Block D and F Elements
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a year agoContributor-Level 10
8.15 For answering this question, we can compare the electronic configuration of standard elements and then write their corresponding oxidation
S. No | Electronic configurations in ground state | Stable oxidation states |
1 | 3d3 Vanadium | +2, +3, +4, +5 |
2 | 3d5 Chromium | +3, +4, +6 |
3 | 3d5 Manganese | +2, +4, +6, +7 |
4 | 3d8 Nickel | +1, +2, +3, +4 |
5 | 3d4 | 3d4 configuration is not stable at ground state |
New answer posted
a year agoContributor-Level 10
8.14 The elements in the first-half of the transition series exhibit many oxidation states with Mn exhibiting a maximum number of oxidation states (+2 to +7). The stability of +2 oxidation state increases with the increase in atomic number. This happens as more electrons are getting filled in the d-orbital.
However, Sc ( [Ar] 3d14s2) does not show +2 oxidation state, instead, it loses all the three valence electrons to form Sc3+. The +3 oxidation state of Sc is very stable as it attains stable configuration.
For Mn ( [Ar] 3d54s2), +2 oxidation state is very stable because after losing two electrons, it attains stable half-filled str
New answer posted
a year agoContributor-Level 10
8.13 The oxidation states displayed by the first half of the first row of transition metals are given in the table below.
METALS | Sc ( [Ar] 3d14s2) | Ti ( [Ar] 3d24s2) | V ( [Ar] 3d34s2 ) | Cr ( [Ar] 3d54s1) | Mn ( [Ar] 3d54s2) |
OXIDATION STATES |
| +2 | +2 | +2 | +2 |
+3 | +3 | +3 | +3 | +3 | |
| +4 | +4 | +4 | +4 | |
|
| +5 | +5 | +6 | |
|
|
| +6 | +7 |
Except for Sc, all others metals display +2 oxidation state. This is because as the atomic number increases, the number of electrons in the valence shell increases. +2 oxidation state is attained by the loss of the two 4s electrons by these metals. As the number of electron increases, the possibility of an ion with +2 oxidation state being stable (by attaining half-filled structure) also increases. Finally, Mn2+ ions have half-filled structure and are very stable.
New answer posted
a year agoContributor-Level 10
8.12 Electronic configuration of Mn2+ is [Ar]183d5 and Electronic configuration of Fe2+ is [Ar]18 3d6 . It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in (+2) state has a half-filled stable configuration, whereas the Fe in +3 oxidation state has partially filled subshells, which are relatively unstable. This is the reason Mn2+ shows resistance to oxidation to Mn3+. Also, Fe2+ has 3d6 configuration and by losing one electron, it attains half- filled stable Hence, Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state.
Mn+2→
Manganese has the atomic no. 25 and its electronic config
New question posted
a year agoNew answer posted
a year agoContributor-Level 10
8.11 (i) The atomic number of Cr is 24 and the electronic configuration is [Ar] 3d54s1. When 3 electrons are removed, it becomes Cr3+. The electronic configuration of Cr3+: 1s2 2s2 2p6 3s2 3p6 3d3 Or [Ar]3d3
(ii) The atomic number of Pm is 61 and the electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f5 5s2 5p6 6s2 or [Xe] 4f56s2. When 3 electrons are removed, it becomes Pm+3, having the electronic configuration,
Pm3+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f4
Or, [Xe]544f4
(iii) The atomic number of Cu is 29 and has the electronic configuration of [Ar] 3d104s1. On removing one electron, the Cu+
New answer posted
a year agoContributor-Level 10
8.10 The 5f orbitals (in actinoids) have a poorer shielding effect than 4f orbitals (in lanthanoids). Thus, the effective nuclear charge experienced by electrons in outer shells in case of actinoids is much more that experienced by lanthanoids.
As the effective nuclear charge experienced is high, the electrons are attracted with much force, hence the size of the atom decreases. Hence, actinoid contraction is greater from element to element than lanthanoid contraction.
New answer posted
a year agoContributor-Level 10
8.9 The stability in aqueous condition depends on the hydration energy of the ions when they bond to the water molecules. And, the hydration energy is the amount of heat released as an ionic substance is dissolved and its constituent ions are hydrated or surrounded by water molecules.
Now, in Cu2+ and Cu+ ion, Cu2+ has a greater charge density than the Cu+ ion and so forms much stronger bonds releasing more energy. Therefore, in an aqueous medium, Cu2+ ion is more stable than Cu+ ion. This is because the energy required to remove one electron from Cu+ to Cu2+, is compensated by the high hydration energy of Cu2+.
New answer posted
a year agoContributor-Level 10
8.8 Given, Z=atomic number = 27 = [Ar]3d7 4s2
⇒ M2+ = [Ar]3d7
Hence, 3 unpaired electrons are present.
The spin only magnetic moment μ= √n(n+2)
Where n is the number of unpaired electrons.
Hence, μ =√(3*(3+2))
μ = √15 BM or μ = 3.87 BM
New answer posted
a year agoContributor-Level 10
8.7 The following reactions are involved when Cr2+ and Fe2+ act as reducing
Cr2+ = Cr3+ ( E? Cr3+/ Cr2+ = - 0.41 V)
Fe2+⇒ Fe3+ (E? Fe3+/ Fe2+ = +0.77 V)
Since, Cr has less potential value, Cr2+ gets oxidised easily than Fe2+. Therefore, Cr2+ is a better reducing agent that Fe3+.
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