Classification of elements & Periodicity in properties Test

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Classification of elements & Periodicity in properties Test - 2

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Question 1/9
1 / -0

The increasing order of ionic mobilities is/ are given by

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A
Na⁺< Mg²⁺< Al³⁺

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B
K⁺< Na⁺< Li⁺

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C
Al³⁺< Mg²⁺< Na⁺

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D
Li⁺< Na⁺< K⁺

Solutions

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 2/9
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Which of the following pair of ions have the same number of unpaired electrons?

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A
Co³⁺ and Cr³⁺

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B
Mn²⁺and Fe³⁺

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C
Ti²⁺and Ni²⁺

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D
Mn²⁺and Fe²⁺

Solutions

Mn²⁺ and Fe³⁺ have 5 unpaired electrons each; Ti²⁺ and Ni²⁺ have 2 unpaired electrons each.
Question 3/9
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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

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A
33 and 7

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B
25 and 7

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C
33 and 15

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D
25 and 15

Solutions

Atomic number of the given element =15;

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

period = 4th
Question 4/9
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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.

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

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

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

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

Solutions

Both assertion and reason are correct and reason is the correct explanation of assertion.
Question 5/9
1 / -0

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

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A
sixth period

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B
fourth period

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C
fifth period

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D
third period

Solutions

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

⇒ n = 4, hence it should be placed in 4^thperiod.
Question 6/9
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which of the following is the correct sequence of the ionic radii.

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A
Br^–> Cl^–> S^2–>O^2–> F^–

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B
none of these

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C
Br^–> S^2–> Cl^–> O^2–> F^–

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D
Br^–> S^2–> Cl^–> F^–> O^2–

Solutions

Radii of anions carrying same charge decrease from left to right in a period and increase down the group.
Question 7/9
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Match the elements listed in column I with the period/group listed in column II

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A
A-q; B-p, r; C-p, s; D-p

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B
A-s; B-q, s; C-p, r; D-p

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C
A-r; B-p, s; C-p, q; D-q

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D
A-p; B-p, q; C-r, s; D-r

Solutions

(A) At. No. 47: [Kr]4d⁹ 5s² ;group : 9 + 2 = 11^th; period: 5^th

(B) For lowest At. No. having (n - 1)d¹⁰ ns² np³ configuration, n - 1 =3

Þ n = 4

Hence, group: 10 + 5 = 15^th; period: 4^th

(C) At. No.34:[Ar] 3d¹⁰4s²4p⁴;Group:10 + 6=16^th;period = 4^th

(D) For lowest At. No. having (n -1) d¹⁰ns¹ configuration, n - 1=3

Þ n = 4

Group = 10+1 = 11^th; Period = 4^th
Question 8/9
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The correct order of radii is

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A
F^–<O^2–<N^3–

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B
N<Be<B

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C
Na<Li<K

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D
Fe³⁺< Fe²⁺< Fe⁴⁺

Solutions

The size of isoelectronic anions decreases with the increased nuclear charge.
Question 9/9
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The second ionization energies of the following atoms are such that

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A
C>N>O>F

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B
O>N>F>C

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C
O>F>N>C

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D
F>O>N>C

Solutions

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⁺.

Question 1/9

Na+< Mg2+< Al3+

K+< Na+< Li+

Al3+< Mg2+< Na+

Li+< Na+< K+

Question 2/9

Co3+ and Cr3+

Mn2+and Fe3+

Ti2+and Ni2+

Mn2+and Fe2+

Question 3/9

33 and 7

25 and 7

33 and 15

25 and 15

Question 4/9

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.

Question 5/9

sixth period

fourth period

fifth period

third period

Question 6/9

Br–> Cl–> S2–>O2–> F–

none of these

Br–> S2–> Cl–> O2–> F–

Br–> S2–> Cl–> F–> O2–

Question 7/9

A-q; B-p, r; C-p, s; D-p

A-s; B-q, s; C-p, r; D-p

A-r; B-p, s; C-p, q; D-q

A-p; B-p, q; C-r, s; D-r

Question 8/9

F–<O2–<N3–

N<Be<B

Na<Li<K

Fe3+< Fe2+< Fe4+

Question 9/9

C>N>O>F

O>N>F>C

O>F>N>C

F>O>N>C

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⁺