Classification of Elements and Periodicity in Properties Test

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Classification of Elements and Periodicity in Properties Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

Among fluorine and chlorine, the electron affinity of the latter is high. This is due to:

A
high electronegativity of fluorine

B
low dissociation energy of fluorine

C
fluorine repels with the added electron due to its small size

D
small size of chlorine atom

Solution

Fluorine atom, due to small size, repels incoming electron. Hence, among fluorine and chlorine, the electron affinity of chlorine is high. In general, the electron affinities of halogen atoms are high.
Question 2
1 / -0

Which of the following has zero electron affinity?

A
Oxygen

B
Fluorine

C
Nitrogen

D
Neon

Solution

Noble gases or inert gases do not have a tendency to gain an electron. Hence, its electron affinity is zero.
Question 3
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The electron affinity values (kJmol$$^{-1}$$) of three halogens $$X,\ Y $$ and $$Z$$ respectively are $$-349$$, $$-333 $$ and $$-325$$. Then $$X,\ Y $$ and $$Z$$ respectively are :

A
$$F_{2},\ Cl_{2} $$ and $$ Br_{2}$$

B
$$Cl_{2},\ F_{2}$$ and $$Br_{2}$$

C
$$Cl_{2},\ Br_{2}$$ and $$F_{2}$$

D
$$Br_{2},\ Cl_{2}$$ and $$F_{2}$$

Solution

Order of $$EA$$ of halogens is $$Cl>F>Br$$.

Hence, the value of $$EA$$ will be that of $$Cl_2$$, $$F_2$$, $$Br_2$$ respectively.
Question 4
1 / -0

Number of short periods in Mendeleev's periodic table is :

A
$$2$$

B
$$3$$

C
$$4$$

D
$$1$$

Solution

Period number 1, 2 and 3 contain 2, 8 and 8 elements respectively. Other periods contain more than 8 elements. Hence, the first three periods are called as short periods. So, total of 3 short periods are there in Mendeleev periodic table.
Question 5
1 / -0

In a period, elements are arranged strictly in a sequence of:

A
decreasing charge in the nucleus

B
increasing charge in the nucleus

C
constant charge in the nucleus

D
equal charge in the nucleus

Solution

In a period, the elements are arranged strictly in a sequence of increase in atomic number by one unit, i.e., increase in a number of protons by one and thereby increase in charge in the nucleus by one unit.

Option B is correct.
Question 6
1 / -0

Considering the chemical properties, atomic weight of $$Be$$ was corrected based on:

A
electronic configuration

B
valency

C
atomic number

D
both B and C

Solution

The atomic weight of Be was corrected based on:

Atomic weight $$=$$ equivalent weight $$\times$$ valency

From this, we can understand that the atomic weight of $$Be$$ was corrected based on valency.
Question 7
1 / -0

What is the atomic number of another element present in the same group as the element with $$Z = 13$$?

A
$$Z=50$$

B
$$Z=32$$

C
$$Z=49$$

D
$$Z=20$$

Solution

An element having atomic no. 13 is present in the 3rd period. Now, next element will have atomic number $$13 + 18 = 31$$. After that element, another element will have atomic number $$31 + 18 = 49$$.
Question 8
1 / -0

The order of electron affinities of $$N,\ O,\ S\ $$and$$\ Cl$$ are :

A
$$N < O < S < Cl $$

B
$$ O < N < Cl < S$$

C
$$ O < Cl < N <S$$

D
$$O < S < Cl < N$$

Solution

Generally, electron affinity increases as we move from left to right in a period and decreases as we go down in a group. But electron affinities of second-period elements (such as $$N$$, $$O$$) are less negative as compared to corresponding⋅ third-period elements. This is because of the small atomic size of second-period elements.
Hence, the correct order of electron affinities is $$N<O<S<Cl$$
Question 9
1 / -0

Which of the following will have almost positive $$EA_{1}$$?

A
Chlorine

B
Oxygen

C
Magnesium

D
Sulphur

Solution

For alkaline earth metals, filled 's' subshell discourages the addition of an electron. The electron affinity is almost zero. EA is a positive value for the magnesium atom.
Question 10
1 / -0

Beryllium shows diagonal relationship with aluminium. Which of the following similarity is incorrect?

A
$$Be_{2}C$$ like$$ Al_{4}C_{3}$$ yields methane on hydrolysis

B
$$Be$$, like $$Al$$ is rendered passive by $$HNO_{3}$$

C
Be$$ (OH)_{2}$$ like $$Al(OH)_{3}$$ is basic

D
$$Be$$ forms beryllates and $$Al$$ forms aluminate

Solution

Diagonal relationship of Be with Al is because of small size. Be is different from other earth alkaline earth metals. But it resembles in many of its properties with Al. The $$Be(OH)_{2}$$ and $$Al(OH)_{3}$$ are amphoteric in nature. The two structures involve the only movement of electrons and not of atoms or groups, hence these are resonating structures.
Hence, the correct answer is option $$\text{C}$$.

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Classification of Elements and Periodicity in Properties Test - 20

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Classification of Elements and Periodicity in Properties Test - 20

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Question 1

high electronegativity of fluorine

low dissociation energy of fluorine

fluorine repels with the added electron due to its small size

small size of chlorine atom

Solution

Question 2

Oxygen

Fluorine

Nitrogen

Neon

Solution

Question 3

$$F_{2},\ Cl_{2} $$ and $$ Br_{2}$$

$$Cl_{2},\ F_{2}$$ and $$Br_{2}$$

$$Cl_{2},\ Br_{2}$$ and $$F_{2}$$

$$Br_{2},\ Cl_{2}$$ and $$F_{2}$$

Solution

Question 4

$$2$$

$$3$$

$$4$$

$$1$$

Solution

Question 5

decreasing charge in the nucleus

increasing charge in the nucleus

constant charge in the nucleus

equal charge in the nucleus

Solution

Question 6

electronic configuration

valency

atomic number

both B and C

Solution

Question 7

$$Z=50$$

$$Z=32$$

$$Z=49$$

$$Z=20$$

Solution

Question 8

$$N < O < S < Cl $$

$$ O < N < Cl < S$$

$$ O < Cl < N <S$$

$$O < S < Cl < N$$

Solution

Question 9

Chlorine

Oxygen

Magnesium

Sulphur

Solution

Question 10

$$Be_{2}C$$ like$$ Al_{4}C_{3}$$ yields methane on hydrolysis

$$Be$$, like $$Al$$ is rendered passive by $$HNO_{3}$$

Be$$ (OH)_{2}$$ like $$Al(OH)_{3}$$ is basic

$$Be$$ forms beryllates and $$Al$$ forms aluminate

Solution

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