Question

Match the following:

	List (Molecular geometry)		List II (Molecule)
A	Trigonal planar	I	$PCl _{5}$
B	Tetrahedral	II	$SF _{6}$
C	Trigonal bipyramidal	III	$BF _{3}$
D	Octahedral	IV	$CCl _{4}$
		V	$BeCl _{2}$

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 $\rightarrow$ (III) $BF _{3}$, because boron has three $(3)$ valence electrons which are involved in formation of three sigma ( $\sigma$ ) bonds with $3 F$ -atoms and has no lone pair $(l p)$ of electrons.

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

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

(D) Octahedral $\rightarrow$ (II) $SF _{6}$, because sulphur has six valence electrons which are involved in formation of six sigma $(\sigma)$ bonds with $6 F$ -atoms and has no $l p$ of electrons.