Classification of elements & Periodicity in properties Test

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Classification of elements & Periodicity in properties Test - 2

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Question 1
1 / -0

The increasing order of ionic mobilities is/ are given by

A
Na⁺< Mg²⁺< Al³⁺

B
K⁺< Na⁺< Li⁺

C
Al³⁺< Mg²⁺< Na⁺

D
Li⁺< Na⁺< K⁺

Solution

The extent of hydration increases with decrease in ionic size. Thus, extent of hydration varies in the order: Li⁺> Na⁺> K⁺ and so also the sizes of hydrated ions.
Question 2
1 / -0

Which of the following pair of ions have the same number of unpaired electrons?

A
Co³⁺ and Cr³⁺

B
Mn²⁺and Fe³⁺

C
Ti²⁺and Ni²⁺

D
Mn²⁺and Fe²⁺

Solution

Mn²⁺ and Fe³⁺ have 5 unpaired electrons each; Ti²⁺ and Ni²⁺ have 2 unpaired electrons each.
Question 3
1 / -0

The electronic configuration of an element is 1s² 2s² 2p⁶ 3s² 3p³.The atomic number and the group number of the element ‘X’ which is just below the above element in the periodic table are respectively

A
33 and 7

B
25 and 7

C
33 and 15

D
25 and 15

Solution

Atomic number of the given element =15;

group no. = 10 + 5 (valence electrons) =15,

period = 4th
Question 4
1 / -0

Statement-1: The most negative reduction potential values is that of lithium.

Statement-2: Lithium has a high hydration energy value because of its small size.

A
Both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation of Statement-1.

B
Both Statement-1 and Statement-2 are true and Statement-2 is not the correct explanation of Statement-1.

C
Statement-1 is true but Statement-2 is false.

D
Statement-1 is false but Statement-2 is true.

Solution

Both assertion and reason are correct and reason is the correct explanation of assertion.
Question 5
1 / -0

The element with the lowest atomic number and having ground state electron configuration of (n – 1)d¹⁰ns²np³ is placed in the

A
sixth period

B
fourth period

C
fifth period

D
third period

Solution

For lowest atomic number of the element, n - 1 = 3

⇒ n = 4, hence it should be placed in 4^thperiod.
Question 6
1 / -0

which of the following is the correct sequence of the ionic radii.

A
Br^–> Cl^–> S^2–>O^2–> F^–

B
none of these

C
Br^–> S^2–> Cl^–> O^2–> F^–

D
Br^–> S^2–> Cl^–> F^–> O^2–

Solution

Radii of anions carrying same charge decrease from left to right in a period and increase down the group.
Question 7
1 / -0

The correct order of radii is

A
F^–<O^2–<N^3–

B
N<Be<B

C
Na<Li<K

D
Fe³⁺< Fe²⁺< Fe⁴⁺

Solution

The size of isoelectronic anions decreases with the increased nuclear charge.
Question 8
1 / -0

The second ionization energies of the following atoms are such that

A
C>N>O>F

B
O>N>F>C

C
O>F>N>C

D
F>O>N>C

Solution

Half-filled 2p³ subshell of O⁺ is more stable than 2p⁴ subshell of F⁺. Hence, IE₂ of O will be greater than that of F⁺.

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Re-Attempt Weekly Quiz Competition

Classification of elements & Periodicity in properties Test - 2

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Classification of elements & Periodicity in properties Test - 2

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Question 1

Na+< Mg2+< Al3+

K+< Na+< Li+

Al3+< Mg2+< Na+

Li+< Na+< K+

Solution

Question 2

Co3+ and Cr3+

Mn2+and Fe3+

Ti2+and Ni2+

Mn2+and Fe2+

Solution

Question 3

33 and 7

25 and 7

33 and 15

25 and 15

Solution

Question 4

Both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation of Statement-1.

Both Statement-1 and Statement-2 are true and Statement-2 is not the correct explanation of Statement-1.

Statement-1 is true but Statement-2 is false.

Statement-1 is false but Statement-2 is true.

Solution

Question 5

sixth period

fourth period

fifth period

third period

Solution

Question 6

Br–> Cl–> S2–>O2–> F–

none of these

Br–> S2–> Cl–> O2–> F–

Br–> S2–> Cl–> F–> O2–

Solution

Question 7

F–<O2–<N3–

N<Be<B

Na<Li<K

Fe3+< Fe2+< Fe4+

Solution

Question 8

C>N>O>F

O>N>F>C

O>F>N>C

F>O>N>C

Solution

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Na⁺< Mg²⁺< Al³⁺

K⁺< Na⁺< Li⁺

Al³⁺< Mg²⁺< Na⁺

Li⁺< Na⁺< K⁺

Co³⁺ and Cr³⁺

Mn²⁺and Fe³⁺

Ti²⁺and Ni²⁺

Mn²⁺and Fe²⁺

Br^–> Cl^–> S^2–>O^2–> F^–

Br^–> S^2–> Cl^–> O^2–> F^–

Br^–> S^2–> Cl^–> F^–> O^2–

F^–<O^2–<N^3–

Fe³⁺< Fe²⁺< Fe⁴⁺