A metal ‘M’reacts with N₂ to give a compound ‘A’ (M₃N).‘A’ on heating at high temperature gives back ‘M’ and ‘A’;on reacting with H₂O gives a gas B. ‘B’ turns CuSO₄ solution blue on passing through it. M and B can be

Al and NH₃
Li and NH₃
Na and NH₃
Mg and NH₃

Solution

When CH₂ = CH —COOH is reduced with LiAlH₄, the compound obtained will be

CH₂= CH —CH₂OH
CH₃ — CH₂ — CH₂OH
CH₃— CH₂— CHO
CH₃— CH₂— COOH

Solution

The ease of liquefaction of noble gases increases in the order

He < Ne
Xe
Kr
Ar < Kr < Xe < Ne < He

Solution

As size increases, van der Waal's forces of attraction between noble gas atoms also increases. Consequently,ease of their liquefaction increases.

The oxidation number of sulphur in S₈, S₂F₂, H₂S respectively, are

0, +1 and –2
+2, +1 and –2
0, +1 and +2
–2, +1 and –2

Solution

(i) Oxidation state of element in its free state is zero.
(ii) Sum of oxidation states of all atoms in compound is zero.O.N. of S in S₂₈= 0; O.N. of S in S₂F₂= + 1;O.N. of S in H₂S = –2;

In which of the following molecules / ions BF₃, \(NO_{2}^{-}\),\(NH_{2}^{-}\)and H₂O , the central atom is sp² hybridized ?

\(NH_{2}^{-}\) and H₂O
\(NO_{2}^{-}\) and H₂O
BF₃ and \(NO_{2}^{-}\)
\(NO_{2}^{-}\) and \(NH_{2}^{-}\)

Solution

On determining hybridisation from H =1/2 (V + M – C + A).The hybridisation of BF₃, \(NO_{2}^{-}\), \(NH_{2}^{-}\) and H₂O are sp2, sp2, sp3 and sp3 respectively.

Which compound in each of the following pairs is most reactive to the conditions indicated ?

A and C
B and C
A and D
B and D

Solution

In the first case the reaction gives side chain substitution product which is easier in A. In the second case the reaction will proceed by benzyne mechanism for which ortho position w. r. to Cl must have H-atoms.

Work done during isothermal expansion of one mole of an ideal gas from 10 atm. to 1 atm. at 300 K is:

4938.8 joule
4138.8 joule
5744.1 joule
6257.2 joule

Solution

Molecular shapes of SF₄, CF₄ and XeF₄ are

the same, with 2, 0 and 1 lone pairs of electrons respectively
the same, with 1, 1 and 1 lone pairs of electrons respectively
different, with 0, 1 and 2 lone pairs of electrons respectively
different, with 1, 0 and 2 lone pairs of electrons respectively

Solution

Read the passage given below and answer the questions that follow :

A carbonyl compound P, which gives positive iodoform test,undergoes reaction with MeMgBr followed by dehydration togive an olefin Q. Ozonolys is of Q leads to a dicarbonyl compound R, which undergoes intramolecular aldol reaction to give predominantly S.

The structure of the product S is

Solution

Iodoform test of compound P points out that P has– COCH₃ group which shows that it may be either option(a) or (b). Further since the dicarbonyl compound R has at least one α-H atom (as it undergoes aldol reaction) w.r.tone of the carbonyl groups which is possible when R is produced from (b) ; (a) option will give dicarbonyl

Solution

Higher the value of reduction potential higher will be the oxidising power whereas the lower the value of reduction potential higher will be the reducing power.

Which one of the following statements regarding photochemical smog is not correct?

Carbon monoxide does not play any role in photochemical smog formation.
Photochemical smog is an oxidising agent in character.
Photochemical smog is formed through photochemical reaction involving solar energy.
Photochemical smog does not cause irritation in eyes and throat.

Solution

The oxidised hydrocarbons and ozone in presence of humidity cause photochemical smog.Hydrocarbons + O₂, NO₂, NO, O, O₃ ⟶ Peroxides,formaldehyde, peroxyacetylnitrate (PAN), acrolein etc.It is oxidising in nature and causes irritation to eyes,lungs, nose, asthamatic attack and damage plants.

After how many seconds will the concentration of the reactants in a first order reaction be halved, if the decay constant is 1.155 × 10^-3 sec^-1.

Solution

A compound of molecular formula of C₇H₁₆ shows optical isomerism, compound will be

2, 3-Dimethylpentane
2,2-Dimethylpentane
2, 4 - Dimethylpentane
None of these

Solution

A compound is said to exhibit optical isomerism if it atleast contains one chiral carbon atom, which is anatom bonded to 4 different atoms or groups.

Among the ligands NH³, en, CN^– and CO the correct order of their increasing field strength, is:

NH³ < en < CN^- < CO
CN^- < NH³ < CO < en
en < CN^- < NH³ < CO
CO < NH³ < en < CN^-

Solution

The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassium dichromate solution is:

Solution

How many stereoisomers does this molecule have?
CH₃CH = CHCH₂CHBrCH₃

Solution

DIRECTIONS : Read the passage given below and answer the questions that follow :

1 mole of CCl₄ vapours at 77°C occupies a volume of 35 litre. vander Waal’s constants are a = 20.39 L² atm mol^-2 and b = 0.1383 L mol^-1.

At T = 500K, P = 40 atm, the state will be liquid
At T = 300K, P = 50 atm, the state will be gas
At T < 300K, P > 20 atm, the state will be gas
At 300K < T < 500 K, P > 50 atm, the state will be liquid

Solution

Inside the dome showing graph state is liquid.

The rate constant of a reaction is 3.00 × 10³ L mol^-1 sec^-1.The order of this reaction will be:

zero
1
2
3

Solution

The energies E₁ and E₂ of two radiations are 25 eV and 50 eV,respectively. The relation between their wavelengths i.e., ?₁ and ?₂ will be :

?₁ = ?₂
?₁ = 2?₂
?₁ = 4?₂
?₁ = ¹⁄₂ ?₂

Solution

Solution

Bronsted base is a substance which accepts proton.In option (a), H₂O is accepting proton, i.e., acting as abase.

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