Chemistry: Topic-wise Test- 2 - NEET MCQ

for maximum ionic character,
* IE of cation must be less
* EA of anion must be larger
⇒ CS and  F
⇒ (B)

(A) N₂ B.O = 3 B.O α stability
N₂+ B.O = 2.5
Correct Statement
(B) O₂ ⇒ B.O = 2
O₂ ⇒ B.O = 2.5 ⇒ Correct Statement

(D) NO⁺, B.O. = 3
NO, B.O = 2.5
= Correct statement

It is strictly covalent does not shows cationic & anionic form.

Pseudo halides are
CN^–, NC^–, OCN^–, SCN^–, SeCN^–
NCS^–, N₃^–, NCO^–

Fe (OH)₃ is used to remove arsenic impurities.

Non metals have high value of ionization energy while metals have comparatively low value of ionization energy.

In the manufacture of sulphuric acid by contact process, Tyndall box is used to. remove impurities. Solution : Tyndall box is for testing the presence of dust particles.

In equation d, H⁺(aq) is formed from H₂(g). Enthalpy of formation of both entity is considered to be zero. So, ∆rH (∆H_product - ∆H_reactant) is zero.

No. of moles of HCl = 5 millimoles
No. of moles of NaOH = 5 millimoles
Mass of solution mixed = 50 gm+50 gm=100 gm
ΔH=−cmΔT(c=4.18 kJKg⁻¹)
⇒ ΔH=−4.18×0.1×3
⇒ ΔH=−1.254 kJ (For 5 millimoles of water formed)
For 1 mole water = −1.254/5×10⁻³
=−2.5×10²kJ

The correct answer is option C
Ideally, the enthalpy of neutralization should be - 57.3 K.J + 1.5 K.J = - 55.8 KJ.
But it is - 56.1 KJ.
∴ Energy used for neutralization 
= 57.3 - 56.1 = 1.2 KJ
∴ Percent ionization of weak acid

∴ % of weak acid in solution = 20%

∆H given in the question is for one mole of C (g). If 6 gm of diamond and graphite are burnt in oxygen then the C (diamond) will first convert to graphite and then it will form CO₂. While C (graphite) will directly form CO₂. So due to the conversion of diamond into graphite, we will get extra heat. And since we have 6gm (0.5 mol) of diamond, so the heat released will be 0.5×1.9 kJ or 0.95 kJ more than the second case.

According to Hess law, the thermal effects of a reaction depends upon the initial and final conditions of the reacting substances. It does not depend upon the immediate steps.

According to Hess law, the thermal effects of a reaction depends upon the initial and final conditions of the reacting substances. It does not depend upon the immediate steps.

For the reaction of 1 mole of H⁺ and OH^-,we have 13.6 kcal energy released. In H₂SO₄, we have 2 moles of H⁺. So for its complete neutralisation, we need 2 moles of NaOH. 
So in the end, 2 moles of H⁺ reacts with2 moles of OH^- and 13.6 2 = 27.4 kcal energy is released.

Product 1-Does not have any chiral carbon so only one product
Product 2-Two chiral centres so RR, RS and SR possible
Product 3-Two chiral centres so RR, RS and SR possible
Product 4-Two chiral centres but plane of symmetry exists, so only two isomers.

For boiling point, we have to consider both branching and Molecular mass. In 4  bromopentane molecular mass is nearly the same as compared to 3 chloro pentane but we have 3,3-dichloropentane extended into 2 directions so the boiling point of 3,3-dichloropentane will be more and the other order will be followed by option C.

This is the case of allylic substitution. The radical will delocalise itself to two locations. And so, we will get two different positions for radical substitution.

There is also a possibility at C₆, but it is not in our option.

The statements (B) and (C) are true regarding free radical iodination of the alkane.
(B) Direct iodination of alkane with iodine in presence of light is impractical.
During this reaction, Hl, a strong reducing agent is obtained as a byproduct, which catalyzes the reverse reaction thereby preventing direct iodination of alkane.
(C) Iodination of alkane can be achieved successfully using an oxidizing agent catalyst. 
Presence of an oxidizing agent oxidizes unwanted byproduct Hl and enables iodination of alkane to proceed.