CBSE Class 12 Chemistry The P Block Elements Notes Set C

CBSE Class 12 Chemistry notes and questions for The p block Elements Part D. Learning the important concepts is very important for every student to get better marks in examinations. The concepts should be clear which will help in faster learning. The attached concepts made as per NCERT and CBSE pattern will help the student to understand the chapter and score better marks in the examinations. 

Anomalous behaviour of first member of

p-Block Elements

Anomalous behaviour of first element in the p-block elements is attributed to small size, large (charge/radius) ratio, high ionization enthapy, high electronegativity and unavailability of d-orbitals in its valance shell.

Consequences :

1. The first element in p-block element has four valence orbitals i.e. one 2s and three 2p, Hence maximum covalency of the first element in limited to four. The other elements of the p-block elements have vacant d-orbitals in their valence shell, e.g. elements of the third period have nine (9) one 3s, three 3p and five three 3d orbitals. Hence these show maximum covalence greater than four. Following questions can be answered -

(i) Nitrogen (N) does not from pentahalide while P froms PCl5, PF5, and PF–. Why?

(ii) Sulphur (S) forms SF6 but oxygen does not form OF6. Why?

(iii) Though nitrogen forms pentoxide but it does not form pentachloride. Explain. Why?

(iv) Fluorine forms only one oxoacid while other halogens form a number of oxoacids. Why?

(2) The first member of p-block elements displays greater ability to from pπpπbond (s) with itself, (e.g., C = C, C ≡C, N = N, N ≡N) and with the other elements of second period (e.g., C = O, C ≡N, N = O) compared to the subsequent members of the group.

This is because p-orbitals of the heavier members are so large and diffuse that they cannot have effective sideways overlapping. Heavier members can form pπ– dπbonds with oxygen.


Now, the following questions can be explained using the above reasoning-

(i) Nitrogen forms N2 but phosphorus forms P4 at room temperature. Why?

(ii) Oxygen forms O2 but sulphur exists as S8. Why?

(iii) Explain why (CH3)3 P = O is known but (CH3)3 N = O is not known.

3. Due to small size and high electronegativity and presence of lone pair(s) of electrons, elements N, O, F when bonded to hydrogen atom, forms intermolecular hydrogen bonds which are stronger than other intermolcular forces. This results in exceptionally high m.p. and b.p. of the compounds having N–H/O–H/F–H bonds.

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