Match the compounds given in column I with the hybridization and shape given in column II and mark the correct option.

Column I

Column II

(a) XeF6

(i)    distorted octahedral

(b) XeO3

(ii)   square planar

(c) XeOF4

(iii)  pyramidal

(d) XeF4

(iv)  square pyramidal

Code:

(a)    (b)    (c)    (d)

  • Option 1)

    (i)    (iii)    (iv)    (ii)

  • Option 2)

    (i)    (ii)    (iv)    (iii)

  • Option 3)

    (iv)    (iii)    (i)    (ii)

  • Option 4)

    (iv)    (i)    (ii)    (iii)

 

Answers (1)

 

VSEPR Theory -

1.  The shape of the molecule is determined by repulsions between all of the electron pair present in valence shell.

2.  Order of repulsion 

             lone \:pair-Lone\: pair> Lone\:pair-Bond\:pair> Bond\:pair-bond\:pair

3  Repulsion among the bond pair is directly proportional to the bond  order and electronegativity difference between the central atom and the other atom.

-

 

 

Dettermination of shape of molecules using VSEPR Theory -

calculate \:X

X=(No.\;of\:valence\:electrons\:of\:central\:atom\:)+(No.\:of\:other\:atom\:)+(Negative\:charge\:on\:the\:molecule)-(Positive\:charge\:on\:the\:molecule)

-

 

 X_{e}F_{6}\rightarrow distorted octahedron

X_{e}O_{3}\rightarrow pylamidal

X_{e}OF_{4}\rightarrow square pyramidal

X_{e}F_{4}\rightarrow square planal

 


Option 1)

(i)    (iii)    (iv)    (ii)

Correct option

Option 2)

(i)    (ii)    (iv)    (iii)

incorrect option

Option 3)

(iv)    (iii)    (i)    (ii)

incorrect option

Option 4)

(iv)    (i)    (ii)    (iii)

incorrect option

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