Chemistry Test 251

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Chemistry Test 251

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Weekly Quiz Competition

Question 1
4 / -1

A reaction mixture containing H₂, N₂ and NH₃ has partial pressures 2 atm,1 atm and 3 atm respectively at 725 K. If the value of Kp for the reaction, N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is 4.28 × 10⁻⁵ atm⁻² at 725 K, in which direction the net reaction will go ?

A
Forward

B
Backward

C
No net reaction

D
Direction of reaction cannot be predicted

Solution

= 1.125 atm⁻²

Since value of Q_p is larger than K_p (4.28 × 10⁻⁵ atm⁻²), it indicates net reaction will proceed in backward direction.
Question 2
4 / -1

If SO₂ content in the atmosphere is 0.12ppm by volume, pH of rain water is (assume 100% ionisation of acid rain as monobasic acid)

A
5.7

B
5.6

C
5.4

D
2.0

Solution
Question 3
4 / -1

Consider the following reaction

Which of the following statements are correct ?

a. I is the major product of the reaction

b. II is the major product of the reaction

c. Formation of I is in accordance with Saytzeff's rule

d. II is more stable because it is more substituted

A
a,c

B
b,c

C
a,d

D
b,d

Solution
Question 4
4 / -1

A
3 and 4

B
All of these

C
2

D
1 and 3

Solution

The four lobes of orbital are lying along x and y axes, so, density in XY plane can't be zero.

The two lobes of orbital are lying along z axis, and contain a ring of negative charge surrounding the nucleus in xy plane. So, the density in XY plane is non-zero.

2s orbitals have one spherical node, where the electron density is zero.

In 2p_x orbital, both the lobes lie along x axis. Hence,the density in yz plane is zero, thus it is the nodal plane.
Question 5
4 / -1

Assertion (A): Mg²⁺ and Al³⁺ are isoelectronic but the magnitude of ionic radius of Al³⁺ is less than that in Mg²⁺.

Reason (R): The effective nuclear charge on the outermost electrons in Al³⁺ is greater than that in Mg²⁺.

The correct option among the following is

A
A is true, R is true and R is the correct explanation for A.

B
A is true, R is true but R is not the correct explanation for A.

C
A is true but R is false.

D
A is false but R is true.

Solution

Higher the electrostatic attraction between nucleus and valence electron, smaller will be the size of the atom/ion.

Electrostatic attraction is given by Z_eff= Z−S where, Z= the number of protons in the nucleus of an atom or ion (the atomic number) and S= shielding from core electrons.

In case of Al³⁺and Mg²⁺, they are isoelectronic species and thus have same number of electrons as 1s²,2s²,2p⁶ isoelectric series of atoms and ions with different numbers of protons (and thus different nuclear attraction, gives) the relative ionic sizes of each atom or ion with respect to atomic number.

Atomic number, Z_Al= 13 and ZMg = 12.

Thus, Al³⁺ has lower ionic radii than Mg²⁺ due to higher nuclear charge.

Hence, A is true, R is true and R is the correct explanation for A.
Question 6
4 / -1

Electromagnetic radiations having λ = 310 Å are subjected to a metal sheet having work function = 12.8 eV. What will be the velocity of photoelectrons with maximum Kinetic energy.

A
0, no emission will occur

B
2.18 × 10⁶ m/s

C
3.09 × 10⁶ m/sec.

D
8.72 × 10⁶ m/s

Solution
Question 7
4 / -1

Azide ion [N₃^-] exhibits an (N−N) bond order of 2 and may be represented by the resonance structures I,II and III given below:

The most stable resonance structure is

A
I

B
II

C
III

D
All of them have equal stability
Solution
The structure with more delocalization and more charge separation is the most stable.
- Structure I:
  Formal charges: Left = -1, Middle = +1, Right = -1
- Structure II:
  Formal charges: Left = -2, Middle = +1, Right = 0
- Structure III:
  Formal charges: Left = 0, Middle = +1, Right = -2
Conclusion:
- Structure I has more delocalization and charge separation, making it the most stable.
- Structures II and III have equal stability.
Question 8
4 / -1

C—Cl bond in (vinyl chloride) is stabilised in the same way as in

A
benzyl chloride

B
benzoyl chloride

C
chlorobenzene

D
allyl chloride

Solution

Due to delocalisation of π-electrons, (C—Cl) bond is stable and it does not show S_N reactions; Cl directly attached (C=C) bond, i.e. vinyl group.
Question 9
4 / -1

The following equations are balanced atomwise and chargewise.

(i) Cr₂O₇^2- + 8H⁺ + 3H₂O₂ → 2Cr³⁺ + 7H₂O + 3O₂

(ii) Cr₂O₇^2- + 8H⁺ + 5H₂O₂→ 2Cr³⁺ + 9H₂O + 4O₂

(iii) Cr₂O₇^2- + 8H⁺ + 7H₂O₂→ 2Cr⁺ + 11H₂O + 5O₂

The precise equations representing the oxidation of H₂O₂ is/are

A
(i) Only

B
(ii) Only

C
(iii) Only

D
All the three

Solution
Question 10
4 / -1

In the general electronic configuration -

(n - 2)f^1-14 (n - 1)d^0-1 ns², if value of n = 7 the configuration will be -

A
Lanthanides

B
Actinides

C
Transition elements

D
None

Solution

The correct answer is Option B.

General electronic configuration is given:

(n − 2)f^1-14(n − 1)d⁰¹ns² , where(n=7)

5f^1-146d⁰¹7s²

The seventh period (n=7) is similar to the sixth period with the successive filling up of the 7s, 5f, 6d and 7p orbitals and includes most of the man-made radioactive elements. This period will end at the element with atomic number 118 which would belong to the noble gas family. Filling up of the 5f orbitals after actinium (Z=89) gives the 5f-inner transition series known as the Actinide Series.