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Class 12th Haloalkanes and Haloarenes

10.1 Write structures of the following compounds: (i) 2-Chloro-3-methylpentane (ii) 1-Chloro-4- ethylcyclohexane (iii) 4-tert. Butyl-3-iodoheptane (iv) 1,4-Dibromobut-2-ene (v) 1-Bromo-4-sec. butyl- 2-methylbenzene.

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Vishal Baghel

Contributor-Level 10

10.1 From the name of the compound it is clear that the parent ring is pentane and chloro and methyl groups are attached in the straight chain at 2nd and 5th position respectively. Hence, the structure is as follows:

From the name of the compound given it is clear that the parent group is hexane with 2 attachments namely chloro and ethyl groups at 1 and 4 positions respectively. Hence, the structure is as follows: -

From the name of the compound given it is clear that the parent group is heptane with tertiary butyl and iodine groups attached at 4 and 3 positions respectively. Hence, the structure is as follows: -

From the name of the comp

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Class 12th Pandit Prithi Nath College

What is the minimum percentage required in class 12th cbse to get admission in your college in bcom?

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Class 12th Block D and F Elements

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

Example	Magnetic moment
[K₄ [Mn(CN)₆]	2.2
[Fe(H₂O)₆]^2+.	5.3
K₂ [MnCl₄]	5.9

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Payal Gupta

Contributor-Level 10

8.48 To calculate magnetic moment of the complex species, we use the spin formula:

μ =√n(n+2) BM

When n= 1

⇒ μ = √1(1+2)

⇒ u= √3⇒

u=1.73 BM

When n=2

⇒ μ = √2(2+2)

⇒ μ = √8

⇒ μ = 2.83 BM

When n= 3

⇒ μ = √3(3+2)

⇒ μ = 15

⇒ μ = 3.87 BM

When n = 4

⇒ μ = √ 4(4+2)

⇒ μ = √24

⇒ μ = 4.899 BM

When n= 5

⇒ μ = √5(5+2)

⇒ μ = √35

⇒ μ = 5.92 BM

1. [K₄ [Mn(CN)₆]

⇒μ = 2.2 BM (given)

We can see from the above calculation that the given value(2.2) is close to n=1. It means that it has only one unpaired electron Also in this complex Mn is in +2 oxidation state,i.e., as Mn²⁺.Thus when CN^- ligands approach Mn²⁺ ion, The electrons in 3d do not pair up.

The atomic number of Manganese (Mn) is Z = 25

The electronic configuration of ₂₅Mn= [Ar] 3d⁵ 4s²

And, the electronic configuration of Mn²⁺=[Ar] 3d⁵

Thus CN^- is a strong ligand.

The hybridization involved is d²sp³ forming inner orbital oc

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8.48 To calculate magnetic moment of the complex species, we use the spin formula:

μ =√n(n+2) BM

When n= 1

⇒ μ = √1(1+2)

⇒ u= √3⇒

u=1.73 BM

When n=2

⇒ μ = √2(2+2)

⇒ μ = √8

⇒ μ = 2.83 BM

When n= 3

⇒ μ = √3(3+2)

⇒ μ = 15

⇒ μ = 3.87 BM

When n = 4

⇒ μ = √ 4(4+2)

⇒ μ = √24

⇒ μ = 4.899 BM

When n= 5

⇒ μ = √5(5+2)

⇒ μ = √35

⇒ μ = 5.92 BM

1. [K₄ [Mn(CN)₆]

⇒μ = 2.2 BM (given)

The atomic number of Manganese (Mn) is Z = 25

The electronic configuration of ₂₅Mn= [Ar] 3d⁵ 4s²

And, the electronic configuration of Mn²⁺=[Ar] 3d⁵

Thus CN^- is a strong ligand.

The hybridization involved is d²sp³ forming inner orbital octahedral complex

2. [Fe(H2O)₆]^2+.

⇒ μ = 5.3 BM (given)

We can see from the above calculation that the given value(5.3) is close to n = 4. It means that it has four unpaired electrons. In this complex, Fe is in +2 state, i.e., as Fe²⁺. This means that the electrons in 3d do not pair up when the ligands, H₂O molecules approach.

The atomic number of Iron(Fe)is Z = 26

The electronic configuration of ₂₆Fe= [Ar] 3d⁶ 4s²

And, the electronic configuration of Fe²⁺=[Ar] 3d⁶

Thus H₂O is a weak ligand. To accommodate the electrons donated by six H₂O molecules, the hybridization will be sp³d². Hence it will be an outer orbital octahedral complex.

3. K₂ [MnCl₄]

⇒μ = 5.9 BM (given)

We can see from the above calculation that the given value(5.9) is close to n=5. It means that it has five unpaired electrons

In this complex, Mn is in +2 state, i.e., as Mn²⁺. Hence, we can say that Cl- is a weak ligand and does not cause the pairing of electrons.

The atomic number of Manganese (Mn) is Z = 25

The electronic configuration of ₂₅Mn= [Ar] 3d⁵ 4s²

And, the electronic configuration of Mn²⁺=[Ar] 3d⁵

Thus, the hybridization involved will be sp³ and the complex will be tetrahedral complex

Being Lewis bases (those who donate electrons) the ligands with less electronegativity will be stronger. Therefore, in general halogen or oxygen donors (F^-, Cl^-,Br^-, H₂O) are weak field ligands and the ones in which carbon or nitrogen atom is the donor (eg-CN^-,CO,NH₃) are strong field ligands.

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Class 12th Block D and F Elements

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

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Class 12th Block D and F Elements

8.47 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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Payal Gupta

Contributor-Level 10

8.47 The given statement is true as explained below:

1. Atomic radii of the heavier transition elements (4d and 5d series) are larger than those of the corresponding elements of the first transition series through those of 4d and 5d series are very close to each (Lanthanoid contraction)

2. Due to stronger intermetallic bonding (M-M bonding), the melting and boiling points of heavier transition elements are greater than those of the first transition series

3. The ionization enthalpies of 5d series are higher than the corresponding elements of 3d and 4d

4. The heavier transition elements form low spin complexes whereas the elements of t

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Class 12th Block D and F Elements

8.47 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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Payal Gupta

Contributor-Level 10

8.47 The given statement is true as explained below:

2. Due to stronger intermetallic bonding (M-M bonding), the melting and boiling points of heavier transition elements are greater than those of the first transition series

3. The ionization enthalpies of 5d series are higher than the corresponding elements of 3d and 4d

4. The heavier transition elements form low spin complexes whereas the elements of t

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Class 12th Block D and F Elements

8.46 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).

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Payal Gupta

Contributor-Level 10

8.46

S.no	Ion	Configuration	Number of 3d electrons	No. of unpaired Electrons	3d orbitals
1	Ti²⁺	3d²	2	2	^t22g e⁰g
2	V²⁺	3d³	3	3	^t32g e⁰ g
3	Cr³⁺	3d³	3	3	^t32g e⁰ g
4	Mn²⁺	3d⁵	5	5	^t32g e² g
5	Fe²⁺	3d⁶	6	4	^t42g e² g
6	Fe³⁺	3d⁵	5	5	^t32g e² g
7	Co²⁺	3d⁷	7	3	^t52g e² g

Ni²⁺

3d⁸

^t62g e² g

Cu²⁺

3d⁹

^t62g e³g

Note: In an octahedral field, the d-orbitals split into two sets of orbitals, the set of orbitals ( d_xy, d_yz, d_xz) with lower energy is called t_2g and the set of orbitals (d_x2-y2 and d_z2) with higher energy is called e_g

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Class 12th Block D and F Elements

8.45 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:

(i) Electronic configurations

(ii) Oxidation states

(iii) Ionisation enthalpies

(iv) Atomic sizes.

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Payal Gupta

Contributor-Level 10

8.45 (i) Electronic configuration:

In the first transition series, 3d orbitals are progressively filled while in the second and third transition series, 4d and 5d orbitals are filled. However the first series shows only two exceptions Cr and Cu, both have a single electron in the 4s orbital ( 3d⁵ 4s¹, 3d¹⁰ 4s¹) but the second series shows more exceptions. Similarly, third series elements show exceptions. Thus in the same vertical column, in a number of series, the electronic configuration of the three series are not similar at all.

(ii) Oxidation states:

The number of oxidation states shown by the elements in the middle of each ser

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8.45 (i) Electronic configuration:

(ii) Oxidation states:

The number of oxidation states shown by the elements in the middle of each series is maximum and minimum at the extreme ends. However, the first row elements differ from the second and third-row elements in the fact that for all the first row elements, +2 and +3 states are important. For second and third row elements, higher oxidation states are more important than those of the first row elements.

(iii) Ionisation enthalpies:

The first ionization enthalpies in each series generally increase gradually as we move from left to right through some exceptions are observed in each series. The first ionization enthalpies of some elements in the second (4d) series are higher while some of them have a lower value than the elements of 3d series in the same vertical column.

However the first ionization enthalpies of third (5d) series are higher than those of 3d and 4d series. This is because of weak shielding of nucleus by 4f electrons in the 5d series

(iv) Atomic sizes:

In general, ions of the same charge or atoms in a given series show progressively decrease in radius with increasing atomic number though the decrease is quite small. But the size of the atoms of the 4d series is larger than the corresponding elements of the 3d series whereas those of corresponding elements of the 5d series are nearly the same as those of4d series because of Lanthanoid contraction

Note: Lanthanoid contraction is the regular decrease (contraction) in the atomic and ionic radii of lanthanoids with increasing atomic number

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Class 12th Block D and F Elements

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

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Payal Gupta

Contributor-Level 10

8.44 Promethium (Pm) has atomic number 61. Hence electronic configuration of Promethium is [Xe]⁵⁴4f⁵ 5d⁰ 6s²

Protactium (Pa) has atomic number 91. Hence electronic configuration of Protactium is [Rn]⁸⁶ 5f² 6d¹ 7s²

Mendelevium (Md) has atomic number 101. Hence electronic configuration of Mendelevium is [Rn]⁸⁶ 5f¹⁴ 6d⁰ 7s²

Meitnerium (Mt) has atomic number 109. Hence electronic configuration of Meitnerium is [Rn]⁸⁶ 5f¹⁴ 6d⁷ 7s²

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Class 12th Block D and F Elements

8.43 Compare the chemistry of the actinoids with that of lanthanoids with reference to: (i) electronic configuration (ii) oxidation states and (iii) chemical reactivity.

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Payal Gupta

Contributor-Level 10

8.43

Lanthanoids	Actinoids
1.The electronic configuration of lanthanoids is [Xe]⁵⁴ 4f^1-14	1.The electronic configuration of actinoids is [Rn]86 5f1-14 6d0-1 7s2
2. They show limited oxidation states (+2, +3, 4) out of which +3 is most common. This is because of the large energy gap between 4f and 5d subshells.	2. Actinoids show a large number of oxidation states (+3, +4, +5, +6, +7) because of the small energy gap between 5f, 6d, and 7s subshells.
3. The first few members of the series are quite reactive, almost like calcium with increasing atomic no. their behaviour becomes similar to that of aluminium	3.They are highly reactive metals especially in the finely divided state. When they are added to boiling water they give a mixture of oxide and hydride.

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