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weqwewe [10]
3 years ago
11

Indicate the electron pair geometry and the molecular geometry for each of the six compounds. Compound Electron pair geometry Mo

lecular geometry CO 2 CO2 linear linear BF 3 BF3 SO 2 SO2 trigonal planar bent SiCl 4 SiCl4 tetrahedral tetrahedral PF 3 PF3 tetrahedral trigonal pyramidal OF 2 OF2 tetrahedral bent
Chemistry
1 answer:
marusya05 [52]3 years ago
5 0

Answer:

CO2

Electron pair geometry- Linear

Molecular geometry- Linear

BF3

Electron pair geometry - Trigonal planar

Molecular geometry- trigonal planar

SO2

Electron geometry - Trigonal planar

Molecular geometry - bent

SiCl4

Electron geometry- tetrahedral

Molecular geometry - tetrahedral

PF3

Electron pair geometry - tetrahedral

Molecular geometry- trigonal pyramidal

OF2

Electron pair geometry- tetrahedral

Molecular geometry- bent

Explanation:

Considering the molecule CO2, there are two regions of electron density in the molecule positioned at an angle of 180 degrees from each other hence the molecule is linear.

For BF3, the three bond pairs are arranged at the corners of a triangle to give a trigonal planar geometry at a bond angle of 120 degrees.

SO2 has two bonding groups and one lone pair giving three regions of electron density and a trigonal planar electron pair geometry. Due to the distortion to geometry caused by the presence of a lone pair, the molecule is bent.

For SiCl4, the four bonding groups are arranged at the corners of a regular tetrahedron hence it is tetrahedral both in electron pair geometry and in molecular geometry.

PF3 molecule has four regions of electron density corresponding to tetrahedral electron pair geometry. The presence of the lone pair leads to a trigonal pyramidal molecular geometry.

For OF2, there are four regions of electron density around the central oxygen atom. Two bond pairs and two lone pairs leads to a tetrahedral electron pair geometry but a bent molecular geometry is observed due to the two lone pairs.

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Phosphorus-32 is radioactive and has a half life of 14.3 days. What percentage of a sample would be left after 12.6 days
nadya68 [22]

Answer:

There is 54.29 % sample left after 12.6 days

Explanation:

Step 1: Data given

Half life time = 14.3 days

Time left = 12.6 days

Suppose the original amount is 100.00 grams

Step 2: Calculate the percentage left

X = 100 / 2^n

⇒ with X = The amount of sample after 12.6 days

⇒ with n = (time passed / half-life time) = (12.6/14.3)

X = 100 / 2^(12.6/14.3)

X = 54.29

There is 54.29 % sample left after 12.6 days

8 0
2 years ago
Some versions of the periodic table show hydrogen at the top of Group 1A(1) and at the top of Group 7A(17). What properties of h
lawyer [7]

Answer:Hydrogen is placed such because it exhibits some similar characteristics of both group1 and group VII elements.

Explanation:

The reason why hydrogen is similar to group 1 metals:

#It has same valence electron and inorder achieve octet state it can lose that electron and forms H+ ion

#It acts as a good reducing agent similar to group1 metals

#It can also halides

Similarity to halogens:

#hydrogen can also gain one electron to gain noble gas configuration. It can combine with other non metals to form molecules with covalent bonding.

#It exists as diatomin molecule,H2

#Have the same electronegativity nature

#its reaction with other metal

8 0
3 years ago
What is the volume of 45.6g of silver if the density of silver is 10.5g/mL? A. 4.34mL B. 479mL C. 0.23mL
nirvana33 [79]

Explanation:

first you get moles of silver

n=m/M

hence you add no of moles to this equation

c=nv

v=n/c

8 0
3 years ago
Give the symbol for the element with the following orbital diagram:
wel

Answer: 0%

Explanation:

You got a bad grade

5 0
3 years ago
A buffer solution is composed of 1.00 mol of acid and 2.25 mol of the conjugate base. If the p K a of the acid is 4.90 , what is
Gemiola [76]

<u>Answer:</u> The pH of the buffer is 5.25

<u>Explanation:</u>

Let the volume of buffer solution be V

We know that:

\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution}}

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{conjugate base}]}{[acid]})

We are given:

pK_a = negative logarithm of acid dissociation constant of weak acid = 4.90

[\text{conjugate base}]=\frac{2.25}{V}

[acid]=\frac{1.00}{V}

pH = ?

Putting values in above equation, we get:

pH=4.90+\log(\frac{2.25/V}{1.00/V})\\\\pH=5.25

Hence, the pH of the buffer is 5.25

4 0
3 years ago
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