The d and f Block Elements
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27. While filling up of electrons in the atomic orbitals, the 4s-orbital is filled before the 3d-orbital but reverse happens during the ionization of the atom. Explain why?
27. While filling up of electrons in the atomic orbitals, the 4s-orbital is filled before the 3d-orbital but reverse happens during the ionization of the atom. Explain why?
27. As per (n + l) rule, 4s has lower energy than 3d-orbital.
3d−n+l = 3+2 = 5
4s - n + l = 4 + 0 = 4
So, 4s are filled first.
After filling of electrons, 4s-orbital moves beyond 3d-orbital and 4s electrons are loosely held by the nucle
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26. The halides of transition elements become more covalent with increasing oxidation state of the metal. Why?
26. The halides of transition elements become more covalent with increasing oxidation state of the metal. Why?
26. As the positive charge of the ion increases or we can say that oxidation state of a transition element increases, its size decreases and as per Fajan's rule, more the charge on the metal ion, more is its tendency to form covalent compounds because positively charged cation attracts the electron
Upvote 25. E of Cu is +0.34V while that of Zn is −0.76V. Explain.
25. E of Cu is +0.34V while that of Zn is −0.76V. Explain.
25. The sum of sublimation energy and ionisation enthalpy to oxidise cu (s) to Cu2+ is so highly that it is not compensated by the hydration enthalpy of Cu. Due to this, the Eof Cu is positive.
While in case if Zn, the E value is negative or more negative than the expected value because wh
Upvote 24. The second and third rows of transition elements resemble each other much more than they resemble the first row. Explain why?
24. The second and third rows of transition elements resemble each other much more than they resemble the first row. Explain why?
24. As an effect of lanthanide contraction, the second and third rows of transition elements resemble each other. For example, zirconium and hafnium have similar radius i.e. 160pm and 159pm respectively. Due to this similarity in their size, they show similar physical and chemical properties.&n
Upvote 23. A solution of KMnO4 on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?
23. A solution of KMnO4 on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?
23. KMnO4 act as an oxidising agent however it’s activity as oxidising agent is influenced by the pH of the solution i.e. acidic, basic or neutral solution.
K2Cr2O7 + 2KOH → 2K2CrO4 + H2O
(Orange) &nbs
Upvote 22. When orange solution containing Cr2O72− ion is treated with an alkali, a yellow solution is formed and when H+ ions are added to yellow solution, an orange solution is obtained. Explain why this happens?
22. When orange solution containing Cr2O72− ion is treated with an alkali, a yellow solution is formed and when H+ ions are added to yellow solution, an orange solution is obtained. Explain why this happens?
22. This is due to the inter conversion of dichromate (orange) to chromate ion (yellow).
Cr2O72- CrO42-
(orange) (yellow)
Upvote 21. Explain why does colour of KMnO4 disappear when oxalic acid is added to its solution in acidic medium?
21. Explain why does colour of KMnO4 disappear when oxalic acid is added to its solution in acidic medium?
21. 2MnO4- + 16H+ + 5C2O42- → 2Mn2 + + 8H2O + 10CO2
KMnO4 oxidises the oxalic acid to CO2 and reduces itself to Mn2+ state. Mn2+ is colourless that’s why it seems that KMnO4 has been disappeared.
Upvote 20. Although +3 oxidation state is the characteristic oxidation state of lanthanoids but cerium shows +4 oxidation state also. Why?
20. Although +3 oxidation state is the characteristic oxidation state of lanthanoids but cerium shows +4 oxidation state also. Why?
20. Ce = 4f1 5d1 6s2
Ce3 + = 4f1 5d0 6s0
As in Ce3+ ,
4f has only single electron. Cerium holds the capacity to lose this electron also and attain 4f0 configuration which is more stable.
Ce4+ = 4f0 5d0 6s0
Upvote 19. Although Zr belongs to 4d and Hf belongs to the 5d transition series but it is quite difficult to separate them. Why?
19. Although Zr belongs to 4d and Hf belongs to the 5d transition series but it is quite difficult to separate them. Why?
19. As an effect of lanthanoid contraction, zirconium and hafnium have similar radius of 160 pm and 159 pm respectively. Due to this similarity in their size, they show similar physical and chemical properties.
Lanthanoid contraction: Because the elements in Row 3 have 4f electrons. These
Upvote 18. Ionisation enthalpies of Ce, Pr and Nd are higher than Th, Pa and U. Why?
18. Ionisation enthalpies of Ce, Pr and Nd are higher than Th, Pa and U. Why?
18. Ce, Pr and Nd belong to the lanthanide series whereas Th, Pa and U belong to the actinides family.
When electrons start accommodating the 4f and 5f orbitals, the 5f electrons penetrate less into the inner core. They are more effectively shielded nuclei in comparison to 4f-electrons in
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