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uysha [10]
1 year ago
13

If the symbol X represents a central atom, Y represents outer atoms, and Z represents lone pairs on the central atom, the struct

ure A central X atom has two lone pairs. Two Y atoms are attached to X with single bonds, could be abbreviated as XY2Z2.
Classify these structures by the hybridization of the central atom.

HYBRIDIZATIONS:
1. sp
2. sp2
3. sp3
4. sp3d
5. sp3d2

ABREVIATIONS:
1) XY2
2) XY3
3) XY4
4) XY5
5) XY6
6) XY2Z3
7) XY2Z
8) XY3Z2
9) XY4Z
10) XY4Z2
11) XY3Z
12) XY5Z
13) XY2Z2
Chemistry
1 answer:
MA_775_DIABLO [31]1 year ago
5 0

Answer:

1. sp  = XY₂

2. sp²  = XY₂Z, XY₃

3. sp3³ = XY₄, XY₂Z₂, XY₃Z

4. sp³d  = XY₅, XY₂Z₃, XY₃Z₂, XY₄Z

5. sp³d² = XY₆, XY₄Z₂, XY₅Z

Explanation:

this is quite dicey, so it should be looked into carefully.

we would classify each of the abbreviation according to their  hybridization and it electron domain.

⇒ sp hybridization = XY₂

in this, we can see that the central atom X is bonded to two outer atoms Y.

this makes the no of hybrid orbitals and the no of sigma bonds both 2.

electron domain = 2.

⇒ sp² hybridization = XY₂Z, XY₃

Here we can see the central atom X bonded with three outer atoms Y in XY₂Z and in XY₃. For XY₂Z molecule, the no of sigma bonds is 2 and the no of hybrid orbitals is 3. While for XY₃ molecule, the no of sigma bonds is 3 while the no of hybrid orbital is 3.

electron domain = 3.

⇒ sp³ hybridization = XY₄, XY₂Z₂, XY₃Z

for XY₄ molecule, the central atom X is bonded with four outer atoms Y. It has 4 numbers of both the sigma and orbital atoms.

In XY₂Z₂, the central atom X is bonded to 2 outer atoms Y, and has 2 lone pairs Z. From this, the no of hybrid orbitals is 4 and the no of sigma bonds is 2, with 2 lone pairs causing the sp³ hybridization.

⇒ sp³d hybridization = XY₅, XY₂Z₃, XY₃Z₂, XY₄Z

for all the molecules listed above, the sum of both the lone pairs and the outer atoms both give a total of 5, hence have the sp³d structure, viz;

XY₂Z₃:

total electron domain = 2+3 = 5  

XY₃Z₂:

total electron domain = 2+3 = 5

XY₄Z:

total electron domain = 1+4 = 5

⇒ sp³d² hybridization = XY₆, XY₄Z₂, XY₅Z

Same thing goes for the above molecule, where the sum of both the outer atoms and the lone pairs gives a total of 6 as can be seen in the example below.

XY4Z2:

total electron domain = 2+4 = 6

cheers, i hope this helps.

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Answer:

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Answer :

(1) The hybridization of central atom beryllium in BeCl_2  is, sp

(2) The hybridization of central atom nitrogen in NO_2  is, sp^2

(3) The hybridization of central atom carbon in CCl_4  is, sp^3

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Explanation :

Formula used  :

\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]

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N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

Now we have to determine the hybridization of the following molecules.

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\text{Number of electrons}=\frac{1}{2}\times [2+2]=2

The number of electron pair are 2 that means the hybridization will be sp and the electronic geometry of the molecule will be linear.

(2) The given molecule is, NO_2

\text{Number of electrons}=\frac{1}{2}\times [4]=2

If the sum of the number of sigma bonds, lone pair of electrons and odd electrons present is equal to three then the hybridization will be, sp^2.

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(3) The given molecule is, CCl_4

\text{Number of electrons}=\frac{1}{2}\times [4+4]=4

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(4) The given molecule is, XeF_4

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Bond pair electrons = 4

Lone pair electrons = 6 - 4 = 2

The number of electrons are 6 that means the hybridization will be sp^3d^2 and the electronic geometry of the molecule will be octahedral.

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