Chemistry: Topic-wise Test- 2 - NEET MCQ

Na and Cl

Cs and F

Cs and l

Na and F

two sigma and two pi bonds

three sigma and one pi bonds

two sigma and one pi bonds

two sigma and thee pi bonds

BF₄^-

SO₄^2-

XeF₄

PH₄⁺

O₂F₂

H₂O₂

S₂Cl₂

H₂C₂

sp³d and T shaped

sp²d ² and tetragonal

sp3d and bent

none of these

CsBr₃ Cs⁺ + Br₃^–

I₄O₉ I³⁺ + (IO₃^–)₃

AgBrO₃ Ag⁺ + BrO₃^–

I₂O₄ IO₂^– + IO₂⁺

CNO^–

RCOO^–

OCN^–

N₃^–

Mg₃N₂

MgO

Mg₂O₃

MgCl₂

NaCl, Cl₂, AgCl

NaBr, Br2, NaBr

NaCl, Cl₂, Ag₂SO₄

Na₂CO₃, CO₂, Ag₂CO₃

small size

their being liquids

high melting point

high boiling point

to remove arsenic impurity

to remove dust particles

to detect colloidal impurity

to remove moisture

higher ionisation enthalpies

non-metals readily form cations

lower covalent charecter of the compounds formed between them

lower ionisation enthalpies

Cl₂

Br₂

I₂

F₂

 convert SO2 and SO3

test the presence of dust particles
 

filter dust particles

 remove impurities
 

H₂(g) + I₂(g) → 2HI (g)

HCl (aq) + NaOH (aq) → NaCl(aq) + H₂O (l)

C(s) + O₂(g) → CO₂(g)

N₂(g) + 3H₂(g) → 2NH₃(g)

 H₂(g) → 2H⁺(g) + 2e^-

2H(g) + aq → 2H⁺(aq) + 2e^-

2H(g) → 2H⁺(g) + 2e^-

 H₂(g) + aq → 2H⁺(aq) + 2e^-

Hess's law

 Law of conservation of energy

Henry's law

Joule's principle

Initial concentration of reactants

Final condition of the reacting substance

Intermediate states of a reaction

Initial and final conditions of the reacting substances

Hydrogen abstraction by halogen radical in the propagation step is exothermic in both chlorination and bromination

Hydrogen abstraction by halogen radical in propagation step is exothermic in chlorination but endothermic in bromination

A single halogen radical may bring about halogenation of thousands of alkane molecules

Continuous source of energy is required to continue the free radical halogenation reaction

IV < III < II < I 

I < II < IV < III 

I < II < III < IV

II < I < IV < III

It occurs very rapidly due to very small value of bond enthalpy of l₂

Direct iodination of alkane with l₂ in the presence of light is impractical

Iodination of an alkane can be achieved successfully using an oxidising agent catalyst

Presence of some HCI activate the free radical iodination of alkane

NaNH₂ is a very strong base

NaNH₂ abstract H from less hindered carbon

terminal bromo alkene is formed in first step of elimination reaction

terminal alkyne forms salt with NaNH₂, phased out of equilibrium

Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

Statement I is correct but Statement II is incorrect

Statement II is incorrect but Statement I incorrect

Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

Statement I is correct but Statement II is incorrect

Statement II is incorrect but Statement I incorrect

Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

Statement I is correct but Statement II is incorrect

Statement II is incorrect but Statement I incorrect