Classification of Elements and Periodicity in Properties Test

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Classification of Elements and Periodicity in Properties Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following will have lowest electron affinity?

A
Nitrogen

B
Oxygen

C
Carbon

D
Sulphur

Solution

The electron affinity for oxygen is lowest. As we move from left to right in a period, electron affinity becomes more negative. Argon (a noble gas) has large positive value for electron affinity.
Question 2
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Why do noble gases have positive electron gain enthalpy?

A
It is difficult to add an electron due to small size.

B
It is difficult to add an electron due to high electronegativity.

C
It is difficult to add an electron due to stable configuration.

D
It is difficult to add an electron due to high electron affinity.

Solution

Due to the stable configuration, they do not accept an electron and hence they have large positive electron gain enthalpy.

An electron has to enter the next higher principal quantum level which leads to a very unstable electronic configuration.

Hence, the correct option is $$\text{C}$$
Question 3
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Which of the following is not a periodic property for the elements?

A
Electronegativity

B
Atomic size

C
Occurrence in nature

D
Ionization energy

Solution

Occurrence in nature is not a periodic property for the elements.
Electronegativity, electron affinity, atomic size, metallic character, redox potential and ionisation enthalpy are periodic properties for the elements.
Question 4
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Few values of enthalpies are given below:
$$O = - 141 \ kJ\ mol^{-1}$$, $$F = - 328\ kJ \ mol^{-1}$$
$$S = - 200 \ kJ \ mol^{-1}$$, $$CI = - 349 \ kJ \ mol^{-1}$$
What do these values show?

A
Ionisation enthalpy

B
Bond enthalpy

C
Electron gain enthalpy

D
Hydration enthalpy
Question 5
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The electronic states $$X$$ and $$Y$$ of an atom are depeicted below:
$$X : 1s^{2} 2s^{2} 2p^{6} 3s^{1}$$
$$Y: 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{1}$$
Which of the following statements is not correct?

A
$$X$$ represents an alkali metal.

B
Energy is required to change $$X$$ into $$Y$$

C
$$Y$$ represents ground state of the element.

D
Less energy is required to remove an electron from $$X$$ than from $$Y$$

Solution

Electronic configuration for given elements $$X$$ $$= 1s^22s^22p^63s^1, Y=1s^22s^22p^63s^23p^64s^1$$
shows that they belong to the same group (group 11, alkali metals) and $$Y$$ is bigger in size than $$X$$. Thus, the distance between the nucleus and valence shell in $$Y$$ is greater as compared to $$X$$ and therefore $$Y$$ experience less effective nuclear charge and needs less energy to remove the valence electron than that from $$X$$ atom.
Question 6
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Which of the following gases is most predominating in the Sun?

A
Hydrogen

B
Nitrogen

C
Ozone

D
Helium

Solution

The most predominat gas is sum is hydrogen. The percentage of abundance of elements in sum is as follows:
Hydrogen : $$91.2\%$$
Helium :$$8.7\%$$
Oxygen : $$0.078\%$$
Question 7
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Which of the following arrangements represents the correct order of electron gain enthalpy?
(considering only magnitude of the electron gain enthalpy)

A
$$Cl > F > S > O$$

B
$$O > Cl > F >S$$

C
$$F > S > O > Cl$$

D
$$S >F > O > Cl$$

Solution

The order of electron gain enthalpies (kJ/mol) is Cl (-349)< F (-328) < S (-200) < O (-141). But if we neglect negative sign and consider magnitude, then Cl>F>S>O.
Electron gain enthalpy becomes more negative across a period while it becomes less negative in a group. However, electron gain enthalpy of O or F is less negative than that of S or Cl. When an electron is added to n = 2 (in case of O or F), the repulsion is more than when it is added to n = 3 (in case of S or CI). an n=2 level is smaller than the n=3 level.
Question 8
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Why is the electron gain enthalpy of O or F less than that of S or CI?

A
O and F are more electronegative than S and Cl.

B
When an electron is added to O or F, it goes to a smaller (n = 2) level and suffers more repulsion than the electron in S or Cl in larger level (n = 3).

C
Adding an electron to 3p-orbital leads to more repulsion than 2p-orbital.

D
Electron gain enthalpy depends upon the electron affinity of the atom.

Solution

The electron gain enthalpy of O or F less than that of S or CI and is due to smaller size of O/F the inter-electronic repulsion in compact 2p-orbitals is much more than the repulsion in 3p-orbitals of S/Cl hence, the incoming new electron feels less attraction in O/F and more attraction in S/Cl. Due to this the addition becomes difficult in O/F and electron gain enthalpy becomes less than S/Cl.
option B
Question 9
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Choose the correct answer form the alternatives given. $$Ca^{2+}$$ is isoelectronic with:

A
$$Na$$

B
$$4Mg^{2+}$$

C
$$Ba^{2+}$$

D
$$Ar$$

Solution

The electronic configuration of Ca(Z=20) is $$ { Ca } =\ { 1s }^{ 2 }{ 2s }^{ 2 }2p^{ 6 }3s^{ 2 }{ 3p }^{ 6 }4{ s }^{ 2 } $$
while The electronic configuration of $$ { { Ca }^{ 2+ } } $$ is $$ { { Ca }^{ 2+ } }=\ { 1s }^{ 2 }{ 2s }^{ 2 }2p^{ 6 }3s^{ 2 }{ 3p }^{ 6 } $$ = Ar
The no.of electrons in $$ { { Ca }^{ 2+ } } $$ is 18 after the loss of two electrons. So it has noble gas configuration of Ar.
Question 10
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The main reason for showing anomalous properties of the first member of a group in s or p-block is :

A
maximum chemical reactivity

B
maximum electronegativity and different configurations

C
small size, large charge/radius ratio

D
tendency to form multiple bonds

Solution

The main reason for showing anomalous properties of the first member of a group in s or p-block is small size, large charge/radius ratio. For example, in the first group, Li shows anomalous behavior due to small size, large charge/radius ratio( high polarising power).

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Re-Attempt Weekly Quiz Competition

Classification of Elements and Periodicity in Properties Test - 48

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Classification of Elements and Periodicity in Properties Test - 48

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SHARING IS CARING

Question 1

Nitrogen

Oxygen

Carbon

Sulphur

Solution

Question 2

It is difficult to add an electron due to small size.

It is difficult to add an electron due to high electronegativity.

It is difficult to add an electron due to stable configuration.

It is difficult to add an electron due to high electron affinity.

Solution

Question 3

Electronegativity

Atomic size

Occurrence in nature

Ionization energy

Solution

Question 4

Ionisation enthalpy

Bond enthalpy

Electron gain enthalpy

Hydration enthalpy

Question 5

$$X$$ represents an alkali metal.

Energy is required to change $$X$$ into $$Y$$

$$Y$$ represents ground state of the element.

Less energy is required to remove an electron from $$X$$ than from $$Y$$

Solution

Question 6

Hydrogen

Nitrogen

Ozone

Helium

Solution

Question 7

$$Cl > F > S > O$$

$$O > Cl > F >S$$

$$F > S > O > Cl$$

$$S >F > O > Cl$$

Solution

Question 8

O and F are more electronegative than S and Cl.

When an electron is added to O or F, it goes to a smaller (n = 2) level and suffers more repulsion than the electron in S or Cl in larger level (n = 3).

Adding an electron to 3p-orbital leads to more repulsion than 2p-orbital.

Electron gain enthalpy depends upon the electron affinity of the atom.

Solution

Question 9

$$Na$$

$$4Mg^{2+}$$

$$Ba^{2+}$$

$$Ar$$

Solution

Question 10

maximum chemical reactivity

maximum electronegativity and different configurations

small size, large charge/radius ratio

tendency to form multiple bonds

Solution

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