nd carbon atom through a single bond and also to a nitrognaipublishers.com atom with a triple bond. The outer carbon atom is linked to three hydrognaipublishers.com atoms. The nitrognaipublishers.com atom has one lone pair that electrons. The carbon atom in CH3 is: trigonal planar linear trigonal pyramidal tetrahedral bnaipublishers.comt The nitrile carbon atom ( C≡N ) is: trigonal pyramidal bnaipublishers.comt tetrahedral trigonal planar linear

Answer:

The CH₃ carbon atom has tetrahedral geometry and also the nitrile carbon atom has linear geometry

Explanation:

According come VSEPR theory, in one acetonitrile:

The outer carbon atom CH₃, which is associated to 3 hydrognaipublishers.com atoms and also one cnaipublishers.comtral carbon atom, has 4 bond pairs of electrons and zero lone pair. Therefore, outer carbon atom CH₃ has a TETRAHEDRAL geometry.

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The nitrile carbon atom (C≡N), which creates a solitary bond with the external CH₃ carbon atom and triple bond with the nitrognaipublishers.com atom, has two bond pairs and also zero lone pairs.

Therefore, nitrile carbon atom has a linear geometry.


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A 3.3 g sample of salt hydrognaipublishers.com lead carbonate is added to a systems of acetic mountain weighing 10.3 g. The two substances react, rel
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Answer:

1.73g that CO2.

Explanation:

We"ll begin by writing the balanced equation for the reaction. This is givnaipublishers.com below:

NaHCO3 + CH3COOH → CH3COONa + H2O + CO2

Next us shall identify the masses the NaHCO3 and CH3COOH the reacted and the fixed of CO2 developed from the balanced equation. This is portrayed below:

Molar mass of NaHCO3 = 23 + 1 + 12 + (16x3) = 84g/mol

Mass the NaHCO3 from the balanced equation = 1 x 84 = 84g

Molar fixed of CH3COOH = 12 + (3x1) + 12 + 16 + 16 + 1 = 60g/mol

Mass that CH3COOH indigenous the balanced equation = 1 x 60 = 60g

Molar mass of CO3 = 12 + (2x16) = 44g/mol

Mass the CO2 native the balanced equation = 1 x 44 = 44g

From the well balanced equation above,

84g of NaHCO3 reacted through 60g of CH3COOH to develop 44g of CO2.

Next, us shall determine the limiting reactant that the reaction. This is illustrated below:

From the well balanced equation above,

84g that NaHCO3 reacted with 60g of CH3COOH.

Therefore, 3.3g that NaHCO3 will certainly react through = (3.3 x 60)/84 = 2.36g the CH3COOH.

From the over illustration, we deserve to see that just 2.36g that CH3COOH the end of 10.3g givnaipublishers.com reacted fully with 3.3g that NaHCO3. Therefore, NaHCO3 is the limiting reactant if CH3COOH is the overfill reactant.

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Finally, deserve to determine the mass of CO2 created during the reaction.

In this instance the limiting reactant will certainly be used because it will produce the mass productivity of CO2 as all of it were provided up in the reaction. The limiting reactant is NaHCO3 and the massive of CO2 developed is acquired as presented below:

From the balanced equation above,

84g the NaHCO3 reacted to produce 44g that CO2.

Therefore, 3.3g of NaHCO3 will certainly react to create = (3.3 x 44)/84 = 1.73g of CO2.