Chemistry

Periodic Classification and Elements

1,176 Questions

Periodic classification organizes chemical elements based on atomic structure, periodic laws, and elemental properties. It covers historical classification systems, electron configurations, and periodic trends. Test takers preparing for chemistry sections will find these questions crucial for mastering fundamental element properties and atomic interactions.

Periodic table structureAtomic orbitalsElement identificationPeriodic laws

Periodic Classification and Elements Questions

Multiple choice chemistry classification of elements- the periodic table periodic trends in physical properties properties and trend trends in periodic table electronic configuration and valency electron configuration

Electronegativity of $F$ on Pauling's scale is 4.0. The value on Mulliken's scale is:

  1. $11.2$
  2. $14.4$
  3. $16.8$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mulliken proposed that the arithmetic mean of the first ionization energy (I) and the electron affinity (Eea) should be a measure of the tendency of an atom to attract electrons.
$X= \dfrac{E _i + E _{Ea}}{2}$
Electronegativity of $F$ on Mulliken's scale is hence $11.2$

Multiple choice chemistry classification of elements- the periodic table periodic trends in physical properties properties and trend trends in periodic table electronic configuration and valency electron configuration

Electronegativity of F on Pauling scale is 4.0. Its value on Mulliken scale is__________.

  1. 12.35

  2. 11.0

  3. 15.2

  4. 14.2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Electronegativity of F on Pauling scale is 4.0. Its value on Mulliken scale is 11.2
$\displaystyle \chi (Pauling) = 0.34 \chi (Mulliken) -0.2 $
$\displaystyle 4.0 = 0.34\chi (Mulliken) -0.2 $
$\displaystyle \chi (Mulliken) = 12.35 $

Multiple choice chemistry classification of elements- the periodic table periodic trends in physical properties properties and trend trends in periodic table electronic configuration and valency electron configuration

An element $X$ has $IP = 1681$ kJ/mole and $EA =-333$ kJ/mole then its electronegativity is:

  1. $(1681 + 333) / 544$
  2. $(1681 - 333 )/ 544$
  3. $(1681 + (-333)) / 2$
  4. $\dfrac{208\sqrt{1681+333}}{544}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Robert S. Mulliken proposed that the arithmetic mean of the first ionization energy ($E i$) and the electron affinity ($E _{ea}$) should be a measure of the tendency of an atom to attract electrons. As this definition is not dependent on an arbitrary relative scale, it has also been termed absolute electronegativity, with the units of kilojoules per mole or electron volts.
                                        <img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/44e7db9c2cb1be249c807e9435c4d11788ea2efc" class="mwe-math-fallback-image-inline" alt="\chi =(E{\rm {i}}+E_{\rm {ea}})/2\,">

Multiple choice chemistry classification of elements- the periodic table periodic trends in physical properties properties and trend trends in periodic table electronic configuration and valency electron configuration

The ${ Z } { eff }$ for
3d electron of Cr
4s electron of Cr
3d electron of ${ Cr }^{ 3+ }$
3s electron of ${ Cr }^{ 3+ }$ are _________ respectively.

  1. 4.6, 2.95, 4.95, 8.05

  2. 4.95, 2.05, 4.6, 8.05

  3. 4.6, 2.95, 5.3, 12.75

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Cr ${ 1s }^{ 2 }\quad { 2s }^{ 2 }\quad { 2p }^{ 6 }\quad { 3s }^{ 2 }\quad { 3p }^{ 6 }\quad { 3d }^{ 5 }\quad { 4s }^{ 1 }$

${ 24 }^{ - }$
Using later's rule:
for $3de^{ - }\quad { Z } _{ eff }=24-(4\times 0.35)-(18\times 1)=4.6$
for $4se^{ - }\quad { Z } _{ eff }=24-(0\times 0.35)-(13\times 0.85)-(10\times 1)=2.95$
$Cr^{ 3+ }\quad ({ 1s }^{ 2 }\quad { 2s }^{ 2 }\quad { 2p }^{ 6 }\quad { 3s }^{ 2 }\quad { 3p }^{ 6 }\quad { 3d }^{ 3 })$
for $3de^{ - }\quad { Z } _{ eff }=24-(2\times 0.85)-18=5.3$
for $3se^{ - }\quad { Z } _{ eff }=24-(7\times 0.35)-(8\times 0.85)-(2\times 1)=12.75$
                                         ${ s }^{ 1 }p^{ 6 }$           $2({ s }^{ 2 }p^{ 6 })$                    ${ 1s }^{ 2 }$

Multiple choice chemistry periodicity periodic trends in physical properties properties and trend trends in periodic table electronic configuration and valency electron configuration

The screening effect of 'd' electrons is : 

  1. much more than s-electrons

  2. equal to s-electrons

  3. equal to p-electrons

  4. much less than s-electrons
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The screening effect (shielding) depends on the shape of the orbitals. s-orbitals are closest to the nucleus and provide the most shielding, while d-orbitals are more diffuse and have poor shielding ability.

Multiple choice ionic bond atomic structure and chemical bonding chemistry

Sodium atom with electronic configuration 2, 8, 1 achieves stable electronic configuration by ----------  one electron from its outer shell.

  1. loosing

  2. gaining

  3. covalency

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Sodium has unstable configuration $1s^2 2s^2p^6 3s^1$ and it achieves stable configuration by loosing one electron as $Na^+ - 1s^2 2s^2p^6$

Multiple choice ionic bond atomic structure and chemical bonding chemistry

In the atom of an element X, 6 electrons are present in the outermost shell. If it acquires noble gas configuration by accepting requisite number of electrons, then what would be the charge on the ion so formed?

  1. 1

  2. 2

  3. 3

  4. 4
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Electronic configuration of noble gas is $\mathrm{(2,8)}$. It has $\mathrm{8}$ electron in its outermost orbitals.
Element X has $\mathrm{6}$ electrons in its outermost shell. So, it requires $\mathrm{2}$ more electrons to attain the noble gas configuration.
Hence, option $\mathrm{B}$ is the correct answer.

Multiple choice intrinsic and extrinsic semiconductors types of semiconductors electronic devices semiconductor electronics: materials, devices and simple circuits physics

The number density of electrons is equal to the number density of holes in

  1. Intrinsic semiconductors

  2. Extrinsic semiconductors

  3. both of them

  4. none of them
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In an intinsic semiconductor number density of electrons and holes are always equal. As the number of electrons generated in the conduction band is equal to the number of holes generated in the valance band.

At absolute zero temperature valance band is completely filled and conduction band is completely empty.

Multiple choice chemistry occurrence of carbon compounds in nature importance of carbon covalent bonding in carbon compounds carbon and its forms

An element which is an essential constituent of all organic compounds belongs to the_____group. 

  1. 5th

  2. 8th

  3. 14th

  4. 12th
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Carbon is an essential element for all organic compound. It belongs to ${ 14 }^{ th }$ group. Its atomic number is $6$. Methane which is one the very famous organic compound consists of Carbon and hydrogen.

Thus all organic compounds contain carbon and hydrogen of which carbon belongs to the group $14$.

Multiple choice chemistry classification of elements- the periodic table introduction to periodic table necessity of classification need for classification

The elements posses stable electronic configuration are called as __________.

  1. Transition metals

  2. Non-transition metals

  3. Noble gases

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Noble gases have stable configuration of $ns^2 np^6$ with their octet completely filled which, gives them stability. Helium has exception.

Multiple choice chemistry classification of elements- the periodic table introduction to periodic table necessity of classification need for classification

According to classification of elements most of non-metals are placed in ________ block.

  1. $s$ - block
  2. $p$ - block
  3. $d$ - block
  4. $f$ - block
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The $p$ - block is the area of the periodic table containing columns $3A$ to column $8A$ (columns$ 13-18$), including noble gas, all non metals, metals and metalloids. 

Multiple choice chemistry classification of elements- the periodic table introduction to periodic table necessity of classification need for classification

After the classification of elements study of elements become____________.

  1. difficult

  2. easy

  3. both A and B

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 Elements with similar chemical properties generally fall into the same group in the periodic table, and the elements in the same period tend to have similar properties.Thus, it is relatively easy to predict the chemical properties of an element if one knows the properties of the elements around it.