Question

determine the energy required to dissociate a NACL molecule into NA and CL atom given that electron affinity of CL is 3.70 cv the first ionization potential of NA is 5.14 ev and separation of NACL at equilibrium is 0.236 nm



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Answer 1

Answer:

Quantum mechanics has been extraordinarily successful at explaining the structure and bonding in molecules, and is therefore the foundation for all of chemistry. Quantum chemistry, as it is sometimes called, explains such basic questions as why H2O molecules exist, why the bonding angle between hydrogen atoms in this molecule is precisely 104.5°, and why these molecules bind together to form liquid water at room temperature. Applying quantum mechanics to molecules can be very difficult mathematically, so our discussion will be qualitative only.

As we study molecules and then solids, we will use many different scientific models. In some cases, we look at a molecule or crystal as a set of point nuclei with electrons whizzing around the outside in well-defined trajectories, as in the Bohr model. In other cases, we employ our full knowledge of quantum mechanics to study these systems using wave functions and the concept of electron spin. It is important to remember that we study modern physics with models, and that different models are useful for different purposes. We do not always use the most powerful model, when a less-powerful, easier-to-use model will do the job.

Types of Bonds

Chemical units form by many different kinds of chemical bonds. An ionic bond forms when an electron transfers from one atom to another. A covalent bond occurs when two or more atoms share electrons. A van der Waals bond occurs due to the attraction of charge-polarized molecules and is considerably weaker than ionic or covalent bonds. Many other types of bonding exist as well. Often, bonding occurs via more than one mechanism. The focus of this section is ionic and covalent bonding.

Ionic bonds

The ionic bond is perhaps the easiest type of bonding to understand. It explains the formation of salt compounds, such as sodium chloride, NaCl. The sodium atom (symbol Na) has the same electron arrangement as a neon atom plus one 3s electron. Only 5.14 eV of energy is required to remove this one electron from the sodium atom. Therefore, Na can easily give up or donate this electron to an adjacent (nearby) atom, attaining a more stable arrangement of electrons. Chlorine (symbol Cl) requires just one electron to complete its valence shell, so it readily accepts this electron if it is near the sodium atom. We therefore say that chlorine has a large electron affinity, which is the energy associated with an accepted electron. The energy given up by the chlorine atom in this process is 3.62 eV. After the electron transfers from the sodium atom to the chlorine atom, the sodium atom becomes a positive ion and the chlorine atom becomes a negative ion. The total energy required for this transfer is given by

Etransfer=5.14eV−3.62eV=1.52eV.

The positive sodium ion and negative chloride ion experience an attractive Coulomb force. The potential energy associated with this force is given by

Explanation:

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Answer 2

Answer:

The energy required to dissociate a $\mathrm{NaCl}$ molecule into $\mathrm{Na}$ and $\mathrm{Cl}$ atom is $-4.66$ or $4.66 \mathrm{eV}$.

Explanation:

Given: The electron affinity of $\mathrm{Cl}$ is $3.70\mathrm{eV}$, the first ionization potential of $\mathrm{Na}$ is $5.14 \mathrm{eV}$, and the separation of $\mathrm{Na} \mathrm{Cl}$ at equilibrium is $0.236 \mathrm{~nm}$.

To Find: The energy required to dissociate a $\mathrm{NaCl}$ molecule into $\mathrm{Na}$ and $\mathrm{Cl}$ atom.

Solution:

$\text{Electrical potential energy} =\frac{-K e^{2}}{r}$ $=-\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{0.236 \times 10^{-9}} J$

$\mathrm{Cl} = 3.70\mathrm{eV}$

$\mathrm{Na} = 5.14\mathrm{eV}$

Now, determine the energy required to dissociate $\mathrm{NaCl}$ by the below given formula,

$\mathrm{NaCl} = \text {electrical potential energy} + \text {electron affinity} + \text {ionization potential}$

         $=\frac{-K e^{2}}{r} + \mathrm {Cl} + \mathrm {Na}$

         $=-\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{0.236 \times 10^{-9}} J+(-3.7 \mathrm{eV})+5.14 \mathrm{eV}$

         $=-97.627 \times 10^{-20} J+1.44 \mathrm{eV}$

As we know that, $1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}$

So,  $-97.627 \times 10^{-20} \mathrm{~J}=-6.1 \mathrm{eV}$

                                     $=-6.10 \mathrm{eV}+1.44 \mathrm{eV}$

                                     $=-4.66 \mathrm{eV}$

Hence, the energy required to dissociate a $\mathrm{NaCl}$ into $\mathrm{Na}$ and $\mathrm{Cl}$ atoms is $-4.66 \mathrm{eV}$ or $4.66 \mathrm{eV}$.