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2 years ago

For the first ionization energy for an N2 molecule, what molecular orbital is the electron removed from

Chemistry

1 answer:

olga55 [171]2 years ago

7 0

Answer:

p orbital.

Explanation:

Valence electrons are the electrons in an atom holding the very last orbital which is used in chemical bonding with other elements. Their existence could define the chemical properties of that atom.

During the first energy in ionization of an N2 molecule the molecular orbital from which the electron could be extracted is the only one with the highest energy level. Nitrogen has its outermost orbital (p) containing three valence electrons. Each orbital is only half filled, and thus it is unstable Thus, the electron mission must have been removed from p orbital.

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What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.15 M buffer solution at pH 5.0? Not

Leni [432]

Answer:

Acetic acid 0,055M and acetate 0,095M.

Explanation:

It is possible to prepare a 0,15M buffer of acetic acid/acetate at pH 5,0 using Henderson-Hasselblach formula, thus:

pH = pka + log₁₀ [A⁻]/[HA] -Where A⁻ is acetate ion and HA is acetic acid-

Replacing:

5,0 = 4,76 + log₁₀ [A⁻]/[HA]

1,7378 = [A⁻]/[HA] (1)

As concentration of buffer is 0,15M, it is possible to write:

[A⁻] + [HA] = 0,15M (2)

Replacing (1) in (2):

1,7378[HA] + [HA] = 0,15M

2,7378[HA] = 0,15M

[HA] = 0,055M

Thus, [A⁻] = 0,095M

That means you need acetic acid 0,055M and acetate 0,095M to obtain the buffer you need.

i hope it helps!

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3 years ago

What type of particles move to create electricity

yKpoI14uk [10]

Answer:

Electrons

Explanation:

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2 years ago

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Neporo4naja [7]

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2 years ago

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A cube of an unknown metal measures 0.200 cm on one side. The mass of the cube is 52 mg. Which of the following is most likely t

Margarita [4]

Answer:

Option E. Zirconium

Explanation:

From the question given above, the following data were obtained:

Length of side (L) of cube = 0.2 cm

Mass (m) of cube = 52 mg

Name of the unknown metal =?

Next, we shall determine the volume of the cube. This can be obtained as follow:

Length of side (L) of cube = 0.2 cm

Volume (V) of the cube =?

V = L³

V = 0.2³

V = 0.008 cm³

Next, we shall convert 52 mg to g. This can be obtained as follow:

1000 mg = 1 g

Therefore,

52 mg = 52 mg × 1 g / 1000 mg

52 mg = 0.052 g

Thus, 52 mg is equivalent to 0.052 g.

Next, we shall determine the density of the unknown metal. This can be obtained as follow:

Mass = 0.052 g.

Volume = 0.008 cm³

Density =?

Density = mass / volume

Density = 0.052 / 0.008

Density of the unknown metal = 6.5 g/cm³

Comparing the density of the unknown metal i.e 6.5 g/cm³ with those given in table in the above, we can conclude that the unknown metal is zirconium

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3 years ago

What geometry does VSEPR predict for the central atom in NCl3

Archy [21]

Answer:

Tetrahedral electron geometry and trigonal pyramidal molecular geometry.

Explanation:

The Lewis structure is shown in Figure 1.

The central N atom has three bonding pairs and one lone pair, for four electron groups.

VSEPR theory predicts a tetrahedral electron geometry with bond angles of 109.5°.

We do not count the lone pair in determining the molecular shape.

The molecular geometry is trigonal pyramidal (see Figure 2).

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3 years ago