Consider the thioformaldehyde molecule.

a. what is the central atom? enter its chemical symbol.
b. how many lone pairs are around the central atom?
c. what is the ideal angle between the carbon-hydrogen bonds?
d. compared to the ideal angle, you would expect the actual angle between the carbon-hydrogen bonds to be

a. C

b. 0

c. 120°

d. lower

Explanation:

In the attached picture, we can see the Lewis structure of thioformaldehyde.

a. What is the central atom? Enter its chemical symbol.

The central atom is carbon (C).

b. How many lone pairs are around the central atom?

As we can see in the Lewis structure, there are no lone pairs of electrons around carbon.

c. What is the ideal angle between the carbon-hydrogen bonds?

According to the VESPR theory, C is bonded to 3 ligands and has no lone pairs, so the molecular geometry is trigonal planar. The ideal bond angle is 120°.

d. Compared to the ideal angle, you would expect the actual angle between the carbon-hydrogen bonds to be ...

The repulsion due to the double bond  C=S is stronger than the one due to the single bonds C-H. Then, the H-C-H real angle is lower than the ideal one (according to experimental results, it is around 116°).

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answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.