Classification of Elements and Periodicity in Properties Test

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Classification of Elements and Periodicity in Properties Test - 27

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Weekly Quiz Competition

Question 1
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Consider the following statements and arrange in the order of true/ false as given in the codes.
$$S_1: Na_2O_2 < MgO < ZnO < P_4O_{10}$$: Acidic property.
$$S_2$$: Na < Si < Mg < Al: First ionisation energy.
$$S_3$$: F > Cl > Br : Electron affinity.

A
TTT

B
TTF

C
TFF

D
TFT

Solution

$$S_1: Na_2O_2 < MgO < ZnO < P_4O_{10}:$$ as non-metallic character increases the acidic character increases, $$MgO, Na_2O$$ are basic, ZnO amphoteric and $$P_4O_{10}$$ acidic.
$$S_2$$: Mg has a higher first ionization energy than Na due to a small size and higher nuclear charge. Mg has higher than Al because of $$ns^2$$ configuration (has extra stability and high electron penetration power of s-subshell electrons) and Si has higher than Al because of higher nuclear charge and small size. $$IE_1:Na=496, Al=577, Mg=737$$ and Si=786 kJ/mole.
$$S_3$$: There is more interelectronic repulsion in 2p-subshell of fluorine than chlorine (3p). So extra electron will be added easily in 3p-subshell of chlorine as compared to 2p-subshell of fluorine. Down the group electron affinity values generally decreases with increasing atomic number due to increase in atomic size. So $$Cl > F > Br$$.
Question 2
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The $$100^{th}$$ element is named in honor of:

A
Einstein

B
Bohr

C
Fermi

D
Curie

Solution

The $${ 100 }^{ th }$$ element is 'Fermium' . It is named in honor of Scientist Fermi
Question 3
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Structures of nuclei of three atoms A, B, and C are given below-

A has 90 protons and 146 neutrons
B has 92 protons and 146 neutrons
C has 90 protons and 148 neutrons

Based on the above data, which of these atoms are isotopes and which are isobars?

A
A and C are isotopes; B and C are isobars

B
A and B are isotopes; A and C are isobars

C
B and C are isobars; A and B are isobars

D
A and C are isotopes; A and B are isobars

Solution

$$^{236}_{90}A,\quad ^{238}_{92}B,\quad ^{238}_{90}C$$

A and C are isotopes because they have the same atomic number.
B and C are isobars have the same mass number.
Hence, the correct option is A.
Question 4
1 / -0

A neutral atom of an element has a nucleus with nuclear charge $$11$$ times and mass $$23$$ times that of hydrogen. Write the electronic configuration of the element?

A
$$2, 1$$

B
$$2, 8, 1$$

C
$$2, 8$$

D
$$2, 8, 8, 3$$

Solution

Since the charge of the nucleus is $$11$$ times that of the hydrogen atom, the number of electrons is $$11$$ and electronic configuration is $$2, 8, 1$$.

Hence, Option (B) is correct.
Question 5
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There are four elements 'p', 'q', 'r' and 's' having atomic numbers Z-1, Z, Z+1 and Z+2 respectively. If the element 'q' is an inert gas, select the correct answers from the following statements.
(i) 'p' has most negative electron gain enthalpy in the respective period.
(ii) 'r' is an alkali metal.
(iii) 's' exists in +2 oxidation state.

A
(i) and (ii) only

B
(ii) and (iii) only

C
(i) and (iii) only

D
(i), (ii) and (iii)

Solution

As 'q' is a noble gas, p, r, and s having atomic number Z-1, Z+1 and Z+2 should belong to halogen, alkali metal, and an alkaline earth metal respectively.

As halogen has one electron less than stable noble gas configuration it has a greater tendency to accept an additional electron forming anion.

Alkaline earth metal having valence shell configuration $$ns^2$$ exists in the +2 oxidation state.

Thus all the statements are true.
Question 6
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An atom of an element has one electron in the valence shell. It can be represented as :

A
$$_{6}A^{12}$$

B
$$_{19}A^{39}$$

C
$$_{13}A^{27}$$

D
$$_{12}A^{24}$$

Solution

The atomic no. of $$A_{6}^{12}$$ is 6. Its electronic configuration is $$1s^2 2s^2 2p^2$$. It has 4 electrons in valence shell.

The atomic no. of $$A_{19}^{39}$$ is 19. Its electronic configuration is $$1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$$. It has 1 electron in valance shell.

The atomic no. of $$A_{13}^{27}$$ is 13. Its electronic configuration is $$1s^2 2s^2 2p^6 3s^2 3p^1$$. It has 3 electrons in valence shell.

The atomic no. of $$A_{12}^{24}$$ is 12. Its electronic configuration is $$1s^2 2s^2 2p^6 3s^2$$. It has 2 electrons in valence shell.

Therefore, the correct answer is option B.
Question 7
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The atomic number of Uut is:

A
113

B
114

C
108

D
115

Solution

$$Un-1; Tri-3$$. The symbols for these are $$U, T,$$ respectively. Hence for $$Uut$$ it becomes 113. The symbols above mentioned are capital only if they are the first letter otherwise symbols would be small alphabets.

Hence, the correct option is $$\text{A}$$
Question 8
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Which one of the following statements is incorrect?

A
Greater is the nuclear charge, greater is the electron gain enthalpy

B
Nitrogen has almost zero electron gain enthalpy

C
Electron gain enthalpy decreases from flourine to iodine in the group

D
Chlorine has highest electron gain enthalpy

Solution

The first electron affinity of chlorine is greater than that of flourine and order is Cl > F > Br > I.
Nitrogen has almost zero electron gain enthalpy due to stable half filled configuration.
Question 9
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In which of the following process energy is absorbed?

A
$$Cl+e^{\displaystyle-}\longrightarrow Cl^{\displaystyle-}$$

B
$$O^{\displaystyle-}+e^{\displaystyle-}\longrightarrow O^{\displaystyle2-}$$

C
$$O+e^{\displaystyle-}\longrightarrow O^{\displaystyle-}$$

D
$$Na^{\displaystyle+}+e^{\displaystyle-}\longrightarrow Na$$

Solution

$$O(g)+e^{\displaystyle-}\longrightarrow O^{\displaystyle-}(g)$$; $$\Delta H=-142\:kJ/mol$$
$$O^{\displaystyle-}(g)+e\longrightarrow O^{\displaystyle2-}(g)$$; $$\Delta H=844\:kJ/mol$$
In processes $$B$$, energy is released.
Question 10
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Directions For Questions

The amount of energy required to remove the most loosely bound electron from an isolated gaseous atom is called as first ionization energy $$(IE_1)$$. Similarly, the amount of energies required to knock out second, third etc. electrons from the isolated gaseous cation are called successive ionization energies and $$IE_3 > IE_2 > IE_1$$.
(i) Nuclear charge (ii) Atomic size (iii) Penetration effect of the electrons (iv) Shielding effect of the inner electrons and (v) electronic configurations (exactly half filled & completely filled configurations are considered extra stable) affect the ionization energies.
On the other hand, the amount of energy released when a neutral isolated gaseous atom accepts an extra electron to form gaseous anion is called electron affinity.
$$O(g)+e^-\xrightarrow {Exothermic} O^-(g); \Delta H_{eg}=-141 kJ mol^{-1}....(a)$$
$$O^-(g)+e^-\xrightarrow {Endothermic} O^{2-}(g); \Delta H_{eg}=+780 kJ mol^{-1} ....(b)$$
In (b) the energy has to be supplied for the addition of the second electron due to electrostatic repulsion between an anion and extra electron (same charged species). The electron affinity of an element depends upon (i) atomic size (ii) nuclear charge & (iii) electronic configuration. In general, ionization energy and electron affinity increase as the atomic radii decrease and nuclear charge increases across a period. In general, in a group, ionization energy and electron affinity decrease as the atomic size increases.
The members of the third period have some higher (e.g. $$S$$ and $$Cl$$) electron affinity values than the members of the second period (e.g. $$O$$ and $$F$$) because second-period elements have very small atomic size. Hence there is a tendency of electron-electron repulsion, which results in less evolution of energy in the formation of the corresponding anion.

...view full instructions

Considering the elements $$F,\ Cl,\ O$$ and $$N$$, the correct order of their electron affinity values is:

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

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

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

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

Solution

On moving left to right in a period electron affinity increases due to increase in electronegativity (tendency to attract electron) and on moving down, as the atomic radii increases, therefore electron affinity decreases.
So from the element $$F,\ Cl,\ O,\ S$$
The size of $$Cl$$ is having highest electron affinity value than $$F$$ as the size of $$F$$ being small and highly electronegative,it tends to repel the incoming electron hence has low affinity than $$Cl$$ but have a higher affinity value than $$S$$ as sulphur has a large atomic radius compared to $$Cl$$ but $$O$$ having more electronegativity tends to repel the incoming electron hence has high electron affinity than $$O$$. Hence the order will be :
$$Cl > F >S > O$$

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Classification of Elements and Periodicity in Properties Test - 27

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Classification of Elements and Periodicity in Properties Test - 27

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Question 1

TTT

TTF

TFF

TFT

Solution

Question 2

Einstein

Bohr

Fermi

Curie

Solution

Question 3

A and C are isotopes; B and C are isobars

A and B are isotopes; A and C are isobars

B and C are isobars; A and B are isobars

A and C are isotopes; A and B are isobars

Solution

Question 4

$$2, 1$$

$$2, 8, 1$$

$$2, 8$$

$$2, 8, 8, 3$$

Solution

Question 5

(i) and (ii) only

(ii) and (iii) only

(i) and (iii) only

(i), (ii) and (iii)

Solution

Question 6

$$_{6}A^{12}$$

$$_{19}A^{39}$$

$$_{13}A^{27}$$

$$_{12}A^{24}$$

Solution

Question 7

113

114

108

115

Solution

Question 8

Greater is the nuclear charge, greater is the electron gain enthalpy

Nitrogen has almost zero electron gain enthalpy

Electron gain enthalpy decreases from flourine to iodine in the group

Chlorine has highest electron gain enthalpy

Solution

Question 9

$$Cl+e^{\displaystyle-}\longrightarrow Cl^{\displaystyle-}$$

$$O^{\displaystyle-}+e^{\displaystyle-}\longrightarrow O^{\displaystyle2-}$$

$$O+e^{\displaystyle-}\longrightarrow O^{\displaystyle-}$$

$$Na^{\displaystyle+}+e^{\displaystyle-}\longrightarrow Na$$

Solution

Question 10

$$F > Cl > O > S$$

$$F > O > Cl > S$$

$$Cl > F > S > O$$

$$O > F > S > Cl$$

Solution

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