Periodic Properties Test

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Periodic Properties Test - 9

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Weekly Quiz Competition

Question 1
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Which of the following electronic configurations corresponds to the element with the largest negative electron gain enthalpy?

A
1s², 2s², 2p⁵

B
1s², 2s², 2p⁶

C
1s², 2s², 2p⁶, 3s², 3p⁵

D
1s², 2s², 2p⁶, 3s¹

Solution

Electron gain enthalpy becomes more negative across a period as we move from left to right. Electron gain enthalpy becomes less negative down a group. However, adding an electron to the 2p-orbital leads to greater repulsion than adding an electron to the larger 3p-orbital. Hence, the element with the greatest negative electron gain enthalpy is chlorine with 1s², 2s², 2p⁶, 3s², 3p⁵ electronic configuration.
Question 2
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Which of the following orders is not correct according to the property written against it?

A
Al³⁺ < Mg²⁺ < Na⁺ < F^- ⇒ increasing ionic size

B
B < C < N < O ⇒ increasing first ionisation enthalpy

C
I < Br < F < Cl ⇒ increasing electron gain enthalpy (with negative sign)

D
Li < Na < K < Rb ⇒ increasing metallic radius

Solution

Nitrogen has half-filled stable configuration, so its ionisation enthalpy is greater than that of oxygen. So, the correct order of increasing first ionisation enthalpy is: B < C < O < N.
Question 3
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The correct order of decreasing electronegativity values among the elements I - beryllium, II - oxygen, III - nitrogen, and IV - magnesium is

A
II > III > I > IV

B
I > II > III > IV

C
III > IV > II > I

D
II > III > IV > I

Solution

Electronegativity decreases as we move down the group and increases as we move from left to right in the periodic table. Therefore, the correct increasing order of electronegativity is:
II (oxygen) > III (nitrogen) > I (beryllium) > IV (magnesium)

Question 4

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Column I	Column II
(a) He	(i) High negative electron gain enthalpy
(b) Cl	(ii) Strongest oxidising agent
(c) F	(iii) Strongest reducing agent
(d) Li	(iv) Highest ionisation enthalpy

The correct match of the contents in column I with those in column II is

a - iii, b - i, c - ii, d - iv

a - iv, b - iii, c - ii, d - i

a - ii, b - iv, c - i, d - iii

a - iv, b - i, c - ii, d - iii

Solution

Helium has a high ionization enthalpy because of its stable electronic configuration.
Chlorine has the highest value of electron gain enthalpy (higher than even Fluorine which is more electronegative) as owing to its larger size, it can accommodate the incoming electron better.
F₂ is the strongest oxidising agent among the elements.
Lithium metal is a strong reducing agent.

Column I	Column II
(a) He	(iv) Highest ionisation enthalpy
(b) Cl	(i) High negative electron gain enthalpy
(c) F	(ii) Strongest oxidising agent
(d) Li	(iii) Strongest reducing agent

Question 5
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Which of the following statements is/are correct?

A
1^st and 2^nd IEs of nitrogen are more than 1^st and 2^nd IEs of oxygen.

B
IE of oxygen is greater than IE of nitrogen.

C
IE₂ of oxygen is greater than IE₂ of nitrogen.

D
All of the above

Solution

IE₂ of oxygen is greater than IE₂ of nitrogen because after the removal of one electron, oxygen has half-filled electronic configuration, which is more stable. IE₁ of nitrogen is greater than that of oxygen.
Question 6
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Consider the ions K⁺, S^2-, Cl^- and Ca²⁺.

Which of the following is the correct order of ionic radii for the given species?

A
Ca²⁺ > K⁺ > Cl^- > S^2-

B
Cl^- > S^2- > K⁺ > Ca²⁺

C
Ca²⁺ > Cl^- > K⁺ > S^2-

D
S^2- > Cl^- > K⁺ > Ca²⁺

Solution

For the isoelectronic species, the ionic radii decrease with increase in the nuclear charge.
So, the correct order of radii is S^2- > Cl^- > K⁺ > Ca²⁺.
Question 7
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Identify the correct order in which the ionic radii of the following ions increase:

I. F^-
II. Na⁺
III. N^3-

A
III, I, II

B
I, II, III

C
II, III, I

D
II, I, III

Solution

These are all isoelectronic species. Among isoelectronic species, radius decreases with increase in the effective nuclear charge.
Hence the correct order of size is:
Na⁺ < F^- < N^3-
Question 8
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Which of the following processes requires the highest energy?

A
Al (g) → Al⁺ (g) + e^-

B
Al²⁺ (g) → Al³⁺ (g) + e^-

C
Al⁺ (g) → Al²⁺ (g) + e^-

D
Energy required is the same in all the three processes.

Solution

Ionization energy increases with the increase in the effective nuclear charge.
IE₁ < IE₂ < IE₃
Question 9
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Which of the following isoelectronic ions has the lowest ionisation energy?

A
K⁺

B
Ca²⁺

C
Cl⁻

D
S²⁻

Solution

Ionisation energy decreases with decrease in the nuclear charge.
So, S²⁻will have the lowest ionisation energy.

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Periodic Properties Test - 9

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Periodic Properties Test - 9

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Question 1

1s2, 2s2, 2p5

1s2, 2s2, 2p6

1s2, 2s2, 2p6, 3s2, 3p5

1s2, 2s2, 2p6, 3s1

Solution

Question 2

Al3+ < Mg2+ < Na+ < F- ⇒ increasing ionic size

B < C < N < O ⇒ increasing first ionisation enthalpy

I < Br < F < Cl ⇒ increasing electron gain enthalpy (with negative sign)

Li < Na < K < Rb ⇒ increasing metallic radius

Solution

Question 3

II > III > I > IV

I > II > III > IV

III > IV > II > I

II > III > IV > I

Solution

Question 4

a - iii, b - i, c - ii, d - iv

a - iv, b - iii, c - ii, d - i

a - ii, b - iv, c - i, d - iii

a - iv, b - i, c - ii, d - iii

Solution

Question 5

1st and 2nd IEs of nitrogen are more than 1st and 2nd IEs of oxygen.

IE of oxygen is greater than IE of nitrogen.

IE2 of oxygen is greater than IE2 of nitrogen.

All of the above

Solution

Question 6

Ca2+ > K+ > Cl- > S2-

Cl- > S2- > K+ > Ca2+

Ca2+ > Cl- > K+ > S2-

S2- > Cl- > K+ > Ca2+

Solution

Question 7

III, I, II

I, II, III

II, III, I

II, I, III

Solution

Question 8

Al (g) → Al+ (g) + e-

Al2+ (g) → Al3+ (g) + e-

Al+ (g) → Al2+ (g) + e-

Energy required is the same in all the three processes.

Solution

Question 9

K+

Ca2+

Cl−

S2−

Solution

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1s², 2s², 2p⁵

1s², 2s², 2p⁶

1s², 2s², 2p⁶, 3s², 3p⁵

1s², 2s², 2p⁶, 3s¹

Al³⁺ < Mg²⁺ < Na⁺ < F^- ⇒ increasing ionic size

1^st and 2^nd IEs of nitrogen are more than 1^st and 2^nd IEs of oxygen.

IE₂ of oxygen is greater than IE₂ of nitrogen.

Ca²⁺ > K⁺ > Cl^- > S^2-

Cl^- > S^2- > K⁺ > Ca²⁺

Ca²⁺ > Cl^- > K⁺ > S^2-

S^2- > Cl^- > K⁺ > Ca²⁺

Al (g) → Al⁺ (g) + e^-

Al²⁺ (g) → Al³⁺ (g) + e^-

Al⁺ (g) → Al²⁺ (g) + e^-

K⁺

Ca²⁺

Cl⁻

S²⁻