Page No 236: - Chapter 8 D & F Block Elements Exercise Solutions class 12 ncert solutions Chemistry - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 8.31:

Use Hund’s rule to derive the electronic configuration of Ce³⁺ ion and calculate its magnetic moment on the basis of ‘spin-only’ formula.

Answer:

Magnetic moment can be calculated as:

Where,

n = number of unpaired electrons

The electronic configuration of Ce³⁺ : 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶4f¹

In Ce³⁺, n = 1

∴ μ = 1 ( 1 +2 ) = 3 = 1.732 B.M

Question 8.32:

Name the members of the lanthanoid series which exhibit +4 oxidation state and those which exhibit +2 oxidation state. Try to correlate this type of behavior with the electronic configurations of these elements.

Answer:

The lanthanides that exhibit +2 and +4 states are shown in the given table. The atomic numbers of the elements are given in the parenthesis.

Tb after forming Tb⁴⁺ attains a stable electronic configuration of [Xe] 4f^7.Ce after forming Ce⁴⁺ attains a stable electronic configuration of [Xe].

Eu after forming Eu²⁺ attains a stable electronic configuration of [Xe] 4f^7.

Yb after forming Yb²⁺ attains a stable electronic configuration of [Xe] 4f^14.

Question 8.33:

Compare the chemistry of the actinoids with that of lanthanoids with reference to:

(i) electronic configuration

(ii) oxidation states and

(iii) chemical reactivity.

Answer:

Electronic configuration

The general electronic configuration for lanthanoids is [Xe]⁵⁴ 4f^0-14 5d^0-1 6s² and that for actinoids is [Rn]⁸⁶ 5f^1-14 6d^0-1 7s². Unlike 4f orbitals, 5f orbitals are not deeply buried and participate in bonding to a greater extent.

Oxidation states

The principal oxidation state of lanthanoids is (+3). However, sometimes we also encounter oxidation states of + 2 and + 4. This is because of extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d, and 7s levels are of comparable energies. Again, (+3) is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in +3 state than in +4 state.

Chemical reactivity

In the lanthanide series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperatures. Alkalies have no action on these actinoids. In case of acids, they are slightly affected by nitric acid (because of the formation of a protective oxide layer).

Question 8.34:

Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.

Answer:

Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:Question 8.35:

(i) electronic configurations,

(ii) oxidation states,

(iii) ionisation enthalpies, and

(iv) atomic sizes.

Answer:

(i) In the 1^st, 2^nd and 3^rd transition series, the 3d, 4d and 5d orbitals are respectively filled.

We know that elements in the same vertical column generally have similar electronic configurations.

In the first transition series, two elements show unusual electronic configurations:

Similarly, there are exceptions in the second transition series. These are:

There are some exceptions in the third transition series as well. These are:

As a result of these exceptions, it happens many times that the electronic configurations of the elements present in the same group are dissimilar.

(ii) In each of the three transition series the number of oxidation states shown by the elements is the maximum in the middle and the minimum at the extreme ends.

However, +2 and +3 oxidation states are quite stable for all elements present in the first transition series. All metals present in the first transition series form stable compounds in the +2 and +3 oxidation states. The stability of the +2 and +3 oxidation states decreases in the second and the third transition series, wherein higher oxidation states are more important.

For example  are stable complexes, but no such complexes are known for the second and third transition series such as Mo, W, Rh, In. They form complexes in which their oxidation states are high. For example: WCl₆, ReF₇, RuO₄, etc.

(iii) In each of the three transition series, the first ionisation enthalpy increases from left to right. However, there are some exceptions. The first ionisation enthalpies of the third transition series are higher than those of the first and second transition series. This occurs due to the poor shielding effect of 4f electrons in the third transition series.

Certain elements in the second transition series have higher first ionisation enthalpies than elements corresponding to the same vertical column in the first transition series. There are also elements in the 2^nd transition series whose first ionisation enthalpies are lower than those of the elements corresponding to the same vertical column in the 1^st transition series.

(iv) Atomic size generally decreases from left to right across a period. Now, among the three transition series, atomic sizes of the elements in the second transition series are greater than those of the elements corresponding to the same vertical column in the first transition series. However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction.

Question 8.36:

Write down the number of 3d electrons in each of the following ions:

Ti²⁺, V²⁺, Cr³⁺, Mn²⁺, Fe²⁺, Fe³⁺, CO²⁺, Ni²⁺ and Cu²⁺.

Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).

Answer:

Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.Question 8.37:

Answer:

The properties of the elements of the first transition series differ from those of the heavier transition elements in many ways.

(i) The atomic sizes of the elements of the first transition series are smaller than those of the heavier elements (elements of 2^nd and 3^rd transition series).

However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction.

(ii) +2 and +3 oxidation states are more common for elements in the first transition series, while higher oxidation states are more common for the heavier elements.

(iii) The enthalpies of atomisation of the elements in the first transition series are lower than those of the corresponding elements in the second and third transition series.

(iv) The melting and boiling points of the first transition series are lower than those of the heavier transition elements. This is because of the occurrence of stronger metallic bonding (M−M bonding).

(v) The elements of the first transition series form low-spin or high-spin complexes depending upon the strength of the ligand field. However, the heavier transition elements form only low-spin complexes, irrespective of the strength of the ligand field.

Question 8.38:

What can be inferred from the magnetic moment values of the following complex species?

Example Magnetic Moment (BM)

K₄[Mn(CN)₆] 2.2

[Fe(H₂O)₆]²⁺ 5.3

K₂[MnCl₄] 5.9

Answer:

Magnetic moment ( ) is given as .

For value n = 1,  .

For value n = 2,  .

For value n = 3,  .

For value n = 4,  .

For value n = 5,  .

(i) K4[Mn(CN)6]

For in transition metals, the magnetic moment is calculated from the spin-only formula. Therefore,

We can see from the above calculation that the given value is closest to . Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d-orbital.

Hence, we can say that CN⁻ is a strong field ligand that causes the pairing of electrons.

(ii) [Fe(H2O)6]2+

We can see from the above calculation that the given value is closest to . Also, in this complex, Fe is in the +2 oxidation state. This means that Fe has 6 electrons in the d-orbital.

Hence, we can say that H₂O is a weak field ligand and does not cause the pairing of electrons.

(iii) K2[MnCl4]

We can see from the above calculation that the given value is closest to . Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d-orbital.

Hence, we can say that Cl⁻ is a weak field ligand and does not cause the pairing of electrons.