All categories

1 year ago

The highest energy occupied molecular orbital in the C-C bond of the C2 molecule is?

Chemistry

1 answer:

Butoxors [25]1 year ago

5 0

The highest energy occupied molecular orbital in the C-C bond of the C₂ molecule is 2pπ orbitals.

<h3>What is Molecular Orbital Theory?</h3>

According to this theory,

Molecular orbitals are formed by intermixing of atomic orbitals of two or more atoms having comparable energies
The number of molecular orbitals formed is equal to the number of atomic orbitals combined.
The shape of molecular orbitals formed depends on the type of atomic orbitals combined
Only atomic orbitals having comparable energies and the same orientation can intermix
Bonding M.O. is formed by the additive effect of atomic orbitals and thus, has lower energy and high stability.
Antibonding M.O. is formed by the subtractive effect of atomic orbitals and thus, has higher energy and low stability.
Bonding M.O. is represented by $\sigma, \pi, \delta$ while Antibonding M.O. is represented by $\sigma^*, \pi^*, \delta^*\\$

Molecular Orbital Diagram of C₂

Learn more about Molecular Orbital Theory:

brainly.com/question/17371976

#SPJ4

You might be interested in

Fractional crystallization recrystallizes mineral rocks in the reverse order they melted.

schepotkina [342]

Answer: B

Explanation:

I think this is false. I really don't know the explination, but due to me just finishing this class. I'm going to go with B.

3 0

3 years ago

Read 2 more answers

A sample of ethanol (C2H6O) has a mass of 0.2301 g. Complete combustion of this sample causes the temperature of a bomb calorime

Keith_Richards [23]

Answer: $4.994\times 10^{-3}$ moles

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass , occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

Given mass of ethanol = 0.2301

Molar mass of ethanol = 46.07 g/mol

$\text{Number of moles of ethanol}=\frac{0.2301g}{46.07g/mol}=4.994\times 10^{-3}$

Thus there are $4.994\times 10^{-3}$ moles of ethanol are present in the sample.

7 0

4 years ago

Read 2 more answers

A<br> *<br> has no definite (fixed) shape or size.<br> Liquid<br> Solid<br> Gas

Vilka [71]

Answer:

the answer is Gas

Explanation:

hope this helps!

6 0

4 years ago

What is the mass number of C12,C13 and C14

Elena L [17]

Carbon-13 and Carbon-14 are isotopes of the element carbon. All carbon atoms have 6 protons in their nucleus and most have 6 neutrons. Since the mass number consists of # of protons + # of neutrons, Carbon 12 has an atomic mass of 12 ( = 6 protons + 6 neutrons). Carbon-13 atoms have one extra neutron, giving it a total of 7 neutrons. Carbon-14 atoms have two extra neutrons, giving them a total of 8 neutrons. Therefore the mass number of C-12= 12, C-13= 13 and C-14= 14.

3 0

4 years ago

Which of the three has the largest ei1? which of the three has the largest ? 1s22s22p63s23p64s1 1s22s22p63s23p5 1s22s22p63s23p1?

coldgirl [10]

$E_i1$ will be largest for $1s^22s^22p^63s^23p^5$ .

Explanation: Ionization energy is the energy to knock off an electron from a gaseous atom of ion. First ionization energy or $E_i1$ is the energy required to remove 1 loosely held electron from 1 mole of gaseous atoms to produce 1 mole of gaseous ion carrying (+)1 charge.

$M(g)\rightarrow M^+(g)+e^-$

The electrons are filled according to Aufbau's rule and the orbitals which are strongly held to the nucleus follows the order $s>p>d>f$ .

Electron is released from the outermost shell that is from the electrons which are loosely held to the nucleus, this follows the pattern $s$ .

In configurations, $1s^22s^22p^63s^23p^64s^1,1s^22s^22p^63s^23p^5\text{ and } 1s^22s^22p^63s^23p^1$

The loosely held orbital is 4s, therefore electron will be lost from that easily.

Now, in 3p orbital, one configuration has 5 electrons and one has 1 electron.

The configuration having 5 electrons will be more tightly held by the nucleus because it has more electrons that the one having only 1 electron. Hence, the electron will be lost easily from the configuration having $3p^1$ as the valence shell.

Therefore, the configuration $1s^22s^22p^63s^23p^5$ will the largest $E_i1$ .

8 0

3 years ago