
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
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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.
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