Consider a carbon atom that is sp hybridized. Indicate how many of each orbital exist on this carbon atom by sorting each orbita
l type. Consider the outer valence only.
Options include: sp orbitals, p orbitals, s orbitals
Put them in the following categories:
Zero One Two Three
Options include: sp orbitals, p orbitals, s orbitals
Put them in the following categories:
Zero One Two Three
1 answer:
Explanation:
It is known that atomic number of carbon is 6 and its electronic configuration is . This means that in its neutral state it contains 2 electrons in its s-orbital and 2 electrons in its p-orbital.
After excitation there will be one electron present in its s-orbital and three electrons present in p-orbital.
Therefore, after the hybridization there will be in total 2 sp hybrid orbitals, 2 p-orbitals and zero s-orbital.
You might be interested in
The amount, in mg, of CO present in the room will be 191,520 mg.
<h3>Stoichiometric problem</h3>The concentration of the gas in the room is 5.7 x mg/cm3.
The dimension of the room is 3.5 m x 3.0 m x 3.2 m. This is equivalent to 350 cm x 300 cm x 320 cm.
We can obtain the volume of the room as:
350 x 300 x 320 = 33,600,000 cm3
The concentration is in mg/cm3, meaning that it is mass/volume.
Thus:
mass = concentration x volume = 5.7 x mg/cm3 x 33,600,000 cm3
= 191,520 mg
The mass of CO in the room is 191,520 mg
More on stoichiometric problems can be found here: brainly.com/question/14465605
#SPJ1