Classification of Elements and Periodicity in Properties MCQ Quiz - Objective Question with Answer for Classification of Elements and Periodicity in Properties - Download Free PDF

Last updated on May 25, 2025

Latest Classification of Elements and Periodicity in Properties MCQ Objective Questions

Classification of Elements and Periodicity in Properties Question 1:

Which of the following statements are true?
A. Unlike Ga that has a very high melting point, Cs has a very low melting point.
B. On Pauling scale, the electronegativity values of N and Cl are same.
C. Ar, K⁺, Cl⁻, Ca²⁺ and S²⁻ are all isoelectronic species.
D. The correct order of the first ionization enthalpies of Na, Mg, Al, and Si is Si > Al > Mg > Na.
E. The atomic radius of Cs is greater than that of Li and Rb.
Choose the correct answer from the options given below:

  1. A, B and E only
  2. C and E only
  3. C and D only
  4. A, C and E only

Answer (Detailed Solution Below)

Option 2 : C and E only

Classification of Elements and Periodicity in Properties Question 1 Detailed Solution

CONCEPT:

Properties of Elements and Periodic Trends

  • Melting Point:
    • The melting point of elements generally varies based on their metallic bonding and atomic structure.
    • Cesium (Cs) has a very low melting point compared to Gallium (Ga), which has one of the highest among the group 13 elements.
  • Electronegativity:
    • Electronegativity is the ability of an atom to attract electrons in a chemical bond.
    • On the Pauling scale, nitrogen (N) has an electronegativity of 3.0, while chlorine (Cl) has an electronegativity of 3.16.
  • Isoelectronic Species:
    • Isoelectronic species are atoms, ions, or molecules with the same number of electrons.
    • For example, Ar, K⁺, Cl⁻, Ca²⁺, and S²⁻ all have 18 electrons, making them isoelectronic.
  • Ionization Enthalpy:
    • Ionization enthalpy is the energy required to remove an electron from a gaseous atom in its ground state.
    • The correct order of first ionization enthalpies for Na, Mg, Al, and Si is: Si > Al > Mg > Na, due to increasing nuclear charge and electron configuration stability.
  • Atomic Radius:
    • The atomic radius increases down a group and decreases across a period.
    • Cesium (Cs) has a larger atomic radius than both lithium (Li) and rubidium (Rb) due to its position in the periodic table.

EXPLANATION:

  • Statement A: "Unlike Ga that has a very high melting point, Cs has a very low melting point."
    • This statement is False. Gallium (Ga) has a relatively high melting point, while Cesium (Cs) has an almost equal melting point.
      qImage6824503ef142d4ee6347bf8a
  • Statement B: "On Pauling scale, the electronegativity values of N and Cl are same."
    • This statement is incorrect. Nitrogen (N) has an electronegativity of 3.04, and Chlorine (Cl) has an electronegativity of 3.16 on the Pauling scale, so they have different electronegativities.
  • Statement C: "K⁺, Cl⁻, Ca²⁺ and S²⁻ are all isoelectronic species."
    • This statement is true. All these ions (K⁺, Cl⁻, Ca²⁺, and S²⁻) have the same number of electrons (18 electrons), making them isoelectronic species.
  • Statement D: "The correct order of the first ionization enthalpies of Na, Mg, Al, and Si is Si > Al > Mg > Na."
    • This statement is incorrect. The correct order of ionization enthalpies is Si > Al > Mg > Na. Magnesium (Mg) has a higher ionization enthalpy than Aluminum (Al), and Sodium (Na) has the lowest ionization enthalpy due to its larger atomic size and lower nuclear charge compared to others.
  • Statement E: "The atomic radius of Cs is greater than that of Li and Rb."
    • This statement is correct. While Cesium (Cs) has a larger atomic radius than Lithium (Li), it is also larger than Rubidium (Rb). The atomic radius increases as you move down a group in the periodic table, so Li < Na < K < Rb < Cs..

Therefore, the correct answer is: Option 2) C and E only.

Classification of Elements and Periodicity in Properties Question 2:

Arrange the following in increasing order of ionic radii?

C4-, N3, F, O2−

  1. C4 < N3− < O2− < F
  2. N3− < C4 < O2− < F
  3. F< O2−  < C4- < N3−
  4. O2− < F < N3− < C4-
  5. F− < O2− < N3− < C4-

Answer (Detailed Solution Below)

Option 5 : F− < O2− < N3− < C4-

Classification of Elements and Periodicity in Properties Question 2 Detailed Solution

CONCEPT:

Ionic Radii

  • The ionic radius is the measure of an atom's ion in a crystal lattice. It is half the distance between two ions that are barely touching each other.
  • The size of an ion is affected by its charge. As the number of electrons increases (for anions), the ion becomes larger. Conversely, as the number of electrons decreases (for cations), the ion becomes smaller.

EXPLANATION:

  •  C4- :
    • With the lowest nuclear charge ( Z = 6 ), C4-  has the largest ionic radius in this series.
  • N3- :
    • Nitrogen ( Z = 7 ) has a slightly higher nuclear charge than carbon, so N3- has a smaller ionic radius than C4-  .
  • O2- :
    • Oxygen ( Z = 8 ) has a higher nuclear charge than nitrogen, so O2-  has a smaller ionic radius than N3- .
  • F- :
    • Fluorine ( Z = 9 ) has the highest nuclear charge in this series, so F- has the smallest ionic radius.

Increasing Order of Ionic Radii: F- < O2- < N3- < C4- .

Classification of Elements and Periodicity in Properties Question 3:

The correct order of metallic character is

  1. B > Al > Mg > K 
  2. Al > Mg > B > K
  3. Mg > Al > K > B
  4. K > Mg > Al > B

Answer (Detailed Solution Below)

Option 4 : K > Mg > Al > B

Classification of Elements and Periodicity in Properties Question 3 Detailed Solution

CONCEPT:

Metallic Character

  • Metallic character refers to the ability of an element to lose electrons and form positive ions (cations).
  • Elements with higher metallic character are generally found on the left side of the periodic table and towards the bottom.
  • Metallic character increases down a group and decreases across a period from left to right.

EXPLANATION:

  • Potassium (K) is in Group 1, which has the highest metallic character.
    • Magnesium (Mg) is in Group 2, which also has high metallic character but less than Group 1.
    • Aluminum (Al) is in Group 13, which has moderate metallic character.
    • Boron (B) is in Group 13, but it is a metalloid with much lower metallic character compared to Al.
  • Considering the position in the periodic table and the trend of metallic character:
    • K has the highest metallic character.
    • Mg has less metallic character than K but more than Al.
    • Al has less metallic character than Mg but more than B.
    • B has the lowest metallic character among the given elements.

Therefore, the correct order of metallic character is: K > Mg > Al > B.

Classification of Elements and Periodicity in Properties Question 4:

IUPAC name of element having atomic number 108 is 

  1. Unniloctium 
  2. Ununoctium 
  3. Nilniloctinium 
  4. Ununoctinium 

Answer (Detailed Solution Below)

Option 1 : Unniloctium 

Classification of Elements and Periodicity in Properties Question 4 Detailed Solution

CONCEPT:

IUPAC Naming of Elements

  • The IUPAC (International Union of Pure and Applied Chemistry) has a systematic way of naming elements, especially those that are newly discovered or have atomic numbers greater than 100.
  • For elements with atomic numbers 101 and above, a temporary name is derived from the Latin or Greek roots of the digits in the element's atomic number, followed by the suffix "-ium."

EXPLANATION:

- guacandrollcantina.com

Corresponding Prefixes:

Digit Prefix
0 nil
1 un
2 bi
3 tri
4 quad
5 pent
6 hex
7 sept
8 oct
9 enn

  • For the element with atomic number 108:
    • 1 is represented by "Un-" (from the Latin "unus")
    • 0 is represented by "nil-" (from the Latin "nulla")
    • 8 is represented by "oct-" (from the Latin "octo")
  • Combining these roots, the name becomes "Unniloctium."
  • This name is followed by the suffix "-ium" to indicate it is an element.

Therefore, the IUPAC name of the element with atomic number 108 is Unniloctium.

Classification of Elements and Periodicity in Properties Question 5:

The correct sequence which shows decreasing order of the ionic radii of the elements is 

  1. Al3+ > Mg2+ > Na+ > F- > O2- 
  2. Na+ > Mg2+ > Al3+ > O2- > F- 
  3. Na+ > F- > Mg2+ > O2 > Al3+
  4. O2- > F- > Na+ > Mg2+ > Al3+ 

Answer (Detailed Solution Below)

Option 4 : O2- > F- > Na+ > Mg2+ > Al3+ 

Classification of Elements and Periodicity in Properties Question 5 Detailed Solution

CONCEPT:

Ionic Radii

  • The ionic radius is the measure of an atom's ion in a crystal lattice. It is half the distance between two ions that are barely touching each other.
  • The ionic radius varies depending on the ion's charge and the number of electrons present.

EXPLANATION:

  • Ionic radii generally increase as you move down a group in the periodic table and decrease as you move across a period from left to right.
  • Cations (positively charged ions) are smaller than their parent atoms because they lose electrons, resulting in a smaller electron cloud and increased effective nuclear charge.
  • Anions (negatively charged ions) are larger than their parent atoms because they gain electrons, resulting in a larger electron cloud and reduced effective nuclear charge.
  • Form the given ions :- Al3+, Mg2+, Na+, F-, O2-:
    • Al3+: smallest ionic radius due to high positive charge and loss of three electrons.
    • Mg2+: smaller ionic radius than Na+ due to higher positive charge and loss of two electrons.
    • Na+: larger ionic radius than Mg2+ but smaller than neutral Na atom due to the loss of one electron.
    • F-: larger ionic radius than Na+ and anions are generally larger due to gain of electrons.
    • O2-: largest ionic radius due to the gain of two electrons.

Therefore, the correct sequence which shows the decreasing order of the ionic radii of the elements is O2- > F- > Na+ > Mg2+ > Al3+

Top Classification of Elements and Periodicity in Properties MCQ Objective Questions

How many electrons does a hydrogen atom have in its K shell?  

  1. 4
  2. 2

Answer (Detailed Solution Below)

Option 1 : 1 

Classification of Elements and Periodicity in Properties Question 6 Detailed Solution

Download Solution PDF
The correct answer is 1.

Key Point

  • A hydrogen atom, the simplest and lightest of all atoms, consists of only one proton and one electron.
  • The K shell, also known as the first electron shell of an atom, can hold up to two electrons.
  • A hydrogen atom has only one electron.
  • This electron occupies the K shell.

qImage66c0d4c61c70e1c17e39d613

Additional Information

  • Hydrogen is the simplest element with an atomic number of 1.
  • In its atomic form, hydrogen has one proton and one electron.
  • When two hydrogen atoms bond to form a hydrogen molecule, they share their electrons, resulting in a stable H2 molecule.
  • This sharing of electrons is a type of covalent bonding.

Ionisation potential in a period is lowest for-

  1. Alkali metals
  2. Halogens
  3. Inert gases
  4. Alkaline earth metals

Answer (Detailed Solution Below)

Option 1 : Alkali metals

Classification of Elements and Periodicity in Properties Question 7 Detailed Solution

Download Solution PDF

The correct answer is Alkali metals 

Key Points

  • Alkali metals:-
    • Alkali metals have the largest atomic size in their respective periods.
    • This translates to a weaker attraction between the positively charged nucleus and the outermost (valence) electron.
    • As a result, less energy is needed to remove this electron, leading to a lower ionization potential.

Additional Information

  • Alkaline earth metals:-
    • These come after alkali metals in a period and while they are relatively large, they have two valence electrons compared to just one for alkali metals. 
    • This increases the effective nuclear charge experienced by the valence electrons, slightly raising their ionization potential compared to alkali metals.
  • In contrast, halogens are on the rightmost side of the period and have one electron less than a full valence shell. They attract their valence electrons strongly, resulting in high ionization potentials.
  • Inert gases have filled valence shells, achieving a stable configuration. Removing an electron from such a stable configuration requires a lot of energy, resulting in very high ionization potentials.

Which of the following pairs of chemical elements and their symbols is INCORRECT?

  1. Phosphorus - P
  2. Potassium - Po
  3. Iron - Fe
  4. Iodine - I

Answer (Detailed Solution Below)

Option 2 : Potassium - Po

Classification of Elements and Periodicity in Properties Question 8 Detailed Solution

Download Solution PDF

The correct answer is Potassium-Po.

Key Points

  • Potassium:
    • Potassium is a chemical element with the symbol K.
    • The Atomic number of potassium is 19.
    • It is a silvery-white metal that is smooth enough to be sliced with little force with a knife.
    • Potassium reacts rapidly with atmospheric oxygen within seconds of exposure to form flaky, white potassium peroxide.

Additional Information

Phosphorus
  • Phosphorus is a chemical element with atomic number 15.
  • The symbol of phosphorus is P.
  • It was discovered by Hennig Brand in 1669.
Iron
  • Symbol Fe and atomic number = 26.
  • Wrought Iron is the purest form of iron.
  • The Most impure form of Iron is Pig Iron.
Iodine
  • Iodine is a chemical element with the symbol I and atomic number 53. 
  • Iodine is a trace element that is naturally present in some foods, is added to some types of salt, and is available as a dietary supplement.

Among the following, the energy of 2s orbital is lowest in:

  1. K
  2. H
  3. Li
  4. Na

Answer (Detailed Solution Below)

Option 1 : K

Classification of Elements and Periodicity in Properties Question 9 Detailed Solution

Download Solution PDF

Concept:

As the value of Z (atomic number) increases, the energy of orbitals decreases (becomes more negative value)

The energy of 2s orbital is lowest in K (potassium)

Therefore, the order of energy of 2s orbital is H > Li > Na > K

Which of the following is a soft metal that ignites in air and reacts violently with water and has atomic number 37 in the periodic table?

  1. Caesium
  2. Francium
  3. Gallium
  4. Rubidium

Answer (Detailed Solution Below)

Option 4 : Rubidium

Classification of Elements and Periodicity in Properties Question 10 Detailed Solution

Download Solution PDF

The correct option is Rubidium.

Key Points

  • Rubidium is the chemical element with the symbol Rb and atomic number 37.
  • Rubidium is a very soft, whitish-grey metal in the alkali metal group. Rubidium metal shares similarities to Potassium metal and Caesium metal in physical appearance, softness and conductivity.
  • Rubidium was discovered in 1861 by Robert Bunsen and Gustav Kirchhoff, in Heidelberg, Germany.
  • Rubidium reacts violently with water and can cause fires.
  • It is also used in fireworks to give it a purple color.

Additional Information

  • Caesium (or cesium) is a chemical element with the symbol Cs and atomic number 55. Its one of only five elemental metal that are liquid at or near room temperature. In 1860, Robert Bunsen and Gustav Kirchhoff discovered caesium.
  • Francium is a chemical element with the symbol Fr and atomic number 87. It is extremely radioactive and second rarest naturally occuring element. Francium was discovered by Marguerite Perey in France in 1939.
  • Gallium is a chemical element with the symbol Ga and atomic number 31. Discovered by French chemist Paul-Émile Lecoq de Boisbaudran in 1875, Gallium is in group 13 of the periodic table and is similar to the other metals of the group.

which element of group 13 has the atomic number 113 and its electronic configuration is [Rn] 5f146d107s27p1?

  1. Gallium
  2. Indium
  3. Nihonium
  4. Thallium

Answer (Detailed Solution Below)

Option 3 : Nihonium

Classification of Elements and Periodicity in Properties Question 11 Detailed Solution

Download Solution PDF

The correct answer is Nihonium.Key Points

  • The element with atomic number 113 belongs to group 13 of the periodic table, which is also known as the boron group.
  • The electronic configuration [Rn] 5f146d107s27p1 belongs to the element Nihonium, which was officially named in 2016 and is a synthetic element that is not found naturally on Earth.
  • Nihonium is a highly unstable and radioactive element, with a very short half-life of only a few seconds, and its properties are still being studied by scientists.

Additional Information

  • Group 13 elements are characterized by having three valence electrons in their outermost shell, which makes them reactive and able to form compounds with other elements.
  • Gallium is a soft, silvery metal that has a low melting point and is used in a variety of applications, such as in semiconductors, LEDs, and alloys.
  • Indium is a rare, silvery-white metal that is used in electronics, as well as in coatings, alloys, and other industrial applications.
  • Thallium is a toxic, bluish-white metal that is used in electronics, as well as in pesticides, rodenticides, and other chemicals.
  • It is also a radioactive isotope that is used in medical imaging.
  • Thallium is also a group 13 element, but it has a higher atomic number than Nihonium and a different electronic configuration.

In comparison to boron, beryllium has:

  1. Lesser nuclear charge and lesser first ionization enthalpy.
  2. Greater nuclear charge and lesser first ionization enthalpy
  3. Greater nuclear charge and greater first ionization enthalpy
  4. Lesser nuclear charge and greater first ionization enthalpy

Answer (Detailed Solution Below)

Option 4 : Lesser nuclear charge and greater first ionization enthalpy

Classification of Elements and Periodicity in Properties Question 12 Detailed Solution

Download Solution PDF

Concept:

  • Nuclear charge: The nuclear charge is the total charge of all the protons in the nucleus.

Nuclear charge of Boron (B) > Nuclear charge of Be

First ionization enthalpy:

Be = 1s2 2s2 (more stable)

B = 1s2 2s2 2p1

The first ionization of Beryllium is greater than that of boron because beryllium has a stable complete electronic configuration (1s2 2s2) so it requires more energy to remove the first electron from it. Whereas boron has the electronic configuration 1s2 2s2 2p1 which needs lesser energy than that of beryllium to remove the valence electron.

Ionisation energy of Be is greater than B due to ns2 outer electronic configuration.

The ionization energy is the energy required to remove an electron from its orbital around an atom to a point where it is no longer associated with that atom. The ionization energy of an element increases as one moves across the period in the periodic table because the electron is held tighter by the higher effective nuclear charge.

A diagonal relationship is found between which pair of the following elements?

  1. Li and Mg
  2. Be and B
  3. Be and Mg
  4. Na and Mg

Answer (Detailed Solution Below)

Option 1 : Li and Mg

Classification of Elements and Periodicity in Properties Question 13 Detailed Solution

Download Solution PDF

The correct answer is Li and Mg. Key Points

  • Lithium (Li) and Magnesium (Mg) show a diagonal relationship as they are located diagonally opposite to each other in the periodic table.
  • Both Li and Mg have similar atomic radii, electronegativities, and ionization energies due to the similarities in their electronic configurations.
  • A diagonal relationship refers to the similarities in properties between elements that are diagonally opposite to each other in the periodic table.

Additional Information

  • Diagonal relationship is a result of the similarities in the electronic configurations of these elements.
  • Be and B do not show a diagonal relationship as they are not located diagonally opposite to each other in the periodic table.
  • Similarly, Be and Mg do not show a diagonal relationship as they belong to the same group and have different electronic configurations.
  • Na and Mg also do not show a diagonal relationship as they belong to different periods and have different electronic configurations.

In a periodic table, while moving from left to right in a period, number of __________ remains same.

  1. electrons
  2. protons
  3. shells
  4. neutrons

Answer (Detailed Solution Below)

Option 3 : shells

Classification of Elements and Periodicity in Properties Question 14 Detailed Solution

Download Solution PDF

Concept: -

  • Modern Periodic Table has 18 Groups and 7 Periods.
  • It has 118 Elements.
  • Each of the tables has a vertical row called a group. Elements in groups have similar chemical and physical properties because they have the same number of outer electrons.
  • Each of the tables has horizontal rows called periods. During a period, a gradual change in chemical properties occurs from one element to another.
  • The elements are arranged in the modern periodic table on the basis of Mosely's law, which states that the physical and chemical properties of elements are the periodic function of their atomic number.
  • The Structure of it is like a bird Grid.
  • The Modern Periodic table is used to organize all the known elements.
  • Advantages of Modern Periodic Table:- 
    • It is easier to remember the properties of an element if its position in the periodic table is known.
    • It made the study of Chemistry Systematic and Easy.

Modern Periodic Table:-

RRB Group-D 27th Sep 2018 Shift 1 (English) Sunny (Type) Madhu(Dia) D1 utkarsha

Explanation: -

A period in the periodic table represents the principal quantum number.

As Principal quantum numbers stay the same moving left to right in the rows of the periodic table.

So, the shell remains same in a periodic table, while moving from left to right in a period.

Hence, the correct answer is a shell, option 3.

The pair that has similar atomic radii is:

  1. Mn and Re
  2. Ti and Hf
  3. Sc and Ni
  4. Mo and W

Answer (Detailed Solution Below)

Option 4 : Mo and W

Classification of Elements and Periodicity in Properties Question 15 Detailed Solution

Download Solution PDF

Concept:

The pair that has similar atomic radii is Mo and W because Mo and W belong to group-6 and period-5 (4d series), period-6 (5d series) respectively, Due to the lanthanoid contraction, the radius of Mo and W are almost the same i.e. 0.140 nm and 0.141 nm respectively.

Lanthanoid contraction is the steady decrease in the size of the atoms and ions of the rare earth elements with increasing atomic number from lanthanum (atomic number 57) through lutetium (atomic number 71).

Molybdenum (Mo) is a chemical element with the symbol Mo and atomic number 42.

Molybdenum does not occur naturally as a free metal on Earth, it is found only in various oxidation states in minerals. The free element, silvery metal with a grey cast, has the sixth-highest melting point of any element. It readily forms hard, stable carbides in alloys.

Tungsten or wolfram is a chemical element with the symbol W and atomic number 74. Tungsten is a rare metal found naturally on Earth almost exclusively combined with other elements in chemical compounds rather than alone. Its important ores include wolframite and scheelite.
Get Free Access Now
Hot Links: teen patti master teen patti app all teen patti teen patti master game teen patti master old version