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Chemical Bonding and Molecular Structure question

2023 · 30 Jan · Shift 1 · Q13
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Chemical Bonding and Molecular Structure question

2023 · 30 Jan · Shift 1 · Q13

JEE MainChemistryChemical Bonding and Molecular StructureMCQ+4 / −1

Match List I with List II

List I
(molecules/ions)		 List II
(No. of lone pairs of e −^-− on central atom)
A. IF7\mathrm{IF_7}IF7​ I. Three
B. ICl4−\mathrm{ICl}_4^ -ICl4−​ II. One
C. XeF6\mathrm{XeF_6}XeF6​ III. Two
D. XeF2\mathrm{XeF_2}XeF2​ IV. Zero

Choose the correct answer from the options given below :

  1. A
    A - II, B - I, C - IV, D - III
  2. B
    A - IV, B - I, C - II, D - III
  3. C
    A - IV, B - III, C - II, D - I
  4. D
    A - II, B - III, C - IV, D - I
View written solutionFree

Correct answer: C

  1. We need the number of lone pairs on the central atom in each species.

A useful method is:

Steric number=V+M−C+A2\text{Steric number} = \frac{V + M - C + A}{2}Steric number=2V+M−C+A​

where:

  • VVV = valence electrons of central atom
  • MMM = number of attached monovalent atoms
  • CCC = positive charge
  • AAA = negative charge

Then:

Lone pairs=Steric number−number of bonded atoms\text{Lone pairs} = \text{Steric number} - \text{number of bonded atoms}Lone pairs=Steric number−number of bonded atoms
  1. For IF7\mathrm{IF_7}IF7​
  • Central atom: Iodine, V=7V = 7V=7
  • Attached monovalent atoms: M=7M = 7M=7
  • Charge = 0

So,

Steric number=7+72=7\text{Steric number} = \frac{7+7}{2} = 7Steric number=27+7​=7

Bond pairs = 7, hence lone pairs:

7−7=07-7=07−7=0

So, IF7\mathrm{IF_7}IF7​ has zero lone pairs on I.

Thus,

A→IVA \to \text{IV}A→IV
  1. For ICl4−\mathrm{ICl_4^-}ICl4−​
  • Central atom: Iodine, V=7V = 7V=7
  • Attached monovalent atoms: M=4M = 4M=4
  • Charge = −1-1−1

So,

Steric number=7+4+12=6\text{Steric number} = \frac{7+4+1}{2} = 6Steric number=27+4+1​=6

Bond pairs = 4, hence lone pairs:

6−4=26-4=26−4=2

So, ICl4−\mathrm{ICl_4^-}ICl4−​ has two lone pairs on I.

Thus,

B→IIIB \to \text{III}B→III
  1. For XeF6\mathrm{XeF_6}XeF6​
  • Central atom: Xe, V=8V = 8V=8
  • Attached monovalent atoms: M=6M = 6M=6
  • Charge = 0

So,

Steric number=8+62=7\text{Steric number} = \frac{8+6}{2} = 7Steric number=28+6​=7

Bond pairs = 6, hence lone pairs:

7−6=17-6=17−6=1

So, XeF6\mathrm{XeF_6}XeF6​ has one lone pair on Xe.

Thus,

C→IIC \to \text{II}C→II
  1. For XeF2\mathrm{XeF_2}XeF2​
  • Central atom: Xe, V=8V = 8V=8
  • Attached monovalent atoms: M=2M = 2M=2
  • Charge = 0

So,

Steric number=8+22=5\text{Steric number} = \frac{8+2}{2} = 5Steric number=28+2​=5

Bond pairs = 2, hence lone pairs:

5−2=35-2=35−2=3

So, XeF2\mathrm{XeF_2}XeF2​ has three lone pairs on Xe.

Thus,

D→ID \to \text{I}D→I
  1. Final matching
A→IV,B→III,C→II,D→IA\to IV,\quad B\to III,\quad C\to II,\quad D\to IA→IV,B→III,C→II,D→I

This corresponds to Option C.

  1. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.

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