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34kurt
3 years ago
12

For which of the following molecules or ions does the following description apply? "The bonding can be explained using a set of

sp 2 hybrid orbitals, with one of the hybrid orbitals holding a nonbonding pair of electrons."For which of the following molecules or ions does the following description apply? "The bonding can be explained using a set of sp 2 hybrid orbitals, with one of the hybrid orbitals holding a nonbonding pair of electrons." CO2
H2S

O3

CO3^2−

More than one of the above
Chemistry
2 answers:
Marina86 [1]3 years ago
7 0

Answer:

O₃

Explanation:

Consider the molecule CO₂. The carbon is sp hybridized. Carbon has 4 valence electrons and oxygen contributes 2 electrons, 1 for each C=O which indicates that there are 8 electrons around the carbon. Since there are 4 bonds all of them are bond pairs. Each C=O double bond uses 2 bond pairs which are considered as single unit. These two double bond units try to get as far apart as possible making the molecule adopt a linear geometry.

Considering the H₂S molecule both oxygen and sulfur are the in the same group, which means both have a valence of 6. The four valence orbitals of sulfur, one 1 s orbital and three 3p orbitals mix together and forms four sp³ hybridized orbitals. Of the four hybridized orbitals, two overlaps with the 1s orbital of hydrogen forming 2 (S - H) bonds while the other two sp³ orbitals remain on sulfur which has lone pair of electrons. Because of the presence of lone pair, the angle between H-S-H bond is slightly less than the ideal tetrahedral bond angle. Thus, H2S having 2 bonding electron pair and 2 lone pairs has a bent shape.

Considering O₃ and according to the VSEPR theory ozone molecule must have a trigonal – planar geometry. It has a total of 18 valence electrons. From the resonance structure given below it is clear the 4 pairs of electrons exit as bonding pair, sp² or σ- bond and the remaining 10 electrons exit as lone pair. Of the three un- hybridized p orbitals one is anti – bonding and remains empty. In ozone the π bond is distributed between the two bonds, and each receives half a π bond. For this molecule the electron pair geometry is trigonal planar but the molecular geometry is bent. The presence of lone pair exerts slight repulsion on the bonding oxygen atoms and a slight compression of the bond angle greater than 120°.

In carbonate ion, <u>the carbon is sp² hybridized</u>. The carbon has 4 valence electrons and there are four bonds to the oxygen which add another 4 making a total of 8. There are 4 pairs of bonding electrons and no lone pair. Of the 4 bond pairs, 2 pairs are used in forming double bond C=O and 2 bond pairs in forming the two C-O single bonds., Thus CO₃²⁻ adopts a trigonal planar geometry.

Of the two molecules only ozone and carbonate ion, have sp2 hybridized central atoms. In ozone the central atoms have lone pair of electrons the hybridization around is sp². Hence the correct option is O₃

puteri [66]3 years ago
6 0

Answer:

O_{3}

Explanation:

O_{3}  is the only molecule containing sp^2 hybrid orbitals, with one of the hybrid orbitals holding a nonbonding pair of electrons.

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maks197457 [2]

Answer:

a) \Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

b) x=4

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Explanation:

Gas mixture:

n_{Ar}= 4 mol

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n_{Xe}= y mol

n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}

n_{Ne} + n_{Xe}=2*n_{Ar}

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y=8 mol- x

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x_{Ar}=\frac{4 mol}{12 mol}=1/3

x_{Ne}=\frac{x mol}{12 mol}=x/12

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Expression of \Delta G_{mixing}

\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)

\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]

\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

Expression of \Delta G_{max}

\frac{d \Delta G_{mixing}}{dx}=0

\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]

0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)

0=[ln (x/12)-ln ((8-x)/12)

ln (x/12)=ln ((8-x)/12)

x=(8-x)

x=4

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]

\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]

\Delta G_{max}=-2721.9 J/mol

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