Atomic Structure

Step 1: Both ions have 18 electrons.
Step 2: Nuclear charge of Cl^- (17) is less than that of K^+ (19).
Step 3: Lower nuclear charge means weaker attraction, hence larger size.

Step 1: Sodium (Na) and chlorine (Cl) both belong to period 3.

Step 1: Across a period, effective nuclear charge increases.
Step 2: Down a group, atomic size increases.
Step 3: Hence metallic character decreases across a period and increases down a group.

Step 1: Molecular mass of ozone O₃ = 3×16 = 48.
Step 2: Vapour density =molecular mass2. V.D.=(48)/(2)=24

Step 1: Identifying species. The ions F\textsuperscript−, O\textsuperscript2−, and N\textsuperscript3− are isoelectronic.
Step 2: Effect of nuclear charge. Higher nuclear charge results in smaller radius.
Step 3: Ordering radii. F < O²⁻ < N³⁻

Step 1: Atomic size and charge.
The radius of an ion decreases with increasing positive charge. Since has a higher charge than , its radius is smaller. Typically, the radius of is 0.17 Å.
Thus, the correct answer is (1).

Step 1: Identify the electron configuration.
The valence electrons of elements are those in the outermost shell, including the 3d subshell. Phosphorus (P) in the state has no electrons in the 3d subshell, unlike the other elements.
Thus, the correct answer is (4).

Step 1: Understand the concept of screening effect.
The screening effect refers to the ability of inner electrons to shield outer electrons from the attractive force of the nucleus.
Step 2: Conclusion.
The screening effect of -electrons is less than that of -electrons, as -electrons are more effective at shielding. Final Answer:

Step 1: Understand the periodic trend.
Ionization energy increases across a period and decreases down a group. The highest ionization energies are found in the noble gases.
Step 2: Conclusion.
The peaks in the graph correspond to the noble gases, which have the highest first ionization energies. Final Answer:

Step 1: Understand the trend in ionic radii.
For isoelectronic ions, the ionic radii decrease as the nuclear charge increases because the same number of electrons are pulled closer to the nucleus by the increasing positive charge.
Step 2: Conclusion.
The ionic radius decreases significantly from to due to the increasing nuclear charge. Final Answer:

Step 1: Electron gain enthalpy becomes more negative across a period.
Step 2: Due to small size, oxygen and fluorine have less negative values than expected.
Step 3: Correct order (most negative last):

Step 1: Element A has valency 3, element B has valency 2.
Step 2: Cross-multiplying valencies gives formula AB .

Step 1: Atomic number 118 completes a period.
Step 2: Elements at the end of a period are noble gases.

Step 1: Ionization energy increases across a period and decreases down a group.
Step 2: Noble gases have the highest ionization energy.
Step 3: Due to half-filled stability, P > S.

Radial nodes = n-l-1 3s: 3-0-1=2, 2p: 2-1-1=0

Step 1: Half-filled p³ configuration is extra stable.
Step 2: Greater stability implies higher ionisation energy.

Step 1: O₂ has two unpaired electrons → paramagnetic.
Step 2: Bond order: O₂ = 2, O₂⁻ = 1.5

Atomic number 118 belongs to group 18, hence it is a noble gas.

In BF₃, boron has only six electrons in its valence shell.

Step 1: Apply Hess’s law to the Born–Haber cycle: -617 = 161 + 520 + 77 + x - 1047
Step 2: Simplify: -617 = -289 + x
Step 3: x = -328kJ mol⁻1

Step 1: Noble gas Ne has the highest ionization potential.
Step 2: In the same period, Cl has higher ionization potential than P and S.
Step 3: Due to half-filled stability, IE(P) > IE(S).
Step 4: Between Mg and Al, Mg has higher ionization potential because Al has a filled 3s² subshell and a single 3p electron.

Step 1: Formula for number of radial nodes: Radial nodes = n - l - 1 Step 2: 3s: n=3, l=0 ⟹ 3-0-1 = 2 2p: n=2, l=1 ⟹ 2-1-1 = 0

Step 1: Understanding the Question:
We are comparing the sizes of two isoelectronic species (ions with the same number of electrons).

Step 2: Detailed Explanation:
- Chlorine (Atomic number Z = 17) forms Cl by gaining 1 electron. Total electrons = 18.
- Potassium (Atomic number Z = 19) forms K by losing 1 electron. Total electrons = 18.
Both have 18 electrons, but different nuclear charges (number of protons).
- In Cl , 17 protons pull 18 electrons.
- In K , 19 protons pull 18 electrons.
Higher nuclear charge results in a stronger pull on the electrons, making the ion smaller. Therefore, K is smaller than Cl .
Conversely, Cl is bigger than K .

Step 3: Final Answer:
Cl ion is bigger than K ion.

Step 1: Understanding the Question:
The question compares the shielding (screening) ability of electrons in different subshells within the same principle energy level.

Step 2: Detailed Explanation:
The screening effect depends on the shape and penetration of the orbitals.
- s-orbitals are spherical and penetrate closest to the nucleus, providing the maximum screening.
- p-orbitals have a dumb-bell shape and slightly less penetration.
- d and f orbitals have more diffused shapes and poor penetration, leading to poor screening of nuclear charge.
The order of screening effect for a given shell is: .
Therefore, d-electrons provide less screening than p-electrons.

Step 3: Final Answer:
The screening effect of d-electrons is less than p-electrons.