Using VSEPR theory, the molecular shapes are determined based on the number of bonding and lone pairs of electrons around the central atom.

- BrF₅ has five bonding pairs and one lone pair, resulting in a square pyramidal shape (IV).
- H₂O has two bonding pairs and two lone pairs, resulting in a bent shape (III).
- ClF₃ has three bonding pairs and two lone pairs, resulting in a T-shape (I).
- SF₄ has four bonding pairs and one lone pair, resulting in a see-saw shape (II).
Therefore, the correct matching is option (4).

In the first step, sodium acetylide (CH₃–C≡CNa) is formed when the alkyne reacts with sodium metal.

Then, the acetylide reacts with 1-bromopropane (CH₃–CH₂–CH₂–CH₂Br) to give the final product, CH₃–C≡C–CH₂–CH₂–CH₃.
Therefore, A = CH₃–C≡CNa and B = CH₃–CH₂–CH₂–CH₂Br.

In the first step, sodium acetylide (CH₃–C≡CNa) is formed when the alkyne reacts with sodium metal.

In the second step, Chromium trioxide (CrO₃) is used to oxidize the acetylide to an aldehyde. For further functionalization, CrO₂Cl₂ is employed, leading to the final product CH₃–C≡C–CH₂–CH₂–CH₃.
Therefore, reagent A is CrO₃ and reagent B is CrO₂Cl₂.

CH₂=CH−CH₂−Cl is an allyl halide because the halogen is attached to the carbon adjacent to the C=C double bond (sp³ hybridized). However, the Reason (R) incorrectly states that the halogen is attached to an sp² carbon atom.

In allyl halides, the halogen is attached to the allylic position, which is an sp³ hybridized carbon, not sp². Therefore, Assertion (A) is true, but Reason (R) is false.

When a non-volatile solute like naphthalene is added to benzene, it causes a depression in the freezing point due to the lowering of the vapor pressure.

The addition of a solute disrupts the orderly structure of the solvent, making it harder for the solvent molecules to arrange into a solid lattice, thereby lowering the freezing point.

The electronic configurations of the ions are as follows:

- Cr²⁺: [Ar] 3d⁴ (I)
- Mn⁺: [Ar] 3d⁵4s¹ (II)
- Ni²⁺: [Ar] 3d⁸ (III)
- V⁺: [Ar] 3d³4s¹ (IV)
Therefore, the correct matching is option (1).

This is the diazotization reaction, where primary aromatic amines react with sodium nitrite (NaNO₂) in the presence of HCl to form diazonium salts. The diazonium salt then couples with phenol to form a red azo dye.

- Reagent A: NaNO₂ + HCl is used to diazotize the primary amine.
- Reagent B: Phenol is used to couple with the diazonium salt, resulting in the formation of a colored azo compound, confirming the presence of aromatic primary amines.

Using the given data and equations of motion,

Let the initial velocity at time t be u and acceleration be a.
Displacement from t to t+1: s = u * 1 + 0.5 * a * (1)^2 = u + 0.5a = 125 m.
Increase in velocity: Δv = a * 1 = 50 m/s.
From Δv = 50 m/s, we get a = 50 m/s².
Substituting a = 50 m/s² into the displacement equation: u + 0.5 * 50 = 125 ⇒ u = 100 m/s.
The velocity at time t+1: v = u + a = 100 + 50 = 150 m/s.
Distance in the (t+2)th second: s = v * 1 + 0.5 * a * (1)^2 = 150 + 25 = 175 m.

The total initial energy is calculated by considering both capacitors.

The energy stored in the first capacitor: E₁ = 0.5 * C * V².
The energy stored in the second capacitor: E₂ = 0.5 * 2C * (2V)² = 0.5 * 2C * 4V² = 4 * C * V².
Total initial energy = E₁ + E₂ = 0.5CV² + 4CV² = 4.5CV².
When connected, the capacitors share charge and reach a common potential V'.
Total charge initially = CV + 2C * 2V = CV + 4CV = 5CV.
Combined capacitance = C + 2C = 3C.
Final potential V' = Total charge / Combined capacitance = 5CV / 3C = (5/3)V.
Final energy = 0.5 * 3C * (5V/3)² = 0.5 * 3C * (25V²/9) = 25CV²/6 ≈ 4.1667CV².
Energy loss = Initial energy - Final energy = 4.5CV² - 4.1667CV² = 0.3333CV² = (2/6)CV² = (2/3)E.
Therefore, x = 2.

Using conservation of angular momentum and calculating the initial and final kinetic energies of the system,

Moment of inertia of one disc, I = 0.5 * m * r² = 0.5 * 5 kg * (2 m)² = 10 kg·m².
Initial angular momentum, L₁ = I * ω₁ = 10 kg·m² * 10 rad/s = 100 kg·m²/s.
After placing the second disc, total moment of inertia, I₂ = 2 * 10 kg·m² = 20 kg·m².
Conservation of angular momentum: L₁ = I₂ * ω₂ ⇒ 100 = 20 * ω₂ ⇒ ω₂ = 5 rad/s.
Initial kinetic energy, KE₁ = 0.5 * I₁ * ω₁² = 0.5 * 10 * 100 = 500 J.
Final kinetic energy, KE₂ = 0.5 * I₂ * ω₂² = 0.5 * 20 * 25 = 250 J.
Energy dissipated = KE₁ - KE₂ = 500 J - 250 J = 250 J.

For a first-order reaction, using the integrated rate law and given rate data at two different times,

The integrated rate law for a first-order reaction is:
$\ln \left(\frac{\mathrm{[A]}}{\mathrm{[A]}}\right)=-kt$
Using the rates at t = 10 minutes and t = 20 minutes, we can solve for the rate constant k. Once k is known, the half-life t_½ is calculated using:
$t$_½ = 0.693 k Substituting the values, the half-life is found to be 24 minutes.

Using the formula for gravitational potential and gravitational field,

The gravitational potential at a distance r from the center of Earth is given by:
$\phi =\frac{-GM}{r}$
The acceleration due to gravity is:
$g=\frac{G}{r^2}M$

Given φ = −5.12 × 10⁷ J/kg and g = 6.4 m/s², with Earth's radius R = 6400 km = 6.4 × 10⁶ m.
Let the height above the surface be h, so r = R + h.
From the potential formula:
$\phi =\frac{-GM}{R}$
From the acceleration formula:
$g=\frac{G}{R^2}M=\frac{-\phi }{r}$
Solving the equations, we find h = 1600 km.

The total work done by the gas in a cyclic process is equal to the area enclosed by the path on the P-V diagram.

On a P-V diagram, the work done is the integral of P dV over the cycle. The area enclosed by the cyclic path A → B → C → A represents the net work done.
Calculating the area based on the given transformations, the total work done is found to be 200 J.

"Unununnium" corresponds to element 111, which belongs to Group 11 of the periodic table.

The IUPAC systematic naming uses prefixes based on Latin/Greek numerals. "Unununnium" is derived from "un-un-un-nium" indicating element 111.
Element 111 is Roentgenium (Rg), which is placed in Group 11 of the periodic table, alongside copper (Cu), silver (Ag), and gold (Au).

From the 2s and 2p atomic orbitals, we form 8 molecular orbitals in total:

Each atomic orbital (2s and 2p) combines to form bonding and antibonding molecular orbitals.
- 2s orbitals combine to form 2 molecular orbitals (σ2s and σ*2s).
- 2p orbitals combine to form 6 molecular orbitals (σ2p, σ*2p, π2px, π2py, π*2px, π*2py).
Therefore, total molecular orbitals formed = 2 + 6 = 8.

The retardation factor (R_f) is calculated as the ratio of the distance traveled by the compound to the distance traveled by the solvent:

$R$_f = distance_compound distance_solvent = 3.5 5 = 0.7 = 7 × 10 − 1
Therefore, R_f = 7 × 10⁻¹.

E_n = −0.85 eV corresponds to n = 4 in the hydrogen atom. From the 4th level, possible transitions are to n = 3, 2, 1 (3 levels), and from each of those to lower levels, totaling 6 possible transitions.

In hydrogen atom, the energy levels are given by E_n = −13.6 eV / n².
For E_n = −0.85 eV, n = 4.
From n = 4, transitions can be:
4 → 3
4 → 2
4 → 1
3 → 2
3 → 1
2 → 1
Total transitions = 6.

For a first-order reaction, using the integrated rate law and given rate data at two different times, we can find the rate constant k. Then, the half-life (t_1/2) is calculated using t_1/2 = 0.693/k.

Using the rate law for first-order reactions:
$\ln \left(\frac{\mathrm{[A]}}{\mathrm{[A]}}\right)=-kt$
Substituting the given rates at t = 10 min and t = 20 min, solve for k.
Once k is determined, calculate the half-life:
$t$_½ = 0.693 k Substituting the values, the half-life is found to be 24 minutes.

First, convert the area and thickness to compatible units:

- Area = 0.05 m² = 0.05 × (100 cm × 100 cm) = 0.05 × 10,000 cm² = 500 cm².
- Volume = Area × Thickness = 500 cm² × 0.05 cm = 25 cm³.
- Mass = Density × Volume = 7.9 g/cm³ × 25 cm³ = 197.5 g.
- Moles = Mass / Atomic mass = 197.5 g / 108 g/mol ≈ 1.83 mol.
- Number of atoms = Moles × Avogadro's number ≈ 1.83 × 6.022 × 10²³ ≈ 1.1 × 10²⁴ atoms.
- Expressed as 11 × 10²³ atoms.

First, convert the volume to liters: 250 mL = 0.250 L.

Moles required = Molarity × Volume = 0.35 mol/L × 0.250 L = 0.0875 mol.
Mass = Moles × Molar mass = 0.0875 mol × 82.02 g/mol ≈ 7.18 g.
Rounded off, this is approximately 7 g.