CBSE Class 12 Chemistry The P Block Elements Notes Set C

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Revision Notes for Class 12 Chemistry Unit 07 The p-Block Elements

Class 12 Chemistry students should refer to the following concepts and notes for Unit 07 The p-Block Elements in Class 12. These exam notes for Class 12 Chemistry will be very useful for upcoming class tests and examinations and help you to score good marks

Unit 07 The p-Block Elements Notes Class 12 Chemistry

 

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.

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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.

CBSE Class 12 Chemistry notes and questions for The p block Elements Part D

CBSE Class 12 Chemistry notes and questions for The p block Elements Part D

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CBSE Class 12 Chemistry Unit 07 The p-Block Elements Notes

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