Chemistry, 01.07.2019 11:00 foziasif724

Draw the lewis structure of xecl4 showing all lone pairs.

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Answer from: lolirdc

Draw the following structure; replace any sticks with two dots as bonding pairs if necessary.

$:\ddot{\text{Cl}}:\phantom{\text{X}}:\ddot{\text{Cl}}:\\\phantom{:Cl}\backslash\phantom{\text{Xe}}/\\\phantom{:Cl\backslash}\ddot{\text{Xe}}\\\phantom{:Cl}/\ddot{\phantom{Xe}}\backslash\\:{\text{Cl}}:\phantom{e}:{\text{Cl}}:\\ \phantom{:}\;\ddot{\phantom{Cl}} \phantom{:\phantom{e}:}\;\ddot{\phantom{Cl}}$

ExplanationEach neutral xenon atom contains 8 valence electrons in its valence shell. It has already attained an octet as a noble gas element.Each neutral chlorine atom contains 7 valence electrons in its valence shell. It demands one extra electron to form an octet.

Each $\text{Xe}\text{Cl}_4$ xenon tetrachloride molecule contains one xenon atom and four chlorine atoms. It would take $4 \times 1 = 4$ extra electrons to ensure an octet for all atoms in a single $\text{Xe}\text{Cl}_4$ molecule.

Chlorine atoms in a $\text{Xe}\text{Cl}_4$ demands electrons. However, the central xenon atom does not. The chlorine atom would insist sharing electrons with the xenon atom. The xenon atom would end up with forming four single bonds, one with each chlorine atom. It would find more a total of twelve electrons in its valence shell- that's four more than what the xenon atom needs for a complete octet.

The xenon atom has some vacancies in its 5d orbitals (energetically) next its 5p; it would form an "expanded octet." The four extra electrons would end up as lone pairs around the xenon atom, occupying some 5d orbitals.

The central xenon atom ends up with six electron domains, with four bonding pairs and two "lone" nonbonding pairs. By the VSEPR theory, it would demonstrate a square planar molecular geometry. All five atoms would end up on the same plane, with a Cl-Xe-Cl bond angle of 90°. The two nonbonding pairs are located above and beyond the plane right next to the xenon atom. In addition to that, there would be three lone pairs on each of the four chlorine atoms.

Answer from: Quest

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Answer from: Quest

the dissociation of nickel hydroxide is follows:

$ni(oh)_2 > ni^2^+ + 2 oh^-$

$ksp for no(oh)2 = 5.5 * 10^-^1^6$

$5.5 * 10^-^1^6 = s * (2 s)^2 = 4 s^3$

$s^3 = 137.5 * 10^-^1^8$

$s = 5.14 * 10^-^6 = [oh^-]$

the concentration of hydroxide ion is

$5.14 * 10^-^6 = [oh^-]$

the equilibrium constant value of nickel hydroxide is = 6.1 * 10^-^1^6 and thus equilibrium equation can be written as:

$k = [ ni^2^+] * [oh^-]^2$

$6.1 * 10^-^1^6 = [ ni^2] * [5.14 * 10^-^6]^2$

$[ ni^2^+] = 0.23 * 10^-^4 mol /l$

thus, concentration of ni is 0.23 * 10^-^4 mol /l

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Draw the lewis structure of xecl4 showing all lone pairs....

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