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Chemical Bonding and Molecular Structure question

2022 · 29 Jul · Shift 1 · Q5

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Chemical Bonding and Molecular Structure
2022 · 29 Jul · Shift 1 · Q5

Chemical Bonding and Molecular Structure question

2022 · 29 Jul · Shift 1 · Q5

JEE MainChemistryChemical Bonding and Molecular StructureMCQ+4 / −1

Number of lone pairs of electrons in the central atom of

\mathrm{SCl}_{2}, \mathrm{O}_{3}, \mathrm{ClF}_{3}

and

\mathrm{SF}_{6}

, respectively, are :

A
0, 1, 2 and 2
B
2, 1, 2 and 0
C
1, 2, 2 and 0
D
2, 1, 0 and 2

View written solutionFree

Correct answer: B

Find lone pairs on the central atom in each molecule

We use the valence electrons of the central atom and its bonding pattern.

For $\mathrm{SCl_2}$

Central atom: $\mathrm{S}$
Sulfur has $6$ valence electrons.
It forms two single bonds with two chlorine atoms.
Thus, $2$ electron pairs are used in bonding.
Remaining electrons on sulfur: $6 - 2 = 4 \text{ electrons} = 2 \text{ lone pairs}$

So, in $\mathrm{SCl_2}$ , central atom has 2 lone pairs.

For $\mathrm{O_3}$

Central atom: middle oxygen
Oxygen has $6$ valence electrons.
In ozone, the central oxygen is involved in one single bond and one double bond (resonance hybrid overall, but for lone-pair counting this gives the same result).
Total bonding around central oxygen corresponds to $3$ bonding pairs.
One lone pair remains on the central oxygen.

So, in $\mathrm{O_3}$ , central atom has 1 lone pair.

For $\mathrm{ClF_3}$

Central atom: $\mathrm{Cl}$
Chlorine has $7$ valence electrons.
It forms $3$ single bonds with fluorine.
Thus, $3$ electrons of chlorine are used in bonding.
Remaining electrons: $7 - 3 = 4 \text{ electrons} = 2 \text{ lone pairs}$

So, in $\mathrm{ClF_3}$ , central atom has 2 lone pairs.

For $\mathrm{SF_6}$

Central atom: $\mathrm{S}$
Sulfur has $6$ valence electrons.
It forms $6$ single bonds with six fluorine atoms.
All six valence electrons are used in bonding.
Hence, no lone pair remains.

So, in $\mathrm{SF_6}$ , central atom has 0 lone pairs.

Collect the results

For $\mathrm{SCl_2}, \mathrm{O_3}, \mathrm{ClF_3}, \mathrm{SF_6}$ respectively: $2,\ 1,\ 2,\ 0$

This matches Option B.

Check each option

A: $0,1,2,2$ ❌
B: $2,1,2,0$ ✅
C: $1,2,2,0$ ❌
D: $2,1,0,2$ ❌

Therefore, the correct answer is B.

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