Mock Test - 1 - Class 12 MCQ

He is a monoatomic gas, So, it will have equal no. of atoms as that of N₂O₄ molecules

Vander Waal's constant ‘a’ represents, intermolecular attraction. So gas with higher value of ‘a’ can be easily liquefied.
∴ The order of gases will be Y > Z > X

In the above balanced reaction, it can be seen that 3 moles of Cl₂ gives 5 moles of Cl^- and 1 mole of ClO₃^-
∴ 1 mole of Cl will give 5/3 moles of Cl^- and 1/3 moles of ClO₃^-

ZnS crystallizes as a cubic close-packed array of S^2- ions with Zn⁺² ions in tetrahedral holes. If S^2- ions occupy the lattice points of a face-centered cubic unit cell and Zn⁺² ions are packed into every other tetrahedral hole, we get the unit cell of ZnS.

For strong electrolytes, van't Hoff factor 'i' equals to total number of ions in the formula unit.

∴ From the above reaction it can be seen that sodium tetra cyanozincate (II) liberates 3 ions on complete ionisation.

Electrode with least E⁰ has highest reducing power of chemical species on the left.
∴ Order of reducing ability is Zn > Fe²⁺ > Ag > Br^-

Final product of the reaction is methyl amine, intermediate product must be isocyanide as given in the following reaction.

Lindlar’s catalyst gives syn addition of H-atoms & Na in Liq. NH₃ gives anti-addition.
So, A → Cis-but-2-ene & B → Trans-but-2-ene

Elimination of 2⁰ halides is governed by Saytzeff’s rule, so, 2-Butene dominates here.

Glucose is a monosaccharide as it can not be further hydrolyzed.

Lanthanide contraction is the gradual decrease in the atomic and ionic size of lanthanoids with an increase in atomic number.
Causes of lanthanide contraction

• With an increase in the atomic number, the positive charge on nucleus increases by one unit and one more electron enters same 4f subshell. 
• The electrons in 4f subshell imperfectly shield each other. Shielding in a 4f subshell is lesser than in d subshell.
• With the increase in nuclear charge, the valence shell is pulled slightly towards the nucleus. This causes lanthanide contraction.

When H₂O₂ acts as oxidising agent, then H₂O₂ → H₂O + [O]

In reaction with Cu₂S, S is oxidised to SO₂ by nascent oxygen.

Spin multiplicity is 2S+1, where S is total spin

spin multiplicity = 

F^- ion has ability to form H-bonds with HF ; KHF₂ → K⁺ + HF₂^-

When intermolecular forces vanish, water liquid charge to gas, So, mass & moles remain same but volume changes 

According to Heisenberg's Uncertainty Principle, it is impossible to measure position x and momentum p simultaneously.
Using Δx.Δp ⩾ h/4π
as Δx→0, Δp→∞
Here, neither uncertainty can be zero and if the position is zero the uncertainty in momentum becomes infinite.

Disproportionation involves oxidation and reduction of atoms of the same element from same molecular species. Given reaction is cannizarro’s reaction.

Given pH = 3, Volume = 10 ml
∴ [H⁺] = 10^-3 M
No. of moles of H⁺ in 10 ml is ′x′.
= 10^-3 × 10^-3 × 10
=10^-5
No. of [H⁺] gms in 1 mole = 6.023×10²³
Hence, no. of H⁺ ion in 10^-5 moles = 6.023×10²³×10^-5
= 6.023×10¹⁸

It acts as reducing agent as, hydroxy group is oxidised & silver is reduced

Subtracting reaction (1) from reaction (2), We get

∴ Enthalpy change of the reaction is -252 KJ Mol^-1

• Acidic nature of oxides increases with increases in non-metallic character or oxidation state. 
• More the positive oxidation state of a non-metallic oxide more is its acidic nature.

• Propionaldehyde will undergo an aldol condensation reaction when it is treated with dilute NaOH (base).
• The reaction results in the formation of 3-hydroxy-2-methylpentanal.
• This reaction initially involves the formation of the enolate ion which attacks another mole of the aldehyde to form the beta-hydroxyaldehyde product.

The reaction is given as follows:

Reaction of propyne with 2 molecules of HBr to form 2,2 -dibromo propane can be written as:
CH₃​−C≡CH + 2HBr → CH₃​−C(Br)=CH₂​
CH₃​−C(Br)=CH₂ + HBr → CH₃−C(Br)₂−CH₃
The reaction follows Markownikoff's rule.

[Ni(CO)₄ has synergic bonding, CO is a π- acid ligand. Compound has C → Metal, sigma bond & Metal → C, π- bond.

• Chromyl chloride test is done for detecting the presence of Cl- ions.
• The chlorides of silver, lead, mercury and antimony are covalent in nature and thus do not generate Cl- ions and so they do not give the chromyl chloride test also.
• So, Heavy metal chlorides don't give this test because they are not ionic.

 dmg forms stable co-ordination compound with Ni²⁺. So, concentration of metal ion at anode decreases and emf increases.

HIO₄ oxidises following combinations at adjacent positions.

Due to strong electron-withdrawing effect of the - NO₂ group, the nucleophilicity of the - NH₂ is reduced and hence diazotisation becomes difficult.