Home/JEE Main/Chemistry/Chemical Bonding and Molecular Structure/Q-match-list-i-list-iilist-ispecieslist-iishapeai3-ioctahedralbsf6iisee-sawcsf4iiitri-300260Question
ChemistryJEE MainClass 11Medium

Match List-I with List-II.

List-I

(Species)

List-II

(Shape)

(A)

I3-

(I)

Octahedral

(B)

SF6

(II)

See-saw

(C)

SF4

(III)

Trigonal pyramidal

(D)

NH3

(IV)

Linear

Choose the correct answer from the options given below:

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Quick Answer
Option D

A-IV, B-I, C-II, D-III

This question requires matching chemical species with their respective shapes.
Step-by-step solution
1

This question requires matching chemical species with their respective shapes. The shape of a molecule is determined by its hybridization and the number of lone pairs on the central atom, according to VSEPR theory.

Step 1: Determine the shape of I3-.

The central iodine atom in I3- has 7 valence electrons. It forms two single bonds with two other iodine atoms and has 3 lone pairs. The steric number is 2 (bonds) + 3 (lone pairs) = 5. This corresponds to sp3d hybridization. With 3 lone pairs in the equatorial positions and two bond pairs in axial positions, the molecular geometry is linear.

So, A matches with (IV).

Step 2: Determine the shape of SF6.

The central sulfur atom in SF6 has 6 valence electrons. It forms six single bonds with six fluorine atoms and has 0 lone pairs. The steric number is 6 (bonds) + 0 (lone pairs) = 6. This corresponds to sp3d2 hybridization. With 6 bond pairs and no lone pairs, the molecular geometry is octahedral.

So, B matches with (I).

Step 3: Determine the shape of SF4.

The central sulfur atom in SF4 has 6 valence electrons. It forms four single bonds with four fluorine atoms and has 1 lone pair. The steric number is 4 (bonds) + 1 (lone pair) = 5. This corresponds to sp3d hybridization. With 4 bond pairs and 1 lone pair, the molecular geometry is see-saw.

So, C matches with (II).

Step 4: Determine the shape of NH3.

The central nitrogen atom in NH3 has 5 valence electrons. It forms three single bonds with three hydrogen atoms and has 1 lone pair. The steric number is 3 (bonds) + 1 (lone pair) = 4. This corresponds to sp3 hybridization. With 3 bond pairs and 1 lone pair, the molecular geometry is trigonal pyramidal.

So, D matches with (III).

Step 5: Match the options.

  • A - (IV) Linear
  • B - (I) Octahedral
  • C - (II) See-saw
  • D - (III) Trigonal pyramidal

This corresponds to option D.

Correct Answer: (D)

AnswerD·

A-IV, B-I, C-II, D-III

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