NEET Practice Test - 18 - NEET MCQ

Thus Cl in HClO₄ is in highest oxidation state (+7) and hence strongest acid, while HCIO (Cl is in +1 oxidation state) is weakest.

CH₃CI > CH₃CH₂Cl > (CH₃)₂CHCl >(CH₃)₃CCl

Answer: A: NO₂^– and NO₃^– Explanation: To determine the geometry of these ions, we need to understand their hybridization. Let's analyze the hybridization of each ion: - NO₂^–: Nitrite ion has a central nitrogen atom surrounded by two oxygen atoms and one lone pair. The nitrogen atom forms two sigma bonds with the oxygen atoms and has one lone pair. This results in sp² hybridization, and the geometry is bent (approximately 120° bond angles). - NO₃^–: Nitrate ion has a central nitrogen atom surrounded by three oxygen atoms. The nitrogen atom forms three sigma bonds with the oxygen atoms. This results in sp² hybridization, and the geometry is trigonal planar (120° bond angles). - NH₂: This ion has a central nitrogen atom surrounded by two hydrogen atoms and one lone pair. The nitrogen atom forms two sigma bonds with the hydrogen atoms and has one lone pair. This results in sp² hybridization, and the geometry is bent (approximately 120° bond angles). - NH₄ : Ammonium ion has a central nitrogen atom surrounded by four hydrogen atoms. The nitrogen atom forms four sigma bonds with the hydrogen atoms. This results in sp³ hybridization, and the geometry is tetrahedral (approximately 109.5° bond angles). - SCN^–: Thiocyanate ion has a linear geometry with a central carbon atom surrounded by a sulfur atom and a nitrogen atom. The carbon atom forms two sigma bonds with the sulfur and nitrogen atoms. This results in sp hybridization, and the geometry is linear (180° bond angles). From the analysis above, we can see that NO₂^– and NO₃^– have the same sp² hybridization, which explains their similar geometries.

or, r_Cl- = 335.1 -101.0 = 154.1 pm

In bomb calorimeter, there is no expansion in volume, so, work done will be zero. This reaction is exothermic. So, some heat will be evolved which will result in lowering of internal energy. Hence, ΔU < 0 and w = 0

Thus the bromination of 2-methyl butane mainly gives 2- Bromo - 2 - methyl butane

equivalent mass of Cl₂ in reduction half-reaction.

equivalent mass of Cl₂ in oxidation half-reaction.

Thus, equivalent mass in overall reaction

Fe²⁺ present in it is oxidized to Fe³⁺.

So, we have, Fe²⁺ → Fe³⁺ +e

= 392

E_a = activation energy of forward reaction

E_a’ = ...backward...

= (E_a+ΔH)

Thus, E’_a ; > E_a

The correct answer is d. The given relation x/m = pt is not related to the adsorption process. Let's discuss the other options to understand why they are related to the adsorption process: A: x / m = K (p) at very low pressure. - This relation is derived from the Freundlich adsorption isotherm, which describes the relation between the amount of adsorbate adsorbed per unit mass of the adsorbent and the pressure of the adsorbate. - At very low pressure, the Freundlich isotherm becomes linear, and the relation x/m = K(p) holds true. B: x / m = Kp^1/n at intermediate pressure - This relation is the Freundlich adsorption isotherm, a widely used empirical model to describe the adsorption process. - The Freundlich isotherm is given by x/m = Kp^1/n, where x is the amount of adsorbate, m is the amount of adsorbent, K is a constant, p is the pressure of the adsorbate, and n is the heterogeneity factor. C: x/m = K at very high pressure - This relation is derived from the Langmuir adsorption isotherm, which is based on the assumption of monolayer adsorption on a homogeneous adsorbent surface. - At very high pressure, the Langmuir isotherm approaches saturation, and the relation x/m = K holds true, where K is a constant. In summary, options A, B, and C are related to the adsorption process through different adsorption isotherms, while option D is not related to the process.