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Oliga [24]
4 years ago
5

Write out the molecular orbital valence electron configurations for the ground state and first excited state for n2.

Chemistry
1 answer:
Bumek [7]4 years ago
7 0

Answer:

(a) (σ2s)² (σ2*s)² (π2p)⁴ (σ2p)²

(b) (σ2s)² (σ2*s)² (π2p)⁴ (σ2p) (π*2p)

Explanation:

An N atom has five valence electrons, so a molecule of N₂ has 10 valence electrons.

Use the Aufbau Principle to place 10 electrons in the MO diagram for B, C, and N.

1.Ground state

The valence electron configuration of N₂ is

(σ2s)² (σ2*s)² (π2p)⁴ (σ2p)²  

2. First excited state

One of the σ2p electrons is promoted to the next higher level,  π*2p.

The valence electron configuration becomes

(σ2s)² (σ2*s)² (π2p)⁴ (σ2p) (π*2p)

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Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.76
aliina [53]

The given question is incomplete, the complete question is:

Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.763g of oxygen. For a given mass of oxygen, what is the lowest whole-number mass ratio of lead in the two compounds that combines with a given mass of oxygen?

Answer:

The lowest whole-number mass ratio in the two compounds is 1:2.

Explanation:

There is a need to find the mole ratio between lead and oxygen atoms in order to find the whole-number mass ratio of lead in the two compounds. In the first compound, the given mass of lead is 2.98 grams, the molar mass of lead is 207.2 gram per mole.  

The no. of moles can be determined by using the formula,  

moles = mass/molecular mass

moles = 2.98 g/207.2 g/mol

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The mass of oxygen in the compound I is 0.461 grams, the molecular mass of oxygen is 16 gram per mol.  

moles = 0.461 g /16 g/mol

= 0.0288 moles

The ratio between the lead and oxygen in the compound I is 0.0144/0.0288 = 1:2

On the other hand, in the compound II, the mass of lead given is 9.89 grams, therefore, the moles of lead in compound II is,  

moles = 9.89 g / 207.2 g/mol

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The mass of oxygen given in compound II is 0.763 grams, the moles of oxygen present in the compound II is,  

moles = 0.763 g / 16 g

= 0.0477 moles

The ratio between the lead and oxygen in the compound II is, 0.0477 moles lead /0.0477 moles oxygen = 1:1

Hence, of the two compounds, the lowest ratio is found in the compound I, that is, 1:2.  

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3 years ago
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klemol [59]

Answer:

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4, strong nuclear force

5, Gravity and the electromagnetic force have infinite ranges while the nuclear forces have very small ranges.

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