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Chemical Bonding and Molecular Structure question

2021 · 18 Mar · Shift 2 · Q16

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Chemical Bonding and Molecular Structure
2021 · 18 Mar · Shift 2 · Q16

Chemical Bonding and Molecular Structure question

2021 · 18 Mar · Shift 2 · Q16

JEE MainChemistryChemical Bonding and Molecular StructureNumerical+4 / −1

The number of species below that have two lone pairs of electrons in their central atom is

\underline{\hspace{2cm}}

. (Round off to the Nearest Integer).

SF_4

, BF

_4^ -

,

ClF_3

,

AsF_3

,

PCl_5

,

BrF_5

,

XeF_4

,

SF_6

Numerical answer

View written solutionFree

Correct answer: 2

We need to count the species whose central atom has exactly two lone pairs.
Check each species one by one using valence electrons and known shapes:

(i) $SF_4$

Central atom: $S$
Sulfur has $6$ valence electrons.
It forms $4$ bonds with $F$ atoms.
Remaining on sulfur: $1$ lone pair.
So, $SF_4$ has 1 lone pair, not 2.

(ii) $BF_4^-$

Central atom: $B$
Boron has $3$ valence electrons, and the negative charge adds one more effectively.
It forms $4$ bonds with $F$ atoms.
No lone pair on boron.
So, $BF_4^-$ has 0 lone pairs.

(iii) $ClF_3$

Central atom: $Cl$
Chlorine has $7$ valence electrons.
It forms $3$ bonds with $F$ atoms.
Remaining electrons on chlorine give 2 lone pairs.
So, $ClF_3$ has 2 lone pairs.

(iv) $AsF_3$

Central atom: $As$
Arsenic has $5$ valence electrons.
It forms $3$ bonds with $F$ atoms.
Remaining electrons give 1 lone pair.
So, $AsF_3$ has 1 lone pair.

(v) $PCl_5$

Central atom: $P$
Phosphorus has $5$ valence electrons.
It forms $5$ bonds with $Cl$ atoms.
No lone pair remains.
So, $PCl_5$ has 0 lone pairs.

(vi) $BrF_5$

Central atom: $Br$
Bromine has $7$ valence electrons.
It forms $5$ bonds with $F$ atoms.
Remaining electrons give 1 lone pair.
So, $BrF_5$ has 1 lone pair.

(vii) $XeF_4$

Central atom: $Xe$
Xenon has $8$ valence electrons.
It forms $4$ bonds with $F$ atoms.
Remaining electrons give 2 lone pairs.
So, $XeF_4$ has 2 lone pairs.

(viii) $SF_6$

Central atom: $S$
Sulfur has $6$ valence electrons.
It forms $6$ bonds with $F$ atoms.
No lone pair remains.
So, $SF_6$ has 0 lone pairs.

Species with exactly two lone pairs on the central atom are:

$ClF_3$
$XeF_4$

Hence, total number of such species is $2$

Since this is already an integer, rounding to nearest integer also gives: $2$

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