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In order to do hybridization of H2O, we consider the central atom in the molecule. The central atom we look into is oxygen, so we do hybridization in context to oxygen.Â

While doing hybridization of oxygen, the oxygen has 2 sigma bonds + 2 lone pairs of electrons in addition to Hydrogen atom thus it occupies 4 orbitals. As these 4 orbitals are developed, they occupy 1 s and 3 p orbitals. Thus, water exhibits sp3 hybridization.

(Image to be added soon)

Scientists in the past were bewildered by the shapes and types of covalent molecules that did not harmonize with bonding theories of the time. Through much conceptual and experimental work, they formulated the concept of atomic orbital hybridization which offers insight when coupled with other theories of what is noticed in real molecules.

Letâ€™s look into hybridization and formation of H2O based on the Geometry of Hybridization.

The sp3 is oxygen which is hybridized into H2O molecules. Two hybrid orbitals are equipped by lone pairs whereas the two are occupied in bonding with Hydrogen atoms. Seeing that the lone pairs do not contribute towards the geometry of a molecule, hence H2O bears an angular geometry.Â

The combination of (lone pair-lone pair) repulsion is in excess of lone pair-bond pair or (bond pair-bond pair) repulsion. Therefore, the angle between H-O-H is 104.5Â°, which is significantly less than the ideal tetrahedral angle of 109Â°28â€².

Since H_{2}O molecules consist of 2 lone pairs and 2 bond pairs, thus the steric number of the central atom (O) is 2+2 given that the steric no. = lone pairs + bond pairs apart from the odd electron species as well the stereochemically inactive lone pairs.

Such as now, O has sp^{3} hybridization and a steric number 4, tetrahedral arrangement of hybrid orbitals (geometry) and angular shape (visual impression to the human eye when is visible practically by unique techniques like AFM). The measurement of bond angle HO^{H} = 104.5Â°.

Ultimately, the total number of valence shell electrons is 6+1+1 = 8, which is our 4pairs. Of these 4 pairs, two are the lone pair of oxygen and two are the bond pairs with hydrogen forming the â€˜Vâ€™ shape.

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Example 1: Brass alloys have the following compositions:

(i). 30% zn, 70% cu

(ii). 30% cu, 70% zn

(iii). 50% zn, 50% cu

(iv). 40% zn, 60% cu

Solution 1:

Brass is an alloy in proportions of zinc and copper. It actually comprises 66% copper and 34% zinc that can vary slightly in order to obtain varying electrical and mechanical properties. However, since in our question there is no option, the approximate value can be established 30% zinc and 70% copper.

We have considered the estimated value of zinc as 30 and since it is closer to 34 and copper value is closer to 66 (70%), we have got our answer in the option (i).

Thus, the answer is option (i) i.e. 30% zn, 70% cu.

Example 2: Select the correct statement(s) among the following:

A. m= l+2, where â€˜mâ€™ and â€˜lâ€™ are magnetic and azimuthal quantum numbers.

B. The total number of orbitals in a shell having principal quantum number 'n' is n2

C. The total number of subshells in the nth energy level is n.

D. The highest number of electrons in a subshell is provided by the mathematical equation (4l+2).

Solution 2:

The statements (B), (C) & (D) are correct. Let's explore the explanation of the correct statements.

A.Â m=âˆ’l,....âˆ’1,0,+1, +2. +3 .....+l. Therefore, m=2l+1 where m and l represent magnetic and azimuthal quantum numbers. Therefore, when â€˜lâ€™ is equivalent to 3m = âˆ’3,âˆ’2,âˆ’1,0,+1,+2,+3. Hence, there come to be 7 values in a total of m 2l+1= 2(3)+1=7

B. The total number of orbitals in a shell having the principal quantum number 'n' is n2 Therefore, for n=1, the total number of orbitals =12 =1. It comprises 1s orbital. While For n=2, the total number of orbitals =. It basically comprises (2s), (2px), (2py) and (2pz) orbitals. For n=3, the total number of orbitals =. It consists of (3s), (3px), (3py)â€‹, (3pz), (dxy),(dxz),(dyz), (dx2âˆ’y2)and (dZ2)orbitals.

C. The total number of subshells in the nth energy level is â€˜nâ€™. Therefore, for n=1, there is only 1 subshell which exists. It typically consists of [s] orbital. For n=2, there exist 2 subshells. It has [p] and [s] orbitals. For n=3, there are 3 subshells that exist. It consists of [s], [p] and [d] orbitals.

D. The highest number of electrons present in a subshell is provided by the mathematical equation (4l+2). Therefore, [s] subshell having l=0 has (4l+2)=[4{0}+2]=2 electrons. [P] subshell with l=1 is with (4l+2) = (4(1)+2)= 6 electrons. The [d] subshell has l=2 having {4l+2}={4(2)+2)=10 electrons. f subshell has l=3 has (4l+2)=(4(3)+2) =14 electrons.

FAQ (Frequently Asked Questions)

1. What is Meant by the SP^{3} Hybridization of Water?

The valence orbitals of an atom encompassed by a tetrahedral arrangement of lone pairs and bonding pairs having a set of four sp^{3} hybrid orbitals are termed as the hybridization of H_{2}O. The hybrids are an outcome of the blending of one [s] orbital and all three [p] orbitals that generate four similar sp^{3} hybrid orbitals. Refer to the below image to understand well about the mechanism of sp^{3} hybridization in water. If you notice, then you will find that each of these hybrid orbitals is pointing in a different corner of a tetrahedron.Â Â Â Â Â Â Â

(Image to be added soon)

2. Are There Any Criteria to Observe a Particular Type of Hybridization?

Yes. Following are the set of rules that are conjectured in order to establish to understand the type of hybridization in a compound or an ion.

Calculate the total number of lone pairs of electrons

Calculate the number of valence electrons.

Calculate the number of octet or duplex

Evaluate the total number of used orbital = Number of duplex or octet + Number of lone pairs of electrons

In case of no lone pair of electrons, then the H_{2}O hybridization and geometry of orbitals and molecules will be different.