The ionisation potential of a hydrogen atom is –13.6 eV. What will be the energy of the atom corresponding to n = 2.

–3.4 eV
–6.8 eV
–1.7 eV
–2.7 eV

Solution

Energy of an electron E = ^-E₀⁄_n²

For energy level (n = 2)

E = ^-13.6⁄_(2)² = ^-13.6⁄₄ = -3.4 eV.

100 ml solution of an acid of pH= 1, is mixed with 100 ml solution of the same acid of pH = 2. The hydrogen ion concentration of the mixture will be:

5.5 × 10^–2
10^–2
10^–3
10^–4

Solution

The type of glass used in making lenses and prisms is

Flint glass
Jena glass
Pyrex glass
Quartz glass

Read the passage given below and answer the question that follow :
The distribution of electrons among various molecular orbitals is called the electronic configuration of the molecule which provides us the following very important information about the molecule.
(I)Stability of molecule : The molecule is stable if number of bonding molecular orbital electrons (N_b ) is greater than the number of anti bonding molecular orbital electrons(N_a).
(II)Bond order = 12 ( N_b – N_a )A positive bond order means a stable molecule while a negative or zero bond order means an unstable molecule.
(III)Nature of the bond : Bond order 1, 2 and 3 corresponds to single, double and triple bonds respectively.
(IV)Bond length : Bond length decreases as bond order increases.
(V)Magnetic nature : If molecular orbitals in a molecule are doubly occupied, the substance is diamagnetic and if one or more molecular orbitals are singly occupied, it is paramagnetic.

Which of the following statement is incorrect ?

Among \(O_{2}^{+}\), O₂ and \(O_{2}^{-}\)the bond length decreases as \(O_{2}^{-}\) > O₂ > \(O_{2}^{+}\)
He₂ molecule does not exist as the number of electrons in bonding and anti-bonding orbitals are equal.
C₂, \(O_{2}^{2-}\) and Li₂ are diamagnetic
In F₂ molecule, the energy of ?2p_z is more than ?2p_x and ?2p_y

Solution

(i)Bond order ? 1/bond length Bond order \(O_{2}^{+}\) = 2.5, O₂= 2, \(O_{2}^{-}\) = 1.5 So correct order of bond length is \(O_{2}^{-}\) > O₂ > \(O_{2}^{+}\)

Solution

C₂H₅CH₂CHO
C₂H₅CH₂CH₂OH
C₂H₅CH₂OH
C₂H₅CHO

Solution

The electrical charge on a colloidal particle is observed by:

Ultramicroscope
Scattering
Brownian movement
Electrophoresis

Solution

The electrical change on a colloidal particle is observed by electrophoresis. Under electric field, charged particles move in a particular direction.

The region which is greatly affected by air pollution is

Thermosphere
Stratosphere
Troposphere
Mesosphere

Solution

Air pollution greatly affect the troposphere.

Solution

Since water is polar in nature and like dissolves like,the coating must be non polar to polar manner.

Extraction of zinc from zinc blende is achieved by

electrolytic reduction
roasting followed by reduction with carbon
roasting followed by reduction with another metal
roasting followed by self-reduction

Solution

Extraction of Zn from ZnS(Zinc blende) is achieved by roasting followed by reduction with carbon.
2ZnS + 3O₂ ⟶ 2ZnO +2SO₂
ZnO + C ⟶ Zn + CO

The sequence of ionic mobility in aqueous solution is :

K⁺ > Na⁺ > Rb⁺ > Cs⁺
Cs⁺ > Rb⁺ > K⁺ > Na⁺
Rb⁺ > K⁺ > Cs⁺ > Na⁺
Na⁺ > K⁺ > Rb⁺ > Cs⁺

Solution

Smaller the ion more is its ionic mobility in aqueous solution. Ionic radii of the given alkali metals is in the order Na⁺ < K⁺ < Rb⁺ < Cs⁺ and thus expected ionic mobility will be in the order Cs⁺ < Rb⁺ < K⁺ < Na⁺.

However due to high degree of solvation (or hydration)because of lower size or high charge density, the hydrated ion size follows the order Cs⁺ < Rb⁺ < K⁺ < Na⁺ and thus conductivity order is Cs⁺ > Rb⁺ > K⁺ > Na⁺ i.e. option (b) is correct answer.

Nitrogen dioxide cannot be obtained by heating :

KNO₃
Pb(NO₃)₂
Cu(NO₃)₂
AgNO₃

Solution

Only nitrates of heavy metals and lithium decompose on heating to produce NO₂.

Given that bond energies of H – H and Cl –Cl are 430 kJ mol^-1and 240 kJ mol^-1 respectively and ?_fH for HCl is – 90 kJ mol^-1, bond enthalpy of HCl is

380 kJ mol^-1
425 kJ mol^-1
245 kJ mol^-1
290 kJ mol^-1

Solution

Solution

For first order decomposition of N₂O₅, rate law is given by rate, r =k[N₂O₅]
On plotting rate on y-axis and concn[N₂O₅] on x-axis and then comparing it with y = mx + c, we get a line starting from origin with slope'k'. Its positive slope suggests that line is going up. Hence graph shown in option (d) is the correct choice.

Which one of the following complex ions has geometrical isomers ?

(en = ethylenediamine)

[Ni(NH₃)⁵Br]⁺
[Co(NH₃)²(en)²]³⁺
[Cr(NH₃)⁴(en)²]³⁺
[Co(en)₃]³⁺

Solution

The molal elevation constant is the ratio of elevation in boiling point to:

molality
mole fraction
molarity
none

Solution

Molal elevation constant is given by the formula,
K_b = ΔT_b / m , where m is molality and ΔT_b is elevation in boiling point.

A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g. of CO₂. The empirical formula of the hydrocarbon is :

C₂H₄
C₃H₄
C₆H₅
C₇H₈

Solution

The best method to prepare cyclohexene from cyclohexanol is by using

Conc. HCl + ZnCl₂
Conc. H₃PO₄
HBr
Conc. HCl

Solution

Conc. HCl, HBr and conc. HCl + ZnCl₂ all are nucleophiles,thus convert alcohols to alkyl halides. However, conc. H₃PO₄ is a good dehydrating agent which converts an alcohol to an alkene.

The element X [1s², 2s²2p⁶,3s²3p⁵] reacts with Y (atomic no.= 1) to form

XY₂
X₂Y
XY
None

Solution

Which of the following substances acts as an oxidising as well as a reducing agent?

Na₂O
SnCl₂
Na₂O₂
NaNO₂

Solution

In Na₂O, SnCl₂ and Na₂O₂ central atom is either in lowest or highest oxidation state, so it can function either as an oxidising or a reducing agent but not both.However, the oxidation state of N in NaNO₂ is +3 which lies between its highest(+5) and lowest (–3) values.

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