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Svetllana [295]
3 years ago
6

Select the correct hybridization for the central atom based on the electron geometry cocl2 (carbon is the central atom).

Chemistry
1 answer:
Rainbow [258]3 years ago
7 0
In this compound (Phosgene) the central atom (carbon is Sp² Hybridized).

Sp, Sp² and Sp³ can be calculated very simply by doing three steps, 

Step 1:
           Assume triple bond and double bond as one bond and assign s or p to it. In this example carbon double bond oxygen is considered once and let suppose it is s. Now we are having our s.

Step 2:
          Count lone pair of electron, each lone pair counts for s and p. In this case there is no lone pair of electron on carbon, so not included.

Step 3:
          Count single bonds for s and p. As we have already assigned s to the double bond, now one p for one single bond, and other p for the other single bond.

Result:
          So, we counted 1 s for double bond, 1 p for one single and other p for second single bond. As a whole we got,

                                                       Sp²

Practice:
You can practice for hybridization of Oxygen in this molecule. Oxygen has 2 lone pair of electrons. (Hint: Sp² Hybridization)
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Read 2 more answers
Calculate each of the following quantities:<br> (b) Amount (mol) of O atoms in 8.18 g of Ba(NO₃)₂
postnew [5]

The number of moles of O atoms in 8.18 g of Ba(NO₃)₂ is 0.364 mol.

<h3>What is moles?</h3>

"The mole is a SI unit of measurement that is used to calculate the quantity of any substance." It is denoted by mol.

Given that the mass of Ba(NO₃)₂ is  8.18 g

n = m/M

n = moles = ? mol Ba(NO₃)₂

m = mass(g) =  8.18 gram(NO3)2

The molar mass of Ba(NO₃)₂ is  261. 35g/mol.

As given, the moles of oxygen atom in Ba(NO₃)₂ is 6.

So, the moles of oxygen in 8.18 gram will be

Putting the value in the formula:

0.0607 mol Mg(NO3)2 × 6 mol O atoms/mol Mg(NO3)2 = 0.364 mol O atoms to three significant figures.

Thus, the number of moles of O atoms is 0.364 mol.

To learn more about moles, refer to the link:

brainly.com/question/4220551

#SPJ4

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