Classification of Elements and Periodicity in Properties Test

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Classification of Elements and Periodicity in Properties Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

Adsorbed hydrogen by palladium is known as

A
Nascent

B
atomic

C
heavy

D
occluded

Solution

Palladium possesses to a remarkable extent the power of adsorbing or occluding hydrogen. Hence adsorbed hydrogen by palladium is known as occluded hydrogen.
Question 2
1 / -0

The decreasing order of electron affinity is:

$$F> Cl> Br> I$$

A
$$F> Cl> Br> I$$

B
$$Cl> Br> I>F$$

C
$$Cl> Br> F>I$$

D
$$Cl>F> Br> I$$

Solution

Since the atomic size increases down the group, electron affinity generally decreases (At < I < Br < F < Cl). An electron will not be as attracted to the nucleus, resulting in a low electron affinity. However, fluorine has a lower electron affinity than chlorine. This can be explained by the small size of fluorine, compared to chlorine.
Question 3
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The electronic configuration of the element with maximum electron affinity is :

A
$$1s^2,\,2s^2\,2p^3$$

B
$$1s^2,\,2s^2\,2p^5$$

C
$$1s^2,\,2s^2\,2p^6,\,3s^2\,3p^5$$

D
$$1s^2,\,2s^2\,2p^6,\,3s^2\,3p^3$$

Solution

Fluorine is a small atom with a small amount of space available in its 2p orbital. Because of this, any new electron trying to attach to fluorine experiences lower electron affinity from the electrons already living in the element's 2p orbital. Since chlorine's outermost orbital is a 3p orbital, there is more space, and the electrons in this orbital are inclined to share this space with an extra electron. Therefore, chlorine has a higher electron affinity than fluorine, and this orbital structure causes it to have the highest electron affinity of all of the elements.

Chlorine has highest EA and its electronic configuration is $$1s^2,\,2s^2\,2p^6,\,3s^2\,3p^5$$.

Hence, the correct option is $$\text{C}$$
Question 4
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Increasing order of electron gain enthalpy is:

A
$$N < O > Cl < Al$$

B
$$O < N < Al < Cl$$

C
$$Al < N < O < Cl$$

D
$$Cl < N < O < Al$$

Solution

$$Al$$ is a metal so $$Al$$ has lowest electron gain enthalpy.
$$Cl$$ has highest electron gain enthalpy.
And electron gain enthalpy increases along a period.
$$\therefore$$ The correct order is $$Al<N<O<Cl$$.
Question 5
1 / -0

The correct order of increasing electronegativity of halogens is:

A
$$F < Cl < Br < I$$

B
$$I < Br < F < Cl$$

C
$$I < Br < Cl < F$$

D
$$Br < I < F < Cl$$
Question 6
1 / -0

Which of the following has maximum ionization enthalpy ?

A
K

B
Na

C
Mg

D
Be

Solution

We know that the ionization energy decreases down a group and increases from left to right along a period. Hence, beryllium has the maximum ionization enthalpy in this case.
Question 7
1 / -0

Aluminium is diagonally related to :

A
$$Li$$

B
$$Si$$

C
$$Be$$

D
$$B$$

Solution

Aluminium is diagonally related to Beryllium. Both shows similar properties.
Question 8
1 / -0

In which of the following processes is the maximum amount of energy involved ?

A
$$Cl\,\rightarrow\,Cl^-$$

B
$$Br^{-}\,\rightarrow\,Br$$

C
$$F^-\,\rightarrow\,F$$

D
$$I^-\,\rightarrow\,I$$

Solution

As E.A of $$Cl$$ > E.A. of $$F$$, therefore, $$Cl\,\rightarrow\,Cl^-$$ will involve maximum energy.
Question 9
1 / -0

Which of the following species has the highest electron gain enthalpy ?

A
$$F^-$$

B
$$O$$

C
$$O^-$$

D
$$Na^+$$

Solution

Answer : A
In general , electron gain enthalpy becomes more and more negative from left to right in a period. This is due to decreases in size and increase in nuclear charge as the atomic number increases in a period. Both these factors favor the addition of an extra electrons due to higher forces of attraction by the nucleus for the incoming electron.So F- will have the highest electron gain enthalpy.
Question 10
1 / -0

The electronic configurations of four elements are given below. Arrange these elements in the correct order of magnitude (without sign) of their electron affinity.

$$(i)$$ $$2s^2\,2p^5$$
$$(ii)$$ $$3s^2\,3p^5$$
$$(iii)$$ $$2s^2\,2p^4$$
$$(iv)$$ $$3s^2\,3p^4$$

Select the correct answer using the codes given below.

A
$$(i) < (ii) < (iii) < (iv)$$

B
$$(iii) < (iv) < (i) < (ii)$$

C
$$(iii) < (iv) < (ii) < (i)$$

D
$$(ii) < (i) < (iv) < (iii)$$

Solution

On the basis of the electronic configuration the corresponding elements are-
$$i) \ F \ ii)\ Cl\ iii)\ O\ iv)\ S$$

$$Cl$$ has the highest EA. F should have the highest EA but its size can't accommodate the incoming electron easily because of its small size.

So, $$F$$ will come second in order. Then, sulfur will come and then oxygen because of the same size reason.

So, the correct option is $$B$$

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Classification of Elements and Periodicity in Properties Test - 32

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Classification of Elements and Periodicity in Properties Test - 32

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Question 1

Nascent

atomic

heavy

occluded

Solution

Question 2

$$F> Cl> Br> I$$

$$Cl> Br> I>F$$

$$Cl> Br> F>I$$

$$Cl>F> Br> I$$

Solution

Question 3

$$1s^2,\,2s^2\,2p^3$$

$$1s^2,\,2s^2\,2p^5$$

$$1s^2,\,2s^2\,2p^6,\,3s^2\,3p^5$$

$$1s^2,\,2s^2\,2p^6,\,3s^2\,3p^3$$

Solution

Question 4

$$N < O > Cl < Al$$

$$O < N < Al < Cl$$

$$Al < N < O < Cl$$

$$Cl < N < O < Al$$

Solution

Question 5

$$F < Cl < Br < I$$

$$I < Br < F < Cl$$

$$I < Br < Cl < F$$

$$Br < I < F < Cl$$

Question 6

K

Na

Mg

Be

Solution

Question 7

$$Li$$

$$Si$$

$$Be$$

$$B$$

Solution

Question 8

$$Cl\,\rightarrow\,Cl^-$$

$$Br^{-}\,\rightarrow\,Br$$

$$F^-\,\rightarrow\,F$$

$$I^-\,\rightarrow\,I$$

Solution

Question 9

$$F^-$$

$$O$$

$$O^-$$

$$Na^+$$

Solution

Question 10

$$(i) < (ii) < (iii) < (iv)$$

$$(iii) < (iv) < (i) < (ii)$$

$$(iii) < (iv) < (ii) < (i)$$

$$(ii) < (i) < (iv) < (iii)$$

Solution

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