Classification of Elements and Periodicity in Properties Test

Result

Classification of Elements and Periodicity in Properties Test - 62

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Electron gain enthalpy is positive when:

A
$$O^{-}$$ is formed from O

B
$$O^{2-}$$ is formed from $$O^{-}$$

C
$$O^{+}$$ is formed from O

D
electron affinity is always a negative value

Solution

$$O^-(g)+e^-(g) \overset{Endothermic}{\rightarrow}O^{2-}(g);\Delta _{eg}H^{\circleddash} $$ $$= 744 kJ mol^{-1}.$$

Due to electrostatic repulsion on account of the same charge.

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

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

Which one of the following statements is correct?

A
The elements like F, Cl, Br etc having high values of electron affinity act as a strong oxidizing agent.

B
The elements having low values of ionization energies act as strong reducing agent.

C
The formation of $$Be^-(g)$$ from Be(g) is an endothermic process.

D
All of these.

Solution

$$F,\ Cl,\ Br$$ have high value of electron affinity act as a strong oxidizing agent as they have high tendency in accepting electron by oxidizing the other element hence act as strong oxidizing agent.
Elements with low ionization enthalpy mean they can easily lose electron by donating an electron to another element i.e. reduces the other element and hence act as a strong reducing agent.
$$Be(g) + e^- \rightarrow Be^-$$
Since the outermost shell, $$2s$$ of $$Be$$ is filled hence inserting an extra electron requires energy and hence the process is an endothermic process.
Question 3
1 / -0

Which pair of elements has same chemical properties?

A
13, 22

B
3, 11

C
4, 24

D
2, 4

Solution

The pair which belongs to some group i.e., in which both the elements have same outer electronic configuration has same chemical properties.
$$\therefore \,\,\, 3\Rightarrow 1s^2, 2s^1$$
$$11\Rightarrow 1s^2, 2s^1 2p^2, 3s^1$$
Question 4
1 / -0

IP values and EA values of elements A, B, C and D are given in a tabular form. Identify the strongest oxidizing agent and reducing agent among them.
IP Value EA Value
A 450 140
B 225 700
C 300 400
D 700 1035

A
B, D

B
B, A

C
D, A

D
D, B

Solution

Electron affinity is the amount of energy released when an electron is added to a neutral atom or molecule in the gaseous state to form a negative ion. More the Electron affinity stronger is the oxidising agent.

Ionization potential is defined as the amount of energy required to remove an electron from an isolated atom or molecule. Lesser the Ionization potential stronger is the reducing agent.

Oxidizing agent $$\rightarrow$$ gains electrons and reduced.

Reducing agent $$\rightarrow$$ loses electrons and oxidized.

Strongest Oxidizing agent $$\rightarrow$$ D (IP - 700 and EA - 1035)

Strongest Reducing agent $$\rightarrow$$ B (IP - 225 and EA - 700)

Option D is correct.
Question 5
1 / -0

What does 'like dissolves like' mean?

A
That compounds that are from the same period on the periodic table will dissolve each other.

B
Polar solutes dissolve in polar solvents.

C
Nonpolar solutes dissolve in polar solvents.

D
Polar solutes dissolve in nonpolar solvents.

E
That compounds from the same family on the periodic table will dissolve each other.

Solution

"Like dissolves like " means if any substance is polar then polar and ionic substances will be dissolved in it. Example : HCl in water. If the solvent is non-polar then the non-polar substances only can be dissolved in it. Example : hexane in benzene.
Question 6
1 / -0

Which one of the following arrangements represents the correct order of electron gain enthalpy (with negative sign) of the given atomic species?

A
$$Cl < F < S < O$$

B
$$O < S < F < Cl$$

C
$$S < O < Cl < F$$

D
$$F < Cl < O < S$$

Solution

Generally, electron gain enthalpy increases in a period from left to right but decreases in a group on moving down. Therefore, halogens have very high electron affinities.

Fluorine due to its smaller size has unexpectedly low electron gain enthalpy than $$Cl$$.

Similar is shown in case of $$O$$ and $$S$$. Thus, the order of electron gain enthalpy is :

$$O < S < F < Cl$$

Hence, the correct option is B.
Question 7
1 / -0

The decreasing order of electron affinity is:

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

B
$$Br > Cl > F > l$$

C
$$F > Cl > Br > l$$

D
$$Cl > F > Br > l$$

Solution

On moving down the group, electron affinity decreases due to increase in size but $$Cl$$ shows higher electron affinity than fluorine due to vacant d-orbitals.
Question 8
1 / -0

The atomic number of natural radioactive element is:

A
greater than 82

B
lesser than 82

C
not defined

D
at least 92

Solution

The two additional series at the bottom of the periodic table are known as lanthanide and actinide series. The elements present in the actinide series are radioactive elements. Actinide series starts from the element thorium having the atomic number 90. Hence, elements having the atomic number greater than 82 are radioactive.
Question 9
1 / -0

The electron affinity increases on moving from left to right along a period. Arrange the reasons in a proper sequence.
(a) The amount of energy released during the addition of an electrons increases from left to right along a period.
(b) Effective nuclear charge of the elements increases from left to right.
(c) The atomic size of the elements decreases from left to right.
(d) The tendency to gain electrons and form anion increases from left to right.

A
a c b d

B
b c d a

C
c a d b

D
d c b a

Solution

As we know the trends in periodic table
$$\rightarrow $$effective nuclear charge of elements when increase it holds the electron tightly because of these attraction, size of element reduced in a period from left to right. And because of more effective nuclear charge it will attracts or gain electron
and therefore amount of energy released during the addition of an electron increase from left to right along a period.
Question 10
1 / -0

The formation of the oxide ion, O$$^{2-}_{(g)}$$, from oxygen atom requires first an exothermic and then an endothermic step as shown below:

$$O_{(g)} + e^{-} \rightarrow O^{-}_{(g)}; \Delta H^{\circ} = -141 kJ mol^{-1}$$
$$O_{(g)}^{-} + e^{-} \rightarrow O^{2-}_{(g)}; \Delta H^{\circ} = +780 kJ mol^{-1}$$

Thus, process of formation of $$O^{2-}$$ in gas phase is unfavourable even though $$O^{2-}$$ is isoelectronic with neon. It is due to the fact that:

A
oxygen is more electronegative

B
addition of electron in oxygen results in larger size of the ion

C
electron repulsion outweighs the stability gained by achieving noble gas configuration

D
$$O^{-}$$ ion has comparatively smaller size than oxygen atom.

Solution

Due to the small size of $$O$$atom, electron repulsion outweighs the stability gained by achieving noble gas configuration. Due to electron repulsion, a huge amount of energy is required to add an electron to $$O^-$$ ion.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

	IP Value	EA Value
A	450	140
B	225	700
C	300	400
D	700	1035

Classification of Elements and Periodicity in Properties Test - 62

Result

Classification of Elements and Periodicity in Properties Test - 62

Score

-

Rank

-

SHARING IS CARING

Question 1

$$O^{-}$$ is formed from O

$$O^{2-}$$ is formed from $$O^{-}$$

$$O^{+}$$ is formed from O

electron affinity is always a negative value

Solution

Question 2

The elements like F, Cl, Br etc having high values of electron affinity act as a strong oxidizing agent.

The elements having low values of ionization energies act as strong reducing agent.

The formation of $$Be^-(g)$$ from Be(g) is an endothermic process.

All of these.

Solution

Question 3

13, 22

3, 11

4, 24

2, 4

Solution

Question 4

B, D

B, A

D, A

D, B

Solution

Question 5

That compounds that are from the same period on the periodic table will dissolve each other.

Polar solutes dissolve in polar solvents.

Nonpolar solutes dissolve in polar solvents.

Polar solutes dissolve in nonpolar solvents.

That compounds from the same family on the periodic table will dissolve each other.

Solution

Question 6

$$Cl < F < S < O$$

$$O < S < F < Cl$$

$$S < O < Cl < F$$

$$F < Cl < O < S$$

Solution

Question 7

$$I > Br > Cl > F$$

$$Br > Cl > F > l$$

$$F > Cl > Br > l$$

$$Cl > F > Br > l$$

Solution

Question 8

greater than 82

lesser than 82

not defined

at least 92

Solution

Question 9

a c b d

b c d a

c a d b

d c b a

Solution

Question 10

oxygen is more electronegative

addition of electron in oxygen results in larger size of the ion

electron repulsion outweighs the stability gained by achieving noble gas configuration

$$O^{-}$$ ion has comparatively smaller size than oxygen atom.

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.