The only non-metal which is liquid at ordinary temperature is

Hg
Br₂
NH₃
None of the above

Sodium nitroprusside, when added to an alkaline solution of sulphide ions, produces purple colour ion due to the formation of

Na[Fe(H₂O)₅NOS]
Na₂[Fe(H₂O)₅NOS]
Na₃[Fe(CN)₅NOS]
Na₄[Fe(CN)₅NOS]

Solution

Which of the following has highest knocking ?

Olefins
Branchedchain olefins
Straightchain olefins
Aromatic hydrocarbons

Solution

Knocking depends upon the structure of the hydrocarbon and follows the ordern-Alkanes >branched alkanes > straight chain alkenes> branched alkenes> arenes

Which of the following is true regarding hyper conjugation,also known as no-bond resonance?
(i)Like inductive effect it involves donation of electrons through ?-bonds
(ii)Hyper conjugation involves overlapping of filled orbitals with the empty p orbital of the carbocation
(iii)Like resonance, it involves displacement of ? or lone pair of electrons to the carbon bearing positive charge
(iv)It involves delocalisation of ?- and ?- electrons

(ii) and (iv)
(ii)
(iii)
none

Solution

In pyrophosphoric acid, H₂P₂O₇, number of ? and d? – p? bonds are respectively

8 and 2
6 and 2
12 and zero
12 and 2

Solution

A colloidal solution is subjected to an electric field. The particles move towards anode. The coagulation of same sol is studied using NaCl, BaCl₂ and AlCl₃ solutions. The order of their coagulation power should be–

NaCl > BaCl₂ > AlCl₃
BaCl₂ > AlCl₃ > NaCl
AlCl₃ > BaCl₂ > NaCl
BaCl₂ > NaCl > AlCl₃

Solution

As colloidal particles move towards anode so these particles are negatively charged and coagulated by cations of electrolyte.According to Hardy Schulze rule,
Coagulation power ∝ charge of ion
∴ Order of coagulation power is Al³⁺ > Ba^2+> > Na⁺

Solution

The intermediate is carbocation which is destabilised by C = O group (present on α-carbon to the –OH group) in the first three cases. In (d), α–hydrogen is more acidic which can be removed as water. Moreover,the positive charge on the intermediate carbocation is relatively away from the C= O group.

Solution

Two liquids X and Y form an ideal solution. At 300 K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this solution,vapour pressure of the solution increases by 10 mm Hg. Vapour pressure ( in mm Hg) of X and Y in their pure states will be, respectively:

300 and 400
400 and 600
500 and 600
200 and 300

Solution

The following group does not contain a dative bond

–NO₂
-N₂Cl
–NC
all of these

Solution

Dative bond is formed between two atoms when the electrons of shared pair are contributed solely by one of the two atoms and it is directed toward that atom which do not contribute the electrons. It is represented by an arrow on the bond which is directed towards the electron recipient atom.

So, –N⁺ ≣ NCl^- is the only group among the given options which does not contain a dative bond.

The rate constant is doubled when temperature increases from 27°C to 37°C. Activation energy in kJ is

Solution

Solution

The order w.r.t. I₂ is zero because the rate is not dependent on the concentration of I₂.

\(CH_{3} – CH_{2}C \equiv N \overset{X}{\rightarrow} CH_{3}CH_{2}CHO\) The compound X is

SnCl₂/ HCl / H₂O, boil
H₂ / Pd – BaSO₄
LiAIH₄/ ether
NaBH₄/ ether / H₃O⁺

Solution

When excess of SnCl₂ is added to a solution of HgCl₂, a white precipitate turning to grey is obtained. This grey colour is due to the formation of

Hg₂Cl₂
Cl₄SnCl
Sn
Hg

Solution

2HgCl₂ + SnCl₂ → \(\underset{White\, ppt}{HgCl_{2}}\) + SnCl₄

Hg₂Cl₂SnCl₂ → \(\underset{grey}{2Hg}\) + SnCl₄

In the first order reaction, the concentration of the reactant is reduced to 25% in one hour. The half life period of the reaction is

2 hr
4 hr
1/2 hr
1/4 hr

Solution

The Ca²⁺ and F–are located in CaF₂ crystal, respectively at face centred cubic lattice points and in

tetrahedral voids
half of tetrahedral voids
octahedral voids
half of octahedral voids

Which of the following is the correct decreasing order of acidic strength of

(i)Methanoic acid (ii)Ethanoic acid
(iii)Propanoic acid (iv)Butanoic acid

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

Solution

An electron releasing substituent (+I) intensify the negative charge on the anion resulting in the decrease of stability and thus decreases the acidity of the acid. Hence acid character decreases as the +I-effect of the alkyl group increases as
\(CH_{3}^{-}<CH_{3}CH_{2}^{-}<CH_{3}CH_{2}CH_{2}^{-}<CH_{3}CH_{2}CH_{2}CH_{2}^{-}\)
Hence the order becomes :(i) > (ii) > (iii) > (iv)

The melting point of lithium (181°C) is just double the melting point of sodium (98°C) because –

down the group, the hydration energy decreases
down the group, the ionization energy decreases
down the group the cohesive energy decreases
None of these

Solution

The atom becomes larger on descending the group, so the bonds becomes weaker (metallic bond), the cohesive force/energy decreases and accordingly melting point also decreases.

If the value of C_p for nitrogen gas is 7 JK^-1 mol^-1, then the value of ?H on heating 28 g of nitrogen gas from 0°C to 100°Cat constant pressure will be :

1200 J
1300 J
1400 J
1500 J

Solution

Which one of the following substances gives a positive charged sol?

Gold
A metal sulphide
Ferric hydroxide
An acidic dye

Solution

Fe(OH)₃ particles absorb Fe³⁺ ions and get peptized to give a positively charged sol

UnAttempted

0

Attempted

0

Correct

0

Wrong