Question

Match the following:

	List (Molecular geometry)		List II (Molecule)
A	Trigonal planar	I	$PC{l}_5$
B	Tetrahedral	II	$S{F}_6$
C	Trigonal bipyramidal	III	$B{F}_3$
D	Octahedral	IV	$CC{l}_4$
		V	$BeC{l}_2$

The correct answer is

• A. A - V , B - I , C - III , D - II
• B. A - V , B - II , C - I , D - IV
• C. A - III , B - V , C - I , D - IV
• D. A - III , B - IV , C - I , D - II

Answer 1

Answer: (D)

(A) Trigonal planar $\to$ (III) $B{F}_3$, because boron has three $(3)$ valence electrons which are involved in formation of three sigma ( $\sigma$ ) bonds with $3F$ -atoms and has no lone pair $(lp)$ of electrons.

(B) Tetrahedral $\to (IV)CC{l}_4$, because carbon has four $(4)$ valence electrons which are involved in formation of four sigma $(\sigma )$ bonds with $4Cl$ -atoms and has no $lp$ of electrons.

(C) Trigonal bipyramidal $\to$ (I) $PC{l}_5$, because phosphorus has five valence electrons which are involved in formation of five sigma $(\sigma )$ bonds with $5Cl$ atoms and has no $lp$ of electrons.

(D) Octahedral $\to$ (II) $S{F}_6$, because sulphur has six valence electrons which are involved in formation of six sigma $(\sigma )$ bonds with $6F$ -atoms and has no $lp$ of electrons.