The polarity of the water molecule aids in dissolving compounds. This is demonstrated by all but which statement?A)The water mol
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<h3><u>1. Lewis Structure</u></h3>There are three N-H bonds and one lone pair on the nitrogen atom in the Lewis structure of NH3. On hydrogen atoms, there are no lone pairs that can only hold two electrons.
The lewis structure is as shown in the diagram.
<h3><u>2. Perspective Drawing</u></h3>A molecule is shown from a viewpoint with its atoms' bonds pointing either in your direction (bolded wedge) or away from you (hash wedge).
The perspective drawing of NH3 is given here
<h3><u>3.Number of atoms bonded to central atom </u></h3>Nitrogen is the central atom in the Lewis structure of NH3, which also contains one lone pair and is bound to the three hydrogen atoms.
<h3><u>4.Number of non-bonding electron pairs on the central atom</u></h3>Three bond pairs and two lone pairings exist. This is due to the fact that nitrogen contains five electrons in its outer shell, three of which are bound to hydrogen atoms and two of which are free.
<h3><u>5. Electronic geometry:</u></h3>The electronic geometry of nitrogen is based on a tetrahedral arrangement of electron pairs, and ammonia likewise possesses four electron pairs. There is only one lone pair because there are only three connected groupings. But since the lone pairs are "invisible," the ammonia has a pyramidal form.
<h3><u>6. Molecular geometry with ideal bond angles</u></h3>The molecular geometry of ammonia it has a trigonal pyramidal or distorted tetrahedral structure.
The bond angle in ammonia is less than the standard 109.5⁰. The bond angle is 107⁰
<h3><u>7.Hybridization of central atom</u></h3>The core nitrogen (N) atom in the NH3 molecule has a steric number of (3 + 1) = 4, which causes Sp3 hybridization.
<h3><u>8. Polarity:</u></h3>Because of its asymmetrical form, a trigonal pyramidal structure, and the different electronegativities of N(3.04) and H(2.2), the NH3 (ammonia) molecule is polar in nature .
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According to Charles' Law the volume of an ideal gas is directly proportional to its absolute temperature in Kelvin keeping the pressure constant.
V∝ T, P is constant
where V, T and P are volume, temperature and pressure
=
where V₁, T₁, V₂ and T₂ are initial volume, initial temperature, final volume and final temperature.