In petrochemical industry alcohols are directly converted to gasoline by passing over heated

Platinum
ZSM-5
Iron
Nickel

Solution

ZSM-5 is a shape selective catalyst. Zeolites are good shape selective catalysts because of the honey comb like structure

Solution

Assuming that Hund’s rule is violated, the bond order and magnetic nature of the diatomic molecule B₂ is:

1 and diamagnetic
0 and dimagnetic
1 and paramagnetic
0 and paramagnetic

Solution

Copper displaces which of the metal from their salt solutions ?

AgNO₃
ZnSO₄
FeSO₄
All of the above

Solution

Cu is more electropositive than Ag therefore, it displaces Ag from their salt solution AgNO₃.

Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag

The unit of rate constant for a zero order reaction is

mol L^-1 s^-1
L mol^-1 s^-1
L² mol^-1 s^-1
s^-1

Solution

Rate = k[A]°
Unit of k = mol L^-1 sec^-1

The first order rate constant for a certain reaction increases from 1.667 × 10^-6 s^-1 at 727ºC to 1.667 × 10^-4 s^-1 at 1571 ºC.The rate constant at 1150ºC, assuming constancy of activation energy over the given temperature range is[Given : log 19.9 = 1.299]

3.911 × 10^-5 s^-1
1.139 × 10^-5 s^-1
3.318 × 10^-5 s^-1
1.193 × 10^-5 s^-1

Solution

In \(O_{2}^{-}\), O₂ and \(O_{2}^{-2}\) molecular species, the total number of anti bonding electrons respectively are

7, 6, 8
1, 0, 2
6, 6, 6
8, 6, 8

Solution

Which of the following terms is not correct for hydrogen ?

Its molecule is diatomic
It exists both as H⁺ and H^- in different chemical compounds
It is the only species which has no neutrons in the nucleus
Heavy water is unstable because hydrogen is substituted by its isotope deuterium

Solution

Heavy water is stable.

Which of the following volume (V) – temperature (T) plots represents the behaviour of one mole of an ideal gas at one atmospheric pressure ?

Solution

Find the volume by either V = RT/P (PV = RT) or P₁V₁ = P₂V₂ and match it with the values given in graph to find correct answer.

Volume of 1 mole of an ideal gas at 273 K and 1 atm pressure is 22.4 L and that at 373 K and 1 atm pressure is calculated as ;

\(V = \frac{RT}{P} = \frac{0.082 \times 373}{1}\) = 30.58 L = 30.6 L

The formula of ferrocene is

[Fe(CN)₆]^4–
[Fe(CN)₆]^3–
[Fe(CO)₅]
[(C₅H₅)₂Fe]

Which of the following is the wrong statement ?

ONCl and ONO^– are not isoelectronic.
O3 molecule is bent
Ozone is violet-black in solid state
Ozone is diamagnetic gas.
All of above

Solution

The basic character of the transition metal monoxides follows the order(Atomic Nos.,Ti = 22, V = 23, Cr = 24,Fe = 26)

TiO > VO >CrO > FeO
VO > CrO> TiO > FeO
CrO > VO > FeO > TiO
TiO > FeO > VO > CrO

Solution

The order of basic character of the transition metal monoxide is TiO > VO > CrO > FeO because basic character of oxides decreases with increase in atomic number.

Solution

Enamine (ene for carbon-carbon double bond and amine for amine group). >C = C – N<

The magnetic moments of Cu(Z = 29), Ti (Z = 22), and Cr (Z = 24) are in the ratio of

1 : √5 : 4
4 : √5 : 1
1 : 2 : 6
1 : √3 : √21

Commercial 10 volume H₂O₂ is a solution with a strength of approximately

Solution

Strength of 10V H₂O₂ = \(\frac{68 \times 10}{22400}\) × 100 = 3.035%

Standard enthalpy of vapourisation Δ_vapH° for water at 100°C is 40.66 kJ mol^-1. The internal energy of vaporisation of water at 100°C (in kJ mol^-1) is : (Assume water vapour to behave like an ideal gas).

+ 37.56
– 43.76
+ 43.76
+ 40.66

Solution

Below, some catalysts and corresponding proceses/reactions are matched.The mismatch is :

[RhCl(PPh₃)₂]: Hydrogenation
TiCl₄ + Al(C₂H₅)₃: Polymerization
V₂O₅: Haber-Bosch process
Nickel-Hydrogenation

Solution

V₂O₅ is used as a catalyst in contact process for the manufacture of SO₃ and hence H₂SO₄. In Haber-Bosch process for the manufacture of NH₃, finely divided Fe+ molybdenum are used.

C₂H₆ (g) + nO₂(g) ⟶ CO₂(g) + H₂O(l)In this equation, the ratio of the coefficients of CO₂ and H₂O is

1 : 1
2 : 3
3 : 2
1 : 3

Solution

The balanced equation is 2C₂H₆ + 7O₂ ⟶ 4CO₂ + 6H₂O.
Ratio of the coefficients of CO₂ and H₂O is 4 : 6 or 2 : 3

The treatment of Cu with dilute HNO₃ gives :

N₂O
NH₂
\(NH_{4}^{+}\)
NO₂

Solution

3Cu+ (dil) 8 HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O

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