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NCERT · Class 12 · Chemistry

Chapter 4: The d & f Block Elements

Practice each question, then open the answer or worked solution.

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Questions

37 questions
Q8.31Written answer

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

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Worked solution
\(\begin{array}{l}Ce \; : \; 1s^2 \; 2s^2 \; 2p^6 \; 3s^2 \; 3p^6 \; 3d^{10} \; 4s^2 \; 4p^6 \; 4d^{10} \; 5s^2 \; 5p^6 \; 4f^1 \; 5d^1 \; 6s^2\end{array} \)
\(\begin{array}{l}Ce^{3+} \; : \; 1s^2 \; 2s^2 \; 2p^6 \; 3s^2 \; 3p^6 \; 3d^{10} \; 4s^2 \; 4p^6 \; 4d^{10} \; 5s^2 \; 5p^6 \; 4f^1\end{array} \)

Magnetic moment can be calculated as

\(\begin{array}{l}\nu = \sqrt{n(n+2)}\end{array} \)

Where,

n = number of unpaired electrons

In Ce3+, n = 1

\(\begin{array}{l}\nu = \sqrt{1(1+2)}\end{array} \)
\(\begin{array}{l}= \sqrt{3)}\end{array} \)

= 1. 732 BM

Q8.32Written answer

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

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+2 +4
Nd(60) Ce (58)
Sm(62) Pr (59)
Eu (63) Nd (60)
Tm(69) Tb (65)
Yb (70) Dy (66)

In the parenthesis are the atomic numbers of the elements given.

Tb, after forming Tb4+, attains a stable electronic configuration of [Xe] 4f7.

Yb, after forming Yb2+, attains a stable electronic configuration of [Xe] 4f14.

Eu, after forming Eu2+, attains a stable electronic configuration of [Xe] 4f7.

Ce, after forming Ce4+, attains a stable electronic configuration of [Xe].

Q8.33Written answer

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

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Atomic number Electronic configuration
61 [Xe]544f55d06s2
91 [Rn]865f26d17s2
101 [Rn]865f136d07s2
109 [Rn]865f146d77s2
Q8.34Written answer

Compare the general characteristics of the first series of transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:
(i) electronic configurations (ii) oxidation states (iii) ionisation enthalpies and (iv) atomic sizes

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(i) In the 1st, 2nd and 3rd 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

Cu (29) = 3d104s1

Cr (24) = 3d54s1

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

Ag(47) = 4d105s1

Pd(46) = 4d10 5s0

Rh(45) = 4d85s1

Ru (44) = 4d75s1

Tc (43) = 4d65s1

Mo (42) = 4d55s1

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

Au (79 ) = 5d10 6s1

Pt (78) = 5d9 6s1

W (74) = 5d4 6s2

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 third transition series, wherein higher oxidation states are more important.

Some stable complexes are

\(\begin{array}{l}{\left [ Fe^{II}(Cn)_6 \right ]}^{4-}\end{array} \)
\(\begin{array}{l}{\left [ Co^{III}(NH_3)_6 \right ]}^{3+}\end{array} \)
\(\begin{array}{l}{\left [ Ti(H_2O)_6 \right ]}^{3+}\end{array} \)

The issue is that 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 2nd transition series whose first ionisation enthalpies are lower than those of the elements corresponding to the same vertical column in the 1st transition series.

(iv) Atomic size generally decreases from left to right across a period. Now, among the three transition series, the 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.

Q8.35Written answer

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

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Worked solution
Metal ion Number of d-electrons Filling of d-orbitals

 
Cu2+ 9
\(\begin{array}{l}t^{6}_{2g}e^{3}_{g}\end{array} \)
Ni2+ 8
\(\begin{array}{l}t^{6}_{2g}e^{2}_{g}\end{array} \)
CO2+ 7
\(\begin{array}{l}t^{5}_{2g}e^{2}_{g}\end{array} \)
Fe3+ 5
\(\begin{array}{l}t^{3}_{2g}e^{2}_{g}\end{array} \)
Fe2+ 6
\(\begin{array}{l}t^{4}_{2g}e^{2}_{g}\end{array} \)
Mn2+ 5
\(\begin{array}{l}t^{3}_{2g}e^{2}_{g}\end{array} \)
Cr3+ 3
\(\begin{array}{l}t^{3}_{2g}\end{array} \)
V2+ 3
\(\begin{array}{l}t^{3}_{2g}\end{array} \)
Ti2+ 2
\(\begin{array}{l}t^{2}_{2g}\end{array} \)
Q8.36Written answer

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

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In many ways, the properties of elements of heavier transition elements differ from those of the first transition series.

(a) 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.

(b) 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.

(c) 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).

(d)  For heavier elements, the higher oxidation states are more common, whereas, for first transition series elements, th +2 and +3 oxidation states are more common.

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

Q8.37Written answer

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

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Worked solution

Magnetic moment  .

\(\begin{array}{l}\nu = \sqrt{n(n+2)}\end{array} \)

For value n = 1,

\(\begin{array}{l}\nu = \sqrt{1(1+2)} = \sqrt{3} = 1.732\end{array} \)

For value n = 2,

\(\begin{array}{l}\nu = \sqrt{2(2+2)} = \sqrt{8} = 2.83\end{array} \)

For value n = 3,

\(\begin{array}{l}\nu = \sqrt{3(3+2)} = \sqrt{15} = 3.87\end{array} \)

For value n = 4,

\(\begin{array}{l}\nu = \sqrt{4(4+2)} = \sqrt{24} = 4.899\end{array} \)

For value n = 5,

\(\begin{array}{l}\nu = \sqrt{5(5+2)} = \sqrt{35} = 5.92\end{array} \)

K4[Mn(CN)6]

\(\begin{array}{l}\sqrt{n(n+2)} = 2.2\end{array} \)

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 n = 1. 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.

[Fe(H2O)6]2+

\(\begin{array}{l}\sqrt{n(n+2)} = 5.3\end{array} \)

Thus, the calculation given above proves that the closest value to n= 4. In addition to this, the +2 oxidation state in this complex is none other than Fe. The d-orbital for Fe would have 6 electrons.

Thus, we can assume that H2O is a weak field ligand and does not induce electron pairing.

K2[MnCl4] 

\(\begin{array}{l}\sqrt{n(n+2)} = 5.9\end{array} \)

From the equation, we can see that the given value is nearest to n = 5. Mn is in the +2 oxidation state in this complex. This implies that in the d-orbital, Mn has 5 electrons.

We may also assume that Cl – is a weak field ligand and does not induce electron pairing.