Classification of Elements and Periodicity in Properties Test

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Classification of Elements and Periodicity in Properties Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

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 metal so Al has the lowest electron gain enthalpy.

Cl has the highest electron gain enthalpy.

And electron gain enthalpy increases along a period.

∴ The correct order is $$Al<N<O<Cl$$.
Question 2
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On the basic of electronic configuration the known elements are grouped in:

A
$$3-$$ block

B
$$2-$$ block

C
$$4-$$ block + lanthanides

D
$$3$$ block + lanthanides + actinides

Solution

The periodic table is a tabular display of the chemical elements, which are arranged by atomic number, electron configuration, and recurring chemical properties. It consist of four different blocks $$s$$, $$p$$, $$d$$ and $$f$$. The $$f$$ block elements are divided into two series, namely lanthanides and actinides. So the elements are grouped in $$3$$ blocks $$+$$ lanthanides $$+$$ actinides.
Question 3
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Which of the outer electronic configuration represents argon?

A
$$ns^2$$

B
$$ns^2 np^2$$

C
$$ns^2 np^4$$

D
$$ns^2 np^6$$

Solution
Question 4
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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
$$ 1s^2, 2s^2 2p^6, 3s^2 3p^3 $$

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

Solution

Answer C
Fluorine is a small atom compared with chlorine 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 $1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{5}$
Question 5
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$$K_2Al(C_2O_4)_3$$ is called:

A
potassium aluminium oxide

B
potassium aluminium $$(III)$$ trioxalate

C
potassium trioxatatoaluminate $$(III)$$

D
potassium tris (oxalato) aluminium

Solution

$${C_2O_4}^{2−}$$ is oxalate ion. In a given complex oxidation state of aluminium is +3. So the IUPAC name is potassium $$trioxalatoaluminate(III)$$.

$$\mathbf{Hence\ the\ correct\ answer\ is\ option\ (C)}$$.
Question 6
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How many unpaired electrons are present in the central metal ion of $$[CoCl_4]^{2-}$$?

A
$$5$$

B
$$4$$

C
$$3$$

D
$$2$$

Solution

In $$[Co{Cl}^{ }_{4}$$$${]}^{-2}$$

Oxidation state of Co is (x-4=-2) i.e. +2.

Electronic configuration of $${Co}^{+2}$$ is [Ar]$${3d}^{7}$$.

As $$Cl$$ is weak ligand electron pairing not takes place.

The no. of unpaired electron present in $${Co}^{+2}$$ is three.

Hence,correct option is (C).
Question 7
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Elements A, B, C, D and E have the following electronic configurations:
(A) $$ 1s^2 , 2s^2 ,2p^1 $$
(B) $$ 1s^2, 2s^22p^6, 3s^2 3p^1 $$
(C) $$ 1s^2, 2s^2 2p^6,3s^3p^3 $$
(D) $$ 1s^2 , 2s^2 2p^6 , 3s^2 3p^5 $$
(E) $$ 1s^2, 2s^2 2p^6 ,3s^23p^6 , 4s^1 $$

The elements which belong to the same group of the periodic table are:

A
A and C

B
A and B

C
A and D

D
A and E

Solution
Question 8
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Which of the following pairs show reverse properties on moving along a period from left to right and from top to bottom in a group

A
Atomic radius and electron gain enthalpy (negative value)

B
Nuclear charge and ionisation enthalpy

C
lonisation enthalpy and electron gain enthalpy (negative value)

D
None of the above.

Solution

Atomic radius increase from top to bottom in a group while decrease from left to right in a period. On the other hand, electron gain enthalpy shows severe trends i.e. decrease from top to bottom in a group and increase from left to right in a period.
Hence, option A is correct.
Question 9
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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 number of energies required to knock out second, third, etc.
(i) Nuclear change (ii) Atomic size (iii) penetration effect of the electrons (iv) shielding effect of the inner electrons and (v) electronic configurations (exactly half filed & completely filled configurations are considered extra stable) affect the ionization energies.
On the other hand, the amount of energy released when a natural isolated gaseous atom accepts an extra electron to form gaseous anion is called electron affinity.$$ O(g)+e^{-}\overset{exothermic}{\rightarrow}O^-(g);\Delta H_{eg}=-141 kJ mol^{-1} $$.....(a)

$$ O(g)+e^{-}\overset{Endothermic}{\rightarrow}O^{2-}(g);\Delta H_{eg}=+780kJmol^{-1} $$......(b)

In the (b) 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(iii) unclear charge & (iii) electronic configuration. In general, ionization energy and electron affinity increases as the atomic radii decrease and nuclear charge increase 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.O and F ) because second-period elements have a very small atomic size. Hence there is a tendency of electron-electron repulsion, which results in less evolution of energy in the formation of corresponding again.

...view full instructions

Which one of the following statement is correct

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

B
The elements having low values of ionisation 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 a high value of electron affinity act as a strong oxidizing agent as they have a 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.
$$Be(g)+e^−→Be^−$$
Since the outermost shell, $$2s$$ of $$Be$$ is filled hence inserting an extra electron requires energy and hence the process is an endothermic process.

Option D is correct.
Question 10
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Consider the following statements
(I) Rutherford name was associated with the development of periodic table.
(II) A metal M having electronic configuration $$ 1 s^{2}$$, $$ 2 p^{6}$$, $$ 3 s^{2}$$ ,$$ 3 p^{6}$$ , $$ 3 d^{10}$$ , $$ 4 s^{1}$$ is d-block element.
(III) Diamond is not an element.
(IV) The electronic configuration of the most electronegative element is $$ 1 s^{2}$$ , $$ 2s^{2}$$ , $$ 2 p^{6}$$

Select the correct option from the given codes

A
i ,ii iv

B
i, ii, iii ,iv

C
ii ,iv

D
i, iii, iv

Solution

(II) The d-block elements have general electronic configuration [noble gas] $$\left ( n-1 \right )d^{1-10}ns^{1-2}$$.It is configuration of copper which belongs to d-block and group number $$11^{th}$$.
(IV) This is the configuration of F (electronegativity = 4.0) which belongs to p-block and group number $$17^{th}$$

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Classification of Elements and Periodicity in Properties Test - 55

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Classification of Elements and Periodicity in Properties Test - 55

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Question 1

N < O < Cl < Al

O < N < Al < Cl

Al < N < O < Cl

Cl < N < O < Al

Solution

Question 2

$$3-$$ block

$$2-$$ block

$$4-$$ block + lanthanides

$$3$$ block + lanthanides + actinides

Solution

Question 3

$$ns^2$$

$$ns^2 np^2$$

$$ns^2 np^4$$

$$ns^2 np^6$$

Solution

Question 4

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

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

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

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

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

Solution

Question 5

potassium aluminium oxide

potassium aluminium $$(III)$$ trioxalate

potassium trioxatatoaluminate $$(III)$$

potassium tris (oxalato) aluminium

Solution

Question 6

$$5$$

$$4$$

$$3$$

$$2$$

Solution

Question 7

A and C

A and B

A and D

A and E

Solution

Question 8

Atomic radius and electron gain enthalpy (negative value)

Nuclear charge and ionisation enthalpy

lonisation enthalpy and electron gain enthalpy (negative value)

None of the above.

Solution

Question 9

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

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

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

All of these

Solution

Question 10

i ,ii iv

i, ii, iii ,iv

ii ,iv

i, iii, iv

Solution

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