
Consider the following compounds from A to E .
A. $Xe{{F}_{n}}$
B. $XeF_{n+1}^{+}$
C. $XeF_{n+1}^{-}$
D. $Xe{{F}_{n+2}}$
E. $XeF_{n+4}^{2-}$
If the value of n is 4, then calculate the value of “p+q” here, ‘p’ is the total number of bond pairs and ‘q’ is the total number of lone pairs on central atoms of compound A to E .
Answer
162.9k+ views
Hint: Firstly , draw the structures of all the given compounds by putting the respective value of n in all the general formulas and then count the bond pairs and lone pairs of electrons individually and then add them. Bond pairs electrons are those electrons which are bonded to any other atom from the central atom while lone pair of electrons are those electrons which are not bonded to any other atom and are free . Xenon has 8 electrons in its valence orbital.
Complete Step by Step Answer:
Xenon is a noble gas element with an atomic number of 54 and the electronic configuration of $[Kr]4{{d}^{10}}5{{s}^{2}}5{{p}^{6}}$ $[Kr]4{{d}^{10}}5{{s}^{2}}5{{p}^{6}}$. In its outer orbitals like s and p it has a total of 8 valence electrons. Although it is a noble gas element still it forms a large number of compounds under specific conditions as it ionization enthalpy resembles with that of oxygen . Here putting the value of n in all the compounds given here we get the compounds structural formula and there respective bond pair and lone pairs as-
$A.Xe{{F}_{n}}=Xe{{F}_{4}}->b.p=4,l.p=2 \\ B.XeF_{n+1}^{+}=XeF_{5}^{+}->b.p=5,l.p=1 \\ C.XeF_{n+1}^{-}=XeF_{5}^{-}->b.p=5,l.p=2 \\ D.Xe{{F}_{n+2}}=Xe{{F}_{6}}->b.p=6,l.p=1 \\ E.XeF_{n+4}^{2-}=XeF_{8}^{2-}->b.p=8,l.p=1 \\ $
The structures of the following compounds are respectively:
As, p = total number of bond pair =4+5+5+6+8=28
and q = total number of lone pair=2+1+2+1+1=7
and so p+q=28+7=35.
Hence, the answer is p = 35.
Note: The structures of the compounds above the charge given at the upper subscript of the denotes the loss or gain of electrons . And thus that will affect the structures and bond pair and lone pair electrons also. Many students ignore the charge and draw general structures which can cause incorrect answers.
Complete Step by Step Answer:
Xenon is a noble gas element with an atomic number of 54 and the electronic configuration of $[Kr]4{{d}^{10}}5{{s}^{2}}5{{p}^{6}}$ $[Kr]4{{d}^{10}}5{{s}^{2}}5{{p}^{6}}$. In its outer orbitals like s and p it has a total of 8 valence electrons. Although it is a noble gas element still it forms a large number of compounds under specific conditions as it ionization enthalpy resembles with that of oxygen . Here putting the value of n in all the compounds given here we get the compounds structural formula and there respective bond pair and lone pairs as-
$A.Xe{{F}_{n}}=Xe{{F}_{4}}->b.p=4,l.p=2 \\ B.XeF_{n+1}^{+}=XeF_{5}^{+}->b.p=5,l.p=1 \\ C.XeF_{n+1}^{-}=XeF_{5}^{-}->b.p=5,l.p=2 \\ D.Xe{{F}_{n+2}}=Xe{{F}_{6}}->b.p=6,l.p=1 \\ E.XeF_{n+4}^{2-}=XeF_{8}^{2-}->b.p=8,l.p=1 \\ $
The structures of the following compounds are respectively:
As, p = total number of bond pair =4+5+5+6+8=28
and q = total number of lone pair=2+1+2+1+1=7
and so p+q=28+7=35.
Hence, the answer is p = 35.
Note: The structures of the compounds above the charge given at the upper subscript of the denotes the loss or gain of electrons . And thus that will affect the structures and bond pair and lone pair electrons also. Many students ignore the charge and draw general structures which can cause incorrect answers.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Types of Solutions

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions

Solutions Class 12 Notes: CBSE Chemistry Chapter 1

NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes

NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry

Electrochemistry Class 12 Notes: CBSE Chemistry Chapter 2
