Q. 25

(a) Account for t

Answer :

(a) (i) Since, Bi is not stable in +5 oxidation state due to inert pair effect so Bi(V) is a strong oxidizing agent i.e. it will oxidize others and itself will get reduced to a lower stable oxidation state like +3.

(ii) H – O – I is a weaker acid than H – O – Cl because Cl is more electronegative than F due to which the polarity of OH bond increases i.e. more the electronegativity of the halogen atom, more will be the tendency to pull electron towards itself and weaker will be the OH bond and hence H+ can easily be removed thereby increasing the acidity.

(iii) Bond angle decreases from H2O to H2S because O is more electronegative than S due to which the electrons of 2 O-H bonds are more towards O and the bond pair – bond pair repulsion increases which further increases the bond angle.

(b) (i) SF4

Geometry- trigonal bipyramidal, shape- see saw

(ii) XeF2

Geometry- trigonal bipyramidal, shape- linear


(i) PCl5 fumes in moisture due to formation of HCl.

PCl5 + 4H2O H3PO4 + 5HCl

(ii) The name of the allotrope of sulphur which is stable at room temperature is Rhombic sulphur ().

(iii) Chlorine water on standing loses its yellow colour due to formation of HCl and HOCl.

Cl2 + H2O HCl + HOCl

(iv) On heating, H3PO3 disproportionates into phosphoric acid and phoshine.

4H3PO3 3 H3PO4 + PH3

(v) 2F2 + 2H2O O2 + 4H+ + 4F-

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