Sample JEE Main Chemistry Mock Test - JEE MCQ

Phenol is more acidic than cyclohexanol and acyclic alcohols because the phenoxide ion is more stable than the alkoxide ion. In an alkoxide ion, such as the one derived from cyclohexanol, the negative charge is localized at the oxygen atom.

For a compound to be ionic there must be:

• large cation
• small anion.

Lesser the number of charge, which is there in MnO.

Principle quantum no. (n) represent shell no. while azimuthal quantum no. (l) represent sub shell no.

• Deuterium is also known as "Heavy Hydrogen".
• A Deuterium atom consists of a proton, a neutron and an electron.
• It is just like the Hydrogen, but only with a. neutron.
• So, electronic configuration is same as that of Hydrogen: 1s¹.

• An electrophile is a reagent attracted to electrons.
• Electrophiles are positively charged or neutral species having vacant orbitals that are attracted to an electron rich centre.
• It participates in a chemical reaction by accepting an electron pair in order to bond to a nucleophile.

• Polarity produced in a molecule due to transmission of unequal sharing/partial displacement of the bonding electron through a chain of atoms in molecules towards more electronegative atom, giving rise to a permanent dipole in a bond.
• It is present in a sigma bond, as opposed to the electromeric effect, which is present on a π-bond. it's called +I effect.

It is achiral as it has two (−CH2OH)(−CH2OH) groups

KO₂ is a superoxide having anion as( O2)^(-1) and so number of valence electrons = odd and hence it is paramagnetic

The IUPAC name of the given coordination complex according to the nomenclature rules is Pentamminenitrito-N-cobalt(III)chloride. Hence the correct option is B.

An octahedral complex has 6 bonds around the central atom. One EDTA molecule complexes with one molecule of calcium cation to form octahedral complex.

This is because pairing in these (5d & 4d series) elements does not depend upon strength of ligand. it takes almost always, forming low spin complex, unlike 3-d series elements , pairing in 3d series elements depends upon strength of ligand.

There are three sigma bonds and three pi bonds showing a hybridization type: sp³ type.

• Solubility of NaCl decreases.
• Because of the increase in Cl- ions from the HCl , common ion effect takes place and ionisation of NaCl decreases reducing its solubility.
• Hence, correct answer is (D).

But-1-ene and cyclobutane exhibits ring chain isomerism.
 

• In carbon family, C and Si show +4 oxidation state. The remaining elements of this group, i.e., Ge, Sn and Pb, however show two oxidation states +2 and +4 due to inert pair effect which arises due to ineffective shielding of the valence s-electrons by the intervening d- or f-electrons.
• Evidently, as the number of d and f-electrons increases down the group from Ge to Pb, the inert pair effect becomes more and more prominant.
• As a result, the stability of +4 oxidation state decreases while that of the +2 oxidation state increases from Ge to Pb.

It is a thermosetting plastic. It is obtained by condensation of ethylene glycol and phthalic acid or glycerol and phthalic acid. When its solution in a suitable solvent is evaporated, it leaves a tough but non-flexible film. It is, therefore, used in the manufacture of paints and lacquers.

• The vant Hoff factor for the non electrolyte is equal to 1 whereas that of non volatile electrolyte is greater than 1.
• Hence all colligative properties will be increased.

As the given reaction is oxidation reaction and lithium has highest oxidation potential, so it has highest tendency for this half reaction.

(A) = CrO₂Cl₂
(B) = Na₂CrO₄
(C) = CrO₅

FeSO₄ (NH₄)₂ SO₄ . 6H₂O + Fe₂ (SO₄)₃
Mohr salt
x mol               (4–x) mol
∵ (Eq) K₂Cr₂O₇ = (Eq) Mohr solt
(Cr⁺⁶ → Cr⁺³)    (Fe⁺² → Fe⁺³)
N × V_(L)                        = mol × n factor
1 × 6 × 0.5                   =  x × 1
x = 3 mol
= 75