Chemistry NCERT Exemplar Solutions Class 12th Chapter Eight

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 nucleus. Hence, electrons are removed first during the process of ionisation.

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 cloud strongly towards itself. 

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 when the electrons are removed from the 4s-orbital. Zn acquires a stable 3d¹⁰ configuration state.

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. 

Lanthanoid contraction: Because the elements in Row 3 have 4f electrons. These electrons do not shield good, causing a greater nuclear charge. This greater nuclear charge has a greater pull on the electrons and result in the decrease in their size and atomic radii.

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)                      (yellow)

 2K₂CrO₄ + H₂SO₄ → K₂Cr₂O₇ + K₂SO₄ + H₂O

(yellow)                                       (orange) 

22. This is due to the inter conversion of dichromate (orange) to chromate ion (yellow). 

Cr₂O₇^2-   CrO₄^2-  

(orange)     (yellow) 

21. 2MnO₄^-  +  16H⁺ +  5C₂O₄^2-  → 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. 

20. Ce  =  4f¹ 5d¹ 6s²

Ce^{3 +}  =  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

Hence it shows 4⁺ oxidation states.

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 electrons do not shield good, causing a greater nuclear charge. This greater nuclear charge has a greater pull on the electrons and result in the decrease in their size and atomic radii. 

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 lanthanides. This leads to the fact that 5f-electrons experience reduced nuclear force of attraction and hence they have Lower ionisation enthalpies than lanthanoids.