The sum of the number of lone pairs of electrons on each central atom in the following species is \[ [\mathrm{TeBr}_6]^{2-}, [\mathrm{BrF}_2]^+, \mathrm{SNF}_3, \text{ and } [\mathrm{XeF}_3]^- \] (Atomic numbers: \(\mathrm{N}=7, \mathrm{~F}=9, \mathrm{~S}=16, \mathrm{Br}=35, \mathrm{Te}=52, \mathrm{Xe}=54\))

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Chemical Bonding and Molecular Structure

The sum of the number of lone pairs of electrons on each central atom in the following species is

[TeBr_{6}]^{2 -}, [BrF_{2}]^{+}, SNF_{3}, and [XeF_{3}]^{-}

(Atomic numbers:

N = 7, F = 9, S = 16, Br = 35, Te = 52, Xe = 54

)

Answer: 6

Solution

[TeBr_{6}]^{2 -}

**: Te has 6 valence electrons + 2 from charge = 8 electrons. With 6 bonds, 8 - 6 = 2 electrons = 1 lone pair.

**

[BrF_{2}]^{+}

**: Br has 7 valence electrons - 1 from charge = 6 electrons. With 2 bonds, 6 - 2 = 4 electrons = 2 lone pairs.

**

SNF_{3}

**: The most stable structure has S as the central atom with a double bond to N and three single bonds to F. S has 6 valence electrons, uses all in bonding, so 0 lone pairs.

**

[XeF_{3}]^{-}

**: Xe has 8 valence electrons + 1 from charge = 9 electrons. With 3 bonds, 9 - 3 = 6 electrons = 3 lone pairs.

Sum = 1 + 2 + 0 + 3 = 6

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Integer answer

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JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Chemical Bonding and Molecular Structure

Solution

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