The sum of the number of lone pairs of electr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) To find the sum of lone pairs on the central atoms:$1.$ In $[TeBr_6]^{2-}$: $Te$ (Group $16$) has $6$ valence electrons. With a $-2$ charge,it has

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) To find the sum of lone pairs on the central atoms:$1.$ In $[TeBr_6]^{2-}$: $Te$ (Group $16$) has $6$ valence electrons. With a $-2$ charge,it has $6 + 2 = 8$ electrons. It forms $6$ single bonds with $Br$. Lone pairs = $(8 - 6) / 2 = 1$.$2.$ In $[BrF_2]^+$: $Br$ (Group $17$) has $7$ valence electrons. With a $+1$ charge,it has $7 - 1 = 6$ electrons. It forms $2$ single bonds with $F$. Lone pairs = $(6 - 2) / 2 = 2$.$3.$ In $SNF_3$: $S$ (Group $16$) is the central atom with $6$ valence electrons. It forms a triple bond with $N$ and $3$ single bonds with $F$. Total electrons used = $3 + 3 = 6$. Lone pairs = $(6 - 6) / 2 = 0$.$4.$ In $[XeF_3]^-$: $Xe$ (Group $18$) has $8$ valence electrons. With a $-1$ charge,it has $8 + 1 = 9$ electrons. It forms $3$ single bonds with $F$. Lone pairs = $(9 - 3) / 2 = 3$.Sum of lone pairs = $1 + 2 + 0 + 3 = 6$.

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

Similar Questions

The linear structure is assumed by

The number of $120^{\circ}$ $Cl-P-Cl$ angles in phosphorus pentachloride is

What is the shape of $I_3^{-}$ ion?

Out of $CHCl_3$,$CH_4$,and $SF_4$,the molecules that do not have regular geometry are

$XH_3 + H^{+} \longrightarrow XH_4^{+} (X = N, P, As, Sb)$. In this process,the maximum increase in bond angle will be in the case of $:-$

Vedclass Test Series

Exam Paper Generator

Online Exam Module