Which of the following is strongest nucleophile –

Br^-
: OH^-
:CN^-
\(c_{2}H_{5}\bar{O}:\)

Solution

The strength of nucleophile depends upon the nature of alkyl group R on which nucleophile has to attack and also on the nature of solvent. The order of strength of nucleophiles follows the order :CN^- > I^- > C₆H₅O^- > OH^- > Br^- > Cl^-

The role of fluorspar ( CaF₂) which is added in small quantities in the electrolytic reducation of alumina dissolved in fused cryolite (Na₃AlF₆) is

As a catalyst
To make the fused mixture very conducting
To increase the temperature of the melt.
To decrease the rate of oxidation of carbon at the anode.

Solution

CaF₂ when added to fused Cryolite, lowers them.p.and increases the conductivity.

Which of the following statements about amorphous solids is incorrect ?

They melt over a range of temperature
They are anisotropic
There is no orderly arrangement of particles
They are rigid and in compressible

Solution

Amorphous solids are isotropic, because these substances show same properties in all directions.

Which of the following can not be isoelectronic?

two different cations
two different anions
cation and anion
two different atoms

Stainless steel contains iron and

Cr + Ni
Cr + Zn
Zn + Pb
C +Cr + Ni

Eka-aluminium and EKa-silicon are known as

Gallium and Germanium
Aluminium and Silicon
Iron and Sulphur
Neutron and Magnesium

Stronger is oxidising agent, more is :

standard reduction potential of that species
the tendency to get it self oxidised
the tendency to lose electrons by that species
standard oxidation potential of that species

Solution

More is E_RP^°, more is the tendency to get itself reduced or more is oxidising power.

The rate of a first order reaction is 1.5 × 10^-2 mol L^-1 min^-1 at 0.5 M concentration of the reactant. The half life of the reaction is

0.383 min
23.1 min
8.73 min
7.53 min

Solution

If 0.5 mol of BaCl₂ is mixed with 0.2 mole of Na₃PO₄, find the maximum amount of Ba₃(PO₄)₂ that can be formed.

1 mole
0.5 mole
0.1 mole
0.01 mole

Solution

Solution

KNO₃ is a strong electrolyte which dissociates into two ions. Therefore, its van’t Hoff factor is 2. Acetic acid (CH₃COOH) is a weak electrolyte, it does not dissociate completely. So, its van’t Hoff factor is less than that of KNO₃
∴ Osmotic pressure of 0.1 M KNO₃ > Osmotic pressure of 0.1 M CH₃COOH
or P₁ > P₂

The geometrical isomerism is shown by

Solution

Geometrical isomerism is obseved when different groups are attached to each of the doubly bonded carbon atom.

Solution

(ΔH –ΔU) for the formation of carbon monoxide (CO) from its elements at 298 K is(R = 8.314J K^-1mol^-1)

–2477.57 J mol^-1
2477.57 J mol^-1
–1238.78 J mol^-1
1238.78 J mol^-1

Solution

C(s)+¹⁄₂O₂(g)→CO(g)

[Δn=1-¹⁄₂=¹⁄₂]

ΔH –ΔU =ΔnRT=¹⁄₂×18.314×298=1238.78 J mol^-1

The existence of two different coloured complexes with the composition of [Co(NH₃)₄Cl₂]⁺ is due to :

linkage isomerism
geometrical isomerism
coordination isomerism
ionization isomerism

Solution

A meta-directing functional group is :

— COOH
OH
—CH3
— Br

Solution

DIRECTIONS : These are Assertion-Reason type questions. Each of these question contains two statements: Statement-1 (Assertion) and Statement-2 (Reason). Answer these questions from the following four options.

Statement – 1 : Nuclide ³⁰Al₁₃ is less stable than ⁴⁰Ca₂₀

Statement – 2 : Nuclide having odd number of protons and neutrons are generally unstable.

Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1
Statement - 1 is True, Statement - 2 is True; Statement - 2 is NOT a correct explanation for Statement - 1
Statement - 1 is True, Statement - 2 is False
Statement - 1 is False, Statement - 2 is True

Solution

It is observed that a nucleus which is made up of even number of nucleons (No. of n & p) is more stable than nuclie which consist of odd number of nucleons. If number of neutron or proton is equal to some numbers i.e., 2, 8, 20, 50, 82, or 126 (which are called magic numbers),then these posses extra stability.

Bauxite ore is made up of Al₂O₃ + SiO₂ + TiO₂ + Fe₂O₃. This ore is treated with conc. NaOH solution at 500 K and 35 bar pressure for few hours and filtered hot. In the filtrate the species present, is/are

NaAl(OH)₄ only
Na2Ti(OH)₆ only
NaAl(OH)₄ and Na2SiO₃
Na₂SiO₃ only

Solution

The bond angle between two hybrid orbitals is 105°. The percentage of s-character of hybrid orbital is between

50 - 55%
9 - 12%
22 - 23%
11 - 12%

Solution

Read the passage given below and answer the question that follow :J.W. Gibbs and H.Von Helmoltz had given two equations which are known as Gibbs-Helmholtz equation. One equation can be expressed in terms of change in free energy (ΔG) and enthalpy(ΔH) while other can be expressed in terms of change in internal energy (ΔE) and work function (ΔW)

Internal energy change at 25ºC is ΔE₁ while at 35ºC is ΔE₂ then –

ΔE₁= ΔE₂
ΔE₂ > ΔE₁
ΔE₁ > ΔE₂
ΔE₁ ≥ ΔE₂

Solution

ΔE = ΔW – Td \(\left (\frac{\Delta w}{dT} \right )\)
From above relation it is clear that with increase in temperature ΔE decreases.So ΔE₁ > ΔE₂

DIRECTIONS: Read the passage given below and answer the question that follow :A hydrocarbon (X) of the formula C₆H₁₂ does not react with bromine water but react with bromine in presence of light, forming compound (Y). Compound (Y) on treatment with Alc. KOH gives compound [Z] which on ozonolysis gives (T) of the formula C₆H₁₀O₂. Compound (T) reduces Tollens reagent and gives compound (W). (W) gives iodoform test and produce compound(U) which when heated with P₂O₅ forms a cyclic anhydride (V)

Solution

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