Which one of the following exist in the oxidation state other than +3?

Solution

Boron (B), aluminium (Al) and gallium (Ga) are present in group 13. They show + 3 oxidation state. While cerium (Ce) is a lanthanoid. It presents in lanthanide series. It shows + 3 and + 4 oxidation states.

Which pair does not show hydrogen isotopes ?

Ortho and para hydrogen
Protium & deuterium
Deuterium & tritium
Tritium and protium

Solution

Ortho and para hydrogens are two forms of hydrogen(₁H¹ ) which differ only in the direction of spin of proton.

The volume of 4 N HCl and 10 N HCl required to make 1 litre of 6 NHCl are

0.75 litre of 10 N HCl and 0.25 litre of 4 N HCl
0.50 litre of 4N HCl and 0.50 litre of 10 N HCl
0.67 litre of 4 N HCl and 0.33 litre of 10 N HCl
0.80 litre of 4N HCl and 0.20 litre of 10 N HCl

Solution

N₁V₁ + N₂V₂ = NV
4x+ 10 (1– x) = 6 × 1; –6x = –4 ; x = 0.67
Thus 0.67 litre of 4N HCl
1 – x= 1– 0.67 = 0.33 litre of 10 N HCl

The number of d-electrons retained in Fe²⁺ (At.no. of Fe = 26) ion is

Solution

Fe⁺⁺ (26 – 2 = 24) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰ 3d⁶ hence no.of delectrons retained is 6. [Two 4s electron are removed]

Which is correct about physical adsorption?

High temperature and high pressure favour adsorption
High temperature and low pressure favour adsorption
Low temperature and high pressure favour adsorption
Low temperature and low pressure favour adsorption

Solution

Physical adsorption involves weak forces, physical in nature with small heat of adsorption. Thus low temperature and high pressure favours physical adsorption.

Which of the following compounds will give a yellow precipitate with iodine and alkali ?

(i)Acetophenone (ii)Acetamide

(iii)Methyl acetate (iv)2-Hydroxypropane

(i), (ii) and (iii)
(i) and (iv)
(ii) and (iv)
(i), (iii) and (iv)

Solution

Which one of the following statements regarding helium is incorrect ?

It is used to produce and sustain powerful superconducting magnets.
It is used as a cryogenic agent for carrying out experiments at low temperatures.
It is used to fill gas balloons instead of hydrogen because it is lighter and non-inflammable.
It is used in gas-cooled nuclear reactors.

Solution

Helium is twice as heavy as hydrogen it is in flammable but not lighter than hydrogen. Helium has the lowest melting and boiling point of any element which makes liquid helium an ideal coolant for many extremely low temperature application such as super conducting magnet and cryogenic research where temperature close to absolute zero are needed. He is used in gas cool edatomic reactors as a heat transfer agent.

Solubility product of silver bromide is 5.0 × 10^-13. The quantity of potassium bromide (molar mass taken as 120 g mol^-1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is

1.2 × 10^-10 g
1.2 × 10^-9 g
6.2 × 10^-5 g
5.0 ×10^-8 g

Solution

Based on lattice energy and other considerations which one of the following alkali metal chlorides is expected to have the highest melting point

LiCl
NaCl
KCl
RbCl

Solution

Although lattice energy of LiCl is higher than NaCl but LiCl is covalent in nature and NaCl ionic there after, the melting point decreases as we move towards NaCl because the lattice energy decreases as a size of alkali metal atom increases (lattice energy α melting point of alkali metal halide)

The product obtained by the reaction of an aldehyde and hydroxylamine is

hydrazone
aldoxime
primary amine
alcohol

The correct order of heat of formation of halogen acids is

HI > HBr >HCl > HF
HF > HCl > HBr > HI
HCl > HF > HBr > HI
HCl > HBr > HF > HI

Solution

On moving from top to bottom of halogen group the bond dissociation energy of hydrogen halides decreases and so the heat of formation of halogen acids also decreases.

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 : [Ti(H₂O)₆]³⁺ is coloured while [Sc(H₂O)₆]³⁺ is colourless.

Statement-1 : d-d transition is not possible in [Sc(H₂O)₆]³⁺.

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

Both statement-1 and statement-2 are true and statement-2 is the correct explanation of statement-1.[Sc(H₂O₆]³⁺ has no unpaired electron in its d subshell and thus d–d transition is not possible whereas[Ti(H₂O)₆]³⁺ has one unpaired electron in its d subshell which gives rise to d–d transition to impart colour

Oxidation state of phosphorus in \(PO_{4}^{3-}\),P₄O₁₀ and \(P_{2}O_{7}^{4-}\) is :

+ 3
+ 2
– 3
+ 5

Solution

In \(PO_{4}^{3-}\) : x + 4(–2) = –3 ⇒ x = +5
In P₄O₁₀

Four successive members of the first row transition elements are listed below with atomic numbers. Which one of them is expected to have the highest E°_{M³⁺/M²⁺} value ?

Cr(Z =24)
Mn(Z = 25)
Fe(Z = 26)
Co(Z = 27)

Solution

Arrange the following compounds in order of increasing solubility
(i)MgF₂
(ii)CaF₂ ,
(iii)BaF₂

(i)< (ii) < (iii)
(ii) < (i) < (iii)
(ii) < (iii) <(ii)
(iii) < (ii) < (i)

Solution

BaF₂ > MgF₂ > CaF₂

For a first order reaction the rate constant is 6.909 min^-1.The time taken for 75% conversion in minutes is

\(\frac{3}{2}log 2\)
\(\frac{2}{3}log 3\)
\(\frac{2}{3}log 2\)
\(\frac{2}{3}log (\frac{3}{4})\)

Solution

Match List – I (substances) with List – II (processes) employed in the manufacture of the substances and select the correct option.

List – I List – II

Substances Processes

(A) Sulphuric acid (i) Haber’s process

(B) Steel (ii) Bessemer’s process

(D) Ammonia (iv) Contact process Options

(iv) (ii) (iii) (i)
(i) (iv) (ii) (iii)
(i) (ii) (iii) (iv)
(iv) (iii) (ii) (i)

Solution

(A)Sulphuric acid (iv)Contact process

(B)Steel (ii)Bessemer’s process

(C)Sodium hydroxide (iii)Leblanc process

(D)Ammonia (i)Haber’s process

An inorganic compound which on heating first melts, then solidifies and liberates O₃ gas, the inorganic compound is

Al₂O₃
KMnO₄
MnO₂
KClO₃

Solution

The Ce (Z = 58) belongs to III rd group of periodic table.If it furnish one α particle to form an element ‘X’, then X belongs to

III^rd group
II^nd group
I^st group
zero group

Solution

₅₈Ce is lanthanide. Lanthanides are from 57 to 71 all present in III^rd group. Hence ₅₈Ce forms ₅₆X element on emission of one α particle with belongs to II^nd group.

Which of the following ions can exhibit optical isomerism–

[Co(NH₃)₄Cl₂]⁺
[Co(NH₃)₂Cl₄]^-
Cis-[Co(en)₂Cl₂]⁺
trans-[Co(en)₂Cl₂]⁺

Solution

cis-[Co(en)₂Cl₂]+: cis-[M(aa)₂b₂] can show optical isomerism.

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