Classification of Elements and Periodicity in Properties Test

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Classification of Elements and Periodicity in Properties Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

The increasing order of ionic radii for given species is,

A
N³⁻ < O²⁻ <F⁻ < Na⁺ < Mg²⁺ < Al³⁺

B
Mg²⁺ < Na⁺ < F⁻< O²⁻ < N³⁻ < Al³⁺

C
Na⁺< F⁻ < O²⁻ < N³⁻ < Al³⁺ < Mg²⁺

D
Al³⁺ < Mg²⁺< Na⁺ < F⁻< O²⁻ < N³⁻
Solution
- In case of a cations,
  (i) there is a decrease in ionic radii of the resulting cation as compared to its neutral parent atom , due to loss of electron /(s)
  (ii) greater the positive charge over the cation more would be decrease in ionic size or radius .
- In case of anion,
  (i) the addition of each new electron (or negative charge) results into an increase in anionic radius.
  (ii) With increase in negative charge over anion the anionic radius also becomes larger.
  Thus, considering as above the correct answer is-
Al³⁺ < Mg²⁺< Na⁺ < F⁻< O²⁻ < N³⁻
Question 2
1 / -0

Al⁽³⁺⁾< Mg⁽²⁺⁾< Na⁺< F⁻< O⁽²⁻⁾< N⁽³⁻⁾ The above can be aptly described as

A
isoelectronic species

B
the same number of electrons (9 electrons).

C
the same number of electrons (8 electrons).

D
the same number of electrons (7 electrons).

Solution

Isoelectronic species are elements or ions that have the same, or equal number of electrons. Although isoelectronic species have the same number of electrons, they are different in their physical and chemical properties.
Question 3
1 / -0

The correct increasing order of radii of following species is ----?

A
I + < I⁻ < I

B
I < I⁺ < I⁻

C
I⁻ < I⁺ < I

D
I⁺ < I < I⁻

Solution

It is because

(i) radius of anion ( anionic radius ) is greater , and

(ii) radius of cation ( cationic radius ) is lesser

than that of its parent neutral atom.
Question 4
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The increasing order of bond dissociation energies of following species is -----?

A
F₂< N₂< Cl₂< O₂

B
N₂< O₂< F₂< Cl₂

C
F₂< Cl₂< O₂< N₂

D
Cl₂< O₂< N₂< F₂

Solution

As bond order increases bond dissociation energy also increases .

Thus N₂

(i) the triple bonded  N₂ with bond order

= 3

has the  highest bond dissociation energy .

(ii) The doubly bonded   O₂ with bond order

= 2

has lesser bond dissociation energy than  N₂

(iii) F₂  and Cl₂ both are single bonded having the same bond order

= 1

but , the electron - electron repulsive force ( e  - e repulsion ) is maximum in a   F₂ molecule due to its small size. This accounts for the lower bond dissociation energy of  F₂ as compared to  Cl₂
Question 5
1 / -0

The typical range of molar enthalpies for the strongest intermolecular (Hydrogen) bonds is

A
( 4 − 25 ) kJ

B
( 200 −300 ) kJ

C
( 300 − 500 ) kJ

D
( 50 −100 ) kJ

Solution

A hydrogen bond is the electrostatic attraction between two polar groups that occurs when a hydrogen (H) atom linked to a small sized highly electronegative atom such as nitrogen (N) , oxygen (O) , or fluorine (F) . Therefore unlike a covalent bond, it has very low dissociation energy with a range of molar enthalpy =

( 4 − 25 ) kJ
Question 6
1 / -0

In terms of period and group where would you locate the element with

Z = 11

in the modern periodic table?

A
6th period 14th group

B
7th period and 13th group

C
7th period and 14th group

D
6th period and 13th group

Solution

Elements with atomic numbers Z = (87−114) are placed in 7th period.Elements with atomic numbers Z = (87−114) are placed in 7th period.

Thus , the element with

Z=114(Flerovium) is placed in 7th period and 14th group of the modern periodic table.
Question 7
1 / -0

For an ionic compound with tetrahedral structure the ratio

r⁺/r⁻

lies in the range of ---- ?

A
0.225 to 0.414

B
0.155 to 0.255

C
0.732 to 1.0

D
0.414 to 0.732

Solution

For tetrahedral structure of a ionic solid ,

r⁺/r⁻lies in the range of 0.225 to 0.414
Question 8
1 / -0

The relative tendency of an atom in a molecule to attract the shared pair of electrons towards itself is termed as its

A
ionisation enthalpy

B
electron gain enthalpy

C
electro-positivivity

D
electro-negativity

Solution

A qualitative measure of the ability of an atom in a chemical compound to attract shared electrons to itself is called electronegativity. Unlike ionization enthalpy and electron gain enthalpy, it is not a measureable quantity. However, a number of numerical scales of electronegativity of elements viz., Pauling scale, Mulliken-Jaffe scale, Allred-Rochow scale have been developed. The electronegativity of any given element is not constant; it varies depending on the element to which it is bound.
Question 9
1 / -0

The order of screening effect of electrons of s, p, d and f orbitals of a given shell of an atom on its outer shell electrons is:

A
s > p > d > f

B
f > d > p > s

C
f > p > s > d

D
p < d < s > f

Solution

Screening effect order follows s>p>d>f, this due to the nature of the orbitals involved. s-orbitals are least diffused and the two electrons in s-orbitals provide greatest shielding of the nuclear charge to the outermost electron. The diffused character increases in the order p>d>f. f-orbitals are so diffused that fourteen electrons inside them provide virtually no screening of the nuclear charge.
Question 10
1 / -0

Which of the following is not an actinoid?

A
Uranium (Z = 92)

B
Terbium (Z = 65)

C
Californium (Z = 98)

D
Curium (Z = 96)

Solution

Terbium (Z=65) has an electronic configuration of [Xe] 4f⁹ 6s². It is a member of 4f-inner transition series which is called lanthanoid series. Rest option are all the members of 5f-inner transition series, also known as the actinoid series i.e. Curium ([Rn] 5f⁷ 6d¹ 7s²) ; Californium ([Rn] 5f¹⁰ 7s₂); and Uranium ([Rn] 5f³ 6d¹ 7s²).

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Classification of Elements and Periodicity in Properties Test - 6

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Question 1

N3− < O2− <F− < Na+ < Mg2+ < Al3+

Mg2+ < Na+ < F− < O2− < N3− < Al3+

Na+< F− < O2− < N3− < Al3+ < Mg2+

Al3+ < Mg2+< Na+ < F−< O2− < N3−

Solution

Question 2

isoelectronic species

the same number of electrons (9 electrons).

the same number of electrons (8 electrons).

the same number of electrons (7 electrons).

Solution

Question 3

I + < I− < I

I < I+ < I−

I− < I+ < I

I+ < I < I−

Solution

Question 4

F2< N2< Cl2< O2

N2< O2< F2< Cl2

F2< Cl2< O2< N2

Cl2< O2< N2< F2

Solution

Question 5

( 4 − 25 ) kJ

( 200 −300 ) kJ

( 300 − 500 ) kJ

( 50 −100 ) kJ

Solution

Question 6

6th period 14th group

7th period and 13th group

7th period and 14th group

6th period and 13th group

Solution

Question 7

0.225 to 0.414

0.155 to 0.255

0.732 to 1.0

0.414 to 0.732

Solution

Question 8

ionisation enthalpy

electron gain enthalpy

electro-positivivity

electro-negativity

Solution

Question 9

s > p > d > f

f > d > p > s

f > p > s > d

p < d < s > f

Solution

Question 10

Uranium (Z = 92)

Terbium (Z = 65)

Californium (Z = 98)

Curium (Z = 96)

Solution

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N³⁻ < O²⁻ <F⁻ < Na⁺ < Mg²⁺ < Al³⁺

Mg²⁺ < Na⁺ < F⁻< O²⁻ < N³⁻ < Al³⁺

Na⁺< F⁻ < O²⁻ < N³⁻ < Al³⁺ < Mg²⁺

Al³⁺ < Mg²⁺< Na⁺ < F⁻< O²⁻ < N³⁻

I + < I⁻ < I

I < I⁺ < I⁻

I⁻ < I⁺ < I

I⁺ < I < I⁻

F₂< N₂< Cl₂< O₂

N₂< O₂< F₂< Cl₂

F₂< Cl₂< O₂< N₂

Cl₂< O₂< N₂< F₂