RBSE Solutions Class 11 Chemistry Chapter 3 Periodic Table

 

Question 1. The basis of Mendeleevs Periodic Law was:
(a) Valency
(b) Atomic weight
(c) Atomic number
(d) Atomic volume
Answer: (b) Atomic weight
In simple words: Mendeleev grouped elements by their atomic weight. He saw that elements with similar properties appeared regularly.

🎯 Exam Tip: Remember that Mendeleev used atomic weight, while the modern periodic law uses atomic number. This is a key difference.

 

Question 3. Which of the following element has highest ionisation enthalpy?
(a) Boron
(b) Carbon
(c) Nitrogen
(d) Oxygen
Answer: (c) Nitrogen
In simple words: Nitrogen has the highest ionization energy here. It is harder to take an electron from nitrogen because its electron shell arrangement is very stable.

🎯 Exam Tip: Ionisation enthalpy generally increases across a period, but there are exceptions, such as nitrogen having a higher value than oxygen due to its stable half-filled configuration.

 

Question 4. The element from atomic number 58 to 71 are called:
(a) Transition elements
(b) Lanthanoids
(c) Actinoids
(d) Alkali Metals
Answer: (b) Lanthanoids
In simple words: Elements from atomic number 58 to 71 are called lanthanoids. They are a special group of elements that are usually found at the bottom of the periodic table.

🎯 Exam Tip: Lanthanoids (atomic numbers 58-71) and Actinoids (atomic numbers 90-103) are the two series of f-block elements.

 

Question 5. Which of the following has highest atomic radius?
(a) F-
(b) Cl-
(c) Br-
(d) I-
Answer: (d) I-
In simple words: The iodide ion (I-) is the biggest. This is because it has more electron layers than the other ions listed.

🎯 Exam Tip: Atomic radius increases down a group and generally decreases across a period. Negative ions (anions) are always larger than their neutral atoms.

RBSE Class 11 Chemistry Chapter 3 Very Short Answer Type Questions

 

Question 2. What is the basis of modern periodic law?
Answer: The modern periodic law is based on the atomic number of elements. It states that the physical and chemical properties of elements are a periodic function of their atomic numbers. This means when elements are arranged by increasing atomic number, their properties repeat regularly.
In simple words: The modern periodic law uses atomic number. It says that element properties repeat when they are lined up by their atomic number.

🎯 Exam Tip: The shift from atomic weight (Mendeleev) to atomic number (Modern Periodic Law) was a crucial development in chemistry.

 

Question 3. What is the name given to elements of 17th group?
Answer: The elements of the 17th group are known as halogens. This group includes elements like fluorine, chlorine, bromine, iodine, and astatine, which are very reactive non-metals.
In simple words: The elements in the 17th group are called halogens.

🎯 Exam Tip: Halogens are known for forming salts easily and are highly reactive due to having seven valence electrons.

 

Question 4. In a periodic table, the penultimate shell of which of the elements is incomplete?
Answer: The penultimate (second to last) shell is incomplete in transition elements. This is because their d-orbitals are being filled, which means both the outermost and the second-to-last shells are not fully complete.
In simple words: Transition elements have an incomplete second-to-last electron shell. This is due to how their d-orbitals fill up with electrons.

🎯 Exam Tip: Transition elements are found in the d-block of the periodic table and are known for forming colored compounds and having variable oxidation states.

 

Question 5. Why there are 14 elements in 4f series?
Answer: There are 14 elements in the 4f series because lanthanides are f-block elements. The f-subshell has seven orbitals, and each orbital can hold two electrons, allowing a maximum of 14 electrons to be filled.
In simple words: The 4f series has 14 elements because f-orbitals can hold a maximum of 14 electrons. This block fills seven f-orbitals, each with two electrons.

🎯 Exam Tip: The 4f series (lanthanides) and 5f series (actinides) each contain 14 elements because of the filling capacity of the f-orbitals.

 

Question 6. What are transuranic elements?
Answer: Transuranic elements are chemical elements that have atomic numbers greater than 92, which is the atomic number of uranium. All of these elements are unstable and decay radioactively. They are not found naturally and are typically made in labs.
In simple words: Transuranic elements are those with atomic numbers larger than 92 (uranium). They are all man-made and not stable.

🎯 Exam Tip: Uranium \( (Z = 92) \) is the heaviest naturally occurring element; all elements beyond it are synthetic and radioactive.

 

Question 8. Element X is placed at first position in 15 group. Write outer most electronic configuration of this element.
Answer: The element placed first in Group 15 is Nitrogen (N), which has an atomic number of 7. Its electronic configuration is \( 1s^2 2s^2 2p^3 \). Therefore, the outermost electronic configuration is \( 2s^2 2p^3 \).
In simple words: The first element in group 15 is Nitrogen. Its outer electron shell has 2 electrons in the s-orbital and 3 electrons in the p-orbital.

🎯 Exam Tip: The group number often indicates the number of valence electrons (e.g., Group 15 elements have 5 valence electrons, which is \(ns^2 np^3\)).

 

Question 9. Write two de-merits of Mendeleev's periodic table.
Answer: Two de-merits of Mendeleev's periodic table are:

• It could not explain the proper position of Hydrogen. Hydrogen behaves like both alkali metals and halogens.
• The position of isotopes was not clearly explained. Isotopes have different atomic masses but the same chemical properties.

In simple words: Two problems with Mendeleev's table were:
- He could not find a clear place for Hydrogen.
- He did not explain where isotopes (elements with same properties but different weights) should go.

🎯 Exam Tip: Mendeleev's table faced challenges with elements like hydrogen and isotopes because it was based on atomic mass, not atomic number.

 

Question 10. Write two demerits of Modern periodic table.
Answer: Two demerits of the Modern Periodic Table are:
(a) Hydrogen resembles both alkali metals and halogens, but it is placed with alkali metals. This placement is still a point of debate.
(b) The lanthanides and actinides are not placed within the main body of the table. They are shown separately at the bottom.
In simple words: Two issues with the Modern Periodic Table are:
- Hydrogen acts like two different types of elements, but it is only placed with one.
- The special lanthanide and actinide elements are not inside the main table but are shown below it.

🎯 Exam Tip: While vastly improved, even the modern periodic table has some unresolved issues regarding the placement of certain elements.

 

Question 11. Write atomic number of element with symbol Uub.
Answer: The atomic number of the element with the symbol Uub (Ununbium, now Copernicium, Cn) is 112. Uub is an IUPAC temporary systematic element name.
In simple words: The element with the symbol Uub has an atomic number of 112.

🎯 Exam Tip: IUPAC systematic names (like Ununbium) are used for newly synthesized elements before a permanent name is officially adopted.

 

Question 12. Which element in a period has highest atomic size?
Answer: In an isoelectronic series (ions with the same number of electrons but different nuclear charges), the atomic size decreases as the positive charge increases. For example, in the series \( Sc^{3+} < Ca^{2+} < K^{+} < Cl^{-} < S^{2-} \), \( S^{2-} \) has the highest size. The reason is that all these ions have 18 electrons. With more positive charge (less electrons compared to protons), the nucleus pulls the electrons more strongly, making the ion smaller.
In simple words: When atoms or ions have the same number of electrons, the one with the smallest positive charge (or most negative charge) will be the largest. This is because the pull from the nucleus is weaker on the outer electrons.

🎯 Exam Tip: For an isoelectronic series, the more negative the charge (or less positive), the larger the radius due to weaker nuclear attraction per electron.

 

Question 14. What is Vander Waal's radius?
Answer: Van der Waals radius is defined as half the internuclear distance between two non-bonded atoms of the same element when they are at their closest possible approach. It represents the non-covalent interactions between atoms. It is often denoted by \( r_v \).

In simple words: It is half the distance when two same atoms are just touching, but not actually bonded together.

🎯 Exam Tip: Van der Waals radius is typically larger than covalent radius for the same element because it measures non-bonded interactions.

 

Question 15. Which of the following has smallest radius - Na+, Mg2+ and Al3+?
Answer: Among \( Na^{+} \), \( Mg^{2+} \), and \( Al^{3+} \), the smallest radius belongs to \( Al^{3+} \). These are isoelectronic ions (all have 10 electrons). As the positive charge on the nucleus increases across a period, the effective nuclear charge becomes stronger. This stronger pull shrinks the electron cloud, leading to a smaller ionic radius.
In simple words: \( Al^{3+} \) has the smallest radius. Even though they all have the same number of electrons, \( Al^{3+} \) has more protons, pulling its electrons in tighter.

🎯 Exam Tip: For isoelectronic species, the ionic radius decreases with an increase in the positive nuclear charge (number of protons).

 

Question 16. Which of the following has smallest radius- cation or parent atom?
Answer: A cation always has a smaller radius than its parent atom. This happens because a cation is formed when an atom loses one or more electrons from its outermost shell. This electron loss typically removes the entire outermost shell, and the remaining electrons are pulled more strongly by the nucleus, making the ion smaller.
In simple words: A cation is smaller than its original atom. When an atom loses electrons to become a cation, it shrinks because the nucleus pulls the remaining electrons closer.

🎯 Exam Tip: Cations are smaller due to electron loss, while anions (formed by electron gain) are larger than their parent atoms.

 

Question 17. How the ionization enthalpy changes on moving left to right in a period?
Answer: When moving from left to right across a period, the ionization enthalpy generally increases. This is because the atomic number increases, meaning more protons are added to the nucleus. As the extra electrons are added to the same main shell, the effective nuclear charge on the valence electrons increases, pulling them more strongly. This makes it harder to remove an electron, thus increasing the ionization enthalpy.
In simple words: Ionization enthalpy usually goes up as you move from left to right in a period. This is because the nucleus gets stronger, holding onto electrons more tightly, so it takes more energy to pull one away.

🎯 Exam Tip: The increasing nuclear charge and constant number of electron shells across a period lead to a stronger attraction for valence electrons.

 

Question 18. Which of the element has lowest ionization enthalpy in alkali metals?
Answer: Among the alkali metals, Caesium (Cs) has the lowest ionization enthalpy (376 kJ/mol). This is because Caesium is the largest alkali metal, and its outermost electron is furthest from the nucleus, experiencing the least attraction.
In simple words: Caesium has the lowest ionization energy among alkali metals. It is a very large atom, so its outermost electron is easy to remove.

🎯 Exam Tip: Ionization enthalpy decreases down a group due to increasing atomic size and greater shielding effect, making it easier to remove the outermost electron.

 

Question 19. Which element has highest electron gain enthalpy in periodic table?
Answer: Chlorine (Cl) has the highest electron gain enthalpy in the periodic table. While electron gain enthalpy generally increases across a period and decreases down a group, fluorine (F) has a lower electron gain enthalpy than chlorine due to its very small size. This small size leads to significant electron-electron repulsion within its compact \( 2p \) subshell, making it less favorable to accept an incoming electron compared to chlorine.
In simple words: Chlorine takes in electrons less easily than fluorine, even though it pulls them strongly. This is because fluorine is very small, and new electrons face a lot of pushing from existing electrons inside. Chlorine is bigger, so it has more space for a new electron.

🎯 Exam Tip: Despite fluorine being more electronegative, chlorine has a higher electron gain enthalpy because its larger atomic size reduces electron-electron repulsion, allowing it to more readily accept an electron.

 

Question 20. Which of the following has highest second ionization enthalpy Na or Mg?
Answer: Sodium (Na) has the highest second ionization enthalpy. This is because after losing its first electron, Na achieves a stable noble gas configuration (like Neon). Removing a second electron from this highly stable, complete electron shell would require a very large amount of energy, which is energetically unfavorable. Magnesium (Mg), after losing its first electron, still has one valence electron, so its second ionization enthalpy is lower than sodium's second ionization enthalpy.
In simple words: Sodium has a much higher second ionization energy than Magnesium. Once sodium loses one electron, it becomes very stable, making it extremely hard to remove a second electron.

🎯 Exam Tip: Removing an electron from a stable noble gas configuration (like \(Na^+\)) always requires significantly more energy than removing an electron from a partially filled or less stable configuration.

 

Question 21. W negative element in a periodic table?
Answer: Electronegativity is a measure of an atom's ability to attract electrons in a chemical bond. According to the Mulliken scale, electronegativity can be calculated using the following formulas:

• If Ionization Energy (IE) and Electron Gain Enthalpy (EGE) are in eV: \( \text{Electronegativity} = \frac{\text{IE} + \text{EGE}}{2} \)
• If IE and EGE are in kJ/mol: \( \text{Electronegativity} = \frac{\text{IE} + \text{EGE}}{540} \)
• The Pauling electronegativity \( (\chi_{Pauling}) \) can be related to the Mulliken electronegativity \( (\chi_{Mulliken}) \) by \( \chi_{Pauling} = \frac{\chi_{Mulliken}}{2.8} \).
• Alternatively, using IE and EGE in eV: \( \chi_{Pauling} = \frac{\text{IE} + \text{EGE}}{5.6} \). This scale helps us compare how strongly different atoms pull electrons in a bond.

In simple words: Electronegativity tells us how strongly an atom pulls electrons in a bond. We can calculate it by using its ionization energy and electron gain enthalpy values.

🎯 Exam Tip: Electronegativity is different from electron gain enthalpy; electronegativity describes an atom's attraction for electrons *within a bond*, while electron gain enthalpy is the energy change when an isolated atom *gains* an electron.

 

Question 23. Write the formula for calculation of percentage ionic character.
Answer: Even in ionic compounds, there's always some degree of covalent character, as explained by Fajan's rules. The percentage of ionic character in a compound that also has covalent character can be calculated using the formula:
\[ \text{Percentage ionic character} = \frac{\text{Observed dipole moment}}{\text{Calculated dipole moment}} \times 100 \]
This formula helps quantify how much an ionic bond truly behaves like a purely ionic bond.
In simple words: To find out how much an ionic compound is truly "ionic," we use a formula. It compares the actual dipole moment with what a fully ionic bond would have.

🎯 Exam Tip: A higher difference in electronegativity between two bonding atoms generally leads to a higher percentage of ionic character.

 

Question 24. What is the Valency?
Answer: Valency is the combining capacity of an element. It tells us how many bonds an atom can form with other atoms.
In simple words: Valency is how many bonds an atom can make with other atoms.

🎯 Exam Tip: Valency is determined by the number of electrons an atom needs to lose, gain, or share to achieve a stable electron configuration.

 

Question 25. What is the valency of alkaline earth metal?
Answer: Alkaline earth metals have a valency of 2. These elements are found in Group 2 of the periodic table, and they all have two electrons in their outermost electron shell. They readily lose these two electrons to form a stable \( +2 \) ion.
In simple words: Alkaline earth metals have a valency of 2. They like to lose two electrons to become stable.

🎯 Exam Tip: Elements in the same group generally have the same valency because they have the same number of valence electrons.

RBSE Class 11 Chemistry Chapter 3 Short Answer Type Questions

 

Question 1. What is the necessity of classification of elements?
Answer: The classification of elements is necessary for several reasons:

• It helps us efficiently study the vast number of elements by grouping those with similar properties.
• It allows us to understand the regular changes in properties that occur across groups and periods.
• It also helps in predicting the properties of new elements.

In simple words: We classify elements to study them better. It helps us see patterns in their properties and understand how they relate to each other.

🎯 Exam Tip: Classification simplifies the study of chemistry by organizing information and highlighting trends, making it easier to predict chemical behavior.

 

Question 2. In transition elements, on moving left to right, there is less change in atomic size. Why?
Answer: While atomic size generally decreases from left to right in a period, this trend is less prominent in transition elements. As electrons are added to the inner d-orbitals across the transition series, the added d-electrons provide some shielding effect, which partially counteracts the increasing nuclear charge. This results in a smaller, less systematic decrease in atomic size. The lanthanoid contraction is a good example where \( 4f \) electrons provide poor shielding, making \( 4d \) and \( 5d \) series elements have similar atomic radii.
In simple words: In transition metals, the atomic size does not change much as you move across. This is because inner electrons block some of the pull from the nucleus, making the size changes less noticeable.

🎯 Exam Tip: The filling of inner d-orbitals and the associated shielding effect cause the atomic radius trend to be less consistent for transition elements compared to main group elements.

 

Question 3. The radius of anion is always larger than parent atom. Why?
Answer: An anion is always larger than its parent neutral atom. The reasons are:

• An anion is formed when a neutral atom gains one or more electrons. This increases the number of electrons in the outermost shell while the nuclear charge (number of protons) remains the same.
• The added electrons increase electron-electron repulsion in the valence shell. This repulsion causes the electron cloud to expand, increasing the size of the ion. Also, the same nuclear charge now has to attract more electrons, reducing the effective nuclear pull per electron.

In simple words: An anion is bigger than its original atom. When an atom gains electrons, the new electrons push each other away, making the electron cloud spread out and the anion larger.

🎯 Exam Tip: Electron-electron repulsion in the outermost shell is the primary reason for the increased size of an anion compared to its neutral atom.

 

Question 4. Cs+ from Cs is obtained easily as compared to Na⁺ from Na. Explain
Answer: It is easier to form \( Cs^{+} \) from \( Cs \) than \( Na^{+} \) from \( Na \). This is because ionization energy decreases as you move down a group in the periodic table due to increasing atomic size. Cesium (Cs) is much larger than Sodium (Na), so its outermost electron is further from the nucleus and experiences less attraction. This means less energy is needed to remove an electron from \( Cs \) to form \( Cs^{+} \).
In simple words: It is easier to make \( Cs^{+} \) from \( Cs \) than \( Na^{+} \) from \( Na \). This is because \( Cs \) is a bigger atom than \( Na \), so its outermost electron is held less tightly and is easier to remove.

🎯 Exam Tip: The larger the atomic size and the greater the shielding effect, the lower the ionization energy, making it easier to form cations.

 

Question 6. Why Alkali metals do not form di-positive ions?
Answer: Alkali metals do not form di-positive ions (e.g., \( M^{2+} \)) because they have only one electron in their outermost energy level. After losing this single valence electron, they achieve a very stable noble gas configuration (an octet of electrons). Removing a second electron from this already stable, fully filled inner shell would require an extremely large amount of energy, which is energetically unfavorable.
In simple words: Alkali metals do not form \( M^{2+} \) ions. They only have one outer electron, and once they lose it, they become very stable. Removing another electron would take too much energy.

🎯 Exam Tip: The stability gained by achieving a noble gas configuration after losing the first valence electron prevents alkali metals from forming higher oxidation states.

 

Question 7. The ionisation enthalpy of noble gases is very high. Explain
Answer: Noble gases have very high ionization enthalpies. This is because they possess a complete and stable valence shell electron configuration (an octet of electrons, or a duplet for Helium). Their outermost electrons are tightly held, and removing an electron from such a stable configuration requires a substantial amount of energy.
In simple words: Noble gases have very high ionization energy. This is because their outer electron shells are already full and very stable, making it very hard to remove an electron.

🎯 Exam Tip: The stability associated with a completely filled valence shell makes noble gases chemically inert and resistant to losing electrons.

 

Question 8. The ionization enthalpy of N is more than Oxygen. Why?
Answer: The ionization enthalpy of Nitrogen (N) is higher than that of Oxygen (O). Nitrogen's electronic configuration is \( 1s^2 2s^2 2p^3 \), which features a half-filled \( 2p \) subshell. A half-filled subshell is particularly stable due to symmetrical electron distribution and reduced electron-electron repulsion. Oxygen's configuration is \( 1s^2 2s^2 2p^4 \), which is less stable than nitrogen's half-filled p-orbital. Therefore, more energy is required to remove an electron from stable nitrogen than from oxygen.
In simple words: Nitrogen's ionization energy is higher than oxygen's. Nitrogen has a very stable half-filled outer electron shell, which makes it harder to pull an electron away from it compared to oxygen.

🎯 Exam Tip: Half-filled and fully-filled electron configurations confer extra stability, leading to anomalies in ionization enthalpy trends across a period.

 

Question 9. Alkaline Earth Metal do not show variable oxidation states. Why?
Answer: Alkaline earth metals (Group 2 elements) do not show variable oxidation states; they always exhibit a \( +2 \) oxidation state. This is because they have two electrons in their outermost shell, and by losing both of these electrons, they achieve a highly stable noble gas configuration. It is energetically favorable to lose both electrons rather than just one, and losing more than two would be extremely difficult.
For example, for Magnesium (Mg), the configuration is \( 1s^2 2s^2 2p^6 3s^2 \). After losing two electrons, it becomes \( Mg^{2+} \), with configuration \( 1s^2 2s^2 2p^6 \).
In simple words: Alkaline earth metals only show a \( +2 \) charge. They have two outer electrons and always lose both of them to become stable, so they don't have other charges.

🎯 Exam Tip: The strong tendency to achieve a stable noble gas configuration by losing exactly two electrons dictates the fixed \( +2 \) oxidation state for alkaline earth metals.

 

Question 11. Explain the reason that electron gain enthalpy of F is less than Cl.
Answer: Although Fluorine (F) is more electronegative than Chlorine (Cl), its electron gain enthalpy is less than that of chlorine. This is because fluorine has a very small atomic size. When an incoming electron attempts to enter the compact \( 2p \) subshell of fluorine, it experiences strong electron-electron repulsion from the existing electrons. Chlorine, being larger, has less electron density and repulsion in its \( 3p \) subshell, allowing it to more easily accommodate an additional electron and release more energy.
In simple words: Fluorine takes in electrons less easily than chlorine, even though it pulls them strongly. This is because fluorine is very small, and new electrons face a lot of pushing from existing electrons inside. Chlorine is bigger, so it has more space for a new electron.

🎯 Exam Tip: Electron-electron repulsion in small atoms like fluorine can make electron gain less favorable, despite high electronegativity.

 

Question 12. Ionisation enthalpy of boron is less than Beryllium. Explain with reason.
Answer: The ionization enthalpy of Boron (B) is less than that of Beryllium (Be). Their electronic configurations are:

• \( \text{Be}: 1s^2 2s^2 \)
• \( \text{B}: 1s^2 2s^2 2p^1 \)

Beryllium has a stable, completely filled \( 2s \) orbital, making it relatively difficult to remove an electron. Boron, on the other hand, has one electron in its \( 2p \) orbital. Electrons in \( p \) orbitals are generally higher in energy and farther from the nucleus compared to \( s \) orbitals in the same shell. This means the \( 2p \) electron in boron is less tightly held and easier to remove, resulting in a lower ionization enthalpy.
In simple words: It takes less energy to remove an electron from Boron than from Beryllium. Beryllium's outer electron shell is full and stable. Boron's outer electron is in a slightly higher energy level and further away, making it easier to take out.

🎯 Exam Tip: Stability gained from fully-filled or half-filled subshells (like \(2s^2\) in Be) can lead to higher ionization enthalpies than expected from general periodic trends.

 

Question 13. Second electron gain enthalpy of Oxygen is greater than first. Explain with reason.
Answer: When the first electron is added to a gaseous oxygen atom, energy is released (exothermic), forming \( O^- \). This is the first electron gain enthalpy \( (\Delta_{eg}H_1 = -14 \, \text{kJ/mol}) \). However, when a second electron is added to the already negatively charged \( O^- \) ion to form \( O^{2-} \), energy must be supplied (endothermic). This is because the incoming electron experiences strong electrostatic repulsion from the existing negative charge of the \( O^- \) ion. To overcome this repulsion, external energy is required, making the second electron gain enthalpy positive \( (\Delta_{eg}H_2 = +780 \, \text{kJ/mol}) \).
\( O(g) + e^- \rightarrow O^-(g) \quad (\Delta H_1 = -14 \, \text{kJ/mol}) \)
\( O^-(g) + e^- \rightarrow O^{2-}(g) \quad (\Delta H_2 = +780 \, \text{kJ/mol}) \)
In simple words: When oxygen gains the first electron, it releases energy. But when it tries to gain a second electron, it needs energy because the existing negative charge pushes the new electron away.

🎯 Exam Tip: While the first electron gain enthalpy can be exothermic, subsequent electron gain enthalpies are usually endothermic due to electron-electron repulsion.

 

Question 14. The bond angle of NH3 is greater than NF3. Explain with reason.
Answer: The bond angle of \( NH_3 \) (ammonia) is greater than that of \( NF_3 \) (nitrogen trifluoride). Both molecules have a pyramidal shape with one lone pair and three bond pairs around the central nitrogen atom.

• In \( NH_3 \), nitrogen is more electronegative than hydrogen, so the shared electron pairs are pulled closer to the nitrogen atom. This increases electron density around the nitrogen and reduces bond pair-bond pair repulsion, allowing the bond angle to be larger (approximately \( 107.8^\circ \)).
• In \( NF_3 \), fluorine is much more electronegative than nitrogen, so the shared electron pairs are pulled closer to the fluorine atoms. This reduces the electron density around the central nitrogen, increasing the lone pair-bond pair repulsion and causing the bond angle to be smaller (approximately \( 102.5^\circ \)).

In simple words: The bond angle in \( NH_3 \) is larger than in \( NF_3 \). In \( NH_3 \), electrons in bonds are closer to nitrogen, so they push less. In \( NF_3 \), electrons are pulled more towards fluorine, which makes the bond angle smaller due to more repulsion around the central nitrogen.

🎯 Exam Tip: The electronegativity difference between the central atom and the surrounding atoms significantly influences bond angles by affecting the electron density distribution around the central atom.

 

Question 2. Define atomic radius. Explain the factors affecting it and its periodicity.
Answer: Atomic radius is defined as the distance from the center of the nucleus to the outermost electron shell of an atom. It indicates the size of an atom. For metals, it's called metallic radius, and for non-metals, it's covalent radius.

Factors affecting atomic radius:

• Effective Nuclear Charge: This is the net positive charge felt by the outermost electrons. As the effective nuclear charge increases, the electrons are pulled closer to the nucleus, causing the atomic radius to decrease.
• Number of Shells: As the number of electron shells (or principal quantum number) increases, the atomic radius increases because electrons are further from the nucleus.
• Shielding or Screening Effect: In atoms with many electrons, inner electrons shield the outer electrons from the full nuclear charge. More shielding reduces the effective nuclear charge on outer electrons, leading to an increase in atomic radius.
• Number of Bonds: For covalent radius, the number of bonds affects it. As the number of bonds between atoms increases, the atomic radius generally decreases (e.g., C-C bond length is longer than C=C).

Variation of atomic radius in a period:
Atomic radius generally decreases from left to right across a period. This happens because, while electrons are added to the same outermost shell, the nuclear charge (number of protons) increases. This stronger nuclear attraction pulls the electron cloud closer to the nucleus, reducing the atomic size.

Variation of atomic radius in a group:
Atomic radius increases when moving down a group. This is because new electron shells are added with each successive element. These additional shells place the outermost electrons further from the nucleus, and the increased shielding effect from inner electrons reduces the effective nuclear charge felt by the valence electrons, causing the atomic size to increase.
In simple words: Atomic radius is the size of an atom. It gets smaller as you go left to right in a row (period) because the nucleus pulls harder. It gets bigger as you go down a column (group) because atoms get more electron layers.

🎯 Exam Tip: Remember the two main trends for atomic radius: decreasing across a period due to increasing nuclear charge, and increasing down a group due to the addition of new electron shells.

 

Question 4. Write applications of electronegativity and electron gain enthalpy.
Answer: Electronegativity is a measure of how strongly an atom in a molecule pulls shared electrons towards itself. Linus Pauling defined it in 1932 as the "power of an atom to attract electrons". This is a key idea in chemistry. Its applications help us understand chemical bonds and properties.
Applications of Electronegativity:
1. Nature of Bond: Electronegativity helps predict the type of bond between atoms.
(a) If atoms A and B have the same electronegativity (\( X_A = X_B \)), then \( X_A - X_B = 0 \). The bond is non-polar covalent, meaning electrons are shared equally. This is shown as A-B.
(b) If atom A is slightly more electronegative than atom B (\( X_A > X_B \), but the difference is small), the bond is polar covalent. Electrons are shared unevenly. This is represented as \( A^{\delta-} - B^{\delta+} \).
(c) If atom A is much more electronegative than atom B (\( X_A > X_B \), with a very large difference), the bond is more ionic or highly polar. This means electrons are mostly transferred from B to A. It is shown as A-B.

2. Percentage of Ionic Character in a Polar Covalent Bond: The difference in electronegativity (\( X_A - X_B \)) helps determine how much a polar covalent bond behaves like an ionic bond. If \( X_A - X_B \) is greater than 1.7, the bond has more than 50% ionic character.
(a) When \( X_A - X_B = 1.7 \), the bond is exactly 50% ionic and 50% covalent.
(b) When \( X_A - X_B < 1.7 \), the bond has more than 50% covalent character, meaning it is less ionic.

Applications of Electron Gain Enthalpy:

• Nature of Bonds: The difference between ionization enthalpy (energy to remove an electron) and electron gain enthalpy (energy gained when adding an electron) helps decide the bond type. If this difference is small compared to lattice energy, an ionic bond forms. If the difference is large, a covalent bond forms.
• Reactivity of Elements: Elements with high electron gain enthalpy are generally more reactive. Halogens, for example, have the highest electron gain enthalpy in a period and are very reactive.
• Oxidizing Tendency: A higher electron gain enthalpy means an element has a stronger tendency to gain electrons, thus increasing its ability to act as an oxidizing agent.

In simple words: Electronegativity shows how strongly an atom pulls electrons, which helps us understand bond types and how ionic or covalent they are. Electron gain enthalpy helps us predict how reactive an element is and its ability to gain electrons.

🎯 Exam Tip: Remember to clearly define both electronegativity and electron gain enthalpy before listing their applications to score full marks.

 

Electron Gain Enthalpy

Electron gain enthalpy is the amount of energy released when a single electron is added to a neutral gaseous atom. This process turns the atom into a negatively charged ion. It is represented by \( \Delta_{eg}H \).
For example: \( A (g) + e^- \rightarrow A^- (g) \); \( \Delta H = \Delta_{eg}H \).
This enthalpy is measured in electron volts per atom or kilojoules per mole. If more energy is released when an electron is added, the electron gain enthalpy is higher. A higher value means the atom is more likely to become a negative ion.

Factors Affecting Electron Gain Enthalpy:

• Effective Nuclear Charge: The pull of the nucleus on outer electrons. If the effective nuclear charge increases, the electron gain enthalpy also increases because the nucleus pulls new electrons more strongly.
• Atomic Size: This enthalpy is inversely related to the size of the atom. As atomic size increases, the outer electrons are farther from the nucleus, so the attraction for new electrons decreases, leading to lower electron gain enthalpy.
• Shielding Effect: Inner electrons shield the outer electrons from the nucleus's full pull. If the shielding effect increases, the effective nuclear charge felt by outer electrons decreases. This results in less attraction for new electrons, and thus, lower electron gain enthalpy.
• Half-filled and Fully-filled Orbitals: Atoms with orbitals that are exactly half-filled or completely filled are very stable. Because of this stability, they do not readily accept new electrons, leading to a low electron gain enthalpy.

Periodicity:

• Across a Period (left to right): Electron gain enthalpy generally increases as you move from left to right in a period. This happens because the atomic size decreases and the ionization enthalpy increases, meaning atoms hold onto electrons more tightly and attract new ones more strongly.
• Down a Group (top to bottom): Electron gain enthalpy decreases as you go down a group. This is mainly due to the increase in atomic size, which reduces the attraction for additional electrons.