CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES MCQs for NEET UG CHEMISTRY — Easy, Page 4

91 Easy

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Which group of the Periodic Table does not contain only metals?

View answer and explanation

Correct option: D

IIIA

Option D (IIIA) is correct because Group IIIA contains metalloids and non-metals in addition to metals.

Solution approach

Correct Answer + Core Reason: IIIA. Group IIIA contains metals, metalloids, and non-metals.

Step-by-Step Solution:

Step 1: Understand the question. The question asks which group does not contain only metals.

Step 2: Recall the composition of each group.
- Group IA (Alkali Metals): Contains only metals.
- Group IIA (Alkaline Earth Metals): Contains only metals.
- Group IB (Coinage Metals): Contains only metals.
- Group IIIA (Boron Group): Contains Boron (metalloid), Aluminum (metal), Gallium (metal), Indium (metal), and Thallium (metal).

Step 3: Identify the group that does not contain only metals. Group IIIA contains a metalloid (Boron).

Step 4: Conclusion: Group IIIA does not contain only metals.

Why other options are wrong:
- Option A: Group IB (Coinage Metals) contains only metals (Copper, Silver, Gold).
- Option B: Group IA (Alkali Metals) contains only metals (Lithium, Sodium, Potassium, Rubidium, Cesium, Francium).
- Option C: Group IIA (Alkaline Earth Metals) contains only metals (Beryllium, Magnesium, Calcium, Strontium, Barium, Radium).

Key Concepts:
- Periodic Table Groups
- Metals, Non-metals, and Metalloids

Memory Aids:
- Remember the general location of metals, non-metals, and metalloids on the periodic table.

Quick Tips:
- Know the general properties and composition of common groups in the periodic table.

92 Easy

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Which of the following properties show gradual decrease with increase in atomic number across a period in the Periodic Table?

View answer and explanation

Correct option: D

Size of atom

Option D (Size of atom) is correct because atomic size generally decreases across a period due to increasing nuclear charge.

Solution approach

Correct Answer + Core Reason: Size of atom. Effective nuclear charge increases across a period, pulling electrons closer and reducing atomic size.

Step-by-Step Solution:

Step 1: Understand the trend of atomic size across a period.

Step 2: Across a period, the number of protons in the nucleus increases, leading to a greater effective nuclear charge.

Step 3: The increased effective nuclear charge pulls the electrons closer to the nucleus.

Step 4: This results in a decrease in the atomic radius or size of the atom.

Step 5: Therefore, atomic size gradually decreases with an increase in atomic number across a period.

Why other options are wrong:
- Option A: Electron affinity generally increases across a period (becomes more negative) as elements have a greater tendency to accept electrons due to increasing effective nuclear charge.
- Option B: Ionisation potential generally increases across a period because it becomes more difficult to remove an electron due to increasing effective nuclear charge.
- Option C: Electronegativity generally increases across a period because the attraction for electrons increases with increasing effective nuclear charge.

Key Concepts:
- Periodic trends: Atomic size, ionization potential, electron affinity, electronegativity
- Effective nuclear charge

Memory Aids:
- Atomic size decreases across a period (think of it being "squeezed" by the increasing nuclear charge).
- Ionization energy, electron affinity, and electronegativity generally increase across a period.

Quick Tips:
- Remember the general trends across a period and down a group in the periodic table.
- Consider the effect of increasing nuclear charge on the valence electrons.

93 Easy

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Which of the following is the smallest in size?

View answer and explanation

Correct option: A

N{a^ + }

Option A (

N{a^ + }

) is the smallest because, among the given isoelectronic species, it has the highest nuclear charge.

Solution approach

Correct Answer + Core Reason: $N{a^ + }$ . Among isoelectronic species, the species with the highest nuclear charge has the smallest size.

Step-by-Step Solution:

Step 1: Identify the isoelectronic species: $N{a^ + }$ , ${F^ - }$ , ${O^{2 - }}$ , and ${N^{3 - }}$ are all isoelectronic, meaning they have the same number of electrons (10).

Step 2: Determine the number of protons (nuclear charge) for each species:
- $N{a^ + }$ : 11 protons
- ${F^ - }$ : 9 protons
- ${O^{2 - }}$ : 8 protons
- ${N^{3 - }}$ : 7 protons

Step 3: Relate nuclear charge to ionic size: For isoelectronic species, the ion with the greater nuclear charge will have a smaller radius because the protons attract the electrons more strongly, pulling them closer to the nucleus.

Step 4: Compare the nuclear charges: $N{a^ + }$ has the highest nuclear charge (11), so it will have the smallest size.

Why other options are wrong:
- **Option B ( ${F^ - }$ ):** ${F^ - }$ has a smaller nuclear charge (9) than $N{a^ + }$ (11), so it is larger.
- **Option C ( ${O^{2 - }}$ ):** ${O^{2 - }}$ has a smaller nuclear charge (8) than $N{a^ + }$ (11), so it is larger.
- **Option D ( ${N^{3 - }}$ ):** ${N^{3 - }}$ has the smallest nuclear charge (7) among the given species, so it is the largest.

Key Concepts:
- Isoelectronic species
- Nuclear charge
- Ionic size

Memory Aids:
- For isoelectronic species, \

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The IUPAC symbol for the element with atomic number 119 would be

View answer and explanation

Correct option: B

uue

Option B (uue) is correct because it follows the IUPAC naming conventions for elements with atomic numbers greater than 100.

Solution approach

Correct Answer + Core Reason: uue. The IUPAC symbol is derived from the digits of the atomic number: 1-un, 1-un, 9-ennium.

Step-by-Step Solution:

Step 1: Identify the digits of the atomic number: 119.

Step 2: Assign the IUPAC prefixes for each digit: 1 = un, 1 = un, 9 = enn.

Step 3: Combine the prefixes: ununenn.

Step 4: Add the suffix -ium: ununennium.

Step 5: Determine the symbol: uue.

Why other options are wrong:
- Option A: unh - Incorrect. \"h\

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What is the name of element with atomic number 105?

View answer and explanation

Correct option: B

Dubnium

Option B (Dubnium) is correct because the element with atomic number 105 is named Dubnium (Db).

Solution approach

Correct Answer + Core Reason: Dubnium. The element with atomic number 105 is Dubnium.

Step-by-Step Solution:

Step 1: Recall the names of elements with higher atomic numbers.

Step 2: Identify the element with atomic number 105. The element with atomic number 105 is Dubnium (Db).

Why other options are wrong:
- Option A: Kurchatovium was a proposed name for element 104, now known as Rutherfordium.
- Option C: Nobelium has atomic number 102.
- Option D: Holmium has atomic number 67.

Key Concepts:
- Element names and atomic numbers

Memory Aids:
- Remember the names and symbols of common elements.

Quick Tips:
- Memorize the names and symbols of elements with higher atomic numbers.

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The elements in which 4 f- orbitals are progressively filled up are called as

View answer and explanation

Correct option: C

Lanthanoids

Option C is correct because lanthanoids are the elements in which the 4f-orbitals are progressively filled.

Solution approach

Correct Answer + Core Reason: Lanthanoids. Lanthanoids are characterized by the filling of 4f orbitals.

Step-by-Step Solution:

Step 1: Define lanthanoids.\nLanthanoids are the elements in the periodic table from lanthanum (La, atomic number 57) to lutetium (Lu, atomic number 71). They are also known as the rare earth elements.

Step 2: Identify the orbital being filled in lanthanoids.\nIn lanthanoids, the 4f orbitals are progressively filled with electrons.

Step 3: Conclusion.\nThe elements in which 4f-orbitals are progressively filled up are called lanthanoids.

Why other options are wrong:
- Option A: Incorrect because actinoids are the elements in which 5f-orbitals are progressively filled.
- Option B: Incorrect because transition elements are the elements in which d-orbitals are progressively filled.
- Option C: Correct because lanthanoids are the elements in which 4f-orbitals are progressively filled.
- Option D: Incorrect because halogens are the elements in Group 17 of the periodic table.

Key Concepts:
- Lanthanoids
- f-orbitals

Memory Aids:
- Remember that lanthanoids are also known as the rare earth elements.

Quick Tips:
- Review the electronic configurations of lanthanoids.

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Elements of a vertical group have same 

View answer and explanation

Correct option: A

Same valence shell electronic configuration

Option A (Same valence shell electronic configuration) is correct because elements in the same group have the same number of valence electrons and thus similar chemical properties.

Solution approach

Correct Answer + Core Reason: Same valence shell electronic configuration. Elements in the same group have the same number of valence electrons.

Step-by-Step Solution:

Step 1: Define a vertical group: A vertical group is a column in the periodic table.

Step 2: Recall the properties of elements in the same group: Elements in the same group have the same number of valence electrons, which leads to similar chemical properties.

Step 3: Relate valence electrons to electronic configuration: The valence shell electronic configuration determines the number of valence electrons.

Why other options are wrong:
- Option B: Atomic size: Atomic size generally increases down a group, so elements in the same group do not have the same atomic size.
- Option C: Atomic volume: Atomic volume generally increases down a group, so elements in the same group do not have the same atomic volume.
- Option D: Atomic mass: Atomic mass increases down a group, so elements in the same group do not have the same atomic mass.

Key Concepts:
- Periodic table groups and their properties.
- Valence electrons and electronic configuration.

Memory Aids:
- Elements in the same group have similar chemical properties due to the same number of valence electrons.

Quick Tips:
- Focus on the valence electrons when comparing elements in the same group.

98 Easy

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Which of the following pairs has both members of the same period of the Periodic Table?

View answer and explanation

Correct option: A

\mathrm{Na} - \mathrm{Cl}

Option A (

\mathrm{Na} - \mathrm{Cl}

) is correct because both Sodium (Na) and Chlorine (Cl) are in the third period of the periodic table.

Solution approach

Correct Answer + Core Reason: $\mathrm{Na} - \mathrm{Cl}$ . Both Sodium and Chlorine are in the same period (Period 3).

Step-by-Step Solution:

Step 1: Identify the period of each element:
- Sodium (Na): Period 3
- Chlorine (Cl): Period 3
- Calcium (Ca): Period 4
- Bromine (Br): Period 4

Step 2: Compare the periods of the elements in each pair.

Step 3: Sodium and Chlorine are both in Period 3.

Why other options are wrong:
- **Option B ( $\mathrm{Na} - \mathrm{Ca}$ ): Sodium is in Period 3, while Calcium is in Period 4.
- Option C ( $\mathrm{Ca} - \mathrm{Cl}$ ): Calcium is in Period 4, while Chlorine is in Period 3.
- Option D ( $\mathrm{Cl} - \mathrm{Br}$ ): Chlorine is in Period 3, while Bromine is in Period 4.

Key Concepts:
- Periodic Table
- Periods

Memory Aids:
- Remember the positions of common elements in the periodic table.

Quick Tips:**
- Knowing the period numbers of common elements helps in answering such questions.

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The elements present in each group form a family because:

View answer and explanation

Correct option: B

They possess common pattern of electronic configuration.

Option B (They possess common pattern of electronic configuration) is correct because elements in the same group have the same number of valence electrons, leading to similar chemical properties.

Solution approach

Correct Answer + Core Reason: They possess common pattern of electronic configuration. Elements in the same group have the same valence electron configuration.

Step-by-Step Solution:

Step 1: Recall the definition of a group in the periodic table.

Step 2: Understand that elements in the same group have the same number of valence electrons.

Step 3: Recognize that similar valence electron configurations lead to similar chemical properties.

Why other options are wrong:
- Option A (They exhibit different chemical properties): This is incorrect; elements in the same group exhibit *similar* chemical properties.
- Option C (They come immediately after noble gases): This is not a general property of all groups.
- Option D (They are very reactive): This is not true for all groups; some groups are very reactive, while others are not.

Key Concepts:
- Periodic Table Groups
- Electronic Configuration
- Valence Electrons
- Chemical Properties

Memory Aids:
- Remember that group number often corresponds to the number of valence electrons (for main group elements).

Quick Tips:
- Focus on the relationship between electronic configuration and chemical properties.

100 Easy

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Difference between S and

{\mathrm{S}^{2 - }}\mathrm{as\;}{\mathrm{S}^{2 - }}

has

View answer and explanation

Correct option: A

Larger radii and larger size

Option A (Larger radii and larger size) is correct because adding electrons to an atom increases electron-electron repulsion, leading to an expansion of the electron cloud.

Solution approach

Correct Answer + Core Reason: Larger radii and larger size. Adding electrons increases electron-electron repulsion, thus increasing size.

Step-by-Step Solution:

Step 1: Consider the neutral sulfur atom (S) and the sulfide ion (S^2-).

Step 2: S has a certain number of protons and electrons. S^2- has the same number of protons but two additional electrons.

Step 3: The addition of electrons increases the electron-electron repulsion within the electron cloud.

Step 4: This increased repulsion causes the electron cloud to expand, resulting in a larger ionic radius and size for S^2- compared to S.

Why other options are wrong:
- Option B: Incorrect. S^2- has a larger size, not a smaller size.
- Option C: Incorrect. S^2- has a larger size, not a smaller size.
- Option D: Incorrect. S^2- has a larger size, not a smaller size.

Key Concepts:
- Ionic radius
- Electron-electron repulsion
- Anions vs. neutral atoms

Memory Aids:
- Anions (negative ions) are always larger than their corresponding neutral atoms.

Quick Tips:
- Remember that adding electrons increases size, while removing electrons decreases size.

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According to Mendeleev’s periodic law, the properties of elements were the periodic functions of

View answer and explanation

Correct option: B

Atomic mass

Option B (Atomic mass) is correct because Mendeleev's periodic law stated that the properties of elements are periodic functions of their atomic masses.

Solution approach

Correct Answer + Core Reason: Atomic mass. Mendeleev's periodic law was based on atomic mass.

Step-by-Step Solution:

Step 1: Recall Mendeleev's periodic law.\nMendeleev's periodic law states that the properties of elements are periodic functions of their atomic masses.

Step 2: Consider the other options.\nAtomic number was not known at the time of Mendeleev's work.\nNumber of electrons is related to atomic number, so it was also not used by Mendeleev.

Why other options are wrong:
- Option A: Atomic number was not the basis of Mendeleev's periodic law.
- Option C: Number of electrons was not the basis of Mendeleev's periodic law.
- Option D: Mendeleev's periodic law was based on atomic mass.

Key Concepts:
- Mendeleev's periodic law
- Atomic mass

Memory Aids:
- Remember that Mendeleev arranged elements by increasing atomic mass.

Quick Tips:
- Know the historical development of the periodic table.

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Eka-silicon is now known as

View answer and explanation

Correct option: C

Germanium

Option C (Germanium) is correct because Mendeleev predicted the existence of Eka-silicon, which was later discovered and named Germanium.

Solution approach

Correct Answer + Core Reason: Germanium. Eka-silicon, predicted by Mendeleev, was later discovered and named Germanium.

Step-by-Step Solution:

Step 1: Recall Mendeleev's predictions. Mendeleev predicted the existence of several elements that were not yet discovered, and he named them using the prefix "eka-" (meaning "one") before the name of the element above them in the periodic table.

Step 2: Identify Eka-silicon. Eka-silicon was the element predicted to be below silicon in the periodic table.

Step 3: Determine the modern name of Eka-silicon. Eka-silicon was later discovered and named Germanium (Ge).

Why other options are wrong:
- Option A (Scandium): Scandium was predicted as Eka-boron.
- Option B (Gallium): Gallium was predicted as Eka-aluminium.
- Option D (Boron): Boron is not related to Eka-silicon.

Key Concepts:
- Mendeleev's periodic table
- Predictions of undiscovered elements
- Eka-silicon

Memory Aids:
- Remember that Eka-silicon is Germanium.

Quick Tips:
- Know the elements that Mendeleev predicted and their modern names.

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First ionisation enthalpy values of first four group 15 elements are given below. Choose the correct value for the element that is a main component of apatite family:

View answer and explanation

Correct option: A

1012 \mathrm{~kJ} \mathrm{~mol}^{-1}

Option A (

1012 \mathrm{~kJ} \mathrm{~mol}^{-1}

) is correct because it corresponds to the ionization enthalpy of Phosphorus, a key component of the apatite family.

Solution approach

Correct Answer + Core Reason: $1012 \mathrm{~kJ} \mathrm{~mol}^{-1}$ . This value corresponds to Phosphorus, a main component of the apatite family.

Step-by-Step Solution:

Step 1: Identify the element that is a main component of the apatite family. Apatite is a group of phosphate minerals, so the element is phosphorus (P).

Step 2: Recall the trend of ionization enthalpy in group 15. Ionization enthalpy generally decreases down the group.

Step 3: Consider the typical ionization enthalpy values for group 15 elements. Nitrogen has a high ionization enthalpy, followed by Phosphorus, then Arsenic, and finally Antimony.

Step 4: Match the given values with the expected order. The value closest to that of Phosphorus is $1012 \mathrm{~kJ} \mathrm{~mol}^{-1}$ .

Step 5: Conclusion: The correct value for Phosphorus is $1012 \mathrm{~kJ} \mathrm{~mol}^{-1}$ .

Why other options are wrong:
- **Option B ( $1402 \mathrm{~kJ} \mathrm{~mol}^{-1}$ ): This value is too high and likely corresponds to Nitrogen, which has a higher ionization enthalpy than Phosphorus.
- Option C ( $834 \mathrm{~kJ} \mathrm{~mol}^{-1}$ ): This value is too low and likely corresponds to Arsenic or Antimony, which have lower ionization enthalpies than Phosphorus.
- Option D ( $947 \mathrm{~kJ} \mathrm{~mol}^{-1}$ ):** This value is close, but $1012 \mathrm{~kJ} \mathrm{~mol}^{-1}$ is a more reasonable value for Phosphorus based on general trends.

Key Concepts:
- Ionization Enthalpy
- Periodic Trends
- Group 15 Elements
- Apatite Composition

Memory Aids:
- Remember the general trend: Ionization enthalpy decreases down a group.

Quick Tips:
- Know the common elements in important minerals and their group affiliations.

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The element having maximum value for its

3^{\text {rd }}

ionisation enthalpy is

View answer and explanation

Correct option: B

\mathrm{Be}

Option B (Be) is correct because after losing two electrons, Be attains a stable noble gas configuration, making the removal of the third electron require significantly more energy.

Solution approach

Correct Answer + Core Reason: Be. Beryllium has the electronic configuration $1s^22s^2$ . After losing two electrons, it attains a stable noble gas configuration, making the third ionization enthalpy very high.

Step-by-Step Solution:

Step 1: Write the electronic configurations of the elements.\nB: $1s^22s^22p^1$ \nBe: $1s^22s^2$ \nAl: $1s^22s^22p^63s^23p^1$ \nC: $1s^22s^22p^2$

Step 2: Consider the third ionization enthalpy.\nFor B: $B \rightarrow B^+ \rightarrow B^{2+} \rightarrow B^{3+}$ (removing a 2p electron)\nFor Be: $Be \rightarrow Be^+ \rightarrow Be^{2+} \rightarrow Be^{3+}$ (removing a 1s electron)\nFor Al: $Al \rightarrow Al^+ \rightarrow Al^{2+} \rightarrow Al^{3+}$ (removing a 3p electron)\nFor C: $C \rightarrow C^+ \rightarrow C^{2+} \rightarrow C^{3+}$ (removing a 2p electron)

Step 3: Analyze the electronic configurations after the second ionization.
$B^{2+}$ : $1s^22s^1$
$Be^{2+}$ : $1s^2$ (Noble gas configuration)
$Al^{2+}$ : $1s^22s^22p^63s^1$
$C^{2+}$ : $1s^22s^0$

Step 4: Determine which element requires the most energy for the third ionization.\nRemoving an electron from the stable $1s^2$ configuration of $Be^{2+}$ requires significantly more energy than removing an electron from the other elements.

Step 5: Final Answer: Be has the maximum value for its 3rd ionization enthalpy.

Why other options are wrong:
- Option A: B: The third ionization of B involves removing a 2s electron, which is less difficult than removing a 1s electron from Be.
- Option C: Al: The third ionization of Al involves removing a 3s electron, which is less difficult than removing a 1s electron from Be.
- Option D: C: The third ionization of C involves removing a 2s electron, which is less difficult than removing a 1s electron from Be.

Key Concepts:
- Ionization enthalpy
- Electronic configuration
- Noble gas configuration

Memory Aids:
- Removing an electron from a noble gas configuration requires a large amount of energy.

Quick Tips:
- Consider the electronic configuration after each ionization step.

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Highest electron affinity among the following is

View answer and explanation

Correct option: B

Chlorine

Option B (Chlorine) is correct because chlorine has the highest electron affinity among the given options due to its smaller size and higher effective nuclear charge compared to sulphur, and fluorine's electron affinity is lower than chlorine's due to interelectronic repulsions in its small 2p orbitals.

Solution approach

Correct Answer + Core Reason: Chlorine. Chlorine has the highest electron affinity among the given options.

Step-by-Step Solution:

Step 1: Define electron affinity: Electron affinity is the energy change that occurs when an electron is added to a neutral gaseous atom to form a negative ion.

Step 2: General trend: Electron affinity generally increases across a period and decreases down a group in the periodic table.

Step 3: Consider the given elements: Fluorine, Chlorine, Sulphur, and Xenon.\nXenon is a noble gas and has a very low (or zero) electron affinity.\nFluorine and Chlorine are halogens. Chlorine has a higher electron affinity than Fluorine due to the small size of Fluorine and interelectronic repulsions.\nSulphur is in the same period as Chlorine but to the left. Therefore, Chlorine has a higher electron affinity than Sulphur.

Step 4: Conclusion: Chlorine has the highest electron affinity.

Why other options are wrong:
- Option A: Fluorine has a smaller size than chlorine, leading to greater interelectronic repulsions, which reduce its electron affinity compared to chlorine.
- Option C: Sulphur is to the left of chlorine in the periodic table, and electron affinity generally increases from left to right.
- Option D: Xenon is a noble gas with a stable electronic configuration and thus has a very low electron affinity.

Key Concepts:
- Electron affinity
- Periodic trends
- Interelectronic repulsions

Memory Aids:
- Electron affinity generally increases across a period and decreases down a group (with exceptions).

Quick Tips:
- Remember the exceptions to the general trends, such as the lower electron affinity of fluorine compared to chlorine.

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The starting element of fifth period is

View answer and explanation

Correct option: B

Rb

Option B (Rb) is correct because Rubidium (Rb) is the first element in the fifth period of the periodic table.

Solution approach

Correct Answer + Core Reason: Rb. Rubidium is the starting element of the fifth period.

Step-by-Step Solution:

Step 1: Recall the elements in the fifth period.\nThe fifth period starts with Rubidium (Rb).

Step 2: Consider the other options.\nK is in the fourth period.\nKr is in the fourth period.\nXe is in the fifth period, but it's at the end, not the start.

Why other options are wrong:
- Option A: K is in the fourth period.
- Option C: Kr is in the fourth period.
- Option D: Xe is in the fifth period, but it is a noble gas at the end of the period.

Key Concepts:
- Periodic table
- Periods
- Elements

Memory Aids:
- Remember the order of elements in the periodic table.

Quick Tips:
- Know the organization of the periodic table.

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In the Periodic Table metals usually used as catalyst belong to

View answer and explanation

Correct option: B

d-block

Option B (d-block) is correct because d-block elements have variable oxidation states and form complexes, making them good catalysts.

Solution approach

Correct Answer + Core Reason: d-block. d-block elements exhibit variable oxidation states and form complexes, making them suitable as catalysts.

Step-by-Step Solution:

Step 1: Recall the properties of elements in different blocks of the periodic table.

Step 2: Consider the role of catalysts. Catalysts are substances that speed up chemical reactions without being consumed in the process. They often provide an alternative reaction pathway with a lower activation energy.

Step 3: Analyze the properties of d-block elements. d-block elements (transition metals) have the following properties that make them good catalysts:
- Variable oxidation states: They can easily change their oxidation states, allowing them to form temporary bonds with reactants.
- Formation of complexes: They can form complexes with reactants, bringing them together in a favorable orientation for reaction.
- Availability of d-orbitals: They have partially filled d-orbitals that can participate in bonding.

Step 4: Explain why other blocks are not as suitable:
- f-block elements (lanthanides and actinides) are generally radioactive and less commonly used as catalysts.
- p-block elements have a wider range of properties, but fewer are used as catalysts compared to d-block elements.
- s-block elements (alkali and alkaline earth metals) are highly reactive and not generally used as catalysts in their elemental form.

Why other options are wrong:
- Option A: f-block is incorrect because f-block elements are generally radioactive and less commonly used as catalysts.
- Option C: p-block is incorrect because p-block elements have a wider range of properties, but fewer are used as catalysts compared to d-block elements.
- Option D: s-block is incorrect because s-block elements are highly reactive and not generally used as catalysts in their elemental form.

Key Concepts:
- Catalysis
- Transition metals (d-block elements)
- Variable oxidation states
- Complex formation

Memory Aids:
- Transition metals are good catalysts.

Quick Tips:
- Remember the properties of transition metals that make them good catalysts: variable oxidation states, complex formation, and availability of d-orbitals.

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Which group of the Periodic Table contains coinage metal?

View answer and explanation

Correct option: B

IB

Option B (IB) is correct because Group IB (or Group 11) of the Periodic Table contains coinage metals like copper, silver, and gold.

Solution approach

Correct Answer + Core Reason: IB. Coinage metals (Cu, Ag, Au) belong to Group IB.

Step-by-Step Solution:

Step 1: Recall the definition of coinage metals: Coinage metals are typically copper (Cu), silver (Ag), and gold (Au).

Step 2: Locate these elements in the periodic table: These elements belong to Group 11, which is also known as Group IB.

Step 3: Conclusion: The correct answer is IB.

Why other options are wrong:
- Option A: IIA: Group IIA (or Group 2) contains alkaline earth metals, not coinage metals.
- Option C: IA: Group IA (or Group 1) contains alkali metals, not coinage metals.
- Option D: None of these: This is incorrect because Group IB contains coinage metals.

Key Concepts:
- Periodic Table groups
- Coinage metals

Memory Aids:
- Remember that copper, silver, and gold are coinage metals and belong to Group IB.

Quick Tips:
- Familiarize yourself with the common group names in the periodic table.

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The element with atomic number 117 if discovered would be placed in

View answer and explanation

Correct option: D

Halogen family

Option D (Halogen family) is correct because element 117 would be in Group 17, the Halogen family.

Solution approach

Correct Answer + Core Reason: Halogen family. Element with atomic number 117 would belong to the Halogen family.

Step-by-Step Solution:

Step 1: The element with atomic number 117 is Tennessine (Ts).

Step 2: Tennessine is located in Group 17 of the periodic table, which is the Halogen family.

Step 3: Halogens are characterized by having 7 valence electrons and a strong tendency to gain one electron to achieve a stable octet.

Why other options are wrong:
- Option A (Noble gas family): Noble gases are in Group 18 and have a full valence shell.
- Option B (Alkali family): Alkali metals are in Group 1 and have one valence electron.
- Option C (Alkaline earth family): Alkaline earth metals are in Group 2 and have two valence electrons.

Key Concepts:
- Periodic Table
- Halogens
- Group Numbers

Memory Aids:
- Remember that Halogens are in Group 17.

Quick Tips:
- Knowing the group numbers and their corresponding families helps in quickly identifying the family of an element.

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Remembering +4 | -1

As per the modern periodic law, the physical and chemical properties of elements are periodic functions of their

View answer and explanation

Correct option: B

electronic configuration

Option B (electronic configuration) is correct because the modern periodic law states that properties are periodic functions of electronic configuration.

Solution approach

Correct Answer + Core Reason: electronic configuration. The modern periodic law is based on electronic configuration.

Step-by-Step Solution:

Step 1: The modern periodic law states that the physical and chemical properties of elements are periodic functions of their atomic number, which determines the electronic configuration.

Step 2: Electronic configuration dictates how an element will interact with other elements, thus determining its chemical properties.

Step 3: Physical properties are also influenced by electronic configuration, such as atomic size and ionization energy.

Why other options are wrong:
- Option A (atomic volume): Atomic volume is a property that varies periodically, but it is a consequence of the electronic configuration, not the basis of the periodic law.
- Option C (atomic weight): The original periodic law was based on atomic weight, but it had inconsistencies. The modern law uses atomic number.
- Option D (atomic size): Atomic size is a periodic property, but it's a result of the electronic configuration, not the fundamental basis of the law.

Key Concepts:
- Modern Periodic Law
- Electronic Configuration
- Periodic Properties

Memory Aids:
- Remember that the modern periodic table is arranged by increasing atomic number, which is directly related to electronic configuration.

Quick Tips:
- Always remember that the electronic configuration is the key to understanding the periodic properties of elements.

111 Easy

111 of 171Question permalink

Application +4 | -1

\mathrm{O}_{2}^{2-}

is isoelectronic with:

View answer and explanation

Correct option: C

F_{2}

Option C (

F_2

) is correct because both

O_2^{2-}

and

F_2

have 18 electrons.

Solution approach

Correct Answer + Core Reason: $F_{2}$ . Both $O_2^{2-}$ and $F_2$ have 18 electrons.

Step-by-Step Solution:

Step 1: Determine the number of electrons in $O_2^{2-}$ . Oxygen has 8 electrons, so $O_2$ has $2 \times 8 = 16$ electrons. The $2-$ charge indicates 2 additional electrons, so $O_2^{2-}$ has $16 + 2 = 18$ electrons.

Step 2: Determine the number of electrons in each of the given options:
- A. $H_2$ : Hydrogen has 1 electron, so $H_2$ has $2 \times 1 = 2$ electrons.
- B. $N_2$ : Nitrogen has 7 electrons, so $N_2$ has $2 \times 7 = 14$ electrons.
- C. $F_2$ : Fluorine has 9 electrons, so $F_2$ has $2 \times 9 = 18$ electrons.
- D. $S$ : Sulfur has 16 electrons.

Step 3: Compare the number of electrons in $O_2^{2-}$ with the number of electrons in each option. $O_2^{2-}$ has 18 electrons, and $F_2$ also has 18 electrons.

Step 4: Conclude that $O_2^{2-}$ is isoelectronic with $F_2$ .

Why other options are wrong:
- Option A: $H_2$ - Incorrect because $H_2$ has only 2 electrons.
- Option B: $N_2$ - Incorrect because $N_2$ has 14 electrons.
- Option D: $S$ - Incorrect because $S$ has 16 electrons.

Key Concepts:
- Isoelectronic Species
- Counting Electrons in Molecules and Ions

Memory Aids:
- Isoelectronic: Same number of electrons.

Quick Tips:
- Remember to account for the charge when counting electrons in ions.
- Count the number of electrons for each species and compare.

112 Easy

112 of 171Question permalink

Remembering +4 | -1

Which element has the highest electronegativity?

View answer and explanation

Correct option: B

O

Option B (O) is correct because oxygen has the highest electronegativity among the given elements.

Solution approach

Correct Answer + Core Reason: O. Oxygen has the highest electronegativity among the given elements.

Step-by-Step Solution:

Step 1: Recall the electronegativity trend in the periodic table.\nElectronegativity generally increases across a period and decreases down a group.

Step 2: Consider the positions of the elements in the periodic table.
- C, O are in the second period.
- Mg, S are in the third period.
- Oxygen is to the right of Carbon in the same period, so it is more electronegative than Carbon.
- Oxygen is above Sulfur, so it is more electronegative than Sulfur.
- Magnesium is in group 2, so it is less electronegative than Oxygen.

Step 3: Conclusion.\nTherefore, Oxygen has the highest electronegativity among the given elements.

Why other options are wrong:
- Option A: Carbon is less electronegative than Oxygen.
- Option C: Magnesium is less electronegative than Oxygen.
- Option D: Sulfur is less electronegative than Oxygen.

Key Concepts:
- Electronegativity
- Periodic trends

Memory Aids:
- Electronegativity increases across a period and decreases down a group.

Quick Tips:
- Remember the general trends of electronegativity in the periodic table.

113 Easy

113 of 171Question permalink

Analyzing +4 | -1

The correct order of acetic strength is:

View answer and explanation

Correct option: A

\mathrm{C}{\mathrm{l}_2}{\mathrm{O}_7} > S{\mathrm{O}_2} > {\mathrm{P}_4}{\mathrm{O}_{10}}

Option A (

\mathrm{C}{\mathrm{l}_2}{\mathrm{O}_7} > S{\mathrm{O}_2} > {\mathrm{P}_4}{\mathrm{O}_{10}}

) is correct because the acidic strength of oxides increases with the electronegativity and oxidation state of the central atom.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{l}_2}{\mathrm{O}_7} > S{\mathrm{O}_2} > {\mathrm{P}_4}{\mathrm{O}_{10}}$ . Acidic strength increases with the electronegativity and oxidation state of the central atom.

Step-by-Step Solution:

Step 1: Determine the oxidation states of the central atoms in each oxide.
$\mathrm{Cl}_2\mathrm{O}_7$ : Cl has an oxidation state of +7.
$\mathrm{SO}_2$ : S has an oxidation state of +4.
$\mathrm{P}_4\mathrm{O}_{10}$ : P has an oxidation state of +5.
$\mathrm{CO}_2$ : C has an oxidation state of +4.
$\mathrm{N}_2\mathrm{O}_5$ : N has an oxidation state of +5.
$\mathrm{SO}_3$ : S has an oxidation state of +6.
$\mathrm{Na}_2\mathrm{O}$ : Na has an oxidation state of +1.\nMgO: Mg has an oxidation state of +2.
$\mathrm{Al}_2\mathrm{O}_3$ : Al has an oxidation state of +3.
$\mathrm{K}_2\mathrm{O}$ : K has an oxidation state of +1.\nCaO: Ca has an oxidation state of +2.

Step 2: Compare the electronegativity of the central atoms.\nElectronegativity order: Cl > S > P > N > C > Al > Mg > Na > K > Ca

Step 3: Determine the acidic strength based on oxidation state and electronegativity.\nGenerally, higher oxidation state and higher electronegativity lead to stronger acidic character.

Step 4: Compare the given options.\nOption A: $\mathrm{C}{\mathrm{l}_2}{\mathrm{O}_7} > S{\mathrm{O}_2} > {\mathrm{P}_4}{\mathrm{O}_{10}}$ : This is correct because Cl has the highest electronegativity and oxidation state.

Why other options are wrong:
- Option B: $\mathrm{C}{\mathrm{O}_2} > {\mathrm{N}_2}{\mathrm{O}_5} > S{\mathrm{O}_3}$ : Incorrect order of acidic strength.
- Option C: $\mathrm{N}{\mathrm{a}_2}\mathrm{O} > MgO > A{\mathrm{l}_2}{\mathrm{O}_3}$ : Incorrect order of basic strength.
- Option D: ${\mathrm{K}_2}\mathrm{O} > CaO > MgO$ : Incorrect order of basic strength.

Key Concepts:
- Acidic strength of oxides
- Oxidation state
- Electronegativity

Memory Aids:
- Higher oxidation state and electronegativity lead to stronger acidic character.

Quick Tips:
- Remember that non-metal oxides are generally acidic, and metal oxides are generally basic.

114 Easy

114 of 171Question permalink

Analyzing +4 | -1

In which of the following species

{2^{nd}}I.E\mathrm{ } < \mathrm{ }{1^{st}}I.E

View answer and explanation

Correct option: D

None

Option D (None) is correct because the second ionization energy is always greater than the first ionization energy for any element or ion.

Solution approach

Correct Answer + Core Reason: None. The second ionization energy is always greater than the first.

Step-by-Step Solution:

Step 1: Define ionization energy.\nIonization energy (IE) is the energy required to remove an electron from an atom or ion in the gaseous phase.

Step 2: Explain why the second ionization energy is always greater than the first.\nAfter removing one electron, the remaining electrons are more tightly bound to the nucleus due to the increased effective nuclear charge. Therefore, it requires more energy to remove a second electron.

Step 3: Evaluate each species.\nBe: Removing the second electron requires more energy than removing the first.\nNe: Removing the second electron requires more energy than removing the first.
$Na^+$ : Removing the second electron requires more energy than removing the first.

Why other options are wrong:
- Option A: $Be$ : This is incorrect because the second IE is always greater than the first.
- Option B: $Ne$ : This is incorrect because the second IE is always greater than the first.
- Option C: $N{a^ + }$ : This is incorrect because the second IE is always greater than the first.

Key Concepts:
- Ionization energy
- Effective nuclear charge

Memory Aids:
- The second ionization energy is always greater than the first.

Quick Tips:
- Remember the definition of ionization energy and the factors affecting it.

115 Easy

115 of 171Question permalink

Analyzing +4 | -1

The type of oxide formed by the element among
Li, Na, Be, Mg, B and Al that has the least atomic
radius is:

View answer and explanation

Correct option: A

\mathrm{A}_2 \mathrm{O}_3

Option A (

\mathrm{A}_2 \mathrm{O}_3

) is correct because among the given elements, Boron has the smallest atomic radius and forms

\mathrm{B}_2 \mathrm{O}_3

. Aluminum also forms

\mathrm{Al}_2 \mathrm{O}_3

.

Solution approach

Correct Answer + Core Reason: $\mathrm{A}_2 \mathrm{O}_3$ . Boron has the least atomic radius among the given elements and forms $\mathrm{B}_2\mathrm{O}_3$ .

Step-by-Step Solution:

Step 1: Identify the element with the least atomic radius.\nAmong Li, Na, Be, Mg, B, and Al, Boron (B) has the smallest atomic radius.

Step 2: Determine the common oxidation state of Boron.\nBoron commonly exhibits an oxidation state of +3.

Step 3: Determine the formula of the oxide formed by Boron.\nBoron (B $^{3+}$ ) and Oxygen (O $^{2-}$ ) combine to form $\mathrm{B}_2\mathrm{O}_3$ .

Step 4: Generalize the formula.\nSince Boron is represented by A, the formula of the oxide is $\mathrm{A}_2\mathrm{O}_3$ .

Step 5: Final Answer: The type of oxide formed is $\mathrm{A}_2\mathrm{O}_3$ .

Why other options are wrong:
- Option B: $\mathrm{AO}_2$ : This formula corresponds to an element with a +4 oxidation state, which is not typical for the given elements.
- Option C: AO: This formula corresponds to an element with a +2 oxidation state, which is typical for Be and Mg, but not the element with the smallest radius.
- Option D: $\mathrm{A}_2\mathrm{O}$ : This formula corresponds to an element with a +1 oxidation state, which is typical for Li and Na, but not the element with the smallest radius.

Key Concepts:
- Atomic radius
- Oxidation states
- Chemical formula determination

Memory Aids:
- Smaller atomic radius generally leads to stronger non-metallic character.

Quick Tips:
- Remember the common oxidation states of the elements.

116 Easy

116 of 171Question permalink

Application +4 | -1

Which pair of the atomic numbers represents s-block elements?

View answer and explanation

Correct option: A

3, 12

Option A is correct because elements with atomic numbers 3 (Lithium) and 12 (Magnesium) have their valence electrons in s-orbitals, making them s-block elements.

Solution approach

Correct Answer + Core Reason: 3, 12. These atomic numbers represent s-block elements because their valence electrons are in s-orbitals.

Step-by-Step Solution:

Step 1: Identify the electronic configuration of the element with atomic number 3 (Lithium): $1s^22s^1$ .

Step 2: Identify the electronic configuration of the element with atomic number 12 (Magnesium): $1s^22s^22p^63s^2$ .

Step 3: S-block elements have their valence electrons in s-orbitals.

Step 4: Lithium has its valence electron in the 2s orbital, and Magnesium has its valence electrons in the 3s orbital.

Why other options are wrong:
- Option B: 6 (Carbon) is a p-block element, and 12 (Magnesium) is an s-block element.
- Option C: 7 (Nitrogen) and 15 (Phosphorus) are both p-block elements.
- Option D: 9 (Fluorine) and 17 (Chlorine) are both p-block elements.

Key Concepts:
- Electronic configuration
- s-block elements
- Valence electrons

Memory Aids:
- Remember that s-block elements are in Groups 1 and 2 of the periodic table.

Quick Tips:
- Determine the electronic configuration to identify the block of the element.

117 Easy

117 of 171Question permalink

Application +4 | -1

The approximate size of the nucleus of

\mathop {\mathop {Ni}\limits_{28} }\limits^{64}

View answer and explanation

Correct option: C

5fm

Option C (5fm) is correct because the nuclear radius is proportional to

A^{1/3}

, where A is the mass number. Using the formula

R = R_0 A^{1/3}

with

R_0 \approx 1.2 fm

, we get approximately 5 fm.

Solution approach

Correct Answer + Core Reason: 5fm. The nuclear radius is given by $R = R_0 A^{1/3}$ , where A is the mass number and $R_0 \approx 1.2 fm$ .

Step-by-Step Solution:

Step 1: Identify the mass number (A) of the nickel isotope: A = 64.

Step 2: Use the formula for nuclear radius: $R = R_0 A^{1/3}$

Step 3: Substitute the values: $R = 1.2 \times (64)^{1/3}$ fm

Step 4: Calculate the cube root of 64: $(64)^{1/3} = 4$

Step 5: Calculate the nuclear radius: $R = 1.2 \times 4 = 4.8$ fm

Step 6: Approximate the nuclear radius to the nearest whole number: $R \approx 5$ fm

Why other options are wrong:
- Option A: 3fm is incorrect. It is likely a miscalculation of the cube root or using an incorrect value for R0.
- Option B: 4fm is incorrect. It is likely a miscalculation of the cube root or using an incorrect value for R0.
- Option D: 2fm is incorrect. It is likely a significant miscalculation of the cube root or using an incorrect value for R0.

Key Concepts:
- Nuclear radius
- Mass number
- Formula for nuclear radius: $R = R_0 A^{1/3}$

Memory Aids:
- Remember that the nuclear radius is proportional to the cube root of the mass number.

Quick Tips:
- Use $R_0 \approx 1.2 fm$ as a standard value for the constant in the nuclear radius formula.
- Estimate the cube root of the mass number to get a quick approximation of the nuclear radius.

118 Easy

118 of 171Question permalink

Remembering +4 | -1

The lanthanide contraction refers to

View answer and explanation

Correct option: A

radius of the series

Option A is correct because lanthanide contraction refers to the gradual decrease in the atomic and ionic radii of the lanthanide elements as the atomic number increases.

Solution approach

Correct Answer + Core Reason: radius of the series. Lanthanide contraction is the decrease in atomic and ionic radii across the lanthanide series.

Step-by-Step Solution:

Step 1: Define lanthanide contraction.\nLanthanide contraction is the gradual decrease in the size of the atoms and ions of the lanthanide series with increasing atomic number.

Step 2: Explain the cause of lanthanide contraction.\nThe lanthanide contraction is caused by the poor shielding of the 4f electrons. As the nuclear charge increases, the 4f electrons are not effective at shielding the outer electrons from the increasing nuclear attraction. This results in a greater effective nuclear charge, which pulls the outer electrons closer to the nucleus, causing a decrease in atomic and ionic radii.

Step 3: Conclusion.\nLanthanide contraction refers to the radius of the series.

Why other options are wrong:
- Option A: Correct because lanthanide contraction refers to the radius of the series.
- Option B: Incorrect because lanthanide contraction is related to the radius, not the valence electrons.
- Option C: Incorrect because lanthanide contraction is related to the radius, not the density.
- Option D: Incorrect because lanthanide contraction is related to the radius, not the electronegativity.

Key Concepts:
- Lanthanide contraction
- Atomic and ionic radii
- Shielding effect

Memory Aids:
- Remember that lanthanide contraction is a consequence of the poor shielding of 4f electrons.

Quick Tips:
- Understand the relationship between lanthanide contraction and the properties of the elements that follow the lanthanides in the periodic table.

119 Easy

119 of 171Question permalink

Remembering +4 | -1

Which one of the elements has the maximum electron affinity?

View answer and explanation

Correct option: B

Cl

Option B (Cl) is correct because chlorine has the highest electron affinity due to its size and effective nuclear charge, which allows it to accept an electron more readily than fluorine, bromine, or iodine.

Solution approach

Correct Answer + Core Reason: Cl. Chlorine has the highest electron affinity due to its size and effective nuclear charge.

Step-by-Step Solution:

Step 1: Define electron affinity: Electron affinity is the energy change that occurs when an electron is added to a neutral gaseous atom.

Step 2: Recall the trends in electron affinity: Electron affinity generally increases across a period and decreases down a group. However, there is an exception in Group 17 (halogens).

Step 3: Analyze the options:\nA. F\nB. Cl\nC. Br\nD. I

Step 4: Explain the exception in Group 17: Fluorine (F) has a smaller size than chlorine (Cl). Due to the small size of F, there is significant electron-electron repulsion when an electron is added. This reduces the electron affinity of F compared to Cl. Therefore, chlorine has the highest electron affinity among the halogens.

Why other options are wrong:
- Option A: F: Fluorine has a smaller size, leading to increased electron-electron repulsion, which lowers its electron affinity compared to chlorine.
- Option C: Br: Bromine is larger than chlorine and has a lower effective nuclear charge, resulting in a lower electron affinity.
- Option D: I: Iodine is the largest of the given halogens and has the lowest effective nuclear charge, resulting in the lowest electron affinity.

Key Concepts:
- Electron Affinity
- Periodic Trends
- Electron-Electron Repulsion

Memory Aids:
- Remember that chlorine has a higher electron affinity than fluorine.

Quick Tips:
- Consider the size and effective nuclear charge when comparing electron affinities.

120 Easy

120 of 171Question permalink

Remembering +4 | -1

The atomic weight of ‘Be’ and ‘In’ were corrected by Mendeleev using the formula

View answer and explanation

Correct option: C

Atomic weight = Equivalent weight

\times

valency

Option C (Atomic weight = Equivalent weight × valency) is correct because Mendeleev used this formula to correct the atomic weights of elements like Be and In.

Solution approach

Correct Answer + Core Reason: Atomic weight = Equivalent weight $\times$ valency. Mendeleev used this formula to correct the atomic weights of Be and In.

Step-by-Step Solution:

Step 1: Recall how Mendeleev corrected atomic weights.\nMendeleev corrected the atomic weights of some elements by re-evaluating their valencies and using the formula: Atomic weight = Equivalent weight $\times$ valency.

Step 2: Consider the other options.
$\sqrt v = a(z - b)$ is related to Moseley's law.\nmvr = $\frac{{nh}}{{2\pi }}$ is Bohr's quantization condition.\nEquivalent weight = Atomic weight $\times$ valency is incorrect.

Why other options are wrong:
- Option A: $\sqrt v = a(z - b)$ is related to Moseley's law, not Mendeleev's correction of atomic weights.
- Option B: $mvr = \frac{{nh}}{{2\pi }}$ is Bohr's quantization condition, not related to Mendeleev's work.
- Option D: Equivalent weight = Atomic weight $\times$ valency is an incorrect formula.

Key Concepts:
- Mendeleev's periodic law
- Atomic weight
- Equivalent weight
- Valency

Memory Aids:
- Remember the formula: Atomic weight = Equivalent weight $\times$ valency.

Quick Tips:
- Know the historical context of the periodic table and how atomic weights were determined.

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