Periodicity In Properties MCQ Quiz in தமிழ் - Objective Question with Answer for Periodicity In Properties - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Ionization enthalpy is the energy required to remove the most loosely bound electron from a gaseous atom in its ground state. The general trend of ionization enthalpy along the period and group is influenced by several factors:

• Across a Period: Ionization enthalpy increases from left to right across a period due to increasing nuclear charge, which pulls the electrons closer to the nucleus, making them harder to remove.

• Down a Group: Ionization enthalpy decreases as we move down a group due to an increase in atomic size and shielding effect, making it easier to remove electrons from the outermost shell.

• Shielding Effect: The inner electrons shield the outer electrons from the full attraction of the nucleus, making the ionization energy lower.

• Atomic Radius: As the atomic radius increases, the distance between the nucleus and the outermost electron increases, lowering the ionization enthalpy.

• Electronic Configuration: Elements with stable electronic configurations (such as noble gases) have much higher ionization enthalpies compared to elements with an unstable configuration.

Explanation: 

• Sodium (Na) has the lowest ionization enthalpy because it is an alkali metal with a single electron in its outermost shell, making it easy to remove.

• Magnesium (Mg) has a higher ionization enthalpy than Na and Al due to a smaller atomic radius , complete 3s orbital ,and a higher nuclear charge, making electron removal more difficult.

• Aluminum (Al) follows Mg, but its ionization energy is slightly lower than expected because of its electron configuration (3p electron is easier to remove than a 3s electron).

• Silicon (Si), Phosphorus (P), and Sulfur (S) have progressively higher ionization enthalpies as their nuclear charge increases across the period.

• Chlorine (Cl) has the highest ionization enthalpy because it is a halogen with a high effective nuclear charge and a small atomic radius.

Conclusion:

The correct order of first ionization enthalpy for Na, Mg, Al, Si, P, S, and Cl is: Na < Al < Mg < Si < S < P < Cl

Concept:

Electron gain enthalpy is the energy change that occurs when an electron is added to a neutral atom in the gas phase to form a negative ion. More negative values indicate a higher tendency to gain an electron. Typically, electron gain enthalpy becomes more negative across a period from left to right due to increasing nuclear charge and becomes less negative down a group owing to increasing atomic size and electron-electron repulsions.

Explanation:

• Sulfur (S): Has a moderately high electron affinity, resulting in a fairly negative electron gain enthalpy. However, it is less negative than those of fluorine and chlorine.

• Phosphorus (P): Has a half-filled orbital which gives it some stability. Adding an electron would disrupt this stability, resulting in a less favorable (less negative) electron gain enthalpy.

• Fluorine (F): Has a very high nuclear charge and small atomic radius, resulting in the most negative electron gain enthalpy among these elements.

• Chlorine (Cl): Has a high nuclear charge and a relatively small atomic radius, resulting in a highly negative electron gain enthalpy, almost as negative as fluorine’s.

Conclusion:

Based on the explanations, the element with the least negative electron gain enthalpy among the given options is: Phosphorus(P).

Concept:

Isoelectronic species are atoms and ions that have the same number of electrons. Examples given are:

• K⁺ (19 protons, 18 electrons)

• Ca²⁺ (20 protons, 18 electrons)

• Cl^- (17 protons, 18 electrons)

• S^2- (16 protons, 18 electrons)

Ionization energy is the amount of energy required to remove an electron from an atom or ion. For isoelectronic species, this is influenced by the nuclear charge (number of protons).

Explanation:

Effective Nuclear charge:

• K⁺: 19 protons pulling on 18 electrons.

• Ca²⁺: 20 protons pulling on 18 electrons (stronger pull than K⁺).

• Cl^-: 17 protons pulling on 18 electrons (weaker pull than K⁺).

• S^2-: 16 protons pulling on 18 electrons (weakest pull among the given species).

The species with the lowest ionization energy will be the one with the fewest protons, as it has the least nuclear charge to hold onto the electrons.

Conclusion:

Therefore, the isoelectronic species with the lowest ionization energy is: S^2- (16 protons, 18 electrons)

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- .

Concept:

Isoelectronic species are atoms and ions that have the same number of electrons but different nuclear charges (atomic numbers). The size of an isoelectronic species is influenced by the effective nuclear charge (Z_eff), which is the net positive charge experienced by electrons in the valence shell. The greater the effective nuclear charge, the more strongly the electrons are pulled towards the nucleus, resulting in a smaller atomic or ionic radius.

Explanation:

• Fluoride ion (F^–): Has 9 protons and 10 electrons.

• Neon atom (Ne): Has 10 protons and 10 electrons.

• Sodium ion (Na⁺): Has 11 protons and 10 electrons.

The increasing nuclear charge (number of protons) among these species causes their sizes to decrease as you go from F^– to Ne to Na⁺.

Order of Effective Nuclear Charge (Z_eff):

The nuclear charge increases from F^– to Ne to Na⁺, resulting in the following order of effective nuclear charge and atomic or ionic sizes:

• F^– (smallest Z_eff, largest size)

• Ne (intermediate Z_eff, intermediate size)

• Na⁺ (largest Z_eff, smallest size)

Order of Sizes

Therefore, the correct order of the sizes of the isoelectronic species from largest to smallest is:

F^– > Ne > Na⁺

Conclusion:

The order of sizes for isoelectronic species F^– > Ne > Na⁺.

Explanation:

Metallic character :

• It specifies the tendency of an element to lose electrons and form positive ions or cations.
• On moving downwards in a group the metallic character increases.
• On moving from left to right in a period the metallic character decreases.
• The given elements Na, Si, Cl, Mg, and Al belongs to the 3^rd period.
• Hence, the arrangement in the order of their decreasing metallic character will be as follows:
• Na > Mg > Al > Si > Cl

Additional InformationPeriodic table:

• It's an arrangement of all the known elements.
• All elements are arranged (from left to right and top to bottom) in order of increasing atomic number and recurring chemical properties.
• All the rows represent periods.
• All the columns represent groups.
• Elements in the same group have the same valence electron configuration and therefore same chemical properties.
• Elements in the same period have an increasing order of valence electrons.
• The elements are arranged in 7 horizontal rows, called periods, and 18 vertical columns, called groups.

CONCEPT:

Quantum Numbers and Electronic Configuration in Lu³⁺

• Quantum numbers are used to describe the position and energy of electrons in an atom.
• The principal quantum number (n) describes the energy level of the electron and the distance from the nucleus.
• The azimuthal quantum number (l) describes the shape of the orbital and can have integer values from 0 to (n-1).

EXPLANATION:

To determine the principal (n) and azimuthal (l) quantum numbers of the valence electrons in tripositive Lutetium (Lu³⁺), we need to first determine the electronic configuration of neutral Lutetium (Lu).

Lu: [Xe] 4f¹⁴ 5d¹ 6s²

Removing three electrons from neutral Lutetium involves the removal of two 6s electrons and one 5d electron:

Lu³⁺: [Xe] 4f¹⁴

• Neutral Lutetium has an atomic number of 71. Its electron configuration is:
• In the tripositive state (Lu³⁺), three electrons are removed. The electrons are generally removed from the outermost orbitals first.
• In Lu³⁺, the valence electrons are in the 4f subshell.
• The principal quantum number (n) for the 4f subshell is 4, and the azimuthal quantum number (l) for the f subshell is 3.

Therefore, the correct set of quantum numbers for the valence electrons in tripositive Lutetium is n = 4 and l = 3

So, the correc answer is Option 4: n = 4 and l = 3

Concept:

The oxidation state of a transition metal refers to the charge it would have if all bonds to different atoms were completely ionic. Transition metals can exhibit a variety of oxidation states, often including very high ones, due to the involvement of d-electrons.

Explanation:

Consider the highest known oxidation states of the given transition metals:

• Palladium (Pd): The highest oxidation state of palladium is +4.

• Osmium (Os): Osmium can reach an oxidation state of +8, as seen in compounds like osmium tetroxide (OsO₄).

• Chromium (Cr): The highest oxidation state of chromium is +6, seen in compounds such as potassium dichromate (K₂Cr₂O₄).

• Manganese (Mn): Manganese can exhibit an oxidation state of +7, as seen in potassium permanganate (KMnO₄).

Conclusion:

The transition metal that exhibits the highest oxidation state is: Osmium (Os) with an oxidation state of +8.

Concept:

The electronic configuration of an element reveals the arrangement of electrons in its atomic orbitals. This information helps to classify the element as a metal, non-metal, metalloid, or inert gas.

• Metals: Metals typically have few electrons in their outermost shell(ns¹ or ns² or (n-1)d^1-10 ns^1-2) and tend to lose electrons to attain a stable configuration. This gives them characteristic properties such as high electrical conductivity, malleability, ductility, and luster. Examples include alkali metals, alkaline earth metals, and transition metals.
• Non-metals: Non-metals usually have more electrons in their outer shells(ns² np^3-6) and tend to gain or share electrons to achieve stability. They are generally poor conductors of heat and electricity, and they often have lower melting and boiling points compared to metals. Common non-metals include carbon, nitrogen, oxygen, etc.
• Metalloids: Metalloids have properties intermediate between metals and non-metals. They can exhibit mixed characteristics and are often semiconductors. Examples include elements like silicon and germanium.
• Inert Gases (Noble Gases): Inert gases have a complete outer electron shell(ns² np⁶), making them very stable and mostly unreactive. These elements are found in Group 18 of the periodic table and include helium, neon, argon, etc.

Explanation:

The electronic configuration provided is: 1s², 2s², 2p⁶, 3s². We can sum the electrons to determine the atomic number:

2 (1s) + 2 (2s) + 6 (2p) + 2 (3s) = 12

This corresponds to an element with atomic number 12, which is Magnesium (Mg).

Alternate Method Given electronic configuration is comparable to ns² electronic configuration which belongs to metallic group.

Magnesium (Mg) belongs to Group 2 of the periodic table and is classified as an alkaline earth metal. Metals are characterized by their ability to lose electrons and form positive ions, their high electrical conductivity, and their malleability and ductility.

Conclusion:

The element with the electronic configuration 1s², 2s², 2p⁶, 3s² is Metal.

Concept:

Oxidation potential is a measure of the tendency of a chemical species to lose electrons and be oxidized. Elements with the highest oxidation potential are typically those that can easily lose an electron to form a positive ion.

Explanation:

The given configurations represent elements from Group 1 of the periodic table (alkali metals), which are known for their high oxidation potentials due to their tendency to lose one electron easily:

• [He]2s¹: This configuration corresponds to Lithium (Li), which has a relatively high Ionisation potential and low oxidation potential among the alkali metals.

• [Ne]3s¹: This configuration corresponds to Sodium (Na), which also has a high oxidation potential, but less than Lithium due to increased electron shielding and atomic size.

• [Ar]4s¹: This is the electronic configuration of Potassium (K). Its oxidation potential is lower than both Lithium and Sodium due to even greater electron shielding and atomic size.

• [Xe]6s¹: This configuration corresponds to Cesium (Cs). Among the alkali metals listed, Cesium has the Highest oxidation potential because it loses an electron the most easily due to its large atomic size and high electron shielding.

Conclusion:

The element with the highest oxidation potential among the given options is: [Xe]6s1.