Ncert Chemistry Class 12 intext questions solutions chapter 4 The d and f Block Elements

The d and f Block Elements intext questions solutions: Class 12 Chemistry Chapter 4 intext questions solutions

Textbook	Ncert
Class	Class 12
Subject	Chemistry
Chapter	Chapter 4
Chapter Name	The d and f Block Elements Class 12 intext questions solutions
Category	Intext questions
Medium	English

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Question 4.1: Silver atom has completely filled d orbitals (4d¹⁰) in its ground state. How can you say that it is a transition element?

Solution 4.1: A transition element is defined as an element that has partially filled d orbitals in one or more of its oxidation states. While a silver atom (Ag) in its ground state has a filled 4d orbital configuration (4d¹⁰), it is still considered a transition metal because it can exhibit variable oxidation states, such as +1 and +2, where the d orbitals can lose electrons.

In the case of silver:

• When silver forms a +1 oxidation state (Ag⁺), it loses one electron from the 5s orbital, while the 4d orbitals remain filled (4d¹⁰).
• If it were to form a +2 oxidation state (Ag²⁺), it would involve the removal of an electron from the 4d orbital as well (4d⁹).

Question 4.2: In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest i.e., 126 kJ mol^-1. Why?

Solution 4.2: The extent of metallic bonding an element undergoes decides the enthalpy of atomization. The more extensive the metallic bonding of an element, the more will be its enthalpy of atomization. In all transition metals (except Zn, electronic configuration: 3d¹⁰4s²), there are some unpaired electrons that account for their stronger metallic bonding. Due to the absence of these unpaired electrons, the inter-atomic electronic bonding is the weakest in Zn and as a result, it has the least enthalpy of atomization.

Question 4.3: Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?

Solution 4.3: Mn (Z = 25) = 3d5 4s2

Manganese (Mn) in the 3d series of transition metals exhibits the largest number of oxidation states, ranging from +2 to +7. This is because manganese has a half-filled 3d subshell (3d⁵ configuration) and a partially filled 4s orbital, allowing it to lose varying numbers of electrons and adopt a wide range of oxidation states. Manganese can show oxidation states ranging from −3 to +7. The most common oxidation states are +2, +4, and +7.

Question 4.4: The E^θ(M²⁺/M) value for copper is positive (+0.34V). What is possibly the reason for this? (Hint: consider its high Δ_aH^o and low Δ_hydH^o)

Solution 4.4: The E^θ(M²⁺/M) value of a metal depends on the energy changes involved in the following:

1. Sublimation: The energy required for converting one mole of an atom from the solid state to the gaseous state.

M(s) → M(g) (△_sH: Sublimation energy)

2. Ionization: The energy required to take out electrons from one mole of atoms in the gaseous state.

M(g) → M²⁺(aq) (△_iH: Ionization energy)

3. Hydration: The energy released when one mole of ions are hydrated.

M²⁺(g) → M²⁺(aq) (△_hydH: Hydration  energy)

Now, copper has a high energy of atomization and low hydration energy. Hence, the E^θ(M²⁺/M) value for copper is positive.

Question 4.5: How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?

Solution 4.5:The irregular variation in the first and second ionization enthalpies of the first transition series is due to:

Proximity of 3d and 4s orbitals: The close energy levels of 3d and 4s orbitals cause variations in the ease of electron removal, contributing to irregular patterns in ionization energies.

Electron configuration: Elements like chromium (Cr) and copper (Cu) have stable half-filled ((3d^5)) and fully-filled ((3d^{10})) configurations, requiring more energy to remove an electron, causing irregularities.

Poor shielding by 3d electrons: 3d electrons do not shield the nuclear charge effectively, leading to an increased attraction of outer 4s electrons and higher ionization enthalpy, though this increase is not uniform due to varying 3d-electron interactions.

Question 4.6: Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

Solution 4.6: The highest oxidation state of a metal is typically exhibited in its oxides or fluorides due to the extremely high electronegativity of oxygen and fluorine. These elements have a strong ability to attract electrons from the metal, allowing the metal to lose more electrons and achieve its highest possible oxidation state. Fluorine is the most electronegative element, and oxygen is highly electronegative as well, which enables them to stabilize the metal cations in their highest oxidation states by forming strong ionic or covalent bonds.

Additionally, oxygen and fluorine can form multiple bonds with the metal, such as in oxides (e.g., \(MnO_4^-\) where manganese is in the +7 state) or fluorides (e.g., VF₅ where vanadium is in the +5 state), further stabilizing these high oxidation states. Less electronegative elements, like chlorine or sulfur, cannot stabilize metals in such high oxidation states as effectively. Therefore, oxides and fluorides are the compounds in which metals exhibit their highest oxidation states.

Question 4.7: Which is a stronger reducing agent Cr²⁺ or Fe²⁺ and why ?

Solution 4.7: Cr²⁺ (Chromium in +2 oxidation state) is a stronger reducing agent than Fe²⁺ (Iron in +2 oxidation state). This can be explained by their respective tendencies to lose electrons and be oxidized to higher oxidation states.

The electron configuration of Cr²⁺ is \([Ar] 3d^4\), and it prefers to lose an electron to achieve a more stable half-filled 3d⁵ configuration, which is highly stable due to exchange energy. This makes Cr²⁺ more readily oxidized to Cr³⁺, giving it strong reducing properties.

On the other hand, Fe²⁺ has the electron configuration \([Ar] 3d^6\). While Fe²⁺ can be oxidized to Fe³⁺, the difference in stability between Fe²⁺ and Fe³⁺ is less significant compared to chromium. The half-filled 3d⁵ configuration of Cr³⁺ is more stable than Fe³⁺’s 3d⁵, making Cr²⁺ more eager to donate electrons and act as a stronger reducing agent than Fe²⁺.

Question 4.8: Calculate the ‘spin only’ magnetic moment of M²⁺_(aq) ion (Z = 27).

Solution 4.8:

Z = 27

\(\Rightarrow [\text{Ar}] \, 3d^7 \, 4s^2\)

\(\therefore \, \text{M}^{2+} = [\text{Ar}] \, 3d^7\)

\(3d^7 = \begin{array}{|c|c|c|c|c|c|c|} \uparrow\downarrow & \uparrow\downarrow & \uparrow\downarrow & \uparrow & \uparrow \end{array}\)

i.e., 3 unpaired electrons

\(\therefore \, n = 3\)

\(\Rightarrow \sqrt{n(n+2)} = \mu\)

\(\Rightarrow \sqrt{3(3+2)} = \mu\)

\(\Rightarrow \sqrt{15} = \mu\)

\(\mu \approx 4 \, \text{BM}\)

Question 4.9: Explain why Cu⁺ ion is not stable in aqueous solutions?

Solution 4.9: In an aqueous medium, Cu2+ is more stable than Cu+. This is because although energy is required to remove one electron from Cu+ to Cu2+, high hydration energy of Cu2+ compensates for it. Therefore, Cu+ ion in an aqueous solution is unstable. It disproportionates to give Cu2+ and Cu.

2Cu²⁺(aq) → Cu²⁺(aq) + Cu(s)

Question 4.10: Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

Solution 4.10: In actinoids, 5f orbitals are filled. These 5f orbitals have a poorer shielding effect than 4f orbitals (in lanthanoids). Thus, the effective nuclear charge experienced by electrons in valence shells in case of actinoids is much more that that experienced by lanthanoids. Hence, the size contraction in actinoids is greater as compared to that in lanthanoids.