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Agata [3.3K]
3 years ago
8

Give the molecular geometry and number of electron groups for sf4. give the molecular geometry and number of electron groups for

. square planar, 6 electron groups square pyramidal, 6 electron groups t-shaped, 5 electron groups octahedral, 6 electron groups seesaw, 5 electron groups
Chemistry
1 answer:
Andrei [34K]3 years ago
3 0

Answer:

5 electron groups, see saw

Explanation:

During the formation of SF4, the sulfur atom usually bonds with each of four fluorine atoms where 8 of valence electrons are used. The four fluorine atoms have 3 lone pairs of electrons in its octet which will further utilize 24 valence electrons. In addition, two electrons are present as a lone pair on the sulfur atom. We can determine sulfur’s hybridization state by counting of the number of regions of electron density on sulphur (the central atom in the molecule). When bonding takes place there is a formation of 4 single bonds to sulfur and it has 1 lone pair. Looking at this, we can say that the number of regions of electron density is 5. The hybridization state is sp3d.

SF4 molecular geometry is seesaw with one pair of valence electrons. The molecule is polar. The equatorial fluorine atoms have 102° bond angles instead of the actual 120° angle. The axial fluorine atom angle is 173° instead of the actual 180° bond angle.

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Explanation:

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1)

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3 0
3 years ago
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A solution is made by dissolving 0.656 mol of nonelectrolyte solute in 869 g of benzene. calculate the freezing point, tf, and b
solmaris [256]
Answer is: the freezing point is 1.63°C and boiling point is 82.01°C.<span>.

1) n(</span><span>nonelectrolyte solute) = 0.656 mol.
</span>m(C₆H₆ - benzene) = 869 g ÷ 1000 g/kg.
m(C₆H₆) = 0.869 kg.<span>
b(solution) = n(</span>nonelectrolyte solute) ÷ m(C₆H₆).<span>
b(solution) = 0.656 mol ÷ 0.869 kg.
b(solution) = 0.754 mol/kg.

2) ΔT = Kf(benzene) · b(solution).
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</span>
3) ΔTb = Kb(benzene) · b(solution).
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</span>
4 0
3 years ago
There are 9.88x1023 molecules of O2 available.
blondinia [14]

Answer:

1.64 moles O₂

Explanation:

Part A:

Remember 1 mole of particles = 6.02 x 10²³ particles

So, the question becomes, how many  '6.02 x 10²³'s are there in 9.88 x 10²³ molecules of O₂?

This implies a division of given number of particles by 6.02 x 10²³ particles/mole.

∴moles O₂ = 9.88 x 10²³ molecules O₂ / 6.02 x 10²³ molecules O₂ · mole⁻¹ = 1.64 mole O₂

_______________

Part B needs an equation (usually a combustion of a hydrocarbon).

7 0
3 years ago
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