The P-Block Elements - Class 12 Medical Chemistry - Extra Questions

Sulphur when warmed with nonmetal in the absence of air, forms a mixture of nonmetallic trisulphide and pentasulphide. Identify the nonmetal and give the reaction.

Phosphorus when warmed with sulphur in absence of air, gives mixture of trisulphide and pentasulphide.
$$P_4+S_8\longrightarrow P_2S_3+P_2S_5$$


$$(CH_3)_3P = O$$ exists but $$(CH_3)_3N = O$$ does not. Give reasons.

Due to absence of d-orbitals in valence shell,cannot form $$d\pi -p\pi $$ multiple bonds ,so nitrogen cannot form $$(CH_3)_3N=0$$.while in phosphorous, it has d-orbitals in its valance shell so, it can from $$d\pi -p\pi $$ multiple bonds ,so $$(CH_3)_3P=0$$  exists


What is polymorphic from of Sulphur?


Allotropes of sulphur
(1) Rhombic sulphur $$(\alpha-sulphur)$$
(2) Monoclinic sulphur $$(\beta-sulphur)$$
Both these forms $$S_8$$ molecules.
(Refer to Image)

1056681_1117596_ans_f4edaab9c29b4040bea19eb41d6f985a.PNG


$$ \underset { (white) }{ P_{ 4 } } +SOCI_{ 2 }-PCI_{ 3 }+SO_{ 2 }+'x' $$   identify the 'x'

The net reaction is 
$$P_4 + 8SOCl_2 \rightarrow 4PCl_3 + S_2Cl_2$$
hence X=$$S_2Cl_2$$


Explain the structure of ammonia.

Ammonia is a covalent atom. 

The particle is shaped because of the overlap of orbitals of three hydrogen atoms and three $$Sp^3$$ hybrid orbitals of nitrogen in the structure as the central atom. 

The fourth sp³ hybrid orbital is involved by a lone pair.

Due to this the shape of ammonia is trigonal pyramidal.

The H-N-H bond edge is 107.3°, which is somewhat not exactly the tetrahedral edge of 109°28.

This is on the grounds that the bond pair-lone pair repulsions push the N-H bonds somewhat inwards.

In solid and liquid states, ammonia is related through hydrogen bonds.


1324444_1322991_ans_d4805ee5f9784e85a7b0075ee14ae75f.webp


A yellow crystalline allotropic form of sulphur (A) on heating at X $$^{ \circ  }{ C }$$ converts to another allotropic sulphur (B) that has needle shape. Identify A, B, X.

The yellow crystalline form allotrope form of sulphur is Rhombic sulphur (A). The temperature required to convert to another form is $$96^oC$$. So $$X=96$$ and the form is monoclinic sulphur (B).


Class 12 Medical Chemistry Extra Questions