Question

describe the structure of and geometry of H2O and NH3 molecules​

Answer 1

Geometry of $H_2O$ and $NH_3$ molecules​ are bent and trigonal pyramidal respectively.

Explanation:

Formula used  :

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

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

1. For $H_2O$

$\text{Number of electrons}=\frac{1}{2}[6+2-0+0]=4$

The number of electrons are 4  that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral. But as there are two atoms around the central oxygen atom, the third and fourth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be bent shape.

2. For $NH_3$

$\text{Number of electrons}=\frac{1}{2}[5+3-0+0]=4$

The number of electrons are 4  that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral. But as there are three atoms around the central oxygen atom, the fourth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be trigonal pyramidal.

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Answer 2

Answer:

Water molecule (H₂0):-

• Shape- Bent
• Geometry- Tetrahedral

Ammonia molecule (NH₃):-

• Shape - Trigonal pyramidal
• Geometry - Distorted tetrahedral.

Explanation:

• The shape of molecules may be predicted using the Valence Shell Electron Pair Repulsion Theory (VSEPR theory).
• The foundation of the VSEPR theory is the idea that all atoms have a repulsion between their pairs of valence electrons, and that atoms naturally organize themselves to reduce this repulsion as much as possible.
• The order of repulsion between electron pairs may be described using the VSEPR theory as follows:

  Lone pair-Lone pair > Lone pair-Bond pair > Bond pair-Bond pair repulsions.

• We may use the following formula to determine the structure of a molecule:

        $n=\frac{(V+M-C+A)}{2}$

Where, V ⇒ No. of valence electrons of the central atom

M ⇒ No. of monovalent atoms attached to central atom

C ⇒ No. of cations

A ⇒ No. of anions

For Water (H₂O) molecule:-

• The oxygen atom, which is the centre atom, is known to have six valence electrons. Thus, V=6.
• We also know that in water molecules, there are two hydrogen atoms connected to the oxygen. Thus, M=2.
• Water molecules don't contain either anion or cation, hence the values of A and C are both zero.
• Now, if we put these numbers into the above equation, we get:

         $n=\frac{(6+2)}{2}=4$

• Thus, there are four no.orbitals or electron pairs involved is 4.
• Since oxygen has six valences, it shares two electrons—one with each hydrogen—and the other four electrons form the two oxygen lone pairs.
• A water molecule therefore consists of two lone pairs and two bond pairs. Because of the repulsion between lone pair-lone pair, it takes on a bent or V-shape.
• Furthermore, the oxygen atom is sp³ hybridised in the water molecule, resulting in a tetrahedral geometry.
• The H-O-H bond angle is 104.5°, however, slightly less than the tetrahedral bond angle of 109° because of the lone-pair lone-pair repulsion.

For NH₃ (Ammonia) molecule:-

• The nitrogen atom, which is the centre atom, is known to have five valence electrons. Thus, V = 5
• We also know that in ammonia molecules, there are three hydrogen atoms connected to the oxygen. Thus, M = 3.
• Ammonia molecules don't contain either anion or cation, hence the values of A and C are both zero.
• Now, if we put these numbers into the above equation, we get:
•          $n=\frac{(5+3)}{2}=4$

• Thus, there are four no.orbitals or electron pairs involved is 4.
• Since nitrogen has five valences, it shares three electrons—one with each hydrogen—and the other two electrons form the one oxygen lone pairs.
• An ammonia molecule therefore consists of one lone pair and three bond pairs. Therefore, ammonia has a trigonal pyramidal shape.  
• Furthermore, the oxygen atom is sp³ hybridised in the water molecule, resulting in a tetrahedral geometry.
• Since the nitrogen atom possesses a single lone pair of non-bonding electrons that repels the bonding orbitals, the bond angle between nitrogen, and hydrogen atoms (H-N-H) is 107°, giving it  structure distorted tetrahedral geometry.

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Answer 3

Answer:

Water molecule (H₂0):-

• Shape- Bent
• Geometry- Tetrahedral

Ammonia molecule (NH₃):-

• Shape - Trigonal pyramidal
• Geometry - Distorted tetrahedral.

Explanation:

• The shape of molecules may be predicted using the Valence Shell Electron Pair Repulsion Theory (VSEPR theory).
• The foundation of the VSEPR theory is the idea that all atoms have a repulsion between their pairs of valence electrons, and that atoms naturally organize themselves to reduce this repulsion as much as possible.
• The order of repulsion between electron pairs may be described using the VSEPR theory as follows:

  Lone pair-Lone pair > Lone pair-Bond pair > Bond pair-Bond pair repulsions.

• We may use the following formula to determine the structure of a molecule:

        $n=\frac{(V+M-C+A)}{2}$

Where, V ⇒ No. of valence electrons of the central atom

M ⇒ No. of monovalent atoms attached to central atom

C ⇒ No. of cations

A ⇒ No. of anions

For Water (H₂O) molecule:-

• The oxygen atom, which is the centre atom, is known to have six valence electrons. Thus, V=6.
• We also know that in water molecules, there are two hydrogen atoms connected to the oxygen. Thus, M=2.
• Water molecules don't contain either anion or cation, hence the values of A and C are both zero.
• Now, if we put these numbers into the above equation, we get:

         $n=\frac{(6+2)}{2}=4$

• Thus, there are four no.orbitals or electron pairs involved is 4.
• Since oxygen has six valences, it shares two electrons—one with each hydrogen—and the other four electrons form the two oxygen lone pairs.
• A water molecule therefore consists of two lone pairs and two bond pairs. Because of the repulsion between lone pair-lone pair, it takes on a bent or V-shape.
• Furthermore, the oxygen atom is sp³ hybridised in the water molecule, resulting in a tetrahedral geometry.
• The H-O-H bond angle is 104.5°, however, slightly less than the tetrahedral bond angle of 109° because of the lone-pair lone-pair repulsion.

For Ammonia (NH₃) molecule:-

• The nitrogen atom, which is the centre atom, is known to have five valence electrons. Thus, V = 5
• We also know that in ammonia molecules, there are three hydrogen atoms connected to the oxygen. Thus, M = 3.
• Ammonia molecules don't contain either anion or cation, hence the values of A and C are both zero.
• Now, if we put these numbers into the above equation, we get:
•          $n=\frac{(5+3)}{2}=4$

• Thus, there are four no.orbitals or electron pairs involved is 4.
• Since nitrogen has five valences, it shares three electrons—one with each hydrogen—and the other two electrons form the one oxygen lone pairs.
• An ammonia molecule therefore consists of one lone pair and three bond pairs. Therefore, ammonia has a trigonal pyramidal shape.  
• Furthermore, the oxygen atom is sp³ hybridised in the water molecule, resulting in a tetrahedral geometry.
• Since the nitrogen atom possesses a single lone pair of non-bonding electrons that repels the bonding orbitals, the bond angle between nitrogen, and hydrogen atoms (H-N-H) is 107°, giving it  structure distorted tetrahedral geometry.

#SPJ3