All categories

3 years ago

In which cases do you expect deviations from the idealized bond angle?

Chemistry

1 answer:

Allisa [31]3 years ago

5 0

Answer:

Since PF3 and SBr2 both have lone pairs of electrons on the central atom, their bond angles should deviate from the idealized bond angle as lone pairs are more repulsive than bonding pairs of electrons. Large atoms, such as Cl or Br, will cause bond angles to deviate from the ideal, and the presence of a multiple bond will cause a deviation as well owing to the increased electron density, so CH3Br should deviate from the idealized bond angle. BCl3 has identical atoms symmetrically surrounding a central atom with no lone pairs on it, so BCl3 should not deviate from the idealized bond angle.

You might be interested in

The normal freezing point of water (H2O) is 0.00 oC and its Kf value is 1.86 oC/m. A nonvolatile, nonelectrolyte that dissolves

adelina 88 [10]

Answer:

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

Explanation:

One colligative property is the freezing point depression due the addition of a solute. The equation is:

ΔT=Kf*m*i

Where ΔT is change in temperature = 0.400°C

Kf is freezing point constant of the solvent = 1.86°C/m

m is molality of the solution (Moles of solute / kg of solvent)

And i is Van't Hoff constant (1 for a nonelectrolyte)

Replacing:

0.400°C =1.86°C/m*m*1

0.400°C / 1.86°C/m*1 = 0.215m

As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:

0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.

The mass of ethylene glycol must be added is:

0.0602 moles * (62.10g / mol) =

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

6 0

3 years ago

What happens when a parking lot is built in a wetlands area

kondaur [170]

It will most likely get flooded at one point in time.

8 0

3 years ago

Read 2 more answers

Write equations that show the processes that describe the first, second, and third ionization energies for a gaseous gadolinium

Inessa [10]

Answer:

Gd(g) → $Gd^{+}(g) + e^{-}$ , $Gd^{+}(g)$ → $Gd^{2+}(g) + e^{-}$ , $Gd^{2+}(g)$ → $Gd^{3+}(g) + e^{-}$

Explanation:

Ionization energy is the energy required to remove an electron from a gaseous atom or ion.

The first ionization energy is the energy required to remove the valence electron(outermost) from a neutral atom:

Gd(g)→ $Gd^{+}(g) + e^{-}$

The second ionization energy is the energy required to remove next/second electron from $x^{+}$ ion. The second ionization energy is always higher than the first:

$Gd^{+}$ (g) → $Gd^{2+}(g) + e^{-}$

The third ionization energy is the energy required to remove third electron from $x^{2+}$ ion:

$Gd^{2+}$ (g) → $Gd^{3+}(g) + e^{-}$

5 0

3 years ago

1 Identify Draw an example of a force acting on an object.

Alisiya [41]

Answer:

Answer below

Explanation:

Just draw a photo of someone pushing an object across a table. Your push is the force acting on the object you're pushing.

5 0

2 years ago

How do the elements in each group differ? How are they similar?

bulgar [2K]

Answer:

All the elements in one group have the same number of valence electrons. ... Since elements in a group have the same number of valence electrons, they behave similarly in chemistry. An example would be the alkali metals (excepting hydrogen. Hydrogen is in this group only because it has one valence electron.

Explanation:

5 0

3 years ago

Read 2 more answers