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serious [3.7K]
3 years ago
6

Select correct examples of linear molecules for five electron groups. Select correct examples of linear molecules for five elect

ron groups.
a. BeCl2

b. B e C l 2

c. H2O

d. XeF2 X e F 2

e. H2S H 2 S
Chemistry
1 answer:
Shalnov [3]3 years ago
7 0

Explanation:

Formula to calculate hybridization is as follows.

                Hybridization = \frac{1}{2}[V+N-C+A]

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

So, hybridization of BeCl_{2} is as follows.

              Hybridization = \frac{1}{2}[V + N - C + A]

                                    = \frac{1}{2}[2 + 2]

                                    = 2

Hybridization of BeCl_{2} is sp. Therefore, BeCl_{2} is a linear molecule. There will be only two electron groups through which Be is attached.

Similarly, hybridization of XeF_{2} is calculated as follows.

         Hybridization = \frac{1}{2}[V + N - C + A]

                                    = \frac{1}{2}[8 + 2]

                                    = 5

Therefore, hybridization of XeF_{2} is sp^{3}d. Therefore, [tex]XeF_{2} is also a linear molecule. Though there are three lone pair of electrons present on a xenon atom and it is further attached with fluorine atoms through two electron pairs. Hence, there are in total five electron groups.

Thus, we can conclude that out of the given options XeF_{2} is the correct examples of linear molecules for five electron groups.

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3 years ago
If 125g of KClO3 is heated, what is the total mass of the products?​
Andrej [43]

Given parameters:

Mass of KClO₃  = 125g

Unknown:

Total mass of the products = ?

When  KClO₃ is heated, it thermally decomposes to KCl and O₂ according to the chemical equation below;

               2KClO₃  →  2KCl + 3O₂

All chemical equations obeys the law of conservation of matter and with this regard, we know that the amount of reactants used is the same as that of the product.

The total mass of the products must give us 125g according to this law of conservation of matter.

Now to find the masses of each product,

  1. Find the number of moles of the given reactant:

     Number of moles  = \frac{mass}{molar mass}

  molar mass of  KClO₃  = 39 + 35.5 + 3(16)  = 122.5g/mol

    So number of moles of KClO₃ = \frac{125}{122.5}  = 1.02moles

    2. Now, using this number of moles, find the number of moles of the products using this value;

   2 moles of KClO₃ produced 2 moles of KCl

  1.02 moles of KClO₃ will also produce 1.02moles of KCl

   2 moles of KClO₃ produced 3 moles of O₂

   1.02 moles of KClO₃ will produce   \frac{1.02 x 3} {2} mole = 1.53 moles of O₂

   3. Now find the masses of each product;

Mass  = number of moles x molar mass

  molar mass of KCl  = 39 + 35.5 = 74.5g/mol

  molar mass of O₂  = 16 x 2  = 32g/mol

  Mass of KCl  = 74.5 x 1.02  = 75.99g

  Mass of O₂  = 32 x 1.53 = 48.96g

Total mass of products = mass of KCl + Mass of O₂ = 75.99g + 48.96g

                                        = 124.95g

This value is approximately the same as that of mass of  KClO₃

 

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