Classification of Elements and Periodicity in Properties Test

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Classification of Elements and Periodicity in Properties Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following configurations has the highest ionization energy?

A
$$1s^22s^22p^1$$

B
$$1s^22s^22p^3$$

C
$$1s^22s^22p^2$$

D
$$1s^22s^22p^4$$

Solution

The electron will move from a lower stationary state to a higher stationary state when the required amount of energy is absorbed by the electron or energy is emitted when the electron moves from a higher stationary state to lower stationary state. The energy change does not continuously take place.

Hence, the correct option is $$\text{B}$$
Question 2
1 / -0

What is the name of the element with atomic number $$105$$?

A
Dubnium

B
Holmium

C
Kurchatovium

D
Nobelium

Solution

The element with atomic number $$105$$ is Dubnium ($$Db$$). Its IUPAC name is Un-nil pentium.
Question 3
1 / -0

Which of the following has highest electron affinity?

A
$$N$$

B
$$O$$

C
$$F$$

D
$$Cl$$

Solution

The electron affinities of some of the elements of second period (i.e. $$N, O, F$$, etc) are however, lower than the corresponding elements (i.e. $$P, S, Cl$$, etc) of the third period. This is due to the reason that the elements of second period have the smallest atomic sizes amongst the elements in their respective groups. As a result, there are considerable electron-electron repulsion with the atom itself and hence, the additional electron is not accepted with the same case as compared to the remaining elements in the same group.
Question 4
1 / -0

Electronic configuration of $$Gd(64)$$ is written as:

A
$$[Xe]4f^{7} 5d^{1} 6s^{2}$$

B
$$[Xe]4f^{8} 6s^{2}$$

C
$$[Xe]4f^{9} 6s^{1}$$

D
$$[Xe]4f^{10}$$

Solution

$$Gd. (64) = [Xe]4f^{7} 5d^{1}$$
All electrons are unpaired, hence greater stability.
Question 5
1 / -0

Considering the element $$F, Cl, O $$ and $$N,$$ the correct order in their chemical reactivity In terms of oxidising property is :

A
$$F > Cl > O > N $$

B
$$F > O > Cl > N$$

C
$$Cl > F > O > N $$

D
$$O > F > N > Cl$$

Solution

In a group oxidising power decreases from top to bottom. as the size increases but when we move from left to right in a period, it increases because site decreases. Therefore among $$F, Cl, O$$ and $$N,$$ the oxidising power decreases in the order.
$$F > O > CI > N $$
Note oxygen is more electronegative than chlorine,
Hence, $$O$$ is stronger oxidising agent than $$Cl. $$
Question 6
1 / -0

Which of the following has the highest electron affinity?

A
$$O^{-}$$

B
$$F^{-}$$

C
$$F_{2}$$

D
$$Cl_{2}$$

Solution

The electron affinity of $$O^{-}$$ is highest as the other given species have complete octet.
Question 7
1 / -0

Which of the following has the highest electron affinity?

A
$$K$$

B
$$O^-$$

C
$$F^-$$

D
$$O$$

Solution

In general, electron affinity increases in a period from left to right. This is due to decrease in size and increase in nuclear charge as the atomic number increases in a period. Potassium is a metallic element that is electron donor $$O^-$$ and $$F^-$$ ions gain stability by taking one electron.

Only $$O$$ (oxygen atoms) has the tendency to gain electron to attain stability.
Question 8
1 / -0

Which of the following represents correct acidity order $$Li_2O, BeO$$ and $$B_2O_3$$?

A
$$Li_2O$$ < $$BeO$$ < $$ B_2O_3$$

B
$$B_2O_3 < BeO < Li_2O$$

C
$$BeO < Li_2O < B_2O_3$$

D
$$BeO < B_2O_3 < Li_2O$$

Solution

The chemical nature of oxides of elements changes across the period as
Basic oxide $$\rightarrow $$ Amphoteric oxide $$\rightarrow $$ Acidic oxide Thus, the correct increasing order of acidity is
$$Li_2O < Be0 < B_2O_3$$
Question 9
1 / -0

Which of the following electronic configuration is not possible?

A
$$2p^{6}$$

B
$$3s^{1}$$

C
$$2p^{5}$$

D
$$3f^{12}$$
Solution
Correct Option: D

Explanation
- Electrons are filled in orbitals in accordance with three rules:
  1. Hund’s rule: It states that each orbital of a subshell should be singly occupied before pairing starts.
  2. Pauli’s exclusion principle: It states that no two electrons in an atom can have all four quantum numbers the same.
  3. Aufbau’s principle: According to this, electrons are filled in orbitals in order of their increasing energy, i.e., orbitals with lower energy must be filled before those having higher energy.
- Number of maximum electrons in each subshell:
  1. $$s= 2$$
  2. $$p= 6$$
  3. $$d= 10$$
  4. $$f = 14$$
- The shell number from which each sub-shell begins
  1. $$s \rightarrow n = 1$$
  2. $$p \rightarrow n = 2$$
  3. $$d \rightarrow n = 3$$
  4. $$f \rightarrow n = 4$$
  Therefore, $$3f$$ is not possible as $$f-subshell$$ begins from $$n=4$$. Only $$4f$$ and further subshells exist.
Hence, the electronic configuration which is not possible is $$3f^{12}$$
Question 10
1 / -0

Which of the following species have the same number of electrons in its outermost as well as penultimate shell?

A
$$Cl^-$$

B
$$O^{2-}$$

C
$$Na^+$$

D
$$Mg^{2+}$$

Solution

$$Cl^-=2, 8, 8$$
$$O^{2-}=2, 8$$
$$Na^+=2, 8$$
$$Mg^{2+}=2, 8$$
$$Cl^-$$ contains same number of electrons in outermost and penultimate shells.

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Re-Attempt Weekly Quiz Competition

Classification of Elements and Periodicity in Properties Test - 46

Result

Classification of Elements and Periodicity in Properties Test - 46

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SHARING IS CARING

Question 1

$$1s^22s^22p^1$$

$$1s^22s^22p^3$$

$$1s^22s^22p^2$$

$$1s^22s^22p^4$$

Solution

Question 2

Dubnium

Holmium

Kurchatovium

Nobelium

Solution

Question 3

$$N$$

$$O$$

$$F$$

$$Cl$$

Solution

Question 4

$$[Xe]4f^{7} 5d^{1} 6s^{2}$$

$$[Xe]4f^{8} 6s^{2}$$

$$[Xe]4f^{9} 6s^{1}$$

$$[Xe]4f^{10}$$

Solution

Question 5

$$F > Cl > O > N $$

$$F > O > Cl > N$$

$$Cl > F > O > N $$

$$O > F > N > Cl$$

Solution

Question 6

$$O^{-}$$

$$F^{-}$$

$$F_{2}$$

$$Cl_{2}$$

Solution

Question 7

$$K$$

$$O^-$$

$$F^-$$

$$O$$

Solution

Question 8

$$Li_2O$$ < $$BeO$$ < $$ B_2O_3$$

$$B_2O_3 < BeO < Li_2O$$

$$BeO < Li_2O < B_2O_3$$

$$BeO < B_2O_3 < Li_2O$$

Solution

Question 9

$$2p^{6}$$

$$3s^{1}$$

$$2p^{5}$$

$$3f^{12}$$

Solution

Question 10

$$Cl^-$$

$$O^{2-}$$

$$Na^+$$

$$Mg^{2+}$$

Solution

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