$$. An axial position is at $$90^{\circ}$$ to three other positions, whereas an equatorial position is at $$90^{\circ}$$ to only two. Therefore a lone pair experiences less repulsion when it occupies an equatorial site.

Hence Case III, in which both lone pairs occupy the equatorial positions, has the minimum total repulsion and is the most stable. Statement (II) claims the most stable structure has lone pairs in axial orbitals, which is opposite to the VSEPR prediction. Therefore Statement II is incorrect.

So, Statement I is correct while Statement II is incorrect. Hence the correct option is Option B.

We need to find the ions with $$n$$ unpaired electrons, where $$n$$ is the number of lone pairs in the equatorial position of the most stable structure of $$ClF_3$$.

In $$ClF_3$$, the central atom Cl has 7 valence electrons, three of which form bonds with F atoms, leaving two lone pairs (4 electrons). The resulting geometry is trigonal bipyramidal (sp$$^3$$d hybridization) with three F atoms and two lone pairs. Since lone pairs repel more strongly and occupy equatorial positions to minimize repulsion, the most stable T-shaped structure has both lone pairs equatorial. Hence $$n = 2$$.

Examining the given ions for two unpaired electrons: $$V^{3+}$$ has the configuration [Ar] 3d$$^2$$ and thus 2 unpaired electrons; $$Ti^{3+}$$ is [Ar] 3d$$^1$$ with 1 unpaired electron; $$Cu^{2+}$$ is [Ar] 3d$$^9$$ with 1 unpaired electron; $$Ni^{2+}$$ is [Ar] 3d$$^8$$ with 2 unpaired electrons (high‐spin); and $$Ti^{2+}$$ is [Ar] 3d$$^2$$ with 2 unpaired electrons.

Therefore, the ions with 2 unpaired electrons are $$V^{3+}$$, $$Ni^{2+}$$, and $$Ti^{2+}$$, corresponding to options A, D, and E. The correct answer is Option 2: A, D and E Only.

We need to find the maximum covalency of a non-metallic Group 15 element 'E' with the weakest E-E bond.

Non-metallic Group 15 elements are: N, P, As.

Among these, the weakest E-E single bond is the N-N bond (bond energy ≈ 160 kJ/mol). This is because nitrogen atoms are small, and the lone pairs on adjacent nitrogen atoms experience strong repulsion, weakening the single bond.

For comparison, P-P bond energy ≈ 200 kJ/mol.

Nitrogen has the electronic configuration $$1s^2 2s^2 2p^3$$.

Since nitrogen is in the 2nd period, it has no d-orbitals available for bonding and cannot expand its octet.

Maximum number of bonds nitrogen can form = 3 covalent bonds + 1 coordinate bond (using the lone pair) = 4.

Example: $$NH_4^+$$ where nitrogen forms 4 covalent bonds.

The answer is Option A: 4.

For the formation of molecular orbitals in homonuclear diatomic molecules (with the internuclear axis along the $$z$$-direction), the combining atomic orbitals must have the same symmetry with respect to the internuclear axis.

The rule is: Orbitals combine only if they have the same symmetry about the molecular axis. Specifically, they must have the same value of $$m_l$$ along the z-axis ($$\sigma$$ for $$m_l = 0$$, $$\pi$$ for $$|m_l| = 1$$, $$\delta$$ for $$|m_l| = 2$$).

A. $$2p_z$$ and $$2p_x$$: $$2p_z$$ has $$m_l = 0$$, $$2p_x$$ has $$|m_l| = 1$$. Different symmetry. Cannot combine. No.

B. $$2s$$ and $$2p_x$$: $$2s$$ has $$m_l = 0$$ ($$\sigma$$), $$2p_x$$ has $$|m_l| = 1$$ ($$\pi$$). Different symmetry. Cannot combine. No.

C. $$3d_{xy}$$ and $$3d_{x^2-y^2}$$: $$3d_{xy}$$ has $$|m_l| = 2$$ ($$\delta$$), $$3d_{x^2-y^2}$$ also has $$|m_l| = 2$$ ($$\delta$$). Same symmetry, but these are on the same atom. For a homonuclear diatomic, the $$3d_{xy}$$ on one atom combines with $$3d_{xy}$$ on the other. The statement says combining $$3d_{xy}$$ with $$3d_{x^2-y^2}$$, which are different orbitals on different atoms with the same $$|m_l|$$ but different orientations. They do not form bonding/antibonding pairs with each other. No.

D. $$2s$$ and $$2p_z$$: Both have $$m_l = 0$$ ($$\sigma$$ symmetry). They can combine to form molecular orbitals. Yes.

E. $$2p_z$$ and $$3d_{x^2-y^2}$$: $$2p_z$$ has $$m_l = 0$$, $$3d_{x^2-y^2}$$ has $$|m_l| = 2$$. Different symmetry. Cannot combine. No.

Only D leads to formation of molecular orbitals.

The correct answer is Option B: D Only.

The resonance energy of a molecule is the extra stabilisation it enjoys due to resonance.
Mathematically,

$$\text{Resonance energy}= \left( \Delta H_f^{\,\text{calculated (one structure)}} \right) - \left( \Delta H_f^{\,\text{experimental}} \right)$$

The question gives the experimental enthalpy of formation of $$X_2Y$$ as
$$\Delta H_f^{\,\text{exp}} = 80\ \text{kJ mol}^{-1}$$

To obtain $$\Delta H_f^{\,\text{calculated}}$$ we choose any one canonical (non-resonating) structure.
A convenient choice for $$X_2Y$$ is

$$X=X - Y\;,$$

i.e. one $$X=X$$ double bond and one $$X=Y$$ double bond.
The enthalpy of formation is obtained from the usual bond-energy relation

$$\Delta H_f = \sum \text{(bond energies of bonds broken)} \;-\; \sum \text{(bond energies of bonds formed)}$$

Step 1: Bonds broken (to obtain free atoms)
We start from the elemental forms:

• 1 mol $$X_2$$ contains one $$X\equiv X$$ triple bond  Energy to break it  = $$E_{X\equiv X}=940\ \text{kJ}$$
• ½ mol $$Y_2$$ contains half a $$Y=Y$$ double bond  Energy to break it  = $$\tfrac12\,E_{Y=Y} =\tfrac12 \times 500 =250\ \text{kJ}$$

Total energy required to break reactant bonds:

$$E_{\text{broken}} = 940 + 250 = 1190\ \text{kJ}$$

Step 2: Bonds formed in the chosen structure of $$X_2Y$$

• One $$X=X$$ double bond  energy released = $$E_{X=X}=410\ \text{kJ}$$
• One $$X=Y$$ double bond  energy released = $$E_{X=Y}=602\ \text{kJ}$$

Total energy released on forming product bonds:

$$E_{\text{formed}} = 410 + 602 = 1012\ \text{kJ}$$

Step 3: Calculated enthalpy of formation (one structure)

$$\Delta H_f^{\,\text{calc}} = E_{\text{broken}} - E_{\text{formed}} = 1190 - 1012 = 178\ \text{kJ mol}^{-1}$$

Step 4: Resonance energy

$$\text{Resonance energy} = \Delta H_f^{\,\text{calc}} - \Delta H_f^{\,\text{exp}} = 178 - 80 = 98\ \text{kJ mol}^{-1}$$

Hence, the magnitude of the resonance energy of $$X_2Y$$ is 98 kJ mol$$^{-1}$$.

We need to count the number of molecules/ions with linear geometry from the given list.

$$SO_2, BeCl_2, CO_2, N_3^-, NO_2, F_2O, XeF_2, NO_2^+, I_3^-, O_3$$

Analyzing each:

1. $$SO_2$$: S has 2 bonding pairs + 1 lone pair = bent geometry. NOT linear.

2. $$BeCl_2$$: Be has 2 bonding pairs, no lone pairs. sp hybridized. LINEAR.

3. $$CO_2$$: C has 2 double bonds, no lone pairs. sp hybridized. LINEAR.

4. $$N_3^-$$: Azide ion. Central N has 2 bonding regions, no lone pairs. sp hybridized. LINEAR.

5. $$NO_2$$: N has 2 bonding pairs + 1 unpaired electron = bent geometry. NOT linear.

6. $$F_2O$$: O has 2 bonding pairs + 2 lone pairs = bent geometry. NOT linear.

7. $$XeF_2$$: Xe has 2 bonding pairs + 3 lone pairs = LINEAR (sp³d hybridization, trigonal bipyramidal with F atoms in axial positions).

8. $$NO_2^+$$: Nitronium ion. N has 2 double bonds, no lone pairs. sp hybridized. LINEAR.

9. $$I_3^-$$: Central I has 2 bonding pairs + 3 lone pairs = LINEAR (similar to XeF₂).

10. $$O_3$$: Central O has 2 bonding regions + 1 lone pair = bent geometry. NOT linear.

Linear molecules/ions: $$BeCl_2, CO_2, N_3^-, XeF_2, NO_2^+, I_3^-$$ = 6

The answer is 6.

We need to find the total number of paramagnetic species from the given list.

A species is paramagnetic if it has one or more unpaired electrons.

1. $$O_2$$ (Oxygen molecule):

MO configuration: $$(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p})^2(\pi_{2p})^4(\pi_{2p}^*)^2$$. The two electrons in $$\pi_{2p}^*$$ are unpaired (one in each degenerate orbital, by Hund's rule). Paramagnetic (2 unpaired electrons).

2. $$O_2^+$$: Remove one electron from $$\pi_{2p}^*$$: 1 unpaired electron. Paramagnetic.

3. $$O_2^-$$: Add one electron to $$\pi_{2p}^*$$: 3 electrons in two degenerate orbitals gives 1 unpaired. Paramagnetic.

4. NO (Nitric oxide): Total 15 electrons. MO config has one unpaired electron in $$\pi_{2p}^*$$. Paramagnetic.

5. $$NO_2$$: Nitrogen dioxide has 23 electrons (odd number), so at least 1 unpaired electron. Paramagnetic.

6. CO: 14 electrons, all paired in MO configuration. Diamagnetic.

7. $$K_2[NiCl_4]$$: Contains $$[NiCl_4]^{2-}$$. $$Ni^{2+}$$ has $$d^8$$ configuration. $$Cl^-$$ is a weak-field ligand, so tetrahedral geometry. In tetrahedral field, $$d^8$$ has 2 unpaired electrons. Paramagnetic.

8. $$[Co(NH_3)_6]Cl_3$$: Contains $$[Co(NH_3)_6]^{3+}$$. $$Co^{3+}$$ has $$d^6$$. $$NH_3$$ is a strong-field ligand (octahedral), so low-spin: all 6 electrons fill $$t_{2g}$$ with 0 unpaired electrons. Diamagnetic.

9. $$K_2[Ni(CN)_4]$$: Contains $$[Ni(CN)_4]^{2-}$$. $$Ni^{2+}$$ has $$d^8$$. $$CN^-$$ is a strong-field ligand, forming square planar geometry. In square planar, $$d^8$$ has 0 unpaired electrons. Diamagnetic.

Count of paramagnetic species: $$O_2, O_2^+, O_2^-, NO, NO_2, K_2[NiCl_4]$$ = 6.

The answer is 6.

The Lewis structure of $$NO_2^-$$ (nitrite ion):

Nitrogen has 5 valence electrons, each oxygen has 6, plus 1 for the negative charge: Total = 5 + 2(6) + 1 = 18 electrons.

Structure: $$[:O-N=O:]^-$$ with a lone pair on N.

Non-bonded electrons (lone pairs):

- Nitrogen: 1 lone pair = 2 electrons

- Oxygen (single bond): 3 lone pairs = 6 electrons

- Oxygen (double bond): 2 lone pairs = 4 electrons

Total non-bonded electrons = 2 + 6 + 4 = 12

The answer is 12.

Statement I claims that since fluorine is more electronegative than nitrogen, the net dipole moment of $$NF_3$$ is greater than $$NH_3$$. However, experimental dipole moments are $$\mu(NH_3) = 1.47$$ D and $$\mu(NF_3) = 0.23$$ D, so $$NF_3$$ has a much lower dipole moment than $$NH_3$$. This is because in $$NH_3$$, the lone pair dipole and the bond dipoles reinforce each other, while in $$NF_3$$ they oppose each other (partially cancelling). Statement I is false.

Statement II asserts that in $$NH_3$$ the orbital dipole due to lone pair and the dipole moments of N-H bonds are in opposite direction, but in $$NF_3$$ they are in the same direction. In reality, the situation is reversed: in $$NH_3$$ the N-H bond dipoles point from H towards N (since N is more electronegative) and the lone pair on nitrogen also contributes a dipole in the same direction (away from nitrogen, towards the lone pair region), so they reinforce each other. In $$NF_3$$ the N-F bond dipoles point from N towards F (since F is more electronegative than N), i.e., away from nitrogen, while the lone pair dipole still points in the direction from the base towards the apex (away from the lone pair), so they oppose each other, causing partial cancellation and a low dipole moment. Statement II is false.

Since both statements are false, the correct answer is Option (2): Both Statement I and Statement II are false.

We need to identify the correct conditions for linear combination of atomic orbitals (LCAO) to form molecular orbitals.

The three essential conditions for LCAO are:

Condition A: Same energy. The combining atomic orbitals must have comparable (similar or same) energies. For example, the 1s orbital of one atom can effectively combine with the 1s orbital of the other atom, but not with a 2s orbital (which has very different energy). This condition ensures effective orbital mixing. This is CORRECT.

Condition B: Minimum overlap. This is INCORRECT. The correct condition is that the atomic orbitals must have maximum overlap. Greater overlap between atomic orbitals leads to stronger bonding (or antibonding) molecular orbitals. Minimum overlap would result in negligible interaction and no effective molecular orbital formation.

Condition C: Same symmetry about the molecular axis. The combining atomic orbitals must have the same symmetry with respect to the molecular (internuclear) axis. For example, a $$2p_z$$ orbital (symmetric about the bond axis) combines with another $$2p_z$$ orbital to form $$\sigma$$ molecular orbitals, and $$2p_x$$ orbitals combine with $$2p_x$$ orbitals to form $$\pi$$ molecular orbitals. A $$2p_z$$ orbital cannot effectively combine with a $$2p_x$$ orbital because they have different symmetries. This is CORRECT.

Condition D: Different symmetry about the molecular axis. This is INCORRECT — it is the opposite of the correct condition (Condition C).

The correct conditions are A and C.

The correct answer is Option (2): A and C only.

We need to determine the shape of a carbocation.

Understand the electronic structure of a carbocation

A carbocation is a positively charged carbon species, such as $$CH_3^+$$. The carbon atom in a carbocation has only 6 electrons (3 bonding pairs) instead of the usual 8. It has:

- 3 bond pairs (from three bonds to hydrogen or other groups)

- 0 lone pairs

- 1 empty (vacant) p-orbital

Determine the hybridisation

With only 3 regions of electron density (3 bond pairs and no lone pairs), the carbon in a carbocation undergoes sp$$^2$$ hybridisation. The three sp$$^2$$ hybrid orbitals form bonds, while the unhybridised p-orbital remains empty and is perpendicular to the plane of the three bonds.

Determine the geometry

According to VSEPR theory and the sp$$^2$$ hybridisation, three bond pairs with no lone pairs arrange themselves at $$120°$$ angles in a plane to minimise repulsion. This gives a trigonal planar geometry.

The bond angles are all $$120°$$, and all three bonds lie in the same plane with the carbon atom at the centre.

The correct answer is Option (2): trigonal planar.

To compare the ionic character of different bonds we use the Pauling concept:

• Ionic character increases with the electronegativity difference $$\Delta\chi = \chi_{\text{more}} - \chi_{\text{less}}$$ between the two bonded atoms.
• A convenient estimate of percentage ionic character is

$$\%\,\text{ionic} = \left(1 - e^{-\frac{(\Delta\chi)^2}{4}}\right)\times100$$

Although we do not need the exact percentages, the formula shows that a larger $$\Delta\chi$$ means a more ionic bond.

The Pauling electronegativity (EN) values needed are
$$\chi_{Li}=1.0,\; \chi_{K}=0.8,\; \chi_{N}=3.0,\; \chi_{S}=2.5,\; \chi_{Cl}=3.0,\; \chi_{O}=3.5,\; \chi_{F}=4.0$$

Now calculate $$\Delta\chi$$ for the representative bond in each molecule:

$$\begin{aligned} N_2: &\; \Delta\chi = 3.0-3.0 = 0 \\[2pt] SO_2\;(S{-}O): &\; \Delta\chi = 3.5-2.5 = 1.0 \\[2pt] ClF_3\;(Cl{-}F): &\; \Delta\chi = 4.0-3.0 = 1.0 \\[2pt] K_2O\;(K{-}O): &\; \Delta\chi = 3.5-0.8 = 2.7 \\[2pt] LiF\;(Li{-}F): &\; \Delta\chi = 4.0-1.0 = 3.0 \end{aligned}$$

Interpretation:

• $$N_2$$ has $$\Delta\chi = 0$$, so it is perfectly covalent.
• $$K_2O$$ and $$LiF$$ have the largest $$\Delta\chi$$ values; their bonds are highly ionic.
• $$SO_2$$ and $$ClF_3$$ tie at $$\Delta\chi = 1.0$$. However, the S-O bonds in $$SO_2$$ possess significant multiple-bond (π-bond) character because of $$p\pi\!-\!d\pi$$ back-bonding and resonance. Multiple bonding increases covalent character and therefore decreases ionic character when compared with a single S-O bond of the same $$\Delta\chi$$. The Cl-F bonds in $$ClF_3$$ are simple single bonds, so their ionic character is slightly higher than that of the S-O bonds.

Combining all the points, the bonds arranged in order of increasing ionic character are

$$N_2 \lt SO_2 \lt ClF_3 \lt K_2O \lt LiF$$

This corresponds to Option C.

Let us analyze each molecule for polarity:

- $$CCl_4$$: Tetrahedral, symmetric, dipole moments cancel. Non-polar.

- $$CO_2$$: Linear, symmetric, dipole moments cancel. Non-polar.

- $$CH_2=CH_2$$: Planar, symmetric. Non-polar.

- $$CHCl_3$$: Tetrahedral but asymmetric (one H, three Cl). Net dipole moment exists. Polar.

The answer is $$CHCl_3$$, which corresponds to Option (4).

The Assertion A states that $$PH_3$$ has a lower boiling point than $$NH_3$$. This statement is correct because boiling point depends on the strength of intermolecular forces. Ammonia ( $$NH_3$$ ) molecules exhibit hydrogen bonding, which is a stronger intermolecular force compared to the van der Waals forces present in phosphine ( $$PH_3$$ ).

The Reason R states that in the liquid state $$NH_3$$ molecules are associated through van der Waals forces, but $$PH_3$$ molecules are associated through hydrogen bonding. This statement is incorrect. In fact, $$NH_3$$ molecules form hydrogen bonds due to the high electronegativity of nitrogen and the presence of lone pair on nitrogen. On the other hand, $$PH_3$$ molecules cannot form hydrogen bonds because phosphorus is less electronegative and its lone pair is less available for hydrogen bonding; thus, $$PH_3$$ molecules are held together primarily by weaker van der Waals forces.

Since Assertion A is correct and Reason R is not correct, the correct choice is Option D: A is correct but R is not correct.

Assertion (A): $$NH_3$$ and $$NF_3$$ both have pyramidal shapes with a lone pair on nitrogen. The dipole moment of $$NH_3$$ is greater than that of $$NF_3$$.

Reason (R): In $$NH_3$$, the orbital dipole due to the lone pair is in the same direction as the resultant dipole moment of the N-H bonds. F is the most electronegative element.

Evaluate Assertion (A)

Both $$NH_3$$ ($$\mu = 1.47$$ D) and $$NF_3$$ ($$\mu = 0.23$$ D) are pyramidal. Indeed, $$\mu_{NH_3} > \mu_{NF_3}$$. Assertion A is true.

Evaluate Reason (R)

In $$NH_3$$: The bond dipoles of N-H point from H to N (since N is more electronegative). The lone pair on N also creates a dipole pointing away from N in the same direction. Both contributions add up, resulting in a large net dipole moment.

In $$NF_3$$: The bond dipoles of N-F point from N to F (since F is more electronegative), which is opposite to the direction of the lone pair dipole. The two effects partially cancel, resulting in a much smaller net dipole moment.

F is indeed the most electronegative element. Reason R is true and correctly explains why $$\mu_{NH_3} > \mu_{NF_3}$$.

Both (A) and (R) are true and (R) is the correct explanation of (A). The answer is Option 1.

The melting point trend in Group 13 is highly irregular.

• Boron (B) has an extremely high melting point because it forms a very strong icosahedral crystalline structure.
• Gallium (Ga) has an unusually low melting point (about 303 K) because it exists as Ga2 molecules in the crystal lattice rather than a standard metallic structure.
• The Trend: B > Al > Tl > In > Ga

Unlike atomic radii, the ionic radii of M+3 ions follow a regular trend. As you move down the group, more shells are added, and the effective nuclear charge felt by the outermost electrons (in an ionic state) allows for a consistent increase in size.

• The Trend: Tl > In > Ga > Al > B.
  

The First Ionization Enthalpy is irregular due to poor shielding by d and f electrons.

• Boron is highest due to its small size.
• Thallium (Tl) is higher than Aluminium and Gallium due to the Lanthanoid Contraction (poor shielding by 4f electrons), which makes the nucleus hold the 6s2 electrons very tightly.
• Gallium is slightly higher than Aluminium due to the d-block contraction.
• The Trend: B > Tl > Ga > Al > In

Normally, atomic radius increases down a group. However, there is a anomaly between Aluminium and Gallium.

• Gallium vs. Aluminium: Gallium follows the 10 elements of the first transition series 3d10. The 3d electrons shield the nuclear charge poorly, causing the outer electrons to be pulled in more strongly. Consequently, the atomic radius of Gallium is slightly smaller than (or nearly equal to) Aluminium.
• The Trend: Tl > In > Al > Ga > B.

We evaluate each statement about hydrogen bonding.

A: "H bonding exists when H is covalently bonded to a highly electronegative atom."

TRUE. Hydrogen bonds form when H is bonded to F, O, or N (highly electronegative atoms).

B: "Intermolecular H bonding is present in o-nitrophenol."

FALSE. In o-nitrophenol, intramolecular H bonding occurs between the -OH and the -NO₂ group in the ortho position, forming a six-membered chelate ring. It is the para-isomer that shows intermolecular H bonding.

C: "Intramolecular H bonding is present in HF."

FALSE. HF shows strong intermolecular hydrogen bonding (not intramolecular). Intramolecular H bonding requires two functional groups within the same molecule.

D: "The magnitude of H bonding depends on the physical state of the compound."

TRUE. In the solid state, molecules are closer and more ordered, allowing stronger H bonding. In the gas phase, molecules are far apart and H bonding is minimal.

E: "H bonding has powerful effect on the structure and properties of compounds."

TRUE. H bonding significantly affects boiling points, solubility, viscosity, and molecular structure (e.g., DNA double helix, protein folding).

Correct statements: A, D, E.

The correct answer is Option (2): A, D, E only.

We need to determine which of the given compounds is the least ionic.

We know from Fajans’ rules that ionic character decreases (and covalent character increases) when the cation is small with high charge (high charge density equals high polarizing power), the anion is large and easily polarizable, and the cation adopts a pseudo-noble gas configuration (such as an 18-electron outer shell) instead of a noble gas configuration (8-electron outer shell), since cations with (n-1)d$$^{10}$$ configurations exhibit higher polarizing power.

With these considerations in mind, we examine each compound in turn.

In BaCl$$_2$$ the Ba$$^{2+}$$ ion is large and has the noble gas configuration [Xe], which results in low polarizing power; consequently, BaCl$$_2$$ is highly ionic.

In AgCl, the Ag$$^{+}$$ ion has the electronic configuration [Kr]4d$$^{10}$$, and despite its +1 charge, its filled d-shell provides strong polarizing power; this causes AgCl to exhibit significant covalent character.

In KCl, the K$$^{+}$$ ion has the noble gas configuration [Ar] and a relatively large radius, leading to low charge density and making KCl one of the most ionic compounds.

In CoCl$$_2$$ the Co$$^{2+}$$ ion has the configuration [Ar]3d$$^7$$ and thus some polarizing power due to its +2 charge and incomplete d-shell, but it is not as strongly polarizing as Ag$$^{+}$$; therefore, CoCl$$_2$$ is more ionic than AgCl yet less so than BaCl$$_2$$ and KCl.

Ranking these compounds by ionic character from most to least ionic gives KCl > BaCl$$_2$$ > CoCl$$_2$$ > AgCl.

AgCl is the least ionic compound because Ag$$^{+}$$ has exceptionally high polarizing power according to Fajans’ rules.

The correct answer is Option (2): AgCl.

The bond angle in a molecule depends mainly on two factors, (i) the total number of electron pairs around the central atom (bond pairs + lone pairs) and (ii) the number of lone pairs, because lone-pair-lone-pair repulsion $$\gt$$ lone-pair-bond-pair $$\gt$$ bond-pair-bond-pair repulsion (VSEPR theory).

Case 1: $$BF_3$$
• Central atom B has 3 valence electrons.
• It forms three $$\sigma$$ bonds with F and has no lone pair  $$\Rightarrow$$  steric number = 3 (AX$$_3$$ type).
• Geometry: trigonal planar.
• Ideal bond angle for trigonal planar = $$120^{\circ}$$, and with no lone pair the actual angle remains $$\approx 120^{\circ}$$.

Case 2: $$PF_3$$
• Central atom P has 5 valence electrons.
• It forms three $$\sigma$$ bonds with F and retains one lone pair  $$\Rightarrow$$  steric number = 4 (AX$$_3$$E type).
• Electron-pair geometry: tetrahedral; molecular shape: trigonal pyramidal.
• Presence of one lone pair compresses the bond angle below the ideal $$109.5^{\circ}$$. Furthermore, highly electronegative F pulls electron density away from the P-F bond pairs, so bond-pair-bond-pair repulsion decreases and the angle contracts further.
• Observed P-F-P angle $$\approx 97^{\circ}$$.

Case 3: $$ClF_3$$
• Central atom Cl has 7 valence electrons.
• It forms three $$\sigma$$ bonds with F and retains two lone pairs  $$\Rightarrow$$  steric number = 5 (AX$$_3$$E$$_2$$ type).
• Electron-pair geometry: trigonal bipyramidal; molecular shape: T-shaped (the two lone pairs occupy two equatorial positions).
• Two lone pairs produce strong repulsions, squeezing the F-Cl-F bond angle well below the ideal $$90^{\circ}$$ of a T shape; experimental value $$\approx 87^{\circ}$$.

Comparing the three angles:
$$\angle F\!-\!Cl\!-\!F \ (\text{in } ClF_3) \approx 87^{\circ} \lt$$
$$\angle F\!-\!P\!-\!F \ (\text{in } PF_3) \approx 97^{\circ} \lt$$
$$\angle F\!-\!B\!-\!F \ (\text{in } BF_3) \approx 120^{\circ}$$

Therefore, the correct increasing order of bond angles is
$$ClF_3 \lt PF_3 \lt BF_3$$.

The order matches Option B.

Count species with pyramidal geometry around the central atom from: $$S_2O_3^{2-}$$, $$SO_4^{2-}$$, $$SO_3^{2-}$$, $$S_2O_7^{2-}$$.

$$S_2O_3^{2-}$$ (thiosulphate):

The central sulfur atom is bonded to 3 oxygen atoms and 1 sulfur atom in a tetrahedral arrangement (similar to $$SO_4^{2-}$$). Not pyramidal.

$$SO_4^{2-}$$ (sulphate):

S is surrounded by 4 oxygen atoms. Geometry: tetrahedral. Not pyramidal.

$$SO_3^{2-}$$ (sulphite):

S has 3 bond pairs (to O atoms) + 1 lone pair. Electron geometry is tetrahedral, but molecular geometry is pyramidal (trigonal pyramidal). This is pyramidal.

$$S_2O_7^{2-}$$ (disulphate / pyrosulphate):

Two $$SO_4$$ tetrahedra share one oxygen. Each S is in a tetrahedral environment. Not pyramidal.

Only 1 species ($$SO_3^{2-}$$) has pyramidal geometry.

The correct answer is Option (4): 1.

We need to identify the incorrect statement about geometrical isomers of 2-butene.

2-Butene (CH$$_3$$CH=CHCH$$_3$$) exists as two geometrical isomers due to restricted rotation around the C=C bond:

(i) cis-2-butene: Both methyl groups are on the same side of the double bond.

(ii) trans-2-butene: The methyl groups are on opposite sides of the double bond.

Evaluate each statement:

(1) "cis- and trans-2-butene are not interconvertible at room temperature": This is CORRECT. The energy barrier for rotation around a C=C bond is very high (~250 kJ/mol), preventing interconversion at room temperature.

(2) "cis- and trans-2-butene are stereoisomers": This is CORRECT. They have the same molecular formula and connectivity but differ in the spatial arrangement of atoms. Geometrical isomers are a type of stereoisomers.

(3) "cis-2-butene has less dipole moment than trans-2-butene": This is INCORRECT. In cis-2-butene, the two methyl groups are on the same side, so their bond dipoles add up, resulting in a net dipole moment. In trans-2-butene, the methyl groups are on opposite sides, and their dipoles cancel out, giving a dipole moment of approximately zero. Therefore, cis-2-butene has a HIGHER dipole moment than trans-2-butene.

(4) "trans-2-butene is more stable than cis-2-butene": This is CORRECT. trans-2-butene is more stable because the methyl groups are farther apart, reducing steric strain.

The correct answer is Option (3).

A general rule for stability is such that: the compound having no charge at all would be the most stable, then an electronegative atom like Oxygen getting a negative charge would be preferrable and an electronegative atom having a positive charge would be least preferrable. Hence, I>II>III

Phosphorus pentachloride is a trigonal-bipyramidal molecule. In a highly polar organic solvent it undergoes auto-ionisation (self-dissociation) according to

$$PCl_5 \;\longrightarrow\; [PCl_4]^+ + [PCl_6]^- \quad -(1)$$

When the solution is treated with a fluoride ion source (for example $$HF$$, $$SbF_5$$, or any fluorinating agent), the incoming fluoride ions attack the halophosphate anion first because the anion already carries an excess negative charge and can accept further halide substitution more easily than the cation.

Step 1 - Replacement of two equatorial chlorides
Fluoride ions substitute two of the equatorial chlorides of $$[PCl_6]^-$$ to give the mixed-halide anion $$[PCl_4F_2]^-$$:

$$[PCl_6]^- + 2\,F^- \;\longrightarrow\; [PCl_4F_2]^- + 2\,Cl^- \quad -(2)$$

The cation produced in reaction $$(1)$$ remains unchanged, so the ionic species present after this first fluorination are

$$[PCl_4]^+[PCl_4F_2]^-$$

Step 2 - Complete fluorination
If further fluoride ions are supplied, all six positions around phosphorus in the anion can be occupied by fluorides, converting $$[PCl_4F_2]^-$$ into the hexafluorophosphate anion $$[PF_6]^-$$:

$$[PCl_4F_2]^- + 4\,F^- \;\longrightarrow\; [PF_6]^- + 4\,Cl^- \quad -(3)$$

The cation is still $$[PCl_4]^+$$, so a second ionic pair appears:

$$[PCl_4]^+[PF_6]^-$$

Thus, on fluorination of $$PCl_5$$ in a polar organic solvent, two distinct ion pairs can be isolated depending on how many fluorides replace chlorides in the anion:

$$[PCl_4]^+[PCl_4F_2]^- \quad\text{and}\quad [PCl_4]^+[PF_6]^-$$

None of the other suggested combinations can be obtained under these conditions because the cationic part remains $$[PCl_4]^+$$ and the neutral covalent molecules $$PF_5$$, $$PF_3$$ or $$PCl_3$$ are not formed in appreciable amounts in a polar medium.

Therefore, the correct choice is:
Option B which is: $$[PCl_4]^+[PCl_4F_2]^-$$ and $$[PCl_4]^+[PF_6]^-$$

Assertion (A): Cis form of alkene is more polar than the trans form.

In the cis form, similar groups are on the same side, so bond dipoles add up, resulting in a net dipole moment. In the trans form, similar groups are on opposite sides, so dipoles cancel. Thus cis is more polar. True.

Reason (R): Dipole moment of trans isomer of 2-butene is zero.

In trans-2-butene, the two methyl groups and two H atoms are on opposite sides of the double bond, causing the dipole moments to cancel. Dipole moment is zero (or nearly zero). True.

The reason correctly explains the assertion — because trans forms have zero (or very low) dipole moment due to cancellation, cis forms are comparatively more polar.

The correct answer is Option (2): Both (A) and (R) are true and (R) is the correct explanation of (A).

Peroxide (peroxo) linkage means the presence of an $$\mathrm{O\!-\!O}$$ single bond inside the molecule.
Therefore, we must write the accepted structures of the given oxo-acids of sulphur and look for an $$\mathrm{O\!-\!O}$$ bond.

Option A : $$H_2S_2O_7$$ (pyrosulphuric acid, oleum)
$$H_2S_2O_7 : \; HO\!-\!S(=O)_2\!-\!O\!-\!S(=O)_2\!-\!OH$$
The two sulphur atoms are bridged by a single oxygen $$\,(S\!-\!O\!-\!S)$$, but there is no $$\mathrm{O\!-\!O}$$ bond.
Hence, it does not contain a peroxide linkage.

Option B : $$H_2S_2O_8$$ (peroxodisulphuric acid, Marshall’s acid)
$$H_2S_2O_8 : \; HO\!-\!S(=O)_2\!-\!O\!-\!\mathbf{O}\!-\!S(=O)_2\!-\!OH$$
A peroxide bridge $$\mathbf{(O\!-\!O)}$$ joins the two $$SO_3$$ groups, so the molecule definitely contains a peroxide linkage.

Option C : $$H_2S_2O_5$$ (disulphurous acid)
An accepted structure is $$HO\!-\!S(=O)\!-\!O\!-\!S(=O)\!-\!OH$$ (or an $$S\!-\!S$$ bonded form in equilibrium).
Neither form possesses an $$\mathrm{O\!-\!O}$$ bond, so no peroxide linkage is present.

Option D : $$H_2SO_5$$ (peroxomonosulphuric acid, Caro’s acid)
$$H_2SO_5 : \; HO\!-\!\mathbf{O}\!-\!S(=O)_2\!-\!OH$$
One hydroxyl oxygen is linked to the sulphur through an $$\mathbf{O\!-\!O}$$ bond; hence a peroxide linkage is present.

Thus, the compounds that contain a peroxide linkage are:

Option B which is: $$H_2S_2O_8$$,  Option D which is: $$H_2SO_5$$

For a molecule to obey the octet rule, every atom (especially the central atom) must possess exactly $$8$$ valence electrons (4 electron pairs) after the Lewis structure is drawn.
Two types of violation are common:

(i) Incomplete octet - the central atom has <8 electrons, e.g. $$BF_3$$, $$BCl_3$$.
(ii) Expanded octet - the central atom has >8 electrons, e.g. $$PF_5$$, $$H_2SO_4$$.
(iii) Odd-electron (free-radical) structures automatically break the octet rule because one atom necessarily has 7 electrons; e.g. $$NO$$, $$NO_2$$.

We now examine the four options, molecule by molecule.

• $$CO_2$$ - each O has 8 electrons, the central C has 8 electrons ⇒ obeys.
• $$C_2H_4$$ (ethene) - both C atoms achieve an octet through a double bond and two C-H σ-bonds ⇒ obeys.
• $$NO$$ - 11 valence electrons, one electron remains unpaired; N has only 7 electrons ⇒ violates.
• $$HCl$$ - Cl has 8 electrons, H needs only 2 ⇒ obeys.

Total obeying = 3 out of 4 ⇒ Option A satisfies the condition.

• $$NO_2$$ - odd number (17) of valence electrons; one O has 7 electrons ⇒ violates.
• $$O_3$$ - 18 electrons; every O ends with 8 electrons ⇒ obeys.
• $$HCl$$ - obeys (see above).
• $$H_2SO_4$$ - S forms six σ-bonds (4 S-O + 2 S-O(H)); S gets 12 electrons ⇒ expanded octet ⇒ violates.

Total obeying = 2 ⇒ Option B does not qualify.

• $$BCl_3$$ - B ends with only 6 electrons ⇒ violates.
• $$NO$$ - violates (see above).
• $$NO_2$$ - violates (see above).
• $$H_2SO_4$$ - violates (expanded octet).

Total obeying = 0 ⇒ Option C does not qualify.

• $$CO_2$$ - obeys (see Case A).
• $$BCl_3$$ - B has 6 electrons ⇒ violates.
• $$O_3$$ - obeys (see Case B).
• $$C_2H_4$$ - obeys (see Case A).

Total obeying = 3 ⇒ Option D satisfies the condition.

Hence, the options containing at least three molecules that follow the octet rule are:
Option A and Option D.

Final Answer: Option A (CO₂, C₂H₄, NO, HCl) and Option D (CO₂, BCl₃, O₃, C₂H₄).

We need to identify which molecule or ion has a square pyramidal shape. To determine the shape, we use VSEPR theory: first count the total electron pairs (bond pairs + lone pairs) around the central atom, then determine the geometry.

Recall: The shape of a molecule depends on the number of bond pairs (BP) and lone pairs (LP) around the central atom. A square pyramidal shape arises when there are 6 electron pairs total: 5 bond pairs and 1 lone pair (derived from octahedral electron geometry).

Option 1: $$[Ni(CN)_4]^{2-}$$

Nickel in this complex is Ni$$^{2+}$$ with electronic configuration $$[Ar] 3d^8$$. CN$$^-$$ is a strong field ligand, so it causes pairing of d-electrons. With 4 ligands and a $$d^8$$ configuration in a strong field, the complex adopts a square planar geometry ($$dsp^2$$ hybridization). This is NOT square pyramidal.

Option 2: $$PCl_5$$

Phosphorus has 5 valence electrons. Each Cl contributes 1 bond pair, giving 5 BP and 0 LP around P. With 5 bond pairs and no lone pairs, the geometry is trigonal bipyramidal ($$sp^3d$$ hybridization). This is NOT square pyramidal.

Option 3: $$BrF_5$$

Bromine has 7 valence electrons. Each F contributes 1 bond pair, giving 5 BP. The remaining electrons on Br: $$7 - 5 = 2$$ electrons = 1 lone pair (LP). So there are 5 BP + 1 LP = 6 total electron pairs around Br.

With 6 electron pairs, the electron geometry is octahedral ($$sp^3d^2$$ hybridization). When one position of the octahedron is occupied by a lone pair, the molecular shape becomes square pyramidal.

Option 4: $$PF_5$$

Phosphorus has 5 valence electrons, and each F contributes 1 bond pair, giving 5 BP and 0 LP. The geometry is trigonal bipyramidal (same as $$PCl_5$$). This is NOT square pyramidal.

Conclusion: Only $$BrF_5$$ has 5 bond pairs and 1 lone pair arranged in an octahedral electron geometry, resulting in a square pyramidal molecular shape.

The correct answer is Option (3): $$BrF_5$$.

We need to find the number of molecules with zero dipole moment among: $$CH_4, BF_3, H_2O, HF, NH_3, CO_2,$$ and $$SO_2$$.

A molecule has zero dipole moment when the individual bond dipoles cancel each other out due to symmetry.

Analysing each molecule:

1. $$CH_4$$ (Methane): Tetrahedral geometry. All four $$C-H$$ bonds are identical and symmetrically arranged. The bond dipoles cancel out completely.

Dipole moment = Zero

2. $$BF_3$$ (Boron trifluoride): Trigonal planar geometry. The three $$B-F$$ bonds are symmetrically placed at $$120°$$ angles. The bond dipoles cancel out.

Dipole moment = Zero

3. $$H_2O$$ (Water): Bent geometry with a bond angle of about $$104.5°$$. The two $$O-H$$ bond dipoles do not cancel due to the bent shape, and there are two lone pairs on oxygen.

Dipole moment = Non-zero

4. $$HF$$ (Hydrogen fluoride): Linear diatomic molecule with a highly polar $$H-F$$ bond. Since there is only one bond, there is nothing to cancel it.

Dipole moment = Non-zero

5. $$NH_3$$ (Ammonia): Trigonal pyramidal geometry with one lone pair on nitrogen. The bond dipoles and the lone pair contribution do not cancel.

Dipole moment = Non-zero

6. $$CO_2$$ (Carbon dioxide): Linear geometry. The two $$C=O$$ bonds are equal and opposite in direction. The bond dipoles cancel perfectly.

Dipole moment = Zero

7. $$SO_2$$ (Sulfur dioxide): Bent geometry with a bond angle of about $$119°$$. The two $$S=O$$ bond dipoles do not cancel due to the bent shape, and there is a lone pair on sulfur.

Dipole moment = Non-zero

The molecules with zero dipole moment are: $$CH_4$$, $$BF_3$$, and $$CO_2$$.

Therefore, the total number of molecules with zero dipole moment = 3.

We need to count the number of compounds with zero dipole moment from the given list.

Key concept: A molecule has zero dipole moment if it is either nonpolar (same atoms bonded) or if its geometry causes the individual bond dipoles to cancel out due to symmetry.

Let us check each compound:

1. HF — Linear, polar bond. Dipole moment $$\neq 0$$.

2. $$H_2$$ — Homonuclear diatomic, nonpolar. Dipole moment = 0.

3. $$H_2S$$ — Bent shape (like water). Dipole moment $$\neq 0$$.

4. $$CO_2$$ — Linear, $$O=C=O$$. Bond dipoles cancel. Dipole moment = 0.

5. $$NH_3$$ — Trigonal pyramidal with lone pair. Dipole moment $$\neq 0$$.

6. $$BF_3$$ — Trigonal planar, symmetric. Bond dipoles cancel. Dipole moment = 0.

7. $$CH_4$$ — Tetrahedral, symmetric. Bond dipoles cancel. Dipole moment = 0.

8. $$CHCl_3$$ — Tetrahedral but asymmetric (one H, three Cl). Dipole moment $$\neq 0$$.

9. $$SiF_4$$ — Tetrahedral, symmetric. Bond dipoles cancel. Dipole moment = 0.

10. $$H_2O$$ — Bent shape. Dipole moment $$\neq 0$$.

11. $$BeF_2$$ — Linear, $$F-Be-F$$. Bond dipoles cancel. Dipole moment = 0.

Compounds with zero dipole moment: $$H_2$$, $$CO_2$$, $$BF_3$$, $$CH_4$$, $$SiF_4$$, $$BeF_2$$ = 6 compounds.

The answer is 6.

Count compounds/species with non-zero dipole moment from the given list.

Analysis of each:

- $$BeCl_2$$: Linear, symmetric → $$\mu = 0$$

- $$BCl_3$$: Trigonal planar, symmetric → $$\mu = 0$$

- $$NF_3$$: Trigonal pyramidal (lone pair on N) → $$\mu \neq 0$$

- $$XeF_4$$: Square planar, symmetric → $$\mu = 0$$

- $$CCl_4$$: Tetrahedral, symmetric → $$\mu = 0$$

- $$H_2O$$: Bent → $$\mu \neq 0$$

- $$H_2S$$: Bent → $$\mu \neq 0$$

- $$HBr$$: Diatomic, polar → $$\mu \neq 0$$

- $$CO_2$$: Linear, symmetric → $$\mu = 0$$

- $$H_2$$: Homonuclear diatomic → $$\mu = 0$$

- $$HCl$$: Diatomic, polar → $$\mu \neq 0$$

Non-zero dipole moment: $$NF_3, H_2O, H_2S, HBr, HCl$$ → 5.

The correct answer is 5.

We need to count how many of the given species have exactly one unpaired electron, using Molecular Orbital Theory (MOT).

Key Concept: In MOT, electrons fill molecular orbitals in order of increasing energy. For diatomic molecules of second-period elements, the filling order depends on whether the species has 14 or fewer electrons (where $$\sigma_{2p}$$ is higher than $$\pi_{2p}$$) or more than 14 electrons (where $$\sigma_{2p}$$ is lower than $$\pi_{2p}$$).

1. $$O_2$$ (16 electrons):

Electronic configuration: $$(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi^*_{2p})^2$$

The last two electrons go into the two degenerate $$\pi^*_{2p}$$ orbitals. By Hund's rule, they occupy different orbitals with parallel spins. This gives 2 unpaired electrons.

Since we need exactly 1 unpaired electron, $$O_2$$ does not qualify.

2. $$O_2^-$$ (17 electrons):

Compared to $$O_2$$, this has one additional electron: $$(\pi^*_{2p})^3$$

The two degenerate $$\pi^*_{2p}$$ orbitals now hold 3 electrons: one orbital has 2 (paired), the other has 1 (unpaired). This gives 1 unpaired electron. This qualifies.

3. $$NO$$ (15 electrons):

Electronic configuration: $$(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi^*_{2p})^1$$

The 15th electron enters one of the $$\pi^*_{2p}$$ orbitals, giving 1 unpaired electron. This qualifies.

4. $$CN^-$$ (14 electrons):

$$CN^-$$ is isoelectronic with $$N_2$$ and $$CO$$ (all have 14 electrons). For species with 14 or fewer electrons, the MO ordering has $$\pi_{2p}$$ below $$\sigma_{2p}$$:

$$(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p})^4 (\sigma_{2p})^2$$

All electrons are paired, giving 0 unpaired electrons. Does not qualify.

5. $$O_2^{2-}$$ (18 electrons):

Compared to $$O_2$$, this has two additional electrons: $$(\pi^*_{2p})^4$$

Both $$\pi^*_{2p}$$ orbitals are now completely filled (2 electrons each), so all electrons are paired. This gives 0 unpaired electrons. Does not qualify.

Counting the species with exactly one unpaired electron:

$$O_2^-$$ and $$NO$$ — that is 2 species.

The answer is 2.

We need to find the number of species that are paramagnetic AND have bond order = 1 from: $$H_2, He_2^+, O_2^+, N_2^{2-}, O_2^{2-}, F_2, Ne_2^+, B_2$$.

Analyze each species.

$$H_2$$: $$\sigma_{1s}^2$$. Bond order = 1. Diamagnetic (all paired). No.

$$He_2^+$$: $$\sigma_{1s}^2 {\sigma^*_{1s}}^{1}$$. Bond order = 0.5. No (BO ≠ 1).

$$O_2^+$$: Bond order = 2.5. No.

$$N_2^{2-}$$: 16 electrons. Same as $$O_2$$. Bond order = 2. No.

$$O_2^{2-}$$: 18 electrons. $$\sigma_{1s}^2{\sigma^*_{1s}}^{2}\sigma_{2s}^2{\sigma^*_{2s}}^{2}\sigma_{2p}^2\pi_{2p}^4{\pi^*_{2p}}^{4}$$. Bond order = (10-8)/2 = 1. All electrons paired. Diamagnetic. No.

$$F_2$$: 18 electrons. Bond order = 1. Diamagnetic. No.

$$Ne_2^+$$: 19 electrons. Bond order = (10-9)/2 = 0.5. No.

$$B_2$$: 10 electrons. $$\sigma_{1s}^2{\sigma^*_{1s}}^{2}\sigma_{2s}^2{\sigma^*_{2s}}^{2}\pi_{2p}^1\pi_{2p}^1$$. Bond order = (6-4)/2 = 1. Two unpaired electrons in $$\pi_{2p}$$. Paramagnetic with bond order 1. Yes!

Count.

Only $$B_2$$ satisfies both conditions = 1 species.

Conclusion.

The number is 1.

Therefore, the answer is 1.

We need to find the sum of bond orders of CO and $$NO^+$$.

We first determine the bond order of CO using Molecular Orbital Theory (MOT). The formula for bond order is:

$$ \text{Bond Order} = \frac{N_b - N_a}{2} $$

where $$N_b$$ = number of electrons in bonding molecular orbitals, and $$N_a$$ = number of electrons in antibonding molecular orbitals.

CO has a total of $$6 + 8 = 14$$ electrons. The molecular orbital configuration for CO (following the order for molecules with atomic number $$\leq 7$$ for the lighter atom) is:

$$ \begin{aligned} &(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2\\ &(\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2 \end{aligned} $$

Counting electrons gives bonding electrons $$N_b = 2 + 2 + 2 + 2 + 2 = 10$$ (from $$\sigma 1s, \sigma 2s, \pi 2p_x, \pi 2p_y, \sigma 2p_z$$) and antibonding electrons $$N_a = 2 + 2 = 4$$ (from $$\sigma^* 1s, \sigma^* 2s$$). Thus,

$$ \text{Bond Order of CO} = \frac{10 - 4}{2} = \frac{6}{2} = 3 $$

Next, we determine the bond order of $$NO^+$$. This species has $$7 + 8 - 1 = 14$$ electrons (nitrogen has 7, oxygen has 8, minus 1 for the positive charge). Since $$NO^+$$ is isoelectronic with CO, it has the same MO configuration:

$$ \begin{aligned} &(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2\\ &(\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2 \end{aligned} $$

Therefore,

$$ \text{Bond Order of } NO^+ = \frac{10 - 4}{2} = \frac{6}{2} = 3 $$

Finally, the sum of bond orders is

$$ \text{Sum of bond orders} = 3 + 3 = 6 $$

The answer is $$\boxed{6}$$.

Find how many of the given species have trigonal bipyramidal (TBP) shape: $$PF_5, BrF_5, PCl_5, [PtCl_4]^{2-}, BF_3, Fe(CO)_5$$.

Method: For each species, count electron domains around the central atom and determine the geometry using VSEPR theory.

1. $$PF_5$$: P has 5 valence electrons, each F contributes 1 bond = 5 bond pairs, 0 lone pairs. Electron geometry: TBP. Molecular shape: Trigonal bipyramidal.

2. $$BrF_5$$: Br has 7 valence electrons, 5 bonds to F uses 5 electrons, leaving 2 electrons = 1 lone pair. Total electron domains = 6 (octahedral electron geometry). Shape: Square pyramidal (not TBP).

3. $$PCl_5$$: Same as $$PF_5$$. P has 5 bond pairs, 0 lone pairs. Shape: Trigonal bipyramidal.

4. $$[PtCl_4]^{2-}$$: Pt is a transition metal with d$$^8$$ configuration ($$Pt^{2+}$$). With 4 $$Cl^-$$ ligands, it adopts square planar geometry (not TBP).

5. $$BF_3$$: B has 3 valence electrons, 3 bonds = 3 bond pairs, 0 lone pairs. Shape: Trigonal planar (not TBP).

6. $$Fe(CO)_5$$: Fe has 8 valence electrons. In $$Fe(CO)_5$$, iron is $$Fe(0)$$ with 5 CO ligands forming 5 bond pairs. This follows the 18-electron rule (8 + 5×2 = 18). The geometry is Trigonal bipyramidal.

Count of TBP species: $$PF_5$$, $$PCl_5$$, $$Fe(CO)_5$$ = 3.

The answer is 3.

We need to find the total number of antibonding molecular orbitals formed from $$2s$$ and $$2p$$ atomic orbitals in a diatomic molecule.

Molecular Orbital Theory: When atomic orbitals combine to form molecular orbitals, each pair of atomic orbitals produces one bonding and one antibonding molecular orbital.

From $$2s$$ atomic orbitals:

Each atom contributes one $$2s$$ orbital. The two $$2s$$ orbitals combine to form:

• $$\sigma_{2s}$$ (bonding)
• $$\sigma^*_{2s}$$ (antibonding) → 1 antibonding MO

From $$2p$$ atomic orbitals:

Each atom contributes three $$2p$$ orbitals ($$2p_x, 2p_y, 2p_z$$). The six $$2p$$ orbitals (3 from each atom) combine to form:

• $$\sigma_{2p_z}$$ (bonding) and $$\sigma^*_{2p_z}$$ (antibonding) → 1 antibonding MO
• $$\pi_{2p_x}$$ (bonding) and $$\pi^*_{2p_x}$$ (antibonding) → 1 antibonding MO
• $$\pi_{2p_y}$$ (bonding) and $$\pi^*_{2p_y}$$ (antibonding) → 1 antibonding MO

Total antibonding MOs from $$2p$$ orbitals = 3

Total antibonding molecular orbitals = 1 (from 2s) + 3 (from 2p) = 4

Therefore, the answer is 4.

In a diatomic molecule, molecular orbitals are formed by the linear combination of atomic orbitals (LCAO). Each pair of atomic orbitals (one from each atom) combines to form two molecular orbitals — one bonding and one antibonding.

From the 2s atomic orbitals:

Each atom contributes one 2s orbital. These two 2s orbitals combine to form:

1. $$\sigma_{2s}$$ (bonding)

2. $$\sigma^*_{2s}$$ (antibonding)

That gives 2 molecular orbitals.

From the 2p atomic orbitals:

Each atom contributes three 2p orbitals ($$2p_x$$, $$2p_y$$, $$2p_z$$). These six atomic orbitals (3 from each atom) combine to form:

1. $$\sigma_{2p_z}$$ (bonding)

2. $$\sigma^*_{2p_z}$$ (antibonding)

3. $$\pi_{2p_x}$$ (bonding)

4. $$\pi^*_{2p_x}$$ (antibonding)

5. $$\pi_{2p_y}$$ (bonding)

6. $$\pi^*_{2p_y}$$ (antibonding)

That gives 6 molecular orbitals.

Total molecular orbitals = 2 + 6 = 8.

The answer is 8.

CH3-CH(OH)-CH(OH)-CH3: This Fischer projection shows chiral centers at C2 and C3. However, there is a horizontal plane of symmetry passing directly between C2 and C3, dividing the molecule into two identical halves. This makes it a meso compound, optically inactive. 

Heptane-3,4,5-triol: The molecule has chiral centers at C3 and C5, and a pseudoasymmetric center at C4. As drawn, there is a vertical plane of symmetry passing straight through C4 and its attached -OH group. The left side (ethyl group and -OH group) perfectly mirrors the right side. Thus, it is a meso compound.

Butanol: There are no chiral carbons (no carbon is attached to four different groups). Hence optically inactive.

1-chlorobutane: There are no chiral carbons. Hence Optically inactive.

1-chloro-3-methylbutane: There are no chiral carbons. Carbon-3 is attached to two identical methyl groups. Hence, optically inactive.

2-chlorobutane: Carbon-2 is bonded to four distinct groups. This makes C2 a chiral center. There is no plane or center of symmetry in the molecule. Hence optically active compound. 

Total Optically active compounds = 0+0+0+0+0+1 = 1

We need to count the non-polar molecules from the given list: HF, H$$_2$$O, SO$$_2$$, H$$_2$$, CO$$_2$$, CH$$_4$$, NH$$_3$$, HCl, CHCl$$_3$$, BF$$_3$$.

A molecule is non-polar if either (a) it has no polar bonds, or (b) it has polar bonds but they are arranged symmetrically so that the dipole moments cancel out.

Analyze each molecule:
(i) HF: Polar bond, asymmetric → Polar
(ii) H$$_2$$O: Polar O-H bonds, bent geometry → dipoles don't cancel → Polar
(iii) SO$$_2$$: Polar S=O bonds, bent geometry → Polar
(iv) H$$_2$$: Same atoms, no dipole → Non-polar ✓
(v) CO$$_2$$: Polar C=O bonds, but linear and symmetric → dipoles cancel → Non-polar ✓
(vi) CH$$_4$$: Polar C-H bonds, but tetrahedral symmetry → dipoles cancel → Non-polar ✓
(vii) NH$$_3$$: Polar N-H bonds, pyramidal geometry → Polar
(viii) HCl: Polar bond, asymmetric → Polar
(ix) CHCl$$_3$$: Asymmetric substitution → dipoles don't cancel → Polar
(x) BF$$_3$$: Polar B-F bonds, but trigonal planar symmetry → dipoles cancel → Non-polar ✓

Count non-polar molecules: H$$_2$$, CO$$_2$$, CH$$_4$$, BF$$_3$$ = 4.

The answer is 4.

To determine if a molecule is aromatic, it must satisfy Hückel's Rule. A compound is aromatic if it is:

1. Cyclic
2. Planar
3. Fully Conjugated (has a p-orbital on every atom in the ring)
4. Contains (4n + 2) pi electrons (where n is an integer: 2, 6, 10, 14...).

Analyzing each statement:

(A) NF₃ has trigonal planar structure: NF₃ has 3 bond pairs + 1 lone pair on N → pyramidal (like NH₃), NOT trigonal planar. Incorrect.

(B) Bond length of N₂ is shorter than O₂: N₂ has bond order 3 (triple bond, length ~1.10 Å). O₂ has bond order 2 (double bond, length ~1.21 Å). So N₂ is shorter. Correct.

(C) Isoelectronic molecules or ions have identical bond order: Isoelectronic species have the same number of electrons and similar electronic configuration, so they have the same bond order. For example, CO, N₂, NO⁺ are isoelectronic with bond order 3. Correct.

(D) Dipole moment of H₂S is higher than H₂O: H₂O has a dipole moment of 1.85 D while H₂S has 0.97 D. H₂O has a higher dipole moment due to the greater electronegativity of O and the bond angle. Incorrect.

Correct statements: B and C.

This matches option 4: (B) and (C) are correct.

Hydrogen bonding depends on the structural arrangement and purity of water. In ice, every water molecule is hydrogen-bonded to exactly four neighbouring water molecules in a rigid, three-dimensional tetrahedral lattice. This maximises the number of hydrogen bonds per molecule.

In liquid water, thermal energy partially disrupts this ordered network. While extensive hydrogen bonding still exists, some bonds are continuously broken and reformed, so the average number of hydrogen bonds per molecule is fewer than in ice.

In impure water, dissolved solutes (ions, molecules, or other impurities) interfere with the hydrogen-bonding network. The impurity particles occupy space and interact with water molecules differently, further reducing the extent of hydrogen bonding compared to pure liquid water.

Therefore, the decreasing order of hydrogen bonding is $$\text{Ice (B)} > \text{Liquid Water (A)} > \text{Impure Water (C)}$$, i.e., $$B > A > C$$.

Hence, the correct answer is Option 2.

We need to identify the correct statements about OF$$_2$$.

(A) Number of lone pairs on oxygen is 2.

Oxygen has 6 valence electrons. In OF$$_2$$, it forms 2 bonds with F, leaving 2 lone pairs. CORRECT. ✓

(B) FOF angle is less than 104.5°.

OF$$_2$$ has 2 bond pairs and 2 lone pairs around O (similar to water). The lone pair-lone pair repulsion reduces the bond angle. The actual FOF angle is about 103°, which is less than 104.5°. CORRECT. ✓

(C) Oxidation state of O is -2.

In OF$$_2$$, fluorine is more electronegative than oxygen, so F has -1 oxidation state. Therefore, O has +2 oxidation state, not -2. INCORRECT.

(D) Molecule is bent V-shaped.

With 2 bond pairs and 2 lone pairs, the molecular geometry is bent (V-shaped). CORRECT. ✓

(E) Molecular geometry is linear.

The molecule is bent, not linear. INCORRECT.

Correct statements: A, B, D.

The correct answer is Option 4: A, B, D only.

Assertion A: Butan-1-ol has higher boiling point than ethoxyethane (diethyl ether).

Reason R: Extensive hydrogen bonding leads to stronger association of molecules.

Analysis of Assertion A:

Butan-1-ol (CH₃CH₂CH₂CH₂OH) has a boiling point of approximately 117.7°C.

Ethoxyethane (CH₃CH₂OCH₂CH₃, diethyl ether) has a boiling point of approximately 34.6°C.

Both have the same molecular formula C₄H₁₀O, but butan-1-ol has a significantly higher boiling point.

Assertion A is True.

Analysis of Reason R:

Butan-1-ol has an -OH group capable of forming extensive intermolecular hydrogen bonds, leading to stronger molecular association and higher boiling point. Ethoxyethane cannot form H-bonds with itself (no O-H or N-H groups).

Reason R is True and correctly explains Assertion A.

The correct answer is Option B: Both A and R are true and R is the correct explanation of A.

We need to determine which diatomic halogen molecule has the highest bond dissociation energy.

Bond dissociation energies of halogens:

The bond dissociation energies follow the general trend:

$$\text{Cl}_2 (242 \text{ kJ/mol}) > \text{Br}_2 (193 \text{ kJ/mol}) > \text{F}_2 (159 \text{ kJ/mol}) > \text{I}_2 (151 \text{ kJ/mol})$$

Explanation:

Generally, bond dissociation energy decreases as we go down the group due to increasing atomic size and decreasing orbital overlap. However, $$\text{F}_2$$ is an exception — it has an anomalously low bond dissociation energy because the small size of fluorine atoms leads to strong lone pair-lone pair repulsion between the closely placed non-bonding electrons.

Therefore, $$\text{Cl}_2$$ has the highest bond dissociation energy among the halogens.

The correct answer is Option 1: Cl$$_2$$.

We need to find the pair where both compounds have a net non-zero dipole moment.

We start by noting that

A molecule has a non-zero dipole moment when the bond dipoles do not cancel due to molecular asymmetry. Symmetric molecules (like linear CO$$_2$$, benzene, para-dichlorobenzene, trans-butene) have zero dipole moments.

Next,

Option 1: 1,4-Dichlorobenzene and 1,3-Dichlorobenzene

1,4-Dichlorobenzene (para): The two C-Cl bond dipoles are equal and opposite, cancelling exactly. Dipole moment = 0.

1,3-Dichlorobenzene (meta): The bond dipoles do not cancel. Dipole moment is non-zero.

Not both non-zero.

Option 2: cis-butene and trans-butene

cis-2-butene: The two CH$$_3$$ groups are on the same side, creating asymmetry. Has a small but non-zero dipole moment.

trans-2-butene: The CH$$_3$$ groups are on opposite sides, and the molecule has a centre of symmetry. Dipole moment = 0.

Not both non-zero.

Option 3: CH$$_2$$Cl$$_2$$ and CHCl$$_3$$

CH$$_2$$Cl$$_2$$ (dichloromethane): Tetrahedral with 2 H and 2 Cl -- asymmetric arrangement. Has a dipole moment of 1.60 D. Non-zero.

CHCl$$_3$$ (chloroform): Tetrahedral with 1 H and 3 Cl -- asymmetric. Has a dipole moment of 1.04 D. Non-zero.

Both have non-zero dipole moments.

Option 4: Benzene and anisidine

Benzene is perfectly symmetric with zero dipole moment.

Not both non-zero.

The correct answer is Option 3: CH$$_2$$Cl$$_2$$, CHCl$$_3$$.

We need to identify the pair where both compounds form polar molecular solids.

We start by noting that

Polar molecular solids are composed of polar molecules held together by dipole-dipole interactions (and possibly hydrogen bonds). The key requirement is that each molecule must have a net non-zero dipole moment.

Next,

Option 1: SO$$_2$$(s), CO$$_2$$(s)

SO$$_2$$ is bent (bond angle ~119 degrees) with a net dipole moment -- it is polar. However, CO$$_2$$ is linear and symmetric, so its two C=O bond dipoles cancel, giving zero net dipole moment. CO$$_2$$ forms a non-polar molecular solid. This pair does not work.

Option 2: SO$$_2$$(s), NH$$_3$$(s)

SO$$_2$$ is polar (bent structure, as explained above). NH$$_3$$ has a pyramidal geometry with a lone pair on nitrogen, giving a net dipole moment of 1.47 D. Both are polar molecules, and both form polar molecular solids. This is the correct pair.

Option 3: MgO(s), SO$$_2$$(s)

MgO is an ionic solid (Mg$$^{2+}$$ and O$$^{2-}$$ ions), not a molecular solid. This pair does not work.

Option 4: HCl(s), AlN(s)

HCl is a polar molecule, but AlN is a covalent network solid (similar to diamond/BN structure), not a molecular solid. This pair does not work.

The correct answer is Option 2: SO$$_2$$(s), NH$$_3$$(s).

We need to identify the correct statement about the compound GaAlCl$$_4$$.

The compound GaAlCl$$_4$$ is an ionic salt with the formula Ga$$^+$$[AlCl$$_4$$]$$^-$$. Aluminum readily forms the stable tetrachloroaluminate anion [AlCl$$_4$$]$$^-$$, in which Al is tetrahedrally coordinated with four Cl atoms using $$sp^3$$ hybridization. Gallium exists as the monovalent cation Ga$$^+$$ due to the inert pair effect, where the 4s$$^2$$ electrons resist participation in bonding.

Now, the oxidation states are: Ga has +1 (not +3), and Al has +3 (since $$x + 4(-1) = -1$$, so $$x = +3$$). On the Pauling electronegativity scale, Ga (1.81) is more electronegative than Al (1.61). Despite being more electronegative, Ga forms the cation because Al has a very strong tendency to form the [AlCl$$_4$$]$$^-$$ complex anion.

Evaluating each option: Option 1 says Ga is coordinated with Cl, which is incorrect since Ga exists as a free cation. Option 2 says Ga is more electronegative than Al and forms the cationic part, which is correct. Option 3 says Cl bonds with both Al and Ga, which is incorrect. Option 4 says Ga has oxidation state +3, which is also incorrect.

Hence, the correct answer is Option 2.

We determine the molecular geometry using VSEPR theory:

A. XeF$$_4$$: Xe has 8 valence electrons. With 4 F atoms, there are 4 bond pairs and 2 lone pairs (6 electron pairs total). Geometry: Square planar → II

B. SF$$_4$$: S has 6 valence electrons. With 4 F atoms, there are 4 bond pairs and 1 lone pair (5 electron pairs total). Geometry: See-saw → I

C. NH$$_4^+$$: N has 5 valence electrons, minus 1 for positive charge = 4 electrons shared with 4 H atoms. 4 bond pairs, 0 lone pairs. Geometry: Tetrahedral → IV

D. BrF$$_3$$: Br has 7 valence electrons. With 3 F atoms, there are 3 bond pairs and 2 lone pairs (5 electron pairs total). Geometry: Bent T-shaped → III

Matching: A-II, B-I, C-IV, D-III

We need to match each molecule/ion with the number of lone pairs on its central atom.

(A) IF$$_7$$:

Iodine has 7 valence electrons. It forms 7 bonds with 7 fluorine atoms, using all its valence electrons for bonding. The hybridization is $$sp^3d^3$$.

Lone pairs on I = $$\frac{7 - 7}{2} = 0$$ (Zero) --> matches IV

(B) ICl$$_4^-$$:

Iodine has 7 valence electrons, plus 1 from the negative charge = 8 electrons available. It forms 4 bonds with Cl atoms, leaving 4 non-bonding electrons.

Lone pairs on I = $$\frac{8 - 4}{2} = 2$$ (Two) --> matches III

The geometry is square planar ($$sp^3d^2$$, with 2 lone pairs in axial positions of the octahedron).

(C) XeF$$_6$$:

Xenon has 8 valence electrons. It forms 6 bonds with F atoms, leaving 2 non-bonding electrons.

Lone pairs on Xe = $$\frac{8 - 6}{2} = 1$$ (One) --> matches II

The geometry is distorted octahedral ($$sp^3d^3$$ hybridization with 1 lone pair).

(D) XeF$$_2$$:

Xenon has 8 valence electrons. It forms 2 bonds with F atoms, leaving 6 non-bonding electrons.

Lone pairs on Xe = $$\frac{8 - 2}{2} = 3$$ (Three) --> matches I

The geometry is linear ($$sp^3d$$ hybridization with 3 lone pairs in equatorial positions of a trigonal bipyramid).

The correct matching is: A-IV, B-III, C-II, D-I.

The correct answer is Option 2.

Covalent character in an ionic compound is assessed using Fajans' rules. Covalent character increases when (i) the cation is small and highly charged, (ii) the anion is large and easily polarisable, and (iii) the cation has a pseudo-noble-gas configuration (e.g., $$Cu^{+}$$ with $$18$$ electrons in the outermost shell) rather than a noble-gas configuration.

Statement A: $$KF > KI;\; LiF > KF$$. Since $$I^{-}$$ is larger and more polarisable than $$F^{-}$$, we have $$KI > KF$$, not $$KF > KI$$. The first part is wrong, so A is incorrect.

Statement B: $$KF < KI;\; LiF > KF$$. As argued above, $$KI > KF$$ because $$I^{-}$$ is more polarisable. Also, $$Li^{+}$$ is much smaller than $$K^{+}$$ and therefore more polarising, so $$LiF > KF$$. Both parts are correct, so B is correct.

Statement C: $$SnCl_{4} > SnCl_{2};\; CuCl > NaCl$$. In $$SnCl_{4}$$, tin is in the $$+4$$ oxidation state, which is smaller and more highly charged than $$Sn^{2+}$$, so $$SnCl_{4}$$ has greater covalent character than $$SnCl_{2}$$. For $$CuCl$$ vs $$NaCl$$, $$Cu^{+}$$ has a pseudo-noble-gas configuration $$(3d^{10})$$ and is smaller than $$Na^{+}$$, giving $$CuCl$$ much greater covalent character. Both parts are correct, so C is correct.

Statement D: $$LiF > KF;\; CuCl < NaCl$$. While $$LiF > KF$$ is correct, $$CuCl < NaCl$$ is wrong (it should be $$CuCl > NaCl$$). So D is incorrect.

Statement E: $$KF < KI;\; CuCl > NaCl$$. Both parts follow directly from the reasoning above and are correct. So E is correct.

The correct statements are B, C, and E.

Hence, the correct answer is Option 3.

The acetylide ion $$C_2^{2-}$$ contains $$2(6) + 2 = 14$$ electrons. Using the MO diagram for homonuclear diatomic molecules without 2s-2p mixing, its electron configuration is $$\sigma_{1s}^2{\sigma^*_{1s}}^{2}\sigma_{2s}^2{\sigma^*_{2s}}^{2}\pi_{2p}^4\sigma_{2p}^2$$, which gives a bond order of $$\frac{10 - 4}{2} = 3$$. All electrons are paired, so $$C_2^{2-}$$ is diamagnetic.

In $$NO^+$$, nitrogen contributes 7 electrons and oxygen contributes 8, with the positive charge removing one electron, for a total of $$7 + 8 - 1 = 14$$ electrons. This makes $$NO^+$$ isoelectronic with $$C_2^{2-}$$. Its bond order is also $$3$$ and it is diamagnetic, matching the properties of the acetylide ion.

The species $$O_2^+$$ (15 electrons, bond order $$2.5$$, paramagnetic), $$N_2^+$$ (13 electrons, bond order $$2.5$$, paramagnetic), and $$O_2^-$$ (17 electrons, bond order $$1.5$$, paramagnetic) do not share the same bond order and magnetic behavior as $$C_2^{2-}$$.

Therefore, the species that matches both the bond order and magnetic property of the acetylide ion is $$NO^+$$. The correct answer is Option 3: $$NO^+$$.

The carbonate ion $$CO_3^{2-}$$ can be represented by three resonance structures, each showing one C=O double bond and two C-O single bonds with different oxygen atoms.

Key points about resonance:

1. Resonance structures are not real, individual structures. They are merely convenient representations used to describe the actual structure.

2. The actual molecule does not switch between these structures. There is no dynamic equilibrium between resonance structures.

3. The resonance structures do not exist for any amount of time individually.

4. The actual structure of $$CO_3^{2-}$$ is a single structure called the resonance hybrid, which is a weighted average of all three resonance structures. In this hybrid, all three C-O bonds are equivalent with a bond order of $$\frac{4}{3}$$, and all bond lengths are equal.

Therefore, the correct statement is: $$CO_3^{2-}$$ has a single structure i.e., resonance hybrid of the above three structures.

We need to find how many species from the given list carry a single lone pair on the central atom Xenon. Xenon has 8 valence electrons, and we count lone pairs by subtracting electrons used in bonding.

For $$XeF_5^+$$: Effective electrons from Xe = $$8 - 1 = 7$$ (charge adjusts by $$-1$$). Five Xe-F bonds use 5 electrons, leaving $$7 - 5 = 2$$ electrons = 1 lone pair. Geometry is square pyramidal with $$sp^3d^2$$ hybridization.

For $$XeO_3$$: Three double bonds use $$3 \times 2 = 6$$ electrons, leaving $$8 - 6 = 2$$ electrons = 1 lone pair. Geometry is trigonal pyramidal with $$sp^3$$ hybridization.

For $$XeO_2F_2$$: Two double bonds with O use 4 electrons and two single bonds with F use 2 electrons, totalling 6. That leaves $$8 - 6 = 2$$ electrons = 1 lone pair. Geometry is see-saw with $$sp^3d$$ hybridization.

For $$XeF_5^-$$: Effective electrons = $$8 + 1 = 9$$. Five bonds use 5, leaving 4 electrons = 2 lone pairs. This does not qualify.

For $$XeO_3F_2$$: Three double bonds use 6 and two single bonds use 2, totalling 8. That leaves $$8 - 8 = 0$$ lone pairs. This does not qualify.

For $$XeOF_4$$: One double bond uses 2 and four single bonds use 4, totalling 6. That leaves $$8 - 6 = 2$$ electrons = 1 lone pair. Geometry is square pyramidal with $$sp^3d^2$$ hybridization.

For $$XeF_4$$: Four single bonds use 4 electrons, leaving $$8 - 4 = 4$$ electrons = 2 lone pairs. This does not qualify.

The species with a single lone pair are $$XeF_5^+$$, $$XeO_3$$, $$XeO_2F_2$$, and $$XeOF_4$$. Hence, the answer is $$4$$.

We need to determine the geometry/shape of each species using VSEPR theory.

A. H$$_3$$O$$^+$$:

Central atom O has 3 bond pairs + 1 lone pair = 4 electron pairs (sp$$^3$$ hybridized).

Shape: Pyramidal (III)

B. Acetylide anion (C$$_2^{2-}$$):

The acetylide anion has a triple bond between two carbon atoms: $$[:C \equiv C:]^{2-}$$

Shape: Linear (II)

C. NH$$_4^+$$:

Central atom N has 4 bond pairs + 0 lone pairs = 4 electron pairs (sp$$^3$$ hybridized).

Shape: Tetrahedral (I)

D. ClO$$_2^-$$:

Central atom Cl has 2 bond pairs + 2 lone pairs = 4 electron pairs (sp$$^3$$ hybridized).

Shape: Bent (IV)

Therefore: A → III, B → II, C → I, D → IV

We compare bond lengths in $$H_2O_2$$ and $$F_2O_2$$ (dioxygen difluoride, FOOF).

For X (O-O bond comparison): In $$H_2O_2$$, the O-O single bond has a length of about 1.47 Angstrom. In $$F_2O_2$$, the highly electronegative fluorine atoms draw electron density away from the O-O bond region, weakening it and increasing it to about 1.58 Angstrom. Since the O-O bond in $$H_2O_2$$ (1.47 Angstrom) is less than in $$F_2O_2$$ (1.58 Angstrom), X = shorter.

For Y (O-H vs O-F bond comparison): The O-H bond length in $$H_2O_2$$ is about 0.97 Angstrom. The O-F bond length in $$F_2O_2$$ is about 1.41 Angstrom. Since 0.97 Angstrom is less than 1.41 Angstrom, the O-H bond is shorter than the O-F bond. Thus Y = shorter.

The correct answer is Option A: X-shorter, Y-shorter.

The setting of cement involves the hydration of calcium aluminates and silicates, which is an exothermic process that occurs rapidly.

Gypsum (CaSO₄·2H₂O) is added to cement to increase the setting time (slow down the setting process). It reacts with tricalcium aluminate (C₃A) to form calcium sulfoaluminate (ettringite), which coats the C₃A particles and prevents their rapid hydration, thereby increasing the setting time.

Without gypsum, cement would set too quickly (flash setting), making it difficult to work with.

The correct answer is Gypsum.

Gypsum (CaSO₄·2H₂O) is added to cement to slow down the process of setting.

Without gypsum, cement would set too rapidly (flash setting) when water is added, making it difficult to work with. Gypsum reacts with tricalcium aluminate (C₃A) in cement to form calcium sulfoaluminate (ettringite), which coats the C₃A particles and delays their rapid hydration.

This gives workers sufficient time to mix, transport, and place the concrete before it hardens.

The correct answer is To slow down the process of setting.

We need to determine the structures of $$\text{BeCl}_2$$ in the solid state, vapour phase, and at very high temperature.

In the solid state, $$\text{BeCl}_2$$ exists as a polymeric chain structure. Each Be atom is tetrahedrally coordinated, bonded to four Cl atoms through bridging chlorine atoms forming a chain polymer.

When $$\text{BeCl}_2$$ is vaporised at moderate temperatures, the polymeric chain breaks down into dimeric units. In the dimer, two BeCl$$_2$$ units are connected through two bridging Cl atoms, with each Be atom having a coordination number of 3 (trigonal planar geometry with two terminal and one bridging Cl).

At very high temperatures, the dimer dissociates completely into monomeric BeCl$$_2$$ molecules. The monomer is a linear molecule (sp hybridisation) with Be bonded to two Cl atoms at 180°.

Solid state → Polymeric, Vapour phase → Dimeric, Very high temperature → Monomeric. The correct answer is Option D: Polymeric, Dimeric, Monomeric.

Element Y forms a cubic close packed (CCP/FCC) arrangement. In a CCP structure, the number of atoms of Y per unit cell is 4, and the number of tetrahedral voids is $$2 \times 4 = 8$$.

Now, element X occupies one-third of the tetrahedral voids:

$$\text{Number of X atoms} = \frac{1}{3} \times 8 = \frac{8}{3}$$

The ratio of X to Y atoms is:

$$X : Y = \frac{8}{3} : 4 = \frac{8}{3} : \frac{12}{3} = 8 : 12 = 2 : 3$$

Hence, the correct answer is X$$_2$$Y$$_3$$.

A cubic solid is made up of two elements X and Y. We need to find the empirical formula.

X is present on every alternate corner and one at the center of the cube.

A cube has 8 corners. Every alternate corner means 4 corners are occupied by X.

Contribution from corners = $$4 \times \dfrac{1}{8} = \dfrac{1}{2}$$

Contribution from body center = $$1$$

Total X per unit cell = $$\dfrac{1}{2} + 1 = \dfrac{3}{2} = 1.5$$

Y is at $$\dfrac{1}{3}$$rd of the total faces.

A cube has 6 faces. $$\dfrac{1}{3} \times 6 = 2$$ face-centered positions are occupied by Y.

Contribution from face centers = $$2 \times \dfrac{1}{2} = 1$$

Total Y per unit cell = $$1$$

X : Y = 1.5 : 1 = 3 : 2

The empirical formula is $$\text{X}_3\text{Y}_2$$ or equivalently $$\text{X}_{1.5}\text{Y}$$.

The correct answer is Option B: $$\text{X}_{1.5}\text{Y}$$.

In a CsCl unit cell, Cs⁺ occupies the body center and Cl⁻ occupies the corners (or vice versa). The ions touch along the body diagonal.

Body diagonal of a cube with edge length $$a$$: $$d = \sqrt{3}a$$

The body diagonal connects opposite corners passing through the center. Along this diagonal:

$$2(r_{Cs^+} + r_{Cl^-}) = \sqrt{3}a$$

$$r_{Cs^+} + r_{Cl^-} = \frac{\sqrt{3}}{2}a$$

This matches option 2.

Let us analyze the structure of each oxide:

A. N₂O₄: The structure is O₂N-NO₂, where two NO₂ groups are joined by a N-N bond → Match with III

B. NO₂: The structure of NO₂ has one $$N=O$$ double bond and one N-O bond with a lone electron on nitrogen → Match with I (1 N=O bond)

C. N₂O₅: The structure is O₂N-O-NO₂, where two NO₂ groups are connected through an oxygen atom, forming a N-O-N bond → Match with II

D. N₂O: The structure is N=N=O (or N≡N→O), which contains a N=N or N≡N bond → Match with IV

The correct matching is: A-III, B-I, C-II, D-IV.

We need to find the correct relationships between the unit cell edge length $$a$$ and the radius of the sphere $$r$$ for face-centred cubic (FCC) and body-centred cubic (BCC) structures.

Face-Centred Cubic (FCC) Structure:

In an FCC unit cell, atoms touch along the face diagonal. Consider one face of the cube:

- The face diagonal passes through the center of the face atom and connects two corner atoms.

- Along this diagonal, we have: half of corner atom + full face atom + half of corner atom = $$r + 2r + r = 4r$$.

- The face diagonal of a cube with edge length $$a$$ has length $$a\sqrt{2}$$ (by the Pythagorean theorem: $$\sqrt{a^2 + a^2} = a\sqrt{2}$$).

Therefore:

This gives us: $$2\sqrt{2}r = a$$.

Body-Centred Cubic (BCC) Structure:

In a BCC unit cell, atoms touch along the body diagonal. The body diagonal connects two opposite corners and passes through the center atom.

- Along the body diagonal: half of corner atom + full center atom + half of corner atom = $$r + 2r + r = 4r$$.

- The body diagonal of a cube with edge length $$a$$ has length $$a\sqrt{3}$$ (by the 3D Pythagorean theorem: $$\sqrt{a^2 + a^2 + a^2} = a\sqrt{3}$$).

Therefore:

FCC: $$2\sqrt{2}r = a$$

BCC: $$4r = \sqrt{3}a$$

The correct answer is Option 4: $$2\sqrt{2}r = a$$ and $$4r = \sqrt{3}a$$.

Molecular formula of dimethyldichlorosilane: $$(CH_3)_2SiCl_2$$.

Molar mass of $$(CH_3)_2SiCl_2$$:
C: $$2 \times 12 = 24 \text{ g}$$
H: $$6 \times 1 = 6 \text{ g}$$
Si: $$1 \times 28 = 28 \text{ g}$$
Cl: $$2 \times 35.5 = 71 \text{ g}$$
Total $$= 24 + 6 + 28 + 71 = 129 \text{ g mol}^{-1}$$.

Moles of dimethyldichlorosilane taken:
$$n = \frac{516 \text{ g}}{129 \text{ g mol}^{-1}} = 4.0 \text{ mol}$$.

Hydrolysis-condensation stoichiometry to give the cyclic tetramer $$X$$ (octamethylcyclotetrasiloxane):
$$4\,(CH_3)_2SiCl_2 + 4\,H_2O \rightarrow [(CH_3)_2SiO]_4 + 8\,HCl$$.

Thus $$4$$ moles of starting material produce $$1$$ mole of $$X$$.
Therefore, theoretical moles of $$X$$ formed = $$\frac{4.0}{4} = 1.0 \text{ mol}$$.

Molar mass of $$X = [(CH_3)_2SiO]_4$$:
One $$(CH_3)_2SiO$$ unit: C$$_2$$H$$_6$$SiO $$= 30 + 28 + 16 = 74 \text{ g mol}^{-1}$$.
Tetramer (4 units): $$4 \times 74 = 296 \text{ g mol}^{-1}$$.

Theoretical mass of $$X$$ = $$1.0 \text{ mol} \times 296 \text{ g mol}^{-1} = 296 \text{ g}$$.

Given yield = $$75\%$$, actual mass obtained:
$$m = 0.75 \times 296 \text{ g} = 222 \text{ g}$$.

Hence, the weight of the tetrameric cyclic product $$X$$ obtained is 222 g.

Find the number of bent-shaped molecules from: N$$_3^-$$, NO$$_2$$, I$$_3^-$$, O$$_3$$, SO$$_2$$.

N$$_3^-$$ (Azide ion): Central N has 2 bond pairs, 0 lone pairs. Geometry: linear. Not bent.

NO$$_2$$ (Nitrogen dioxide): Central N has 2 bond pairs + 1 unpaired electron. Geometry: bent (bond angle ~134°). $$\checkmark$$

I$$_3^-$$ (Triiodide ion): Central I has 2 bond pairs + 3 lone pairs. Geometry: linear. Not bent.

O$$_3$$ (Ozone): Central O has 2 bond pairs + 1 lone pair. Geometry: bent (bond angle ~117°). $$\checkmark$$

SO$$_2$$ (Sulfur dioxide): Central S has 2 bond pairs + 1 lone pair. Geometry: bent (bond angle ~119°). $$\checkmark$$

Bent-shaped molecules: NO$$_2$$, O$$_3$$, SO$$_2$$ = 3 molecules.

The correct answer is 3.

We need to count the number of species with square planar geometry from the given list.

1. XeF$$_4$$: Xe has 4 bond pairs and 2 lone pairs (sp$$^3$$d$$^2$$). The geometry is square planar. ✔

2. SF$$_4$$: S has 4 bond pairs and 1 lone pair (sp$$^3$$d). The geometry is see-saw (trigonal bipyramidal with one equatorial lone pair). ✘

3. SiF$$_4$$: Si has 4 bond pairs and 0 lone pairs (sp$$^3$$). The geometry is tetrahedral. ✘

4. BF$$_4^-$$: B has 4 bond pairs and 0 lone pairs (sp$$^3$$). The geometry is tetrahedral. ✘

5. BrF$$_4^-$$: Br has 4 bond pairs and 2 lone pairs (sp$$^3$$d$$^2$$). The geometry is square planar. ✔

6. [Cu(NH$$_3$$)$$_4$$]$$^{2+}$$: Cu$$^{2+}$$ is a d$$^9$$ ion. With NH$$_3$$ as a strong field ligand, it adopts a square planar geometry (due to Jahn-Teller distortion). ✔

7. [FeCl$$_4$$]$$^{2-}$$: Fe$$^{2+}$$ is d$$^6$$ with Cl$$^-$$ (weak field). This is a tetrahedral complex. ✘

8. [PtCl$$_4$$]$$^{2-}$$: Pt$$^{2+}$$ is a d$$^8$$ ion. All d$$^8$$ complexes of heavy transition metals are square planar. ✔

Count: XeF$$_4$$, BrF$$_4^-$$, [Cu(NH$$_3$$)$$_4$$]$$^{2+}$$, [PtCl$$_4$$]$$^{2-}$$ = 4 species.

The correct answer is 4.

Determine the number of neighbouring water molecules hydrogen-bonded to each water molecule in an ice crystal.

Structure of ice.

In ice, water molecules arrange in a tetrahedral crystal structure. Each oxygen atom is at the centre of a tetrahedron.

Hydrogen bonding in ice.

Each water molecule forms four hydrogen bonds with its neighbours:

Two hydrogen bonds where the molecule acts as a donor (through its two O-H bonds).

Two hydrogen bonds where the molecule acts as an acceptor (through its two lone pairs on oxygen).

This gives each water molecule exactly 4 hydrogen-bonded neighbours, consistent with the tetrahedral arrangement in the ice crystal.

The correct answer is 4.

We need to find the number of molecules or ions that do not have an odd number of electrons.

We count the total electrons in each species:

(A) NO$$_2$$: N has 7 electrons, each O has 8 electrons. Total = 7 + 2(8) = 23 (odd).

(B) ICl$$_4^-$$: I has 53 electrons, each Cl has 17 electrons, plus 1 for the negative charge. Total = 53 + 4(17) + 1 = 122 (even). Not odd.

(C) BrF$$_3$$: Br has 35 electrons, each F has 9 electrons. Total = 35 + 3(9) = 62 (even). Not odd.

(D) ClO$$_2$$: Cl has 17 electrons, each O has 8 electrons. Total = 17 + 2(8) = 33 (odd).

(E) NO$$_2^+$$: N has 7 electrons, each O has 8 electrons, minus 1 for the positive charge. Total = 7 + 2(8) - 1 = 22 (even). Not odd.

(F) NO: N has 7 electrons, O has 8 electrons. Total = 15 (odd).

The species that do NOT have an odd number of electrons are: ICl$$_4^-$$, BrF$$_3$$, and NO$$_2^+$$.

The answer is $$\boxed{3}$$.

IF₅: Iodine has 7 valence electrons. With 5 F atoms bonded, the structure is square pyramidal.

Total electron pairs around I = 5 bond pairs + 1 lone pair = 6 (sp³d² hybridization).

Lone pairs on central atom = 1.

IF₇: Iodine has 7 valence electrons. With 7 F atoms bonded, the structure is pentagonal bipyramidal.

Total electron pairs around I = 7 bond pairs + 0 lone pairs (sp³d³ hybridization).

Lone pairs on central atom = 0.

Sum of lone pairs = 1 + 0 = 1.

1. Clo3-

2.XeF4

hence Highest no. of loan pairs 3

To determine the number of molecules containing only two lone pairs of electrons, we analyze the total lone pairs (on both central and terminal atoms) for each given molecule:

Molecule-by-Molecule Analysis:

• H2O: Oxygen has 6 valence electrons. It forms 2 single bonds with Hydrogen atoms, leaving 4 non-bonding electrons.
  👉 Total Lone Pairs = 2 (on Oxygen)
• N2: Each Nitrogen atom shares 3 electrons to form a triple bond (N≡N), leaving 1 lone pair on each atom.
  👉 Total Lone Pairs = 2 (1 on each N)
• CO: Carbon and Oxygen share a triple bond (:C≡O:).
  👉 Total Lone Pairs = 2 (1 on C, 1 on O)
• NO: Nitric oxide is an odd-electron molecule (11 valence electrons total). It contains a double bond, 2 lone pairs (1 on N, 1 on O), and 1 single unpaired electron.
  👉 Total Lone Pairs = 2 (Note: Often excluded in standard textbook counting due to its radical/unpaired electron nature).
• NH3: Nitrogen forms 3 single bonds with Hydrogen, leaving 1 lone pair on the Nitrogen atom.
  👉 Total Lone Pairs = 1
• CO2: Carbon forms double bonds with both Oxygen atoms (O=C=O). Each Oxygen retains 2 lone pairs.
  👉 Total Lone Pairs = 4
• F2: Two Fluorine atoms share a single bond. Each Fluorine atom has 3 lone pairs.
  👉 Total Lone Pairs = 6
• XeF4: Xenon has 2 lone pairs, and each of the 4 Fluorine atoms has 3 lone pairs ($2 + 4 \times 3$).
  👉 Total Lone Pairs = 14

Summary Table:

Conclusion:

Excluding the odd-electron radical ($\text{NO}$), the molecules containing exactly two lone pairs of electrons are H2O, N2, and CO.

Correct Answer: 3

For a cubic crystal structure, the number of atoms per unit cell is given by:

$$Z = \frac{\rho \times a^3 \times N_A}{M}$$

where $$\rho$$ is density, $$a$$ is edge length, $$N_A$$ is Avogadro's number, and $$M$$ is molar mass.

$$\rho = 3.0$$ g cm$$^{-3}$$

$$a = 300$$ pm = $$3 \times 10^{-8}$$ cm

$$M = 12$$ g mol$$^{-1}$$

$$N_A = 6.02 \times 10^{23}$$ mol$$^{-1}$$

Substituting the known values:

$$a^3 = (3 \times 10^{-8})^3 = 27 \times 10^{-24} \text{ cm}^3$$

$$Z = \frac{3.0 \times 27 \times 10^{-24} \times 6.02 \times 10^{23}}{12}$$

$$Z = \frac{3.0 \times 27 \times 0.602}{12}$$

$$Z = \frac{48.762}{12} = 4.06 \approx 4$$

The number of atoms present in one unit cell is $$4$$ (corresponding to an FCC unit cell).

We need to determine how many of the seven crystal systems can have a body-centred unit cell. The seven crystal systems are Cubic, Tetragonal, Orthorhombic, Hexagonal, Rhombohedral (Trigonal), Monoclinic, and Triclinic.

A body-centred unit cell has lattice points at the corners and one lattice point at the centre of the unit cell. In the Cubic system, a body-centred cubic (BCC) structure exists. In the Tetragonal system, a body-centred tetragonal structure exists. In the Orthorhombic system, a body-centred orthorhombic structure exists.

However, in the Hexagonal system only the simple (primitive) hexagonal structure exists, so no body-centred form is possible. Similarly, in the Rhombohedral system only the primitive rhombohedral structure exists, and in the Monoclinic system only simple and end-centred (base-centred) monoclinic structures exist—both without any body-centred form. Finally, in the Triclinic system only the primitive triclinic structure exists, and it has no body-centred form either.

Therefore, the crystal systems that allow body-centred unit cells are Cubic, Tetragonal, and Orthorhombic, giving a total of $$3$$.

We use VSEPR theory to determine the molecular geometry of each molecule in List-II:

XeF$$_4$$ (I): Xe has 8 valence electrons. Four are used for bonding with F atoms, leaving 2 lone pairs. Total electron pairs = 6 (octahedral arrangement). With 4 bonding pairs and 2 lone pairs (in trans positions), the shape is square planar.

SF$$_4$$ (II): S has 6 valence electrons. Four are used for bonding with F atoms, leaving 1 lone pair. Total electron pairs = 5 (trigonal bipyramidal arrangement). With 4 bonding pairs and 1 lone pair, the shape is see-saw (distorted tetrahedron).

ClF$$_3$$ (III): Cl has 7 valence electrons. Three are used for bonding with F atoms, leaving 2 lone pairs. Total electron pairs = 5 (trigonal bipyramidal arrangement). With 3 bonding pairs and 2 lone pairs (both in equatorial positions), the shape is T-shaped.

BF$$_3$$ (IV): B has 3 valence electrons. All three are used for bonding with F atoms, leaving 0 lone pairs. Total electron pairs = 3. The shape is trigonal planar.

Matching: A (T-shaped) → III (ClF$$_3$$), B (Trigonal planar) → IV (BF$$_3$$), C (Square planar) → I (XeF$$_4$$), D (See-saw) → II (SF$$_4$$).

The answer is Option D: A-(III), B-(IV), C-(I), D-(II).

A molecule's bond becomes stronger when an electron is removed if that electron was in an antibonding molecular orbital, since removing it increases the bond order.

Bond order = $$\frac{1}{2}$$(bonding electrons - antibonding electrons)

Let us analyze each molecule:

(A) NO: Total electrons = 15. Electronic configuration ends with one electron in $$\pi^*_{2p}$$ (antibonding). Removing this antibonding electron increases bond order from 2.5 to 3. Bond becomes stronger.

(B) $$N_2$$: Total electrons = 14. All bonding MOs are filled up to $$\sigma_{2p}$$, no antibonding electrons in the highest occupied MO. The last electron is in $$\sigma_{2p}$$ (bonding). Removing it decreases bond order from 3 to 2.5. Bond becomes weaker.

(C) $$O_2$$: Total electrons = 16. Has 2 electrons in $$\pi^*_{2p}$$ (antibonding). Removing one antibonding electron increases bond order from 2 to 2.5. Bond becomes stronger.

(D) $$C_2$$: Total electrons = 12. Last electrons are in $$\pi_{2p}$$ (bonding). Removing a bonding electron decreases bond order from 2 to 1.5. Bond becomes weaker.

(E) $$B_2$$: Total electrons = 10. Last electrons are in $$\pi_{2p}$$ (bonding). Removing a bonding electron decreases bond order from 1 to 0.5. Bond becomes weaker.

Only NO and $$O_2$$ have their highest occupied electrons in antibonding orbitals, so only A and C become stronger upon electron removal.

The correct answer is Option C: A, C only.

Analyze each statement about $$PCl_5$$.

Statement A: In $$PCl_5$$, phosphorus undergoes $$sp^3d$$ hybridization.

This is correct. Phosphorus has 5 bond pairs and uses one 3s, three 3p, and one 3d orbital for $$sp^3d$$ hybridization.

Statement B: The geometry of $$PCl_5$$ is trigonal bipyramidal.

This is correct. With 5 bond pairs and no lone pairs, the geometry is trigonal bipyramidal.

Statement C: $$PCl_5$$ has two axial bonds stronger than three equatorial bonds.

This is incorrect. In a trigonal bipyramidal structure, the axial bonds are longer and weaker than the equatorial bonds. The axial P-Cl bond length is 219 pm while the equatorial P-Cl bond length is 204 pm. This is because axial bonds experience greater repulsion from three equatorial bond pairs at 90°, whereas equatorial bonds have only two axial bonds at 90°.

Statement D: The three equatorial bonds of $$PCl_5$$ lie in a plane.

This is correct. The three equatorial bonds are coplanar, forming angles of 120° with each other.

The incorrect statement is Option C.

The answer is $$\boxed{\text{Option C}}$$.

We need to match each xenon compound with its correct hybridization and geometry.

(A) $$XeO_3$$:

Xenon has 8 valence electrons. In $$XeO_3$$, Xe forms 3 double bonds with oxygen and has 1 lone pair. Total electron domains = 4 (3 bonding + 1 lone pair).

Hybridization: $$sp^3$$, Shape: Pyramidal → matches (II)

(B) $$XeF_2$$:

In $$XeF_2$$, Xe forms 2 bonds with fluorine and has 3 lone pairs. Total electron domains = 5 (2 bonding + 3 lone pairs).

Hybridization: $$sp^3d$$, Shape: Linear → matches (I)

(C) $$XeOF_4$$:

In $$XeOF_4$$, Xe forms 1 double bond with O and 4 bonds with F, plus 1 lone pair. Total electron domains = 6 (5 bonding + 1 lone pair).

Hybridization: $$sp^3d^2$$, Shape: Square pyramidal → matches (IV)

(D) $$XeF_6$$:

In $$XeF_6$$, Xe forms 6 bonds with fluorine and has 1 lone pair. Total electron domains = 7 (6 bonding + 1 lone pair).

Hybridization: $$sp^3d^3$$, Shape: Distorted octahedral → matches (III)

The correct matching is: A - II, B - I, C - IV, D - III

Hence, the correct answer is Option C.

We need to match each compound with its molecular shape using VSEPR theory.

(A) BrF₅

Br has 7 valence electrons. With 5 F atoms bonded, there are 5 bond pairs and 1 lone pair.

Hybridization: $$sp^3d^2$$ (6 electron pairs)

Shape: Square pyramidal (II)

(B) [CrF₆]³⁻

Cr in [CrF₆]³⁻ has an oxidation state of +3. With 6 F atoms bonded, there are 6 bond pairs and 0 lone pairs.

Hybridization: $$sp^3d^2$$ or $$d^2sp^3$$

Shape: Octahedral (IV)

(C) O₃ (Ozone)

The central O atom has 2 bonding regions (one double bond and one coordinate bond to the terminal O atoms) and 1 lone pair.

Hybridization: $$sp^2$$

Shape: Bent (I)

(D) PCl₅

P has 5 valence electrons. With 5 Cl atoms bonded, there are 5 bond pairs and 0 lone pairs.

Hybridization: $$sp^3d$$

Shape: Trigonal bipyramidal (III)

Matching:

• A - II (Square pyramidal)
• B - IV (Octahedral)
• C - I (Bent)
• D - III (Trigonal bipyramidal)

Therefore, the correct answer is Option C.

We need to arrange $$CaF_2$$, $$CaCl_2$$, $$CaBr_2$$, and $$CaI_2$$ in increasing order of covalent character.

Applying Fajans' Rules:

According to Fajans' rules, the covalent character of an ionic compound increases when:

1. The cation has a small size and high charge (greater polarizing power)

2. The anion has a large size and high charge (greater polarizability)

Here, the cation ($$Ca^{2+}$$) is the same in all compounds. The difference lies in the anions.

Comparing the anions by size:

$$F^- < Cl^- < Br^- < I^-$$

As the size of the anion increases, it becomes more easily polarized by the cation. A more polarized anion leads to greater sharing of electron density, which means greater covalent character.

Therefore, the increasing order of covalent character is:

$$CaF_2 < CaCl_2 < CaBr_2 < CaI_2$$

$$A < B < C < D$$

The correct answer is Option B.

We are given the following statements:

Assertion A: Zero orbital overlap is an out of phase overlap.

Reason R: It results due to different orientation/direction of approach of orbitals.

When two atomic orbitals approach each other, they can overlap in three ways: positive overlap (in-phase, leading to bonding), negative overlap (out-of-phase, leading to antibonding), and zero overlap. Zero overlap occurs when the positive and negative contributions to the overlap integral cancel each other out exactly, giving a net overlap of zero.

This zero overlap is indeed classified as an out-of-phase overlap because the regions of constructive overlap are exactly cancelled by regions of destructive overlap. For instance, when a $$p_x$$ orbital approaches an $$s$$ orbital along the y-axis, the positive lobe of the p-orbital overlaps with the s-orbital constructively on one side, while the negative lobe overlaps destructively on the other side, resulting in net zero overlap.

The reason this happens is precisely because of the orientation or direction of approach of the orbitals. When orbitals approach in a direction where their symmetries do not match for effective overlap, the result is zero net overlap. So Reason R correctly explains why zero overlap (an out-of-phase overlap) occurs.

Hence, both Assertion A and Reason R are true, and R is the correct explanation of A. The correct answer is Option A.

Let us classify each oxide:

(A) $$Cl_2O_7$$ is the anhydride of perchloric acid ($$HClO_4$$). Chlorine is a non-metal, and $$Cl_2O_7$$ is an acidic oxide (IV).

(B) $$Na_2O$$, sodium being an alkali metal, reacts with water to form NaOH (a strong base), so $$Na_2O$$ is a basic oxide (II).

(C) $$Al_2O_3$$ is a classic example of an amphoteric oxide (I), reacting with both acids and bases: $$Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O$$ and $$Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O$$.

(D) $$N_2O$$ (nitrous oxide, laughing gas) is a neutral oxide (III) that does not react with acids or bases to form salts.

The correct matching is (A) - (IV), (B) - (II), (C) - (I), (D) - (III).

Hence, the correct answer is Option B.

We need to match items in List-I with those in List-II. For A, $$\Psi_{MO} = \Psi_A - \Psi_B$$ represents subtraction of atomic orbitals; since destructive interference yields an anti-bonding molecular orbital, A corresponds to III.

Next, for B, $$\mu = Q \times r$$ is the formula for dipole moment, where Q is the charge and r is the distance between the charges; hence, B matches with I.

Moving on to C, $$\frac{N_b - N_a}{2}$$ is the formula for bond order, where $$N_b$$ is the number of bonding electrons and $$N_a$$ is the number of anti-bonding electrons; thus, C pairs with IV.

Finally, for D, $$\Psi_{MO} = \Psi_A + \Psi_B$$ corresponds to constructive interference of atomic orbitals, giving a bonding molecular orbital, so D matches with II. From the above matchings, the correct matching is A-III, B-I, C-IV, D-II, and the correct answer is Option C.

An electron deficient molecule is one in which the central atom has fewer than 8 electrons in its valence shell (incomplete octet).

Let us check each molecule:

$$PH_3$$: Phosphorus has 5 valence electrons, forms 3 bonds with H and has 1 lone pair = 8 electrons around P. Not electron deficient.

$$B_2H_6$$: Boron has only 3 valence electrons. In $$B_2H_6$$, each boron is involved in 4 bonds but through 3-centre 2-electron bonds. Each B effectively has only 6 electrons in its valence shell. Electron deficient.

$$CCl_4$$: Carbon has 4 valence electrons, forms 4 bonds = 8 electrons around C. Not electron deficient.

$$NH_3$$: Nitrogen has 5 valence electrons, forms 3 bonds and has 1 lone pair = 8 electrons around N. Not electron deficient.

$$LiH$$: This is an ionic compound ($$Li^+$$ and $$H^-$$). It is not considered electron deficient in this context as $$H^-$$ has a complete 1s shell and $$Li^+$$ has a complete 1s shell.

$$BCl_3$$: Boron has 3 valence electrons, forms 3 bonds = only 6 electrons around B. Electron deficient.

The electron deficient molecules are $$B_2H_6$$ and $$BCl_3$$, giving a count of 2.

The correct answer is Option C.

In case of NH3, N has lone pair due to which intermolecular hydrogen bond strength increased, and methane has simple 4 covalent bond with hydrogen N and hence hydrogen bond would be least and HCN has tripple bond which eventually lead to have high hydrogen bond strength than methane.

We need to identify which pair contains one odd-electron molecule and one expanded octet molecule.

An odd-electron molecule is one that has an odd total number of valence electrons, resulting in an unpaired electron. An expanded octet molecule is one where the central atom has more than 8 electrons in its valence shell, which is possible for elements in Period 3 and beyond that can use d-orbitals.

Let us examine each molecule. NO (nitric oxide): Nitrogen has 5 valence electrons, oxygen has 6, giving a total of 11 (odd). So NO is an odd-electron molecule. $$H_2SO_4$$: Sulfur is the central atom bonded to 4 oxygen atoms (two with OH groups). Sulfur has 6 valence electrons and forms bonds with 4 oxygens. In the structure with two S=O double bonds, sulfur has 12 electrons around it — this is an expanded octet. $$BCl_3$$: Boron has 3 valence electrons and forms 3 bonds with Cl, giving only 6 electrons around B — this is an incomplete (electron-deficient) octet, not an expanded octet. $$SF_6$$: Sulfur has 12 electrons around it (6 bonds) — this is also an expanded octet. But $$SF_6$$ is not paired with an odd-electron molecule in the same option as needed.

Looking at Option B: NO (odd-electron molecule) and $$H_2SO_4$$ (expanded octet molecule). This is the correct pairing.

Hence, the correct answer is Option B.

We need to determine if CH₄, NH₄⁺, and BH₄⁻ are isoelectronic and whether they all have tetrahedral structures. CH₄ has C with 6 electrons and 4 H atoms contributing 4 electrons, for a total of 10 electrons. NH₄⁺ has N with 7 electrons and 4 H atoms contributing 4 electrons minus 1 for the positive charge, also totaling 10 electrons. BH₄⁻ has B with 5 electrons and 4 H atoms contributing 4 electrons plus 1 for the negative charge, again totaling 10 electrons. Therefore, all three species are isoelectronic with 10 electrons.

Each species features a central atom bonded to 4 hydrogen atoms with no lone pairs on the central atom, which corresponds to sp³ hybridisation and a tetrahedral geometry. Consequently, CH₄, NH₄⁺, and BH₄⁻ all adopt tetrahedral structures.

The answer is Option B: They are isoelectronic and all have tetrahedral structures.

We need to analyze two statements about magnetic properties of substances.

Statement I: $$O_2$$, $$Cu^{2+}$$, and $$Fe^{3+}$$ are weakly attracted by magnetic field and are magnetized in the same direction as the magnetic field.

Analysis of Statement I:

Substances that are weakly attracted by a magnetic field and magnetized in the same direction as the field are paramagnetic.

• $$O_2$$ has two unpaired electrons in its antibonding molecular orbitals — it is paramagnetic.
• $$Cu^{2+}$$ has the configuration $$[Ar]3d^9$$ with 1 unpaired electron — it is paramagnetic.
• $$Fe^{3+}$$ has the configuration $$[Ar]3d^5$$ with 5 unpaired electrons — it is paramagnetic.

All three species are paramagnetic. Statement I is correct.

Statement II: $$NaCl$$ and $$H_2O$$ are weakly magnetized in opposite direction to the magnetic field.

Analysis of Statement II:

Substances that are weakly magnetized in the opposite direction to the magnetic field are diamagnetic.

• $$NaCl$$ — all electrons are paired ($$Na^+$$ and $$Cl^-$$ both have noble gas configurations) — it is diamagnetic.
• $$H_2O$$ — all electrons are paired — it is diamagnetic.

Both substances are diamagnetic. Statement II is correct.

The correct answer is Option A: Both Statement I and Statement II are correct.

$$SF_4$$ has sulfur with 6 valence electrons forming 4 bonds with fluorine atoms and 1 lone pair, giving a total of 5 electron pairs. The electron geometry is trigonal bipyramidal.

In a trigonal bipyramidal arrangement, the lone pair occupies the equatorial position to minimize repulsion: in this position it has 2 interactions at 90° (with the two axial bond pairs) and 2 interactions at 120° (with the other equatorial bond pairs), whereas in the axial position it would have 3 interactions at 90° (with all three equatorial bond pairs). Since lone pair - bond pair repulsions at 90° are strongest, minimizing the number of 90° repulsions is preferred.

With the lone pair equatorial, it makes 90° angles only with the 2 axial fluorine atoms, the other 2 equatorial atoms being at 120°, resulting in two lone pair - bond pair repulsions at 90°.

Hence, the correct answer is Option A: Equatorial position with two lone pair - bond pair repulsions at 90°.

We need to find the number of lone pairs on the central atom in each of $$SCl_2$$, $$O_3$$, $$ClF_3$$, and $$SF_6$$.

$$SCl_2$$: Sulfur is the central atom with 6 valence electrons. It forms 2 bonds with Cl atoms, using 2 electrons. The remaining 4 electrons form 2 lone pairs. So the central atom has 2 lone pairs.

$$O_3$$: The central oxygen atom has 6 valence electrons. In ozone, the central oxygen forms a double bond with one oxygen and a single bond (plus coordinate bond character) with the other. Using the Lewis structure, the central oxygen has 2 bonding domains and 1 lone pair. So the central atom has 1 lone pair.

$$ClF_3$$: Chlorine is the central atom with 7 valence electrons. It forms 3 bonds with F atoms, using 3 electrons. The remaining 4 electrons form 2 lone pairs. The shape is T-shaped (from a trigonal bipyramidal arrangement with 2 lone pairs in equatorial positions). So the central atom has 2 lone pairs.

$$SF_6$$: Sulfur is the central atom with 6 valence electrons. It forms 6 bonds with F atoms, using all 6 valence electrons (and expanding the octet). There are 0 lone pairs on sulfur.

The lone pairs are therefore 2, 1, 2, and 0 respectively.

Hence, the correct answer is Option B.

We need to find the correct order of bond orders of $$C_2^{2-}$$, $$N_2^{2-}$$, and $$O_2^{2-}$$.

Since each carbon atom contributes 6 electrons, the total electron count in $$C_2^{2-}$$ is:

$$2(6) + 2 = 14$$ electrons.

Similarly, nitrogen contributes 7 electrons each, so for $$N_2^{2-}$$:

$$2(7) + 2 = 16$$ electrons,

and oxygen contributes 8 electrons each, giving for $$O_2^{2-}$$:

$$2(8) + 2 = 18$$ electrons.

For $$C_2^{2-}$$ (14 electrons), which is isoelectronic with $$N_2$$, the molecular orbital configuration is:

$$(\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\pi_{2p})^4 (\sigma_{2p})^2$$

This configuration gives bonding electrons = $$2 + 2 + 4 + 2 = 10$$ and anti-bonding electrons = $$2 + 2 = 4$$. From the above,

$$\text{Bond Order} = \frac{10 - 4}{2} = \frac{6}{2} = 3$$

Now, for $$N_2^{2-}$$ (16 electrons), which is isoelectronic with $$O_2$$, the configuration becomes:

$$(\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi_{2p}^*)^2$$

Here, bonding electrons = $$2 + 2 + 2 + 4 = 10$$ and anti-bonding electrons = $$2 + 2 + 2 = 6$$. Hence,

$$\text{Bond Order} = \frac{10 - 6}{2} = \frac{4}{2} = 2$$

Finally, for $$O_2^{2-}$$ (18 electrons), the peroxide ion, the configuration is:

$$(\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi_{2p}^*)^4$$

This yields bonding electrons = $$2 + 2 + 2 + 4 = 10$$ and anti-bonding electrons = $$2 + 2 + 4 = 8$$, so

$$\text{Bond Order} = \frac{10 - 8}{2} = \frac{2}{2} = 1$$

From the above calculations, $$C_2^{2-}$$ has bond order 3, $$N_2^{2-}$$ has bond order 2, and $$O_2^{2-}$$ has bond order 1, giving the order:

$$C_2^{2-} > N_2^{2-} > O_2^{2-}$$

Therefore, the correct option is Option A: $$C_2^{2-} > N_2^{2-} > O_2^{2-}$$.

We need to find the number of lone pairs on the central atom and the shape of BrF₃.

Bromine (Br) has 7 valence electrons, and each F atom contributes 1 electron. Total electron pairs around Br = $$\frac{7 + 3 \times 1}{2}$$. Br forms 3 bonds with F atoms, using 3 electrons, leaving 7 − 3 = 4 electrons that form 2 lone pairs. Thus, there are 3 bond pairs and 2 lone pairs, for a total of 5 electron pairs, corresponding to a trigonal bipyramidal electron geometry.

In this arrangement, the 2 lone pairs occupy equatorial positions to minimize repulsion, while the 3 F atoms occupy the 2 axial and 1 equatorial positions. This gives a bent T-shape molecular geometry.

Number of lone pairs on central atom = 2, Shape = bent T-shape. The answer is Option C: 2, bent T-shape.

We need to evaluate the Assertion and Reason statements.

Assertion A: Boron is unable to form $$BF_6^{3-}$$. This statement is TRUE. Since boron has the electronic configuration 1s² 2s² 2p¹, it possesses only four orbitals in its valence shell (one 2s and three 2p). Therefore, its maximum coordination number is 4 and it cannot accommodate six fluorine atoms. For instance, boron can form $$BF_4^-$$ with coordination number 4 but not $$BF_6^{3-}$$ with coordination number 6, as it lacks d-orbitals in the second shell.

Reason R: Size of B is very small. This statement is TRUE because boron is indeed a very small atom.

From the above, the actual reason boron cannot form $$BF_6^{3-}$$ is the absence of d-orbitals in its valence shell, which restricts its coordination number to 4. Although the small size of boron is correct, it does not explain Assertion A; the unavailability of d-orbitals is the proper explanation.

Therefore, the correct answer is Option B: Both A and R are true but R is not the correct explanation of A.

We need to identify the incorrect statement about imperfections (defects) in solids.

Option A: "Schottky defect decreases the density of the substance."

In a Schottky defect, equal numbers of cations and anions are missing from their lattice sites, creating vacancies. Since atoms are missing but the volume remains approximately the same, the density decreases.

This statement is correct.

Option B: "Frenkel defect does not alter the density of the substance."

In a Frenkel defect, an ion leaves its normal lattice site and occupies an interstitial site. No atom leaves the crystal — it just moves to a different position within the same crystal. So the total number of atoms and the volume remain the same, meaning the density does not change.

This statement is correct.

Option C: "Interstitial defect increases the density of the substance."

In an interstitial defect, extra atoms or ions occupy the interstitial spaces (voids) in the crystal. This adds extra mass without significantly changing the volume, so the density increases.

This statement is correct.

Option D: "Vacancy defect increases the density of the substance."

In a vacancy defect, some atoms are missing from their lattice positions. This means the crystal has fewer atoms than the ideal structure while maintaining roughly the same volume. Therefore, the density decreases, not increases.

This statement is incorrect.

Hence, the correct answer is Option D.