JEE Chemical Bonding & Molecular Structure Questions

We need to identify the correct combinations of molecule/ion, shape, and hybridisation.

A. $$IF_3$$ → T-shaped ($$sp^3d$$)

I has 7 valence electrons, bonds with 3 F atoms, leaving 2 lone pairs.

Electron geometry: trigonal bipyramidal ($$sp^3d$$), molecular geometry: T-shaped. CORRECT ✓

B. $$IF_5$$ → Square pyramidal ($$sp^3d^2$$)

I has 7 valence electrons, bonds with 5 F atoms, leaving 1 lone pair.

Electron geometry: octahedral ($$sp^3d^2$$), molecular geometry: square pyramidal. CORRECT ✓

C. $$IF_7$$ → Pentagonal bipyramidal ($$sp^3d^3$$)

I has 7 valence electrons, bonds with 7 F atoms, no lone pairs.

Hybridisation: $$sp^3d^3$$, geometry: pentagonal bipyramidal. CORRECT ✓

D. $$ClO_4^-$$ → Square planar ($$sp^3d$$)

Cl has 7 valence electrons + 1 from charge = 8. It bonds with 4 oxygen atoms.

$$ClO_4^-$$ has a tetrahedral geometry with $$sp^3$$ hybridisation, not square planar with $$sp^3d$$. INCORRECT ✗

The correct combinations are A, B and C only, which corresponds to Option 2.

Two p-block elements X and Y form trifluorides $$XF_3$$ and $$YF_3$$. $$XF_3$$ is a Lewis acid and $$YF_3$$ is a Lewis base. Find their hybridizations.

A trifluoride that is a Lewis acid must be electron-deficient. The classic example is $$BF_3$$ (boron trifluoride), where boron has only 6 electrons in its valence shell and can accept a lone pair.

A trifluoride that is a Lewis base must have a lone pair to donate. The classic example is $$NF_3$$ (nitrogen trifluoride), where nitrogen has a lone pair it can donate.

Boron in $$BF_3$$: 3 bond pairs, 0 lone pairs. Geometry: trigonal planar.

Hybridization: $$sp^2$$

Nitrogen in $$NF_3$$: 3 bond pairs, 1 lone pair. Geometry: trigonal pyramidal.

Hybridization: $$sp^3$$

The hybridizations are $$sp^2$$ and $$sp^3$$ respectively.

The correct answer is Option 2: $$sp^2$$ and $$sp^3$$.

We use the Born-Haber cycle to determine the lattice enthalpy of LiF.

The formation of LiF from its elements is broken down into several processes: sublimation of 

Li(s) to Li(g) with $$\Delta H_{\text{sub}} = +155$$ kJ/mol; 

dissociation of 

$$\frac{1}{2}F_2(g) \rightarrow F(g)$$ with $$\frac{1}{2}\Delta H_{\text{diss}} = \frac{150}{2} = +75$$ kJ/mol; 

ionization of 

Li(g) to Li⁺(g) with $$IE = +520$$ kJ/mol; 

electron gain by 

F(g) to F⁻(g) with $$\Delta H_{eg} = -313$$ kJ/mol; 

lattice formation 

Li⁺(g) + F⁻(g) → LiF(s) with enthalpy $$U$$ to be determined.

By Hess’s law, the standard enthalpy of formation $$\Delta H_f$$ equals the sum of these individual steps, so that 

$$\Delta H_f = \Delta H_{\text{sub}} + \frac{1}{2}\Delta H_{\text{diss}} + IE + \Delta H_{eg} + U$$.

Substituting the known values yields 

$$-594 = 155 + 75 + 520 + (-313) + U$$, which simplifies to 

$$-594 = 155 + 75 + 520 - 313 + U = 437 + U$$ and hence 

$$U = -594 - 437 = -1031 \text{ kJ/mol}$$.

The magnitude of the lattice enthalpy of LiF is $$|U| = \mathbf{1031}$$ kJ/mol.

Intermolecular interaction energies are obtained by applying Coulomb’s law to the various charge distributions that can exist on (or be induced in) molecules - permanent charges, permanent dipoles, induced dipoles, etc. The distance dependence and, in some cases, the temperature dependence of the average energy decide whether a given statement is true or false.

Option A
• Potential energy between two point charges (say $$q_1$$ and $$q_2$$) is $$U_{qq}= \dfrac{q_1q_2}{4\pi\varepsilon_0\,r} \propto \dfrac{1}{r}$$.
• Potential energy between a point charge $$q$$ and a point dipole of moment $$\mu$$ (the dipole oriented so that the interaction is maximum) is $$U_{q\mu}= -\dfrac{q\mu}{4\pi\varepsilon_0\,r^{2}} \propto \dfrac{1}{r^{2}}$$.
Because $$1/r^{2}$$ tends to zero faster than $$1/r$$ when $$r\rightarrow\infty$$, the charge-dipole energy approaches zero more rapidly than the charge-charge energy. Option A claims the opposite, hence it is false.

Option B
For two freely rotating permanent dipoles, the orientation-averaged (thermal average) interaction energy has Keesom form
$$\langle U_{\mu\mu}\rangle = -\dfrac{2\mu_1^{2}\mu_2^{2}}{3(4\pi\varepsilon_0)^{2}k_{\mathrm B}T\,r^{6}} \propto -\dfrac{1}{T\,r^{6}}.$$
Thus the energy falls off as $$1/r^{6}$$, not $$1/r^{3}$$. Therefore Option B is false.

Option C
Dipole-induced dipole interaction (Debye interaction): a permanent dipole $$\mu$$ induces a dipole $$\alpha \,E$$ in a neighbouring polarisable molecule (where $$\alpha$$ is its polarisability and $$E$$ the electric field of the permanent dipole). The orientation-averaged energy is
$$\langle U_{\mu\text{-ind}}\rangle = -\dfrac{\mu^{2}\alpha}{(4\pi\varepsilon_0)^{2}\,2\,r^{6}},$$
which contains no temperature term. Hence it is independent of temperature. Option C is true.

Option D
Even molecules with no permanent dipole can attract each other through London dispersion forces. A momentary (instantaneous) dipole in one molecule induces a dipole in the other, giving an average interaction energy
$$\langle U_{\text{disp}}\rangle = -\dfrac{3h\nu\,\alpha_1\alpha_2}{4(4\pi\varepsilon_0)^{2}\,r^{6}},$$
which is always attractive. Hence non-polar molecules do attract one another despite lacking permanent dipoles. Option D is true.

Therefore the correct statements are:
Option C (dipole-induced dipole energy is temperature-independent) and
Option D (non-polar molecules attract each other via dispersion forces).

Final answer: Option C and Option D.

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$$^{-1}$$.

• Experimental Dipole Moment ($$\mu_{\text{exp}}$$):
  $$\mu_{\text{exp}} = 1.2\text{ D} = 1.2 \times 10^{-18}\text{ esu}\cdot\text{cm}$$
• Calculated Dipole Moment ($$\mu_{\text{cal}}$$ for $$100\%$$ ionic character):
  Assuming a full electronic charge ($$e = 4.8 \times 10^{-10}\text{ esu}$$) and bond distance ($$d = 1\text{ \AA} = 10^{-8}\text{ cm}$$):
  $$\mu_{\text{cal}} = q \times d = (4.8 \times 10^{-10}\text{ esu}) \times (10^{-8}\text{ cm}) = 4.8 \times 10^{-18}\text{ esu}\cdot\text{cm}$$
• Fractional Charge:
  $$\text{Fractional charge} = \frac{\mu_{\text{exp}}}{\mu_{\text{cal}}} = \frac{1.2 \times 10^{-18}}{4.8 \times 10^{-18}} = \frac{1.2}{4.8} = 0.25 = 2.5 \times 10^{-1}$$

Conclusion:

Comparing our result with the required format ($$\text{Blank} \times 10^{-1}\text{ esu}$$), the numerical value for the blank space is $$2.5$$.

Answer: 2.5

1. Complete Octet (8 Electrons)

These central atoms follow the standard rule by sharing electrons until they reach a stable count of 8.

• $$\text{CO}_2$$ (Carbon): Carbon has 4 valence electrons and forms two double bonds with Oxygen. $$4 + 4 = 8$$ electrons.
• $$\text{CH}_4$$ (Carbon): Carbon forms 4 single bonds with Hydrogen. $$4 \times 2 = 8$$ electrons.
• $$\text{SiF}_4$$ (Silicon): Like Carbon, Silicon is in Group 14 and forms 4 single bonds. $$4 \times 2 = 8$$ electrons.
• $$\text{C}_2\text{H}_6$$ (Carbons): Each Carbon forms 3 bonds with Hydrogen and 1 bond with the other Carbon. $$4 \times 2 = 8$$ electrons.
• $$\text{CHCl}_3$$ & $$\text{CBr}_4$$ (Carbon): Carbon forms 4 single bonds with halogens. $$4 \times 2 = 8$$ electrons.

2. Exceptions: Incomplete Octet (< 8 Electrons)

These central atoms are stable even though they lack a full shell of 8 electrons.

• $$\text{BF}_3$$ (Boron): Boron is in Group 13. It shares its 3 valence electrons to form 3 single bonds.
  • Total: 6 electrons.
• $$\text{BeF}_2$$ (Beryllium): Beryllium is in Group 2. It shares its 2 valence electrons to form 2 single bonds.
  • Total: 4 electrons.

3. Exceptions: Expanded Octet (> 8 Electrons)

Elements in Period 3 or below have empty $$d$$-orbitals that allow them to hold more than 8 electrons.

• $$\text{H}_2\text{SO}_4$$ (Sulfur): Sulfur forms 6 bonds (two double bonds with Oxygen, two single bonds with -OH).
  • Total: 12 electrons.
• $$\text{PCl}_5$$ (Phosphorus): Phosphorus forms 5 single bonds with Chlorine.
  • Total: 10 electrons.
• $$\text{ClO}_2$$ (Chlorine): Chlorine is bonded to two oxygens and retains lone pairs/unpaired electrons. In many Lewis representations, it exceeds 8 to reduce formal charge.

4. Exceptions: Odd-Electron Molecules

• $$\text{NO}_2$$ (Nitrogen): Nitrogen has 5 valence electrons. Because 5 is an odd number, it is impossible to pair every electron.
  • Total: Nitrogen ends up with 7 valence electrons (one unpaired electron). This makes it a free radical.

We need to find the number of molecules with bond order 2 from: $$C_2, O_2, Be_2, Li_2, Ne_2, N_2, He_2$$.

Using Molecular Orbital Theory, the bond order is calculated as $$\text{Bond Order} = \frac{N_b - N_a}{2}$$ where $$N_b$$ is the number of bonding electrons and $$N_a$$ is the number of antibonding electrons.

Since $$C_2$$ (12 electrons) has configuration $$(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\pi_{2p})^4$$, the bond order is $$(8-4)/2 = 2$$. ✓

Now $$O_2$$ (16 electrons) has configuration $$(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p})^2(\pi_{2p})^4(\pi^*_{2p})^2$$, so the bond order is $$(10-6)/2 = 2$$. ✓

For $$Be_2$$ (8 electrons), the bond order is $$(4-4)/2 = 0$$. ✗

For $$Li_2$$ (6 electrons), the bond order is $$(4-2)/2 = 1$$. ✗

For $$Ne_2$$ (20 electrons), the bond order is $$(10-10)/2 = 0$$. ✗

For $$N_2$$ (14 electrons), the bond order is $$(10-4)/2 = 3$$. ✗

For $$He_2$$ (4 electrons), the bond order is $$(2-2)/2 = 0$$. ✗

Therefore the molecules with bond order 2 are $$C_2$$ and $$O_2$$.

This gives the answer $$\boxed{2}$$.

We need to find the total number of species from the given list that have exactly one unpaired electron. We use Molecular Orbital Theory (MOT) to determine the electronic configurations.

Recall: The molecular orbital filling order for diatomic molecules is:

For $$Z \leq 7$$ (like N, C): $$\sigma_{1s}, \sigma^*_{1s}, \sigma_{2s}, \sigma^*_{2s}, \pi_{2p_x} = \pi_{2p_y}, \sigma_{2p_z}, \pi^*_{2p_x} = \pi^*_{2p_y}, \sigma^*_{2p_z}$$

For $$Z > 7$$ (like O): $$\sigma_{1s}, \sigma^*_{1s}, \sigma_{2s}, \sigma^*_{2s}, \sigma_{2p_z}, \pi_{2p_x} = \pi_{2p_y}, \pi^*_{2p_x} = \pi^*_{2p_y}, \sigma^*_{2p_z}$$

1. $$N_2$$ (14 electrons):

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

All electrons are paired. Unpaired electrons = 0. Not selected.

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

Configuration: ...$$(\sigma_{2p_z})^2 (\pi_{2p})^4 (\pi^*_{2p})^2$$

The two electrons in $$\pi^*_{2p}$$ occupy one each of $$\pi^*_{2p_x}$$ and $$\pi^*_{2p_y}$$ by Hund's rule. Unpaired electrons = 2. Not selected.

3. $$C_2^-$$ (13 electrons):

$$C_2$$ has 12 electrons; adding 1 gives 13. Configuration: $$(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p})^4 (\sigma_{2p_z})^1$$

Unpaired electrons = 1. Selected.

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

$$O_2$$ has 16 electrons; adding 1 gives 17. Configuration: ...$$(\pi^*_{2p})^3$$

Three electrons in two degenerate $$\pi^*$$ orbitals: 2 in one, 1 in the other. Unpaired electrons = 1. Selected.

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

$$O_2$$ has 16 electrons; adding 2 gives 18. Configuration: ...$$(\pi^*_{2p})^4$$

Both $$\pi^*$$ orbitals are fully filled. Unpaired electrons = 0. Not selected.

6. $$H_2^+$$ (1 electron):

Configuration: $$(\sigma_{1s})^1$$

One electron in the bonding orbital. Unpaired electrons = 1. Selected.

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

C has 6 electrons, N has 7, plus 1 for the negative charge = 14 electrons. This is isoelectronic with $$N_2$$.

Configuration is the same as $$N_2$$: all electrons paired. Unpaired electrons = 0. Not selected.

8. $$He_2^+$$ (3 electrons):

Two He atoms contribute 4 electrons; removing 1 gives 3. Configuration: $$(\sigma_{1s})^2 (\sigma^*_{1s})^1$$

Unpaired electrons = 1. Selected.

Summary: Species with exactly 1 unpaired electron: $$C_2^-$$, $$O_2^-$$, $$H_2^+$$, $$He_2^+$$.

The total count is $$\boxed{4}$$.

We need to count the species with the central atom using sp$$^2$$ hybrid orbitals from the given list.

sp$$^2$$ hybridisation occurs when the central atom has 3 electron domains (regions of electron density), giving a trigonal planar arrangement (or bent if lone pairs are present).

Analyze each species:

(i) $$NH_3$$: N has 3 bond pairs + 1 lone pair = 4 electron domains → sp$$^3$$. ✗

(ii) $$SO_2$$: S has 2 bond pairs + 1 lone pair = 3 electron domains → sp$$^2$$. ✓

(iii) $$SiO_2$$: Each Si is bonded to 4 O atoms in a tetrahedral arrangement → sp$$^3$$. ✗

(iv) $$BeCl_2$$: Be has 2 bond pairs + 0 lone pairs = 2 electron domains → sp. ✗

(v) $$C_2H_4$$: Each C has 3 electron domains (2 C-H + 1 C=C) → sp$$^2$$. ✓

(vi) $$C_2H_2$$: Each C has 2 electron domains (1 C-H + 1 C≡C) → sp. ✗

(vii) $$BCl_3$$: B has 3 bond pairs + 0 lone pairs = 3 electron domains → sp$$^2$$. ✓

(viii) $$HCHO$$: C has 3 electron domains (2 C-H + 1 C=O) → sp$$^2$$. ✓

(ix) $$C_6H_6$$: Each C has 3 electron domains → sp$$^2$$. ✓

(x) $$BF_3$$: B has 3 bond pairs + 0 lone pairs → sp$$^2$$. ✓

(xi) $$C_2H_4Cl_2$$: Each C has 4 electron domains → sp$$^3$$. ✗

Count: SO$$_2$$, C$$_2$$H$$_4$$, BCl$$_3$$, HCHO, C$$_6$$H$$_6$$, BF$$_3$$ = 6 species.

The answer is 6.

Analysis of Statement I

Statement I states that dipole moment is a vector quantity and is represented by a small arrow with its tail on the negative centre and head pointing towards the positive centre.

Although dipole moment is indeed a vector quantity, the direction of the dipole moment vector is stated incorrectly.

By convention, the dipole moment vector points from the positive centre to the negative centre, i.e., the tail is at the positive end and the head points towards the negative end.

Hence, Statement I is incorrect.

Analysis of Statement II

Statement II states that the crossed arrow of the dipole moment symbolizes the direction of the shift of charges in the molecule.

This is correct. The crossed arrow represents the direction of electron density shift, pointing from the partial positive charge $$\delta^+$$ towards the partial negative charge $$\delta^-$$.

Hence, Statement II is correct.

Conclusion

• Statement I is incorrect.
• Statement II is correct.

Therefore, the correct answer is

$$\boxed{\text{Option B: Statement I is incorrect but Statement II is correct.}}$$

Explanation

For molecules up to $$N_2$$, the molecular orbital energy order is

For $$O_2$$ and beyond, the order is

For $$N_2$$,

The HOMO is $$\sigma_{2p_z}$$ and it contains two paired electrons.

Therefore, the number of unpaired electrons in the HOMO is

$$0$$

For $$N_2^+$$, one electron is removed from the HOMO.

$$N_2^+=(\sigma_{2p_z})^1$$

The HOMO contains one unpaired electron.

Therefore,

$$1$$

unpaired electron is present in the HOMO.

For $$O_2$$,

$$(\pi_{2p_x}^{})^1(\pi_{2p_y}^{})^1$$

The HOMO consists of the degenerate $$\pi^{*}$$ orbitals. According to Hund's rule, the two electrons occupy separate orbitals.

Therefore, the number of unpaired electrons in the HOMO is

$$2$$

For $$O_2^+$$, one electron is removed from the $$\pi^{*}$$ level.

Only one unpaired electron remains in the HOMO.

Therefore, the number of unpaired electrons is

$$1$$

Hence, for

$$N_2,\ N_2^+,\ O_2,\ O_2^+$$

the numbers of unpaired electrons in the HOMO are

$$0,\ 1,\ 2,\ 1$$

Hence, the correct answer is

$$0,\ 1,\ 2,\ 1$$

The bond order is calculated by:

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

where,

$$N_b$$ is number of bonding electrons

$$N_a$$ is number of antibonding electrons

For, $$O_2^{2-}$$

Total electrons = 8+8+2 = 18 electrons

MO Configuration:
$$\sigma_{1s}^2 \ \sigma_{1s}^{*2} \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ \sigma_{2p_z}^2 \ (\pi_{2p_x}^2 = \pi_{2p_y}^2) \ (\pi_{2p_x}^{*2} = \pi_{2p_y}^{*2})$$
$$N_{b} = 10$$, $$N_{a} = 8$$
$$\text{Bond Order} = \frac{10 - 8}{2} = 1$$

For, $$CO$$
Total Electrons: $6 + 8 = 14\text{ electrons}$
MO Configuration: $$\sigma_{1s}^2 \ \sigma_{1s}^{*2} \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ (\pi_{2p_x}^2 = \pi_{2p_y}^2) \ \sigma_{2p_z}^2$$
$$N_b = 10$$, $$N_a = 4$$
$$\text{Bond Order} = \frac{10 - 4}{2} = 3$$

For, $$\text{NO}^+$$
Total Electrons: $7 + 8 - 1 = 14\text{ electrons}$
MO Configuration: Since it is isoelectronic with $\text{CO}$, it shares the same molecular orbital distribution.
$$N_b = 10$$, $$N_a = 4$$
$$\text{Bond Order} = \frac{10 - 4}{2} = 3$$

Hence, Option A is correct.

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 decreases in the order: Ice > Liquid water > Impure water.

Ice has maximum hydrogen bonding with each water molecule forming 4 H-bonds in a crystalline tetrahedral structure. In liquid water, some bonds are broken. In impure water, dissolved impurities disrupt the hydrogen bonding network further.

Therefore, the decreasing order is B > A > C.

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.

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 arrange the bond dissociation energies of E-H bonds in hydrides of group 16 elements (H$$_2$$E).

The group 16 elements in order are O, S, Se, and Te. As we move down the group, the atomic size increases, the E-H bond length increases, and the orbital overlap between E and H decreases. So the E-H bond dissociation energy decreases down the group.

The bond dissociation energies are approximately: O-H at 463 kJ/mol, S-H at 363 kJ/mol, Se-H at 276 kJ/mol, and Te-H at 266 kJ/mol.

Hence, the correct order is

which corresponds to $$A > B > C > D$$.

So, the answer is $$A > B > C > D$$.

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.

Statement I: SO₂ and H₂O both have V-shaped (bent) structures.

SO₂: Central S atom has 2 bond pairs + 1 lone pair → V-shaped ✓

H₂O: Central O atom has 2 bond pairs + 2 lone pairs → V-shaped ✓

Statement I is correct.

Statement II: The bond angle of SO₂ is less than that of H₂O.

SO₂ bond angle ≈ 119° (sp² hybridization with one lone pair)

H₂O bond angle ≈ 104.5° (sp³ hybridization with two lone pairs)

Actually, SO₂ has a LARGER bond angle than H₂O. So Statement II is incorrect.

Statement I is correct but Statement II is incorrect.

This matches option 4.

We need to find the process where the bond order increases AND the paramagnetic character changes to diamagnetic.

Option A: $$\text{O}_2 \to \text{O}_2^+$$

$$\text{O}_2$$: Bond order = 2, paramagnetic (2 unpaired electrons).

$$\text{O}_2^+$$: Bond order = 2.5, paramagnetic (1 unpaired electron).

Bond order increases but remains paramagnetic. Does not satisfy both conditions.

Option B: $$\text{O}_2 \to \text{O}_2^{2-}$$

$$\text{O}_2$$: Bond order = 2, paramagnetic.

$$\text{O}_2^{2-}$$: Bond order = 1, diamagnetic.

Bond order decreases. Does not satisfy.

Option C: $$\text{NO} \to \text{NO}^+$$

$$\text{NO}$$: Bond order = 2.5, paramagnetic (1 unpaired electron in $$\pi^*_{2p}$$).

$$\text{NO}^+$$: Bond order = 3, diamagnetic (the unpaired antibonding electron is removed).

Bond order increases (2.5 $$\to$$ 3) and paramagnetic $$\to$$ diamagnetic. Both conditions satisfied.

Option D: $$\text{N}_2 \to \text{N}_2^+$$

$$\text{N}_2$$: Bond order = 3, diamagnetic.

$$\text{N}_2^+$$: Bond order = 2.5, paramagnetic (1 unpaired electron).

Bond order decreases and diamagnetic $$\to$$ paramagnetic. Opposite of required.

The correct answer is Option C: $$\text{NO} \to \text{NO}^+$$.

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: $$NO^+$$.

We need to determine the physical state and geometry of ClF$$_5$$ at room temperature.

Chlorine has 7 valence electrons. In ClF$$_5$$, Cl forms 5 bonds with F atoms using 5 electrons, leaving 1 lone pair. The total electron pairs around Cl are 5 bond pairs + 1 lone pair = 6 pairs. So the electron geometry is octahedral, and with one lone pair, the molecular geometry is square pyramidal.

Now for the physical state, ClF$$_5$$ is a colourless liquid at room temperature. Interhalogen compounds with higher number of atoms tend to be liquids or solids.

Hence, the correct answer is Option D: Colourless liquid with square pyramidal geometry.

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 $$4$$.

The dipole moment $$\mu$$ of a hetero-atomic bond is the product of the magnitude of charge separation $$q$$ and the bond length $$r$$, i.e. $$\mu = q\,r$$ (expressed in Debye, D).
A larger electronegativity difference increases $$q$$, while a longer bond increases $$r$$. Hence, to compare dipole moments we must examine the combined effect of both factors.

For the tetrahedral molecules $$CH_3X$$ (where $$X = F, Cl, Br, I$$) the three nearly coplanar C-H bonds have dipoles that almost cancel one another. The net molecular dipole therefore points almost entirely along the C-X bond. Consequently,

$$\mu(CH_3X) \approx \mu(\text{C-}X) = q_{\,\text{C-}X}\; r_{\,\text{C-}X} \quad -(1)$$

Let us analyse both factors moving down the halogen group:

Electronegativity difference (controls $$q$$)
$$\Delta\chi(\text{C-}X):\; F \gt Cl \gt Br \gt I$$
Thus $$q_{\,\text{C-}F} \gt q_{\,\text{C-}Cl} \gt q_{\,\text{C-}Br} \gt q_{\,\text{C-}I}$$.

Bond length (controls $$r$$)
C-F (1.39 Å) < C-Cl (1.78 Å) < C-Br (1.94 Å) < C-I (2.14 Å)

Putting the two trends together:

• From F → Cl the decrease in $$q$$ is modest, whereas the increase in $$r$$ is substantial (≈0.39 Å).
• From Cl → Br → I both factors ($$q$$ decreases and $$r$$ increases only slightly) lower the product $$q\,r$$.

Experimental dipole moments corroborate this assessment:

• $$\mu(CH_3F) = 1.85\;\text{D}$$
• $$\mu(CH_3Cl) = 1.87\;\text{D}$$
• $$\mu(CH_3Br) = 1.79\;\text{D}$$
• $$\mu(CH_3I) = 1.64\;\text{D}$$

The slight fall in charge separation from F to Cl is more than compensated by the longer C-Cl bond, giving $$CH_3Cl$$ the highest dipole moment of the four molecules.

Hence, among the compounds listed, the molecule with the greatest dipole moment is $$\mathbf{CH_3Cl}$$.

Option A which is: $$CH_3Cl$$

The van der Waals constant $$a$$ measures the magnitude of intermolecular attractive forces.

Stronger intermolecular forces $$\Rightarrow$$ larger value of $$a$$.

Given gases:

$$Ar,\ CH_4,\ H_2O,\ C_6H_6$$

Analysis:

For argon: $$Ar$$

Being a noble gas, it has very weak intermolecular forces.

Hence it has the smallest value of $$a$$.

For methane: $$CH_4$$

Non-polar molecule with weak van der Waals forces, but stronger than Ar.

For water: $$H_2O$$

Strong hydrogen bonding is present, so intermolecular attraction is much larger.

For benzene: $$C_6H_6$$

Large molecular size and highly polarizable electron cloud produce very strong intermolecular attractions.

Hence benzene has the largest value of $$a$$.

Therefore the increasing order of van der Waals constant $$a$$ is:

$${Ar < CH_4 < H_2O < C_6H_6}$$

Hence, Option D is correct

Assertion A: Isotopes of hydrogen (H, D, T) have almost the same chemical properties because they have the same electronic configuration. However, their rates of reactions differ due to the kinetic isotope effect (heavier isotopes react slower). This is correct.

Reason R: Isotopes of hydrogen have different bond dissociation enthalpies. The H-H bond has a lower dissociation energy than D-D bond because of the difference in zero-point energies (heavier isotopes have lower zero-point energy, making their bonds slightly stronger). This is correct.

The difference in bond dissociation enthalpy directly causes the difference in reaction rates (kinetic isotope effect), so R correctly explains A.

Both A and R are correct and R is the correct explanation of A.

This matches option 2.

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.

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.

The formula $$A_{0.95}O$$ indicates a metal deficiency defect. For every oxygen ion present in the crystal, the number of metal ions is only $$0.95$$ times the number expected in a perfectly stoichiometric lattice.

The lattice shown contains 20 oxide ions $$O^{2-}$$.

Therefore, the expected number of metal ions is

$$20\times 0.95=19$$

Since there are 20 cation lattice sites but only 19 cations, the structure must contain one cation vacancy.

The total negative charge contributed by the oxide ions is

$$20\times (-2)=-40$$

For electrical neutrality, the total positive charge must be

$$+40$$

Let the number of $$A^{3+}$$ ions be $$x$$ and the number of $$A^{2+}$$ ions be $$y$$.

Since the total number of cations is 19,

$$x+y=19$$

Since the total positive charge must be 40,

$$3x+2y=40$$

Substituting

$$y=19-x$$

into the charge balance equation,

$$3x+2(19-x)=40$$

$$3x+38-2x=40$$

$$x=2$$

Therefore,

$$y=19-2=17$$

Thus, the lattice must contain

$$2\ A^{3+}\ ions$$

and

$$17\ A^{2+}\ ions$$

along with

$$1$$

cation vacancy.

The given diagram shows exactly one vacant cation site, seventeen cations of one type, and two cations of the higher oxidation state. Hence, it satisfies both the stoichiometric requirement

$$A_{0.95}O$$

and the condition of electrical neutrality.

Hence, the given diagram correctly represents the $$A_{0.95}O$$ lattice.

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: $$\text{X}_{1.5}\text{Y}$$.

For every cubic lattice the density, edge length and packing efficiency depend on three basic data: number of atoms per unit cell ($$Z$$), atomic radius ($$r$$) and molar mass ($$M$$).
We tabulate the standard relations first:

Face-centred cubic (fcc, metal $$x$$):
  • $$Z_x = 4$$     • contact along a face diagonal ⇒ $$\sqrt{2}\,L_x = 4\,r_x$$ ⇒ $$r_x = \dfrac{L_x}{2\sqrt{2}}$$.
Body-centred cubic (bcc, metal $$y$$):
  • $$Z_y = 2$$     • contact along a body diagonal ⇒ $$\sqrt{3}\,L_y = 4\,r_y$$ ⇒ $$r_y = \dfrac{\sqrt{3}}{4}\,L_y$$.
Simple cubic (sc, metal $$z$$):
  • $$Z_z = 1$$     • contact along an edge ⇒ $$L_z = 2\,r_z$$ ⇒ $$r_z = \dfrac{L_z}{2}$$.

The question gives the inter-relations

$$r_z = \frac{\sqrt{3}}{2}\,r_y,\qquad r_y = \frac{8}{\sqrt{3}}\,r_x,\qquad M_z = \frac{3}{2}\,M_y,\qquad M_z = 3\,M_x.$$

From the last two equalities

$$M_x = \frac{1}{3}M_z,\qquad M_y = \frac{2}{3}M_z \;\;\Longrightarrow\;\; \frac{M_x}{M_y} = \frac12.$$

1. Comparison of edge lengths

Using $$r_y = \dfrac{8}{\sqrt{3}}\,r_x$$ in $$L_y = \dfrac{4\,r_y}{\sqrt{3}}$$:
$$L_y = \frac{4}{\sqrt{3}}\left(\frac{8}{\sqrt{3}}\,r_x\right) = \frac{32}{3}\,r_x.$$
Using $$r_z = \frac{\sqrt{3}}{2}\,r_y = \frac{\sqrt{3}}{2}\left(\frac{8}{\sqrt{3}}\,r_x\right)=4\,r_x$$ in $$L_z=2\,r_z$$:
$$L_z = 8\,r_x.$$
For fcc, $$L_x = 2\sqrt{2}\,r_x.$$

Numerically (in units of $$r_x$$):
$$L_x = 2.828\,r_x,\qquad L_y = 10.667\,r_x,\qquad L_z = 8\,r_x.$$ Therefore $$L_y \gt L_z \gt L_x$$ and hence Option B ($$L_y \gt L_z$$) is correct, while Option C ($$L_x \gt L_y$$) is incorrect.

2. Packing efficiencies

Standard values are: fcc = 74 %, bcc = 68 %, sc = 52.4 %.
Thus$$\text{fcc (}x\text{)} \gt \text{bcc (}y\text{)} \gt \text{sc (}z\text{)},$$so Option A is correct.

3. Density comparison of $$x$$ and $$y$$

For any cubic lattice $$\rho = \dfrac{Z\,M}{N_A\,L^3}.$$ Taking the ratio $$ \frac{\rho_x}{\rho_y} = \frac{Z_x M_x L_y^{\,3}}{Z_y M_y L_x^{\,3}} = \left(\frac{4}{2}\right) \left(\frac{M_x}{M_y}\right) \left(\frac{L_y}{L_x}\right)^{3} = 2\left(\frac12\right)\left(\frac{L_y}{L_x}\right)^{3} = \left(\frac{L_y}{L_x}\right)^{3}. $$ We already have $$\dfrac{L_y}{L_x} = \dfrac{32/3}{2\sqrt{2}} = \dfrac{16}{3\sqrt{2}} \approx 3.772,$$ so $$\frac{\rho_x}{\rho_y} \approx 3.772^{3} \approx 53.7 \gt 1.$$ Hence $$\rho_x \gt \rho_y,$$ making Option D correct.

Final verdict:
Correct statements: Option A, Option B, Option D.

Greenhouse gases are those that absorb and emit infrared radiation, contributing to the greenhouse effect.

A. Water vapour: The most abundant greenhouse gas. Yes.

B. Ozone (O₃): A greenhouse gas that absorbs infrared radiation. Yes.

C. I₂ (Iodine): Molecular iodine is not a greenhouse gas. It does not significantly absorb infrared radiation in the atmosphere. No.

D. Molecular hydrogen (H₂): H₂ is a homonuclear diatomic molecule with no permanent dipole moment, so it does not absorb infrared radiation. No.

The greenhouse gases are A and B only.

This matches option 3.

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 $$4r = \sqrt{3}a$$.

First write the rule for deciding hybridisation.
For a given central atom, the hybridisation is decided by its steric number (SN).
$$\text{SN}= \text{(number of } \sigma \text{-bonds)} + \text{(number of lone pairs)}$$

• SN = 4  →  $$sp^3$$ hybridisation (tetrahedral electron geometry).
• SN = 5  →  $$sp^3d$$ hybridisation.
• SN = 6  →  $$sp^3d^2$$ hybridisation, etc.

Analyse each species one by one.

Case 1: $$[I_3]^+$$
Central iodine forms two $$I-I$$ $$\sigma$$-bonds and possesses two lone pairs (confirmed by a total valence-electron count of 20).
SN = $$2 + 2 = 4$$ ⇒ $$sp^3$$.

Case 2: $$[SiO_4]^{4-}$$
Silicon is bonded to four $$O^{2-}$$ ions through four $$\sigma$$-bonds and has no lone pair.
SN = $$4 + 0 = 4$$ ⇒ $$sp^3$$.

Case 3: $$SO_2Cl_2$$ (thionyl chloride)
Central sulphur has two $$S-Cl$$ single bonds and two $$S=O$$ double bonds. Each double bond contributes one $$\sigma$$-bond.
Number of $$\sigma$$-bonds = 4, lone pairs = 0.
SN = $$4 + 0 = 4$$ ⇒ $$sp^3$$.

Case 4: $$XeF_2$$
Xe has two $$\sigma$$-bonds and three lone pairs.
SN = $$2 + 3 = 5$$ ⇒ $$sp^3d$$ (not $$sp^3$$).

Case 5: $$SF_4$$
Four $$\sigma$$-bonds + one lone pair → SN = 5 ⇒ $$sp^3d$$ (not $$sp^3$$).

Case 6: $$ClF_3$$
Three $$\sigma$$-bonds + two lone pairs → SN = 5 ⇒ $$sp^3d$$ (not $$sp^3$$).

Case 7: $$Ni(CO)_4$$
Nickel forms four metal-carbon $$\sigma$$-bonds with no lone pair involved in the hybrid orbitals.
SN = $$4 + 0 = 4$$ ⇒ $$sp^3$$ (tetrahedral complex).

Case 8: $$XeO_2F_2$$
Xe has four $$\sigma$$-bonds (two Xe-O, two Xe-F) and one lone pair.
SN = $$4 + 1 = 5$$ ⇒ $$sp^3d$$ (not $$sp^3$$).

Case 9: $$[PtCl_4]^{2-}$$
Square-planar $$[PtCl_4]^{2-}$$ uses $$dsp^2$$ hybridisation (not $$sp^3$$).

Case 10: $$XeF_4$$
Four $$\sigma$$-bonds + two lone pairs → SN = 6 ⇒ $$sp^3d^2$$ (not $$sp^3$$).

Case 11: $$SOCl_2$$ (thionyl chloride)
Sulphur has two $$S-Cl$$ $$\sigma$$-bonds, one $$S=O$$ double bond (one $$\sigma$$), and one lone pair.
SN = $$3 + 1 = 4$$ ⇒ $$sp^3$$.

Species with $$sp^3$$ hybridised central atom:
$$[I_3]^+,\; [SiO_4]^{4-},\; SO_2Cl_2,\; Ni(CO)_4,\; SOCl_2$$

Total number = 5.

Hence the required answer is 5.

Explanation

To determine the total number of $$\pi$$-bonds, first draw the structures of the given oxoacids of sulphur.

Each $$S=O$$ double bond contains one $$\pi$$-bond.

Therefore, the total number of $$\pi$$-bonds is equal to the total number of $$S=O$$ double bonds present in the molecules.

For peroxodisulphuric acid, $$H_2S_2O_8$$

The structure contains:

• One $$S-O-O-S$$ peroxide linkage
• Two $$S-OH$$ bonds
• Four $$S=O$$ double bonds

Hence, the number of $$\pi$$-bonds in $$H_2S_2O_8$$ is

$$4$$

For pyrosulphuric acid, $$H_2S_2O_7$$

The structure contains:

• One $$S-O-S$$ bridge
• Two $$S-OH$$ bonds
• Four $$S=O$$ double bonds

Hence, the number of $$\pi$$-bonds in $$H_2S_2O_7$$ is

$$4$$

Therefore, the total number of $$\pi$$-bonds present in both molecules is

$$4+4=8$$

Hence, the correct answer is

$$8$$

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.

Ozone $$(O_3)$$ contains three oxygen atoms. Since oxygen belongs to Group 16, each oxygen atom contributes 6 valence electrons.

Total valence electrons in ozone:

$$3 \times 6 = 18 \text{ electrons}$$

The Lewis structure of ozone contains one double bond and one single bond between the oxygen atoms:

$$O = O - O \leftrightarrow O - O = O$$

The bonding electrons account for:

$$1 \text{ double bond} + 1 \text{ single bond} = 4 + 2 = 6 \text{ electrons}$$

Therefore, the remaining non-bonding electrons are:

$$18 - 6 = 12 \text{ electrons}$$

These 12 electrons are present as lone pairs.

For the double-bonded terminal oxygen:

$$4 \text{ non-bonding electrons} = 2 \text{ lone pairs}$$

For the central oxygen:

$$2 \text{ non-bonding electrons} = 1 \text{ lone pair}$$

For the single-bonded terminal oxygen:

$$6 \text{ non-bonding electrons} = 3 \text{ lone pairs}$$

Hence, total lone pairs in ozone are:

$$2 + 1 + 3 = 6$$

Therefore, the total number of lone pairs present in $$O_3$$ is

$$\boxed{6}$$

We need to determine which of the given species have a linear shape.

1. XeF₂: Xe has 8 valence electrons (including 2 bond pairs + 3 lone pairs). Geometry: trigonal bipyramidal electron pair, linear molecular shape. ✓

2. I₃⁺: Total valence electrons = 3(7) − 1 = 20. Central I: 2 bond pairs + 2 lone pairs → bent. ✗

3. C₃O₂ (carbon suboxide): Structure: O=C=C=C=O. All carbons are sp hybridized. Linear. ✓

4. I₃⁻: Total valence electrons = 3(7) + 1 = 22. Central I: 2 bond pairs + 3 lone pairs → sp³d, linear. ✓

5. CO₂: O=C=O. Carbon is sp hybridized. Linear. ✓

6. SO₂: S has 2 bond pairs + 1 lone pair → bent. ✗

7. BeCl₂: Be has 2 bond pairs, no lone pairs → sp, linear. ✓

8. BCl₂⁻: B has 3 + 1 = 4 valence electrons. With 2 bonds: 2 bond pairs + 1 lone pair → sp², bent. ✗

Linear species: XeF₂, C₃O₂, I₃⁻, CO₂, BeCl₂ = 5

The correct answer is 5.

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. $$ClO^{3-}$$

$$\text{}$$

2.$$XeF_4$$

hence Highest no. of loan pairs is 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

Given species:

$$PF_5,\; BrF_4^{-},\; IF_5,\; BrF_5,\; XeOF_4,\; ICl_4^{-}$$

For $$PF_5$$,

Central atom phosphorus has: $$5 \text{ bond pairs and }0 \text{ lone pair}$$

Geometry: $$\text{Trigonal bipyramidal}$$

Hence, not square pyramidal.

For $$BrF_4^{-}$$,

Total electron pairs around Br: $$4 \text{ bond pairs } + 2 \text{ lone pairs}$$

Geometry: $$\text{Square planar}$$

Hence, not square pyramidal.

For $$IF_5$$,

Total electron pairs around I:$$5 \text{ bond pairs } + 1 \text{ lone pair}$$

Electronic geometry: $$\text{Octahedral}$$

Shape:$$\text{Square pyramidal}$$

For $$BrF_5$$,

Total electron pairs around Br: $$5 \text{ bond pairs } + 1 \text{ lone pair}$$

Shape: $$\text{Square pyramidal}$$

For $$XeOF_4$$,

Total electron pairs around Xe: $$5 \text{ bond pairs } + 1 \text{ lone pair}$$

Shape: $$\text{Square pyramidal}$$

For $$ICl_4^{-}$$,

Total electron pairs around I: $$4 \text{ bond pairs } + 2 \text{ lone pairs}$$

Shape: $$\text{Square planar}$$

Species having square pyramidal structure:

$$IF_5,\; BrF_5,\; XeOF_4$$

Therefore, number of such species:$${3}$$

We use the density formula for a crystal unit cell:

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

where $$Z$$ is the number of formula units per unit cell, $$M$$ is the molar mass, $$N_A$$ is Avogadro's number, and $$a$$ is the edge length.

We have $$d = 4.0$$ g cm$$^{-3}$$, $$M = 56 + 16 = 72$$ g mol$$^{-1}$$ (for FeO), $$a = 5.0$$ $$\text{\AA}$$ $$= 5.0 \times 10^{-8}$$ cm, and $$N_A = 6.0 \times 10^{23}$$ mol$$^{-1}$$.

Now, $$a^3 = (5.0 \times 10^{-8})^3 = 125 \times 10^{-24} = 1.25 \times 10^{-22}$$ cm$$^3$$.

So $$N_A \times a^3 = 6.0 \times 10^{23} \times 1.25 \times 10^{-22} = 75$$.

From the density formula:

$$Z = \frac{d \times N_A \times a^3}{M} = \frac{4.0 \times 75}{72} = \frac{300}{72} \approx 4.17$$

Rounding to the nearest integer gives $$Z = 4$$.

So, the answer is $$4$$.

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 $$3$$.

Pyrophosphoric acid is: $$H_4P_2O_7$$

Its structure contains:

- Two phosphorus atoms
- One $$P-O-P$$ linkage
- Four $$P-OH$$ bonds
- Two $$P=O$$ bonds

Structure analysis:

Sigma bonds:

From four $$O-H$$ bonds: $$4$$

From four $$P-O$$ single bonds attached to $$OH$$: $$4$$

From one $$P-O-P$$ bridge: $$2$$

(one sigma bond from each $$P-O$$)

From two $$P=O$$ bonds: $$2 \sigma$$

Total sigma bonds: $$4+4+2+2=12$$

Pi bonds:
Each $$P=O$$ contributes one $$\pi$$ bond.

Thus: $$2 \pi \text{ bonds}$$

Therefore the ratio of sigma and $$\pi$$ bonds is: $${12:2}$$ or $${6:1}$$

In a hexagonal close-packed (hcp) structure, for every atom there are 2 tetrahedral voids and 1 octahedral void, giving a total of 3 voids per atom.

The number of atoms in 0.02 mol is $$N = 0.02 \times N_A = 0.02 \times 6.02 \times 10^{23} = 1.204 \times 10^{22}$$.

The total number of voids is $$3N = 3 \times 1.204 \times 10^{22} = 3.612 \times 10^{22} = 36.12 \times 10^{21}$$.

Rounding to the nearest integer, the total number of voids is $$\boxed{36} \times 10^{21}$$.

Sodium crystallizes in a BCC lattice with unit cell edge length $$a = 4$$ Angstrom. We need to find the radius of the sodium atom.

First, recall the relationship between atomic radius and edge length in BCC.

In a body-centred cubic (BCC) structure, atoms touch along the body diagonal. The body diagonal of a cube with edge length $$a$$ has length $$\sqrt{3}a$$. Along this diagonal, there are 4 atomic radii (the corner atom contributes $$r$$, the body-centre atom contributes $$2r$$, and the opposite corner contributes $$r$$):

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

Next, solve for $$r$$.

$$ r = \frac{\sqrt{3}a}{4} = \frac{\sqrt{3} \times 4}{4} = \sqrt{3} \approx 1.732\;\text{Angstrom} $$

Now, express in the required form.

$$ r = 1.732\;\text{A} = 17.32 \times 10^{-1}\;\text{A} $$

Rounding to the nearest integer: $$17 \times 10^{-1}$$ A.

The answer is 17.

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.

The strength of a hydrogen bond depends on the electronegativity of the atoms involved (typically $$N$$, $$O$$, or $$F)$$.

• $$CH_{4}$$ (Methane): Carbon is not electronegative enough to form hydrogen bonds. It only exhibits weak London dispersion forces. Therefore, it has the lowest strength (zero).
• $$HCN$$ (Hydrogen Cyanide): While it contains Nitrogen, the hydrogen is bonded to Carbon. $$HCN$$ can form very weak hydrogen bonds, but it primarily interacts through dipole-dipole forces.
• $$NH_{3}$$ (Ammonia): Hydrogen is directly bonded to a highly electronegative Nitrogen atom with a lone pair. This allows for significant intermolecular hydrogen bonding, making it the strongest in this list.

Hence, Option C is correct

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 arrange LiCl, NaCl, KCl, and CsCl in decreasing order of covalent character.

Key Concept: Fajans' Rules

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

• The cation is smaller (higher polarizing power)
• The cation has higher charge
• The anion is larger (more polarizable)

Analysis:

All four compounds have the same anion (Cl$$^-$$) and cations with the same charge (+1). The difference lies in the size of the cation:

$$\text{Li}^+ < \text{Na}^+ < \text{K}^+ < \text{Cs}^+$$

A smaller cation has greater polarizing power, leading to greater covalent character. Therefore, the decreasing order of covalent character is:

$$\text{LiCl} > \text{NaCl} > \text{KCl} > \text{CsCl}$$

$$\text{(A)} > \text{(B)} > \text{(C)} > \text{(D)}$$

Hence, the correct answer is Option C ((A) > (B) > (C) > (D)).

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.

Step 1: Formation of Product 'A'

When boron trifluoride ($$\text{BF}_3$$) reacts with sodium hydride ($$\text{NaH}$$) at around $$450\text{ K}$$, it undergoes a reduction reaction to form diborane ($$\text{B}_2\text{H}_6$$) along with sodium fluoride ($$\text{NaF}$$):

$$2\text{BF}_3 + 6\text{NaH} \xrightarrow{450\text{K}} \text{B}_2\text{H}_6 + 6\text{NaF}$$

Therefore, product 'A' is diborane, $$\text{B}_2\text{H}_6$$.

Step 2: Formation of Product 'B'

Diborane ($$\text{B}_2\text{H}_6$$) is an electron-deficient molecule that easily undergoes cleavage by Lewis bases. When treated with a relatively small, bulky tertiary amine like trimethylamine ($$\text{NMe}_3$$), it undergoes symmetric cleavage to yield a stable Lewis acid-base adduct:

$$\text{B}_2\text{H}_6 + 2\text{NMe}_3 \rightarrow 2\text{BH}_3\cdot\text{NMe}_3$$

Therefore, product 'B' is the monomeric adduct, $$\text{BH}_3\cdot\text{NMe}_3$$.

Hybridization and Geometry Around Boron in Product 'B':

In the adduct $$\text{BH}_3\cdot\text{NMe}_3$$, the boron atom accepts a lone pair of electrons from the nitrogen atom of the trimethylamine molecule via a coordinate (dative) bond.

• The central Boron atom forms 3 covalent bonds with hydrogen atoms ($$\text{B--H}$$) and 1 coordinate bond with the nitrogen atom ($$\text{B}\leftarrow\text{N}$$).
• This gives a total of 4 bond pairs and 0 lone pairs on the boron atom.
• According to the VSEPR theory, a steric number of 4 corresponds to $$\text{sp}^3$$ hybridization.

The spatial arrangement corresponding to $$\text{sp}^3$$ hybridization with zero lone pairs is completely symmetrical, giving it a tetrahedral geometry.

Conclusion:

The coordination of the lone pair from trimethylamine changes the geometry around the boron atom from trigonal planar (in $$\text{BH}_3$$ fragments) to tetrahedral in the final complex.

Answer: Option B — tetrahedral

We need to identify which gas does NOT contribute to the greenhouse effect.

The greenhouse effect occurs when certain gases in the atmosphere absorb and re-emit infrared radiation, trapping heat. A gas must be able to change its dipole moment during molecular vibration to absorb infrared radiation.

Option A: $$H_2O$$ vapour - Water vapour is the most abundant greenhouse gas. It has a bent molecular geometry and a permanent dipole moment, allowing it to absorb infrared radiation effectively. This IS a greenhouse gas.

Option B: $$O_3$$ - Ozone has a bent structure and absorbs infrared radiation. This IS a greenhouse gas.

Option C: $$N_2$$ - Nitrogen is a homonuclear diatomic molecule. It has no permanent dipole moment and its symmetric stretching vibration does not produce a change in dipole moment. Therefore, it does NOT absorb infrared radiation and is NOT a greenhouse gas.

Option D: $$CH_4$$ - Methane has vibrational modes that change the dipole moment, making it a potent greenhouse gas. This IS a greenhouse gas.

The correct answer is Option C: $$N_2$$.

Dioxygen difluoride, $$O_2F_2$$, is an extremely powerful oxidising as well as fluorinating agent. When it is allowed to react with xenon, it oxidises xenon to the +6 oxidation state while simultaneously supplying four fluorine atoms. The compound obtained is the oxy-fluoride $$XeOF_4$$ (called xenon oxytetrafluoride).

A convenient balanced equation for its formation is

$$2\,Xe + 4\,O_2F_2 \;\longrightarrow\; 2\,XeOF_4 + 3\,O_2$$

Let the product $$P$$ be $$XeOF_4$$. To find the number of moles of $$HF$$ produced on complete hydrolysis of one mole of $$P$$, write its hydrolysis reaction:

$$XeOF_4 + 2\,H_2O \;\longrightarrow\; XeO_3 + 4\,HF$$

Checking the atom balance: Xe (1), O (1 + 2 = 3 on both sides), F (4), and H (4) are all balanced, so the equation is correct.

The equation shows that 1 mole of $$XeOF_4$$ yields 4 moles of $$HF$$ on complete hydrolysis.

Hence, the required number of moles of $$HF$$ produced per mole of $$P$$ is 4.

Silver nitrate decomposes on heating according to

$$2\,AgNO_3 \;\xrightarrow{\;\Delta\;}\; 2\,Ag \;+\; 2\,NO_2 \;+\; O_2$$

The two gaseous products are $$NO_2$$ and $$O_2$$. Both are paramagnetic, but the number of unpaired electrons differs:

• $$NO_2$$ has one unpaired electron (odd total electron count).
• $$O_2$$ has two unpaired electrons (the well-known dioxygen case).

Hence, the gas with the larger number of unpaired electrons is $$O_2$$.

Next, write the molecular orbital (MO) configuration of $$O_2$$ (total 16 electrons):

$$\sigma_{1s}^{2}\;\sigma_{1s}^{*\,2}\; \sigma_{2s}^{2}\;\sigma_{2s}^{*\,2}\; \sigma_{2p_z}^{2}\; \pi_{2p_x}^{2}\;\pi_{2p_y}^{2}\; \pi_{2p_x}^{*\,1}\;\pi_{2p_y}^{*\,1}$$

Antibonding orbitals are those marked with an asterisk ($$^*$$). Counting the electrons in these orbitals:

$$\sigma_{1s}^{*} : 2,\; \sigma_{2s}^{*} : 2,\; \pi_{2p_x}^{*} : 1,\; \pi_{2p_y}^{*} : 1$$

Total electrons in antibonding MOs

$$= 2 + 2 + 1 + 1 = 6$$

Therefore, the required number is 6.

We need to match each species with its correct hybridization.

(A) $$[PtCl_4]^{2-}$$:

Pt is in +2 oxidation state ($$Pt^{2+}$$) with electronic configuration $$[Xe]5d^8$$.

$$Pt^{2+}$$ is a $$d^8$$ system. With 4 $$Cl^-$$ ligands (weak field for most metals, but Pt²⁺ being a 5d metal with large crystal field splitting always forms square planar complexes).

Square planar geometry corresponds to $$dsp^2$$ hybridization → matches (III).

(B) $$BrF_5$$:

Br has 7 valence electrons. With 5 bond pairs and 1 lone pair (total 6 electron pairs around Br).

6 electron pairs require $$sp^3d^2$$ hybridization → square pyramidal geometry → matches (IV).

(C) $$PCl_5$$:

P has 5 valence electrons. With 5 bond pairs and 0 lone pairs (total 5 electron pairs around P).

5 electron pairs require $$sp^3d$$ hybridization → trigonal bipyramidal geometry → matches (I).

(D) $$[Co(NH_3)_6]^{3+}$$:

Co is in +3 oxidation state ($$Co^{3+}$$) with electronic configuration $$[Ar]3d^6$$.

$$NH_3$$ is a strong field ligand, so all 6 electrons in $$3d$$ pair up, leaving two empty $$3d$$ orbitals available.

The hybridization is $$d^2sp^3$$ (inner orbital complex) → octahedral geometry → matches (II).

Final matching:

(A) - (III), (B) - (IV), (C) - (I), (D) - (II)

The correct answer is Option B.

We need to count the number of diamagnetic species among: $$N_2, N_2^+, N_2^-, N_2^{2-}, O_2, O_2^+, O_2^-, O_2^{2-}$$.

A species is considered diamagnetic if all electrons are paired (no unpaired electrons).

Considering first the nitrogen species, we use the MO order $$\sigma_{1s}, \sigma^*_{1s}, \sigma_{2s}, \sigma^*_{2s}, \pi_{2p_x} = \pi_{2p_y}, \sigma_{2p_z}, \pi^*_{2p_x} = \pi^*_{2p_y}, \sigma^*_{2p_z}$$. In $$N_2$$ with 14 electrons the configuration is $$...\sigma_{2s}^2, {\sigma^*_{2s}}^{2}, \pi_{2p_x}^2 = \pi_{2p_y}^2, \sigma_{2p_z}^2$$, so all electrons are paired and the molecule is diamagnetic.

In $$N_2^+$$ there are 13 electrons, corresponding to the removal of one electron from $$\sigma_{2p_z}$$, which leaves one unpaired electron and therefore makes it paramagnetic.

Adding an electron to $$N_2$$ gives $$N_2^-$$ with 15 electrons, where the extra electron occupies a $$\pi^*_{2p}$$ orbital, resulting in one unpaired electron and paramagnetism.

In $$N_2^{2-}$$ the two additional electrons occupy $${\pi^*_{2p_x}}^{1}$$ and $${\pi^*_{2p_y}}^{1}$$, producing two unpaired electrons and a paramagnetic species.

Turning to the oxygen species, we adopt the MO order $$\sigma_{2s}, \sigma^*_{2s}, \sigma_{2p_z}, \pi_{2p_x} = \pi_{2p_y}, \pi^*_{2p_x} = \pi^*_{2p_y}, \sigma^*_{2p_z}$$. For $$O_2$$ with 16 electrons the configuration is $$...\sigma_{2p_z}^2, \pi_{2p_x}^2 = \pi_{2p_y}^2, {\pi^*_{2p_x}}^{1} = {\pi^*_{2p_y}}^{1}$$, which has two unpaired electrons and is therefore paramagnetic.

Removing one electron from $$O_2$$ yields $$O_2^+$$ with 15 electrons, leaving one electron in a $$\pi^*$$ orbital and one unpaired electron; this species is paramagnetic.

In $$O_2^-$$ the 17th electron enters $$\pi^*_{2p_x}$$, resulting in the configuration $${\pi^*_{2p_x}}^{2}, {\pi^*_{2p_y}}^{1}$$ and one unpaired electron, so it is paramagnetic.

The peroxide ion $$O_2^{2-}$$ has 18 electrons filling both $$\pi^*_{2p_x}$$ and $$\pi^*_{2p_y}$$ orbitals with paired electrons, making it diamagnetic.

Among the eight species considered, only $$N_2$$ and $$O_2^{2-}$$ have all electrons paired, giving two diamagnetic species.

Hence, the answer is 2.

To determine whether an oxide is paramagnetic or diamagnetic, we check for the presence of unpaired electrons.

Species containing unpaired electrons are paramagnetic.

Species with all electrons paired are diamagnetic.

$$\mathrm{Na_2O}$$

Contains:

$$\mathrm{Na^+ \ and\ O^{2-}}$$

The oxide ion has completely filled orbitals:

$$\mathrm{O^{2-} : 1s^2\ 2s^2\ 2p^6}$$

All electrons are paired.

Hence:

$$\mathrm{Na_2O \ is\ diamagnetic}$$

$$\mathrm{KO_2}$$

Contains the superoxide ion:

$$\mathrm{O_2^-}$$

The superoxide ion possesses one unpaired electron in the antibonding $$\mathrm{\pi^*}$$ orbital.

Hence:

$$\mathrm{KO_2 \ is\ paramagnetic}$$

$$\mathrm{NO_2}$$

Total valence electrons:

$$\mathrm{= 5 + 2(6) = 17}$$

Odd number of electrons ⇒ one unpaired electron.

Hence:

$$\mathrm{NO_2 \ is\ paramagnetic}$$

$$\mathrm{ClO_2}$$

Total valence electrons:

$$\mathrm{= 7 + 2(6) = 19}$$

Odd number of electrons ⇒ one unpaired electron.

Hence:

$$\mathrm{ClO_2 \ is\ paramagnetic}$$

$$\mathrm{NO}$$

Total valence electrons:

$$\mathrm{= 5 + 6 = 11}$$

Odd number of electrons ⇒ one unpaired electron.

Hence:

$$\mathrm{NO \ is\ paramagnetic}$$

$$\mathrm{N_2O}$$

Total valence electrons:

$$\mathrm{= 2(5) + 6 = 16}$$

All electrons are paired.

Hence:

$$\mathrm{N_2O \ is\ diamagnetic}$$

$$\mathrm{SO_2}$$

Total valence electrons:

$$\mathrm{= 6 + 2(6) = 18}$$

All electrons are paired.

Hence:

$$\mathrm{SO_2 \ is\ diamagnetic}$$

$$\mathrm{Cl_2O}$$

Total valence electrons:

$$\mathrm{= 2(7) + 6 = 20}$$

All electrons are paired.

Hence:

$$\mathrm{Cl_2O \ is\ diamagnetic}$$

Paramagnetic species:

$$\mathrm{KO_2,\ NO_2,\ ClO_2,\ NO}$$

Total number of paramagnetic oxides:

$$\boxed{\mathrm{4}}$$

We need to find how many species/ions from the given list have identical bond order using Molecular Orbital theory.

Bond order = $$\frac{N_b - N_a}{2}$$ where $$N_b$$ is the number of bonding electrons and $$N_a$$ is the number of antibonding electrons.

For $$CN^-$$, the total number of electrons is 6 + 7 + 1 = 14, and its molecular orbital configuration is $$(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\pi_{2p})^4(\sigma_{2p})^2$$. Substituting into the bond order formula gives $$\frac{10 - 4}{2} = 3$$.

Since $$NO^+$$ is isoelectronic with $$CN^-$$ (7 + 8 − 1 = 14 electrons), it has the same MO configuration and therefore the same bond order: $$\frac{10 - 4}{2} = 3$$.

For $$O_2$$, which has 16 electrons (8 + 8), the MO configuration is $$(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p})^2(\pi_{2p})^4(\pi^*_{2p})^2$$, yielding a bond order of $$\frac{10 - 6}{2} = 2$$.

When one electron is removed to form $$O_2^+$$ (16 − 1 = 15 electrons), the configuration becomes $$(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p})^2(\pi_{2p})^4(\pi^*_{2p})^1$$, giving a bond order of $$\frac{10 - 5}{2} = 2.5$$.

Removing two electrons to obtain $$O_2^{2+}$$ (16 − 2 = 14 electrons) leads to the configuration $$(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p})^2(\pi_{2p})^4$$, and the bond order is $$\frac{10 - 4}{2} = 3$$.

From the above calculations, $$CN^-$$, $$NO^+$$, and $$O_2^{2+}$$ each have a bond order of 3. Therefore, the number of species with identical bond order is 3.

We need to determine how many of the given species — $$PF_5$$, $$BrF_5$$, $$PCl_3$$, $$SF_6$$, $$ICl_4^-$$, $$ClF_3$$, and $$IF_5$$ — have $$sp^3d^2$$ hybridisation.

We determine the hybridisation of each species by counting the number of electron domains (bond pairs + lone pairs) around the central atom using the steric number.

$$PF_5$$: Phosphorus has 5 bond pairs and 0 lone pairs, giving a steric number of 5. This corresponds to $$sp^3d$$ hybridisation (trigonal bipyramidal geometry). Not $$sp^3d^2$$.

$$BrF_5$$: Bromine has 7 valence electrons; 5 are used for bonding with fluorine, and 1 lone pair remains. The steric number is 5 + 1 = 6, which corresponds to $$sp^3d^2$$ hybridisation (square pyramidal geometry). This is $$sp^3d^2$$.

$$PCl_3$$: Phosphorus has 3 bond pairs and 1 lone pair, giving a steric number of 4. This is $$sp^3$$ hybridisation (trigonal pyramidal). Not $$sp^3d^2$$.

$$SF_6$$: Sulphur has 6 bond pairs and 0 lone pairs, giving a steric number of 6. This is $$sp^3d^2$$ hybridisation (octahedral geometry). This is $$sp^3d^2$$.

$$ICl_4^-$$: Iodine has 7 valence electrons plus 1 from the negative charge = 8. Four are used for bonding with Cl, leaving 2 lone pairs. The steric number is 4 + 2 = 6, which is $$sp^3d^2$$ hybridisation (square planar geometry). This is $$sp^3d^2$$.

$$ClF_3$$: Chlorine has 7 valence electrons; 3 are used for bonding with fluorine, leaving 2 lone pairs. The steric number is 3 + 2 = 5, which is $$sp^3d$$ hybridisation (T-shaped geometry). Not $$sp^3d^2$$.

$$IF_5$$: Iodine has 7 valence electrons; 5 are used for bonding with fluorine, leaving 1 lone pair. The steric number is 5 + 1 = 6, which is $$sp^3d^2$$ hybridisation (square pyramidal geometry). This is $$sp^3d^2$$.

The species with $$sp^3d^2$$ hybridisation are: $$BrF_5$$, $$SF_6$$, $$ICl_4^-$$, and $$IF_5$$. That gives us a count of 4.

Hence, the correct answer is 4.

We need to determine which of the given molecules have a non-zero net dipole moment.

Analyzing each molecule:

1. $$NH_3$$ (Ammonia): Trigonal pyramidal geometry. The three N-H bond dipoles do not cancel out due to the lone pair on nitrogen. Non-zero dipole moment.

2. HCl (Hydrogen chloride): Linear diatomic molecule with different atoms. The bond is polar and there is no symmetry to cancel the dipole. Non-zero dipole moment.

3. $$H_2O$$ (Water): Bent geometry (bond angle ~104.5°). The two O-H bond dipoles do not cancel due to the bent shape. Non-zero dipole moment.

4. $$BeF_2$$ (Beryllium fluoride): Linear geometry (180°). The two Be-F bond dipoles are equal and opposite, so they cancel out. Zero dipole moment.

5. $$BH_3$$ (Borane): Trigonal planar geometry (120°). The three B-H bond dipoles cancel out due to symmetry. Zero dipole moment.

6. $$CCl_4$$ (Carbon tetrachloride): Tetrahedral geometry. The four C-Cl bond dipoles cancel out due to perfect symmetry. Zero dipole moment.

Count: $$NH_3$$, HCl, and $$H_2O$$ have non-zero dipole moments.

The number of molecules with non-zero net dipole moment is 3.

We need to find the value of $$y$$ in the hybridization $$sp^x d^y$$ of phosphorus in $$PF_5$$. Phosphorus (P) is the central atom, with 5 valence electrons, bonded to 5 fluorine atoms and having no lone pairs.

The steric number (the number of bonded atoms plus lone pairs) for phosphorus is 5 + 0 = 5. For a steric number of 5, the hybridization is $$sp^3d$$ and the molecular geometry is trigonal bipyramidal.

In $$sp^3d$$ hybridization, one s-orbital, three p-orbitals, and one d-orbital combine to form five hybrid orbitals, each forming a sigma bond with a fluorine atom.

This hybridization can be denoted as $$sp^3d^1$$, which corresponds to $$sp^x d^y$$ with $$x = 3$$ and $$y = 1$$. Therefore, the correct answer is $$y = 1$$.

We need to find the number of molecules/ions with non-planar structure from the given list.

Let us analyze each species:

1. $$NO_3^-$$

Nitrogen has 3 bond pairs and 0 lone pairs. Hybridization: $$sp^2$$. Shape: Trigonal planar. $$\Rightarrow$$ Planar

2. $$H_2O_2$$

Hydrogen peroxide has an open book structure. The two O-H bonds are in different planes due to the lone pairs on each oxygen. Shape: Non-planar (open book/skewed). $$\Rightarrow$$ Non-planar

3. $$BF_3$$

Boron has 3 bond pairs and 0 lone pairs. Hybridization: $$sp^2$$. Shape: Trigonal planar. $$\Rightarrow$$ Planar

4. $$PCl_3$$

Phosphorus has 3 bond pairs and 1 lone pair. Hybridization: $$sp^3$$. Shape: Trigonal pyramidal. $$\Rightarrow$$ Non-planar

5. $$XeF_4$$

Xenon has 4 bond pairs and 2 lone pairs. Hybridization: $$sp^3d^2$$. Shape: Square planar. $$\Rightarrow$$ Planar

6. $$SF_4$$

Sulphur has 4 bond pairs and 1 lone pair. Hybridization: $$sp^3d$$. Shape: See-saw. $$\Rightarrow$$ Non-planar

7. $$XeO_3$$

Xenon has 3 bond pairs and 1 lone pair. Hybridization: $$sp^3$$. Shape: Trigonal pyramidal. $$\Rightarrow$$ Non-planar

8. $$PH_4^+$$

Phosphorus has 4 bond pairs and 0 lone pairs. Hybridization: $$sp^3$$. Shape: Tetrahedral. $$\Rightarrow$$ Non-planar

9. $$SO_3$$

Sulphur has 3 bond pairs and 0 lone pairs. Hybridization: $$sp^2$$. Shape: Trigonal planar. $$\Rightarrow$$ Planar

10. $$[Al(OH)_4]^-$$

Aluminium has 4 bond pairs and 0 lone pairs. Hybridization: $$sp^3$$. Shape: Tetrahedral. $$\Rightarrow$$ Non-planar

Counting non-planar species:

$$H_2O_2$$, $$PCl_3$$, $$SF_4$$, $$XeO_3$$, $$PH_4^+$$, $$[Al(OH)_4]^-$$ = 6

Hence, the answer is 6.

We need to find the number of species with exactly two lone pairs of electrons on the central atom among $$SF_4$$, $$XeF_4$$, $$CF_4$$, and $$H_2O$$.

Sulfur has 6 valence electrons.

In $$SF_4$$, sulfur forms 4 bonds with fluorine atoms.

Lone pairs on S = $$\frac{6 - 4}{2} = 1$$ lone pair.

The geometry of $$SF_4$$ is see-saw (sp$$^3$$d hybridization) and it has 1 lone pair.

Xenon has 8 valence electrons.

In $$XeF_4$$, xenon forms 4 bonds with fluorine atoms.

Lone pairs on Xe = $$\frac{8 - 4}{2} = 2$$ lone pairs.

The geometry of $$XeF_4$$ is square planar (sp$$^3$$d$$^2$$ hybridization) and it has 2 lone pairs.

Carbon has 4 valence electrons.

In $$CF_4$$, carbon forms 4 bonds with fluorine atoms.

Lone pairs on C = $$\frac{4 - 4}{2} = 0$$ lone pairs.

The geometry of $$CF_4$$ is tetrahedral (sp$$^3$$ hybridization) and it has 0 lone pairs.

Oxygen has 6 valence electrons.

In $$H_2O$$, oxygen forms 2 bonds with hydrogen atoms.

Lone pairs on O = $$\frac{6 - 2}{2} = 2$$ lone pairs.

The geometry of $$H_2O$$ is bent (sp$$^3$$ hybridization) and it has 2 lone pairs.

The species with two lone pairs are $$XeF_4$$ and $$H_2O$$.

Therefore, the number of species with two lone pairs is 2.

A paramagnetic species has one or more unpaired electrons. We use Molecular Orbital Theory (MOT) to determine the electronic configuration and count unpaired electrons for each species.

Key MOT orbital filling order:

For $$B_2$$, $$C_2$$, $$N_2$$ (up to 14 electrons): $$\sigma_{1s} \lt \sigma^*_{1s} \lt \sigma_{2s} \lt \sigma^*_{2s} \lt \pi_{2p_x} = \pi_{2p_y} \lt \sigma_{2p_z} \lt \pi^*_{2p_x} = \pi^*_{2p_y} \lt \sigma^*_{2p_z}$$

For $$O_2$$ and heavier (15+ electrons): $$\sigma_{1s} \lt \sigma^*_{1s} \lt \sigma_{2s} \lt \sigma^*_{2s} \lt \sigma_{2p_z} \lt \pi_{2p_x} = \pi_{2p_y} \lt \pi^*_{2p_x} = \pi^*_{2p_y} \lt \sigma^*_{2p_z}$$

1. $$B_2$$ (total electrons = 5 + 5 = 10):

Configuration: $$\sigma_{1s}^2\, {\sigma^*_{1s}}^{2}\, \sigma_{2s}^2\, {\sigma^*_{2s}}^{2}\, \pi_{2p_x}^1\, \pi_{2p_y}^1$$

The last two electrons enter the two degenerate $$\pi_{2p}$$ orbitals with parallel spins (Hund's rule), giving 2 unpaired electrons.

Bond order = $$\f\frac{6-4}{2} = 1$$. Paramagnetic.

2. $$Li_2$$ (total electrons = 3 + 3 = 6):

Configuration: $$\sigma_{1s}^2\, {\sigma^*_{1s}}^{2}\, \sigma_{2s}^2$$

All electrons are paired. 0 unpaired electrons.

Bond order = $$\f\frac{4-2}{2} = 1$$. Diamagnetic.

3. $$C_2$$ (total electrons = 6 + 6 = 12):

Configuration: $$\sigma_{1s}^2\, {\sigma^*_{1s}}^{2}\, \sigma_{2s}^2\, {\sigma^*_{2s}}^{2}\, \pi_{2p_x}^2\, \pi_{2p_y}^2$$

Both $$\pi_{2p}$$ orbitals are completely filled. 0 unpaired electrons.

Bond order = $$\f\frac{8-4}{2} = 2$$. Diamagnetic.

4. $$C_2^-$$ (total electrons = 12 + 1 = 13):

Configuration: $$\sigma_{1s}^2\, {\sigma^*_{1s}}^{2}\, \sigma_{2s}^2\, {\sigma^*_{2s}}^{2}\, \pi_{2p_x}^2\, \pi_{2p_y}^2\, \sigma_{2p_z}^1$$

The 13th electron enters the $$\sigma_{2p_z}$$ orbital (which comes after $$\pi_{2p}$$ in the filling order for lighter elements). This gives 1 unpaired electron.

Bond order = $$\f\frac{9-4}{2} = 2.5$$. Paramagnetic.

5. $$O_2^{2-}$$ (total electrons = 16 + 2 = 18):

Configuration (using O-type ordering): $$\sigma_{1s}^2\, {\sigma^*_{1s}}^{2}\, \sigma_{2s}^2\, {\sigma^*_{2s}}^{2}\, \sigma_{2p_z}^2\, \pi_{2p_x}^2\, \pi_{2p_y}^2\, {\pi^*_{2p_x}}^{2}\, {\pi^*_{2p_y}}^{2}$$

Both antibonding $$\pi^*$$ orbitals are completely filled. 0 unpaired electrons.

Bond order = $$\f\frac{10-8}{2} = 1$$. Diamagnetic.

6. $$O_2^+$$ (total electrons = 16 - 1 = 15):

Configuration: $$\sigma_{1s}^2\, {\sigma^*_{1s}}^{2}\, \sigma_{2s}^2\, {\sigma^*_{2s}}^{2}\, \sigma_{2p_z}^2\, \pi_{2p_x}^2\, \pi_{2p_y}^2\, {\pi^*_{2p_x}}^{1}$$

One electron is removed from the antibonding $$\pi^*$$ orbital, leaving 1 unpaired electron.

Bond order = $$\f\frac{10-5}{2} = 2.5$$. Paramagnetic.

7. $$He_2^+$$ (total electrons = 2 + 2 - 1 = 3):

Configuration: $$\sigma_{1s}^2\, {\sigma^*_{1s}}^{1}$$

The antibonding orbital has one electron. 1 unpaired electron.

Bond order = $$\f\frac{2-1}{2} = 0.5$$. Paramagnetic.

Summary:

The paramagnetic species are: $$B_2$$, $$C_2^-$$, $$O_2^+$$, and $$He_2^+$$.

Therefore, the number of paramagnetic species is 4.

We need to find the sum of the number of lone pairs on the central atoms of $$XeO_3$$, $$XeOF_4$$, and $$XeF_6$$.

Analyze $$XeO_3$$

Xe is the central atom. Xe has 8 valence electrons. It forms 3 double bonds with oxygen (uses 6 electrons).

Remaining electrons on Xe = 8 - 6 = 2, which gives 1 lone pair.

Geometry: Trigonal pyramidal (sp$$^3$$ hybridization).

Analyze $$XeOF_4$$

Xe has 8 valence electrons. It forms 4 single bonds with F (uses 4 electrons) and 1 double bond with O (uses 2 electrons).

Remaining electrons on Xe = 8 - 6 = 2, which gives 1 lone pair.

Geometry: Square pyramidal with one lone pair (sp$$^3$$d$$^2$$ hybridization).

Analyze $$XeF_6$$

Xe has 8 valence electrons. It forms 6 single bonds with F (uses 6 electrons).

Remaining electrons on Xe = 8 - 6 = 2, which gives 1 lone pair.

Geometry: Distorted octahedral (sp$$^3$$d$$^3$$ hybridization).

Calculate the sum

$$\text{Total lone pairs} = 1 + 1 + 1 = 3$$

The answer is 3.

In the crystal $$Fe_{0.93}O$$, some $$Fe^{2+}$$ ions are missing and replaced by $$Fe^{3+}$$ ions to maintain electrical neutrality.

Let us assume that in the formula $$Fe_{0.93}O$$, let $$x$$ be the number of $$Fe^{3+}$$ ions and $$(0.93 - x)$$ be the number of $$Fe^{2+}$$ ions per formula unit.

Charge balance:

The total positive charge must equal the total negative charge.

Charge from $$Fe^{2+}$$: $$2(0.93 - x)$$

Charge from $$Fe^{3+}$$: $$3x$$

Charge from $$O^{2-}$$: $$-2(1) = -2$$

For electrical neutrality:

$$2(0.93 - x) + 3x = 2$$

$$1.86 - 2x + 3x = 2$$

$$1.86 + x = 2$$

$$x = 0.14$$

So, number of $$Fe^{3+}$$ ions = 0.14

Number of $$Fe^{2+}$$ ions = $$0.93 - 0.14 = 0.79$$

Percentage of $$Fe^{2+}$$ ions:

$$\% Fe^{2+} = \frac{0.79}{0.93} \times 100$$

$$= \frac{79}{93} \times 100$$

$$= 84.95\%$$

$$\approx 85\%$$

Hence, the percentage of $$Fe^{2+}$$ ions is 85 (nearest integer).

Let the number of atoms forming the hcp lattice be $$\mathrm{N}$$.

In a close-packed structure, total tetrahedral voids are $$\mathrm{2N}$$.

Atoms occupy $$\mathrm{\frac{2}{3}}$$ of the tetrahedral voids.

Hence, number of atoms in tetrahedral voids is $$\mathrm{\frac{2}{3} \times 2N = \frac{4N}{3}}$$.

Ratio of lattice atoms to atoms in voids:

$$\mathrm{N : \frac{4N}{3}}$$

Multiplying by 3 gives:

$$\mathrm{3 : 4}$$

Therefore, the formula is $$\mathrm{X_3Y_4}$$.

Total atoms in one formula unit:

$$\mathrm{3 + 4 = 7}$$

Percentage of lattice atoms:

$$\mathrm{\frac{3}{7} \times 100}$$

$$\mathrm{= 42.857\% \approx 43\%}$$

We are given that anion $$B^-$$ forms a cubic close packing (CCP, i.e., FCC) structure, with cation $$A^+$$ occupying all octahedral voids. The ionic radii are $$r_{A^+} = 102$$ pm and $$r_{B^-} = 181$$ pm.

In an FCC unit cell, there are 4 anions per unit cell (8 corner atoms contributing $$\frac{1}{8}$$ each and 6 face-centred atoms contributing $$\frac{1}{2}$$ each: $$8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$$). The number of octahedral voids in an FCC unit cell equals the number of atoms, which is 4. Since $$A^+$$ occupies all octahedral voids, we have 4 cations per unit cell, giving the stoichiometry AB. This is the well-known rock salt (NaCl) structure.

In the rock salt structure, the octahedral voids are located at the edge centres and at the body centre of the cube. Consider one edge of the unit cell: at one end sits a $$B^-$$ anion (at a corner), at the midpoint of the edge sits an $$A^+$$ cation (in the octahedral void), and at the other end sits another $$B^-$$ anion (at the adjacent corner). The cation and anions touch along the edge.

The edge length $$a$$ therefore equals the distance from the centre of one anion through the cation to the centre of the next anion:

$$a = r_{B^-} + 2r_{A^+} + r_{B^-} = 2r_{A^+} + 2r_{B^-}$$

This can also be written as $$a = 2(r_{A^+} + r_{B^-})$$. The factor of 2 appears because both a cation radius and an anion radius span from the edge centre to each corner, and there are two such spans along the full edge.

Now we substitute the given values:

$$a = 2(102 + 181) = 2 \times 283 = 566 \text{ pm}$$

Hence, the correct answer is 566.

We need to find the value of x in the stoichiometry $$A_xB_y$$ after removing two A atoms from opposite faces of a solid AB with ccp arrangement.

In a ccp (FCC) unit cell, A atoms are arranged in cubic close packing:

Corner A atoms: 8 corners × $$\frac{1}{8}$$ = 1

Face-center A atoms: 6 faces × $$\frac{1}{2}$$ = 3

Total A atoms per unit cell = 1 + 3 = 4

B atoms occupy all octahedral sites:

Body center: 1 × 1 = 1

Edge centers: 12 × $$\frac{1}{4}$$ = 3

Total B atoms per unit cell = 1 + 3 = 4

Initial formula: $$A_4B_4$$ = AB

Two A atoms are removed from opposite faces. Each face-center atom contributes $$\frac{1}{2}$$ to the unit cell.

Removing 2 face-center atoms reduces A by: $$2 \times \frac{1}{2} = 1$$

New A atoms = 4 - 1 = 3

B atoms remain = 4 (unchanged)

The formula becomes $$A_3B_4$$.

So $$x = 3$$.

Therefore, the value of x is 3.

We are given that metal M crystallizes in an FCC lattice with edge length $$a = 4.0 \times 10^{-8}$$ cm, and the density $$\rho = 9.03$$ g/cm$$^3$$. We need to find the atomic mass.

For an FCC unit cell, the number of atoms per unit cell is $$Z = 4$$ (since there are 8 corner atoms each contributing $$\frac{1}{8}$$ and 6 face-centered atoms each contributing $$\frac{1}{2}$$, giving $$8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$$).

The relation between density and the unit cell parameters is given by:

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

Rearranging for $$M$$:

Now substituting the values:

So we get:

Rounding to the nearest integer, we get $$M \approx 87$$ g/mol.

Hence, the correct answer is 87.

To begin with, we need to find the distance between $$Na^+$$ and $$Cl^-$$ ions in solid NaCl. Recall that NaCl crystallizes in a face-centered cubic (FCC) structure and that each unit cell contains four formula units of NaCl, so the number of formula units per unit cell is $$Z = 4$$.

Next, the density of a crystal is given by
$$\rho = \frac{Z \times M}{N_A \times a^3}$$
where $$Z = 4$$, $$M = 23 + 35.5 = 58.5$$ g mol$$^{-1}$$ is the molar mass of NaCl, $$N_A = 6.02 \times 10^{23}$$ mol$$^{-1}$$, and $$\rho = 43.1$$ g cm$$^{-3}$$. Rearranging for $$a^3$$ yields
$$a^3 = \frac{Z \times M}{N_A \times \rho} = \frac{4 \times 58.5}{6.02 \times 10^{23} \times 43.1}$$ which in turn gives
$$a^3 = \frac{234}{2.5946 \times 10^{25}} = 9.02 \times 10^{-24} \text{ cm}^3$$ and hence
$$a = (9.02 \times 10^{-24})^{1/3} = 2.08 \times 10^{-8} \text{ cm}.$$

Then, in the NaCl FCC structure, the $$Na^+$$ and $$Cl^-$$ ions alternate along each edge of the unit cell, so the nearest-neighbor distance (distance between adjacent $$Na^+$$ and $$Cl^-$$) is exactly half the edge length of the unit cell:
$$d = \frac{a}{2} = \frac{2.08 \times 10^{-8} \text{ cm}}{2} = 1.04 \times 10^{-8} \text{ cm}.$$

Finally, since $$1 \text{ cm} = 10^{-2} \text{ m}$$, it follows that
$$d = 1.04 \times 10^{-8} \text{ cm} = 1.04 \times 10^{-10} \text{ m}.$$ Expressing this result as $$d = n \times 10^{-10}$$ m gives $$n \approx 1$$ (to the nearest integer).

The correct answer is $$\mathbf{1}$$.

We need to identify which interhalogens from the given list have a square pyramidal structure.

A square pyramidal structure arises when a molecule has the general formula AX$$_5$$E$$_1$$ (5 bonding pairs and 1 lone pair), with sp$$^3$$d$$^2$$ hybridization.

First, consider ClF$$_3$$. Chlorine has 7 valence electrons, forming 3 bond pairs and retaining 2 lone pairs, which gives 5 electron pairs overall. Since the resulting AX$$_3$$E$$_2$$ arrangement corresponds to a T-shaped geometry, ClF$$_3$$ does not adopt a square pyramidal structure.

Next, IF$$_7$$ involves iodine having 7 valence electrons forming 7 bond pairs with no lone pairs, for a total of 7 electron pairs. The AX$$_7$$ arrangement corresponds to a pentagonal bipyramidal shape, so IF$$_7$$ is not square pyramidal.

Turning to BrF$$_5$$, bromine contributes 7 valence electrons, yielding 5 bond pairs and 1 lone pair to make 6 electron pairs. This matches the AX$$_5$$E$$_1$$ pattern and sp$$^3$$d$$^2$$ hybridization, resulting in a square pyramidal structure, so BrF$$_5$$ qualifies.

In contrast, BrF$$_3$$ has bromine with 7 valence electrons but forms only 3 bond pairs and 2 lone pairs—5 electron pairs in total—leading to an AX$$_3$$E$$_2$$ T-shaped geometry and thus not square pyramidal.

Considering the dimer I$$_2$$Cl$$_6$$, each iodine is bonded to four bridging chlorine atoms and retains one lone pair, producing a square planar arrangement around each iodine and giving rise to an edge-shared bioctahedral framework. Consequently, I$$_2$$Cl$$_6$$ is not square pyramidal.

For IF$$_5$$, iodine’s 7 valence electrons form 5 bond pairs plus 1 lone pair, totaling 6 electron pairs. The resulting AX$$_5$$E$$_1$$ configuration with sp$$^3$$d$$^2$$ hybridization yields a square pyramidal structure, so IF$$_5$$ is included.

As for ClF, the simple diatomic molecule has a linear geometry and obviously does not have a square pyramidal shape.

Finally, ClF$$_5$$ involves chlorine with 7 valence electrons forming 5 bond pairs and 1 lone pair, for 6 electron pairs in an AX$$_5$$E$$_1$$ arrangement. Therefore, ClF$$_5$$ also adopts a square pyramidal structure.

From the above analysis, the interhalogens with square pyramidal structures are BrF$$_5$$, IF$$_5$$, and ClF$$_5$$. Therefore, the number of interhalogens with square pyramidal structure = 3.

To identify the linear species, we apply VSEPR theory by counting the steric number (number of bonding pairs + lone pairs) around the central atom of each molecule.

For $$\text{NO}_2$$: nitrogen is the central atom with 17 total valence electrons (5 from N + 6+6 from two O). After forming two N-O bonds, nitrogen has one unpaired electron remaining. The electron geometry is trigonal planar (steric number 3), and the molecular geometry is bent with a bond angle of about 134°. This is not linear.

For $$\text{Cl}_2\text{O}$$: oxygen is the central atom with 20 valence electrons (7+7 from two Cl + 6 from O). Oxygen forms two single bonds with chlorine and has two lone pairs, giving a steric number of 4. The molecular geometry is bent (similar to water), so this is not linear.

For $$\text{O}_3$$: the central oxygen has 18 valence electrons (6+6+6). After forming bonds with the two terminal oxygens, the central oxygen has one lone pair, giving a steric number of 3. The molecular geometry is bent with a bond angle of about 117°. This is not linear.

For $$\text{N}_3^-$$: this ion has 16 valence electrons (5+5+5 from three N + 1 for the negative charge). The central nitrogen forms two double bonds with the terminal nitrogen atoms and has no lone pairs, giving a steric number of 2. With two bonding regions and zero lone pairs, the geometry is linear with a bond angle of 180°.

Therefore, the linear species is $$\text{N}_3^-$$, which is Option D.

The water molecule has two lone pairs on the central oxygen atom and two bond pairs (O-H bonds). According to VSEPR theory, the electron geometry around oxygen is tetrahedral.

In a perfect tetrahedral arrangement, the bond angle would be $$109.5°$$. However, the lone pair-lone pair repulsion is greater than lone pair-bond pair repulsion, which in turn is greater than bond pair-bond pair repulsion. The two lone pairs on oxygen push the bonding pairs closer together, reducing the H-O-H bond angle from the ideal $$109.5°$$ to $$104.5°$$.

Therefore, Assertion A is true — the H-O-H bond angle in water is indeed $$104.5°$$. Reason R is also true — lone pair-lone pair repulsion is higher than bond pair-bond pair repulsion. Moreover, R correctly explains A, because it is precisely this stronger lone pair repulsion that compresses the bond angle from $$109.5°$$ to $$104.5°$$.

The correct answer is Option (4): Both A and R are true, and R is the correct explanation of A.

First, recall the basic idea: a species shows paramagnetism when it possesses one or more unpaired electrons. Diamagnetic species have all electrons paired. For oxides of s-block metals, the magnetic behaviour is governed by the type of oxygen-containing ion present - oxide $$\left(\mathrm O^{2-}\right)$$, peroxide $$\left(\mathrm O_2^{2-}\right)$$ or superoxide $$\left(\mathrm O_2^{-}\right)$$.

We now look at each given oxide.

1. Li2O, CaO, MgO and K2O
All four are simple ionic oxides. In each of them the oxygen is present as the monatomic oxide ion $$\mathrm O^{2-}$$.

The oxide ion has

$$\text{electrons} = 8\;(\text{from O atom}) + 2\;(\text{extra}) = 10$$

This is the same electron count as neon. Every orbital (1s, 2s, 2p) is completely filled, so

$$\text{number of unpaired electrons} = 0$$

Hence $$\mathrm O^{2-}$$ is diamagnetic, and therefore Li2O, CaO, MgO and K2O are all diamagnetic.

2. Na2O2 (peroxide)
This compound contains the peroxide ion $$\mathrm O_2^{2-}$$. Let us use the molecular-orbital (MO) picture of the O2 molecule and then add electrons.

For O2 (16 electrons) the last two electrons occupy the degenerate antibonding $$\pi^\ast_{2p_x}$$ and $$\pi^\ast_{2p_y}$$ orbitals singly, giving two unpaired electrons and paramagnetism. The peroxide ion has two extra electrons:

$$16 + 2 = 18\;\text{electrons}$$

These two extra electrons pair up in the antibonding $$\pi^\ast$$ orbitals, so now

$$\text{number of unpaired electrons} = 0$$

Thus $$\mathrm O_2^{2-}$$ and hence Na2O2 are diamagnetic.

3. KO2 (superoxide)
KO2 contains the superoxide ion $$\mathrm O_2^{-}$$. Starting again from O2 (16 electrons) and adding one extra electron gives

$$16 + 1 = 17\;\text{electrons}$$

With 17 electrons, one of the antibonding $$\pi^\ast$$ orbitals now contains a single unpaired electron:

$$\text{number of unpaired electrons} = 1$$

Because of this unpaired electron, $$\mathrm O_2^{-}$$ is paramagnetic, and therefore KO2 is paramagnetic.

Counting the paramagnetic oxides
Only KO2 is paramagnetic. All the others are diamagnetic.

So, the required number of paramagnetic oxides is

$$1$$

Hence, the correct answer is Option A.

We need to identify which species has unequal bond lengths among the given options.

$$SiF_4$$ has a tetrahedral geometry with all four Si-F bonds equivalent, so all bond lengths are equal. $$XeF_4$$ has a square planar geometry with all four Xe-F bonds equivalent, so all bond lengths are equal. $$BF_4^-$$ has a tetrahedral geometry with all four B-F bonds equivalent, so all bond lengths are equal.

$$SF_4$$ has a see-saw (distorted tetrahedral) geometry based on a trigonal bipyramidal electron arrangement with one lone pair. In this geometry, there are two types of S-F bonds: two axial bonds and two equatorial bonds. The axial bonds are longer than the equatorial bonds because the axial positions experience greater repulsion from the three equatorial electron pairs (two bonding pairs and one lone pair).

Therefore, $$SF_4$$ has unequal bond lengths, and the correct answer is option (2).

The central atom has 2 lone pairs and forms 3 single bonds, giving a total of $$2 + 3 = 5$$ electron pairs around it. According to VSEPR theory, 5 electron pairs arrange themselves in a trigonal bipyramidal geometry to minimize repulsion.

In a trigonal bipyramidal arrangement, lone pairs preferentially occupy the equatorial positions because these positions have more space (only two 90-degree interactions with axial positions, compared to three 90-degree interactions if placed axially). With 2 lone pairs in equatorial positions, the 3 bonding pairs occupy the remaining positions: 1 equatorial and 2 axial.

This gives a T-shaped molecular geometry, where the three bonded atoms form a shape resembling the letter T. A classic example of this geometry is $$ClF_3$$.

Therefore, the shape of the molecule is T-shaped, which is option (C).

We determine the hybridisation of the central atom in each species.

SF$$_4$$: Sulphur has 6 valence electrons; 4 are used for bonding with F, leaving 1 lone pair. Total electron pairs around S = 4 bonding pairs + 1 lone pair = 5, giving $$sp^3d$$ hybridisation (iii).

IF$$_5$$: Iodine has 7 valence electrons; 5 are used for bonding with F, leaving 1 lone pair. Total electron pairs = 5 bonding + 1 lone pair = 6, giving $$sp^3d^2$$ hybridisation (i).

NO$$_2^+$$: This cation has no lone pairs on nitrogen (N loses one electron, giving a linear structure). With 2 double bonds and no lone pairs, the steric number is 2, giving $$sp$$ hybridisation (v).

NH$$_4^+$$: Nitrogen forms 4 equivalent N-H bonds with no lone pairs (the lone pair donated into the extra bond). Total electron pairs = 4, giving $$sp^3$$ hybridisation (iv).

The correct matching is (a)-(iii), (b)-(i), (c)-(v) and (d)-(iv), which corresponds to option (3).

Intermolecular hydrogen bonding is the hydrogen bonding that occurs between different molecules due to the attraction between H attached to highly electronegative atoms (O, N, F) and lone pairs present on another molecule.

(a) shows intramolecular hydrogen bonding because the -OH group and -NO₂ group are present at ortho positions, allowing H-bonding within the same molecule.

Hence, (a) does not show significant intermolecular H-bonding.

(b) contains both -OH and -NH groups and can form strong hydrogen bonding between different molecules.

Hence, (b) shows significant intermolecular hydrogen bonding.

(c) is a sterically hindered phenol due to bulky groups near the -OH group, preventing effective intermolecular H-bonding.

Hence, (c) does not show significant intermolecular hydrogen bonding.

Therefore, the correct answer is:

$$\boxed{(b)}$$

The melting point of an ionic compound depends on the lattice energy, which in turn depends on the charges of the ions and the interionic distance. According to the Born-Landé equation, lattice energy is directly proportional to the product of ionic charges and inversely proportional to the sum of ionic radii.

Comparing LiF and LiCl: Both have the same cation ($$\text{Li}^+$$) and the same charges ($$+1$$ and $$-1$$). The fluoride ion ($$\text{F}^-$$) is smaller than the chloride ion ($$\text{Cl}^-$$), so the interionic distance in LiF is smaller than in LiCl. This means LiF has a higher lattice energy and therefore a higher melting point: $$\text{LiF} > \text{LiCl}$$.

Comparing MgO and NaCl: MgO consists of $$\text{Mg}^{2+}$$ and $$\text{O}^{2-}$$ ions (charges $$+2$$ and $$-2$$), while NaCl consists of $$\text{Na}^+$$ and $$\text{Cl}^-$$ ions (charges $$+1$$ and $$-1$$). The product of charges in MgO ($$2 \times 2 = 4$$) is much greater than in NaCl ($$1 \times 1 = 1$$). Additionally, the ionic radii of $$\text{Mg}^{2+}$$ and $$\text{O}^{2-}$$ are smaller than those of $$\text{Na}^+$$ and $$\text{Cl}^-$$. Therefore, MgO has a significantly higher lattice energy and melting point: $$\text{MgO} > \text{NaCl}$$.

The correct set with both pairs in the correct order is: LiF $$>$$ LiCl and MgO $$>$$ NaCl.

Therefore, the correct answer is Option (4).

London dispersion forces are weak intermolecular attractions that arise from momentary dipoles produced due to the movement of electrons in atoms or non-polar molecules.

The potential (interaction) energy $$E$$ of London forces varies with the separation $$r$$ between two particles according to the relation
$$$E \;\propto\; \dfrac{1}{r^{6}}$$$

Writing the proportionality in the form $$E \;\propto\; r^{\,x}$$ and comparing with $$$E \;\propto\; r^{-6}$$$, we obtain
$$x = -6$$

Hence, the exponent $$x$$ is $$-6$$, which corresponds to Option B.

We need to identify which pairs of species are isostructural, meaning they have the same geometry and shape.

Pair A: $$SO_4^{2-}$$ and $$CrO_4^{2-}$$. In $$SO_4^{2-}$$, sulphur has 4 bond pairs and no lone pairs, giving a tetrahedral shape. In $$CrO_4^{2-}$$, chromium also has 4 bond pairs and no lone pairs, giving a tetrahedral shape. So this pair is isostructural.

Pair B: $$SiCl_4$$ and $$TiCl_4$$. In $$SiCl_4$$, silicon has 4 bond pairs and no lone pairs, giving a tetrahedral shape. In $$TiCl_4$$, titanium also has 4 bond pairs and no lone pairs, giving a tetrahedral shape. So this pair is isostructural.

Pair C: $$NH_3$$ and $$NO_3^-$$. In $$NH_3$$, nitrogen has 3 bond pairs and 1 lone pair, giving a trigonal pyramidal shape. In $$NO_3^-$$, nitrogen has 3 bond pairs and no lone pairs, giving a trigonal planar shape. These are not isostructural.

Pair D: $$BCl_3$$ and $$BrCl_3$$. In $$BCl_3$$, boron has 3 bond pairs and no lone pairs, giving a trigonal planar shape. In $$BrCl_3$$, bromine has 3 bond pairs and 2 lone pairs, giving a T-shaped geometry. These are not isostructural.

Hence, the isostructural pairs are A and B only.

Hence, the correct answer is Option 3.

A magnetic moment of 1.73 BM corresponds to spin-only value for one unpaired electron, since $$\mu = \sqrt{n(n+2)} = \sqrt{1 \times 3} = \sqrt{3} \approx 1.73$$ BM.

For $$O_2^+$$: molecular orbital configuration removes one electron from $$O_2$$. $$O_2$$ has 2 unpaired electrons; $$O_2^+$$ has 1 unpaired electron. So $$\mu = 1.73$$ BM. ✓

For $$O_2^-$$: adds one electron to $$O_2$$. This gives 1 unpaired electron (the extra electron pairs up one of the $$\pi^*$$ electrons). So $$\mu = 1.73$$ BM. ✓

For $$[\text{Cu(NH}_3)_4]Cl_2$$: here Cu is in +2 state (Cu²⁺). Cu²⁺ has configuration [Ar]$$3d^9$$, with 1 unpaired electron. So $$\mu = 1.73$$ BM. ✓

For CuI: here Cu is in +1 state (Cu⁺). Cu⁺ has configuration [Ar]$$3d^{10}$$, which is completely filled with no unpaired electrons. Therefore $$\mu = 0$$ BM, not 1.73 BM.

CuI does not have a magnetic moment of 1.73 BM.

According to Molecular Orbital Theory, a molecule exists only if its bond order is greater than zero. Bond order is calculated as $$\text{B.O.} = \frac{N_b - N_a}{2}$$, where $$N_b$$ is the number of bonding electrons and $$N_a$$ is the number of antibonding electrons.

For $$He_2^-$$ (5 electrons): The configuration is $$(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^1$$. Bond order $$= \frac{3 - 2}{2} = 0.5$$. This species can exist.

For $$O_2^{2-}$$ (18 electrons): The configuration fills up to $$(\pi^* 2p)^4$$. Bond order $$= \frac{10 - 8}{2} = 1$$. This species can exist.

For $$He_2^+$$ (3 electrons): The configuration is $$(\sigma 1s)^2(\sigma^* 1s)^1$$. Bond order $$= \frac{2 - 1}{2} = 0.5$$. This species can exist.

For $$Be_2$$ (8 electrons): The configuration is $$(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2$$. Bond order $$= \frac{4 - 4}{2} = 0$$. With zero bond order, this molecule does not exist.

Therefore, the species that does not exist is $$Be_2$$, which corresponds to option (4).

We begin by reading the two statements carefully. Assertion A says that lithium halides are somewhat covalent in nature, while Reason R claims that lithium possesses high polarisation capability. In order to judge these statements, we must recall the concept of polarising power and Fajans’ rules.

Fajans’ rules state that the covalent character in an ionic bond increases when:

(i) the cation has a small radius and/or high charge,

(ii) the anion has a large radius and/or high charge,

because under such conditions the cation can strongly distort (polarise) the electron cloud of the anion. This distortion leads to sharing of electrons rather than complete transfer, imparting covalent character to the bond.

Now we apply these rules to lithium halides. The lithium ion is $$\text{Li}^+$$. Among all alkali-metal cations (Li $$^+, \text{Na}^+, \text{K}^+, \ldots)$$, $$\text{Li}^+$$ has the smallest ionic radius. Numerically,

$$r(\text{Li}^+) \approx 0.76\ \text{Å}, \quad r(\text{Na}^+) \approx 1.02\ \text{Å}, \quad r(\text{K}^+) \approx 1.38\ \text{Å}$$

and so on. Because the charge on each of these ions is the same $$(+1)$$, a smaller radius means a larger charge density, i.e.

Charge density $$= \frac{+1}{\text{smaller volume}} \ \Longrightarrow \ \text{larger electric field around Li}^+.$$

The polarising power is directly proportional to this charge density. Therefore $$\text{Li}^+$$ shows the highest polarising power among alkali cations.

Let us see the direct consequence for lithium halides. The general reaction for the formation of a lithium halide is

Li $$^+ + \text{X}^- \longrightarrow \text{LiX}, \quad \text{where } \text{X}^- = \text{F}^-,\ \text{Cl}^-,\ \text{Br}^-,\ \text{I}^-.$$

Because $$\text{Li}^+$$ is so small and highly polarising, it pulls the electron cloud of $$\text{X}^-$$ towards itself. This polarisation produces partial sharing of the electrons between Li and X, and that partial sharing manifests as partial covalent character in the Li-X bond. Thus, lithium halides are indeed somewhat covalent instead of being purely ionic.

Therefore Assertion A is true.

Next we evaluate Reason R. We have just argued—using size, charge density and Fajans’ rules—that $$\text{Li}^+$$ has a high polarising power. Hence Reason R is also true.

Finally, we must determine whether R correctly explains A. Since A (lithium halides are covalent) directly follows from the high polarising capability of $$\text{Li}^+$$ as described in R, Reason R is the correct and relevant explanation of Assertion A.

Hence, the correct answer is Option D.

For molecules and ions of oxygen we have to use the Molecular Orbital (MO) theory. According to MO theory, the electronic energy-level sequence for the 2p orbitals in O2 and its ions is

$$\sigma_{2p_z}\,\lt \,\pi_{2p_x}=\pi_{2p_y}\,\lt \,\pi^{*}_{2p_x}=\pi^{*}_{2p_y}\,\lt \,\sigma^{*}_{2p_z}$$

All lower-energy 1s and 2s orbitals are completely filled in every species and will be included in the counting. The bond order of any species is obtained from the formula that MO theory gives:

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

where $$N_b$$ is the total number of electrons in bonding MOs and $$N_a$$ is the total number of electrons in antibonding MOs.

Now we shall count electrons one by one for each species, write the values of $$N_b$$ and $$N_a$$ explicitly, and then apply the formula.

1. The neutral molecule $$O_2$$ (16 electrons)

The filling is

$$\sigma_{1s}^2\,\sigma_{1s}^{*\,2}\,\sigma_{2s}^2\,\sigma_{2s}^{*\,2}\, \sigma_{2p_z}^2\,\pi_{2p_x}^2\,\pi_{2p_y}^2\,\pi^{*1}_{2p_x}\,\pi^{*1}_{2p_y}$$

Counting electrons:

Bonding electrons $$N_b=2+2+2+2+2=10$$ Antibonding electrons $$N_a=2+2+1+1=6$$

Hence

$$\text{B.O.}(O_2)=\dfrac{10-6}{2}=2$$

2. The ion $$O_2^{+}$$ (15 electrons)

Removal of one electron occurs from the highest occupied antibonding orbital, so

$$\pi^{*1}_{2p_x}\,\pi^{*0}_{2p_y}$$

Now

Antibonding electrons $$N_a=2+2+1=5$$ (Bonding electrons remain $$N_b=10$$.)

Thus

$$\text{B.O.}(O_2^{+})=\dfrac{10-5}{2}=2.5$$

3. The ion $$O_2^{-}$$ (17 electrons)

An extra electron goes into the still singly occupied antibonding orbital, giving

$$\pi^{*1}_{2p_x}\,\pi^{*2}_{2p_y}$$

Antibonding electrons $$N_a=2+2+1+2=7$$ (Bonding electrons are still $$N_b=10$$.)

Therefore

$$\text{B.O.}(O_2^{-})=\dfrac{10-7}{2}=1.5$$

4. The ion $$O_2^{2-}$$ (18 electrons)

The added electron pairs up in the remaining antibonding orbital, making both $$\pi^{*}$$ orbitals completely filled:

$$\pi^{*2}_{2p_x}\,\pi^{*2}_{2p_y}$$

Antibonding electrons $$N_a=2+2+2+2=8$$ (Bonding electrons again $$N_b=10$$.)

So

$$\text{B.O.}(O_2^{2-})=\dfrac{10-8}{2}=1$$

We now have all four bond orders:

$$O_2^{+}:2.5,\; O_2:2,\; O_2^{-}:1.5,\; O_2^{2-}:1$$

Clearly the descending order of bond order is

$$O_2^{+} \;\gt \; O_2 \;\gt \; O_2^{-} \;\gt \; O_2^{2-}$$

This matches Option C in the list provided.

Hence, the correct answer is Option C.

The electronic configuration of one $$\text{O}$$ atom is $$1s^2\,2s^2\,2p^4$$ (i.e. six valence electrons).
Therefore the neutral $$\text{O}_2$$ molecule contains $$2\times6 = 12$$ valence electrons.

For molecules from $$\text{N}_2$$ to $$\text{Ne}_2$$ the molecular-orbital energy order is:
$$$\sigma(1s),\;\sigma^*(1s),\;\sigma(2s),\;\sigma^*(2s),\;\sigma(2p_z),\;\pi(2p_x)=\pi(2p_y),\;\pi^*(2p_x)=\pi^*(2p_y),\;\sigma^*(2p_z)$$$.

Filling these orbitals with 12 electrons (Hund’s rule + Pauli principle) gives neutral $$\text{O}_2$$ the configuration
$$$\sigma(1s)^2\, \sigma^*(1s)^2\, \sigma(2s)^2\, \sigma^*(2s)^2\, \sigma(2p_z)^2$$$ $$$\pi(2p_x)^2\, \pi(2p_y)^2\, \pi^*(2p_x)^1\, \pi^*(2p_y)^1$$$.

Bond order is defined as
$$$\text{Bond order} = \frac{(\text{number of bonding electrons}) - (\text{number of antibonding electrons})}{2}$$$.

For neutral $$\text{O}_2$$:
Bonding electrons $$= 8$$ (in $$\sigma(2p_z),\;\pi(2p_x),\;\pi(2p_y)$$).
Antibonding electrons $$= 4$$ (in $$\pi^*(2p_x),\;\pi^*(2p_y)$$).
Thus $$\text{B.O.} = \frac{8-4}{2} = 2$$.

Now consider $$\text{O}_2^-$$ (superoxide ion). It possesses one extra electron, so total valence electrons $$= 12 + 1 = 13$$.

This extra electron enters the next available orbital $$\pi^*(2p)$$, giving the configuration
$$\dots \pi^*(2p_x)^2\,\pi^*(2p_y)^1$$.

Counting again:
Bonding electrons remain $$8$$.
Antibonding electrons become $$5$$.

Hence
$$\text{Bond order} = \frac{8 - 5}{2} = \frac{3}{2} = 1.5$$.

Magnetic behaviour depends on unpaired electrons. In $$\text{O}_2^-$$ the orbitals contain one unpaired electron (in the singly-occupied $$\pi^*$$ orbital). Any species with at least one unpaired electron is paramagnetic.

Therefore the superoxide ion $$\text{O}_2^-$$ has
Bond order = 1.5, Paramagnetic behaviour.

Thus, correct choice: Option B.

The $$I_3^-$$ ion is formed when an iodide ion ($$I^-$$) donates a lone pair to an iodine molecule ($$I_2$$). The central iodine atom in $$I_3^-$$ has 7 valence electrons plus 1 from the negative charge, giving a total of 8 valence electrons around it. Of these, 2 electrons form bonds with the two terminal iodine atoms, and the remaining 6 electrons form 3 lone pairs.

According to VSEPR theory, the central iodine atom has 5 electron pairs (2 bonding pairs and 3 lone pairs), which corresponds to a trigonal bipyramidal electron geometry. The two bonding pairs occupy the axial positions (which are 180° apart) to minimise lone pair-bonding pair repulsions, while the three lone pairs occupy the equatorial positions.

With the bonding pairs in the axial positions, the molecular shape of $$I_3^-$$ is linear, and the $$I-I-I$$ bond angle is $$180°$$.

Therefore, the correct answer is Option (4): Linear; 180°.

For $$NO_2^-$$ (nitrite ion): Nitrogen has 5 valence electrons, and with the negative charge, nitrogen forms 2 bonds with oxygen and has one lone pair. The electron geometry around nitrogen has 3 regions of electron density (2 bonding + 1 lone pair), giving $$sp^2$$ hybridisation. The bond angle is about 115°.

For $$NO_2^+$$ (nitronium ion): Nitrogen forms two double bonds with two oxygen atoms, giving 2 electron domains around nitrogen with no lone pair. This corresponds to $$sp$$ hybridisation with a linear geometry (180° bond angle).

For $$NH_4^+$$ (ammonium ion): Nitrogen is bonded to 4 hydrogen atoms with no lone pairs, giving a tetrahedral geometry with $$sp^3$$ hybridisation.

Therefore, the hybridisations of nitrogen in $$NO_2^-$$, $$NO_2^+$$ and $$NH_4^+$$ are $$sp^2$$, $$sp$$ and $$sp^3$$ respectively.

A Lewis base is a species that can donate a lone pair of electrons. For a molecule to act as a Lewis base, its central atom must have at least one lone pair of electrons available for donation.

In $$NF_3$$, nitrogen has one lone pair and can donate it, so it acts as a Lewis base. In $$SF_4$$, sulfur has one lone pair available for donation. In $$ClF_3$$, chlorine has two lone pairs and can donate one of them.

However, in $$PCl_5$$, phosphorus uses all five of its valence electrons to form five bonds with chlorine atoms. The phosphorus atom has no lone pairs remaining, meaning it cannot donate any electron pair to act as a Lewis base. In fact, $$PCl_5$$ is an electron-deficient species at phosphorus and tends to act as a Lewis acid instead.

Therefore, the compound that cannot act as a Lewis base is $$PCl_5$$, which is option (B).