Question

With reference to the formation of compounds from atoms by electron transfer-electrovalency, state the basic steps in the conversion of sodium and chlorine atoms to sodium and chloride ions leading to the formation of the compound- sodium chloride [electronic configuration of: Na= 2,8,1 & Cl = 2,8,7]

Answer 1

Answer:

Sodium is a silver-colored metal which is soft enough to cut with a knife. It is an extremely reactive metal, and is always found naturally in ionic compounds, not in its pure metallic form. Pure sodium metal reacts violently (and sometimes explosively) with water, producing sodium hydroxide, hydrogen gas, and heat:

2Na(s) + 2H2O(l) ——> 2NaOH(aq) + H2(g)

Chlorine is a poisonous, yellow-green gas, with a very sharp odor, and was used in gas warfare during World War I.

Sodium and chlorine react with each other, however, to produce a substance that is familiar to almost everyone in the world: sodium chloride, or table salt:

2Na(s) + Cl2(g) ——> 2NaCl(s)

It is easy to see why this reaction takes place so readily when we look at it on an atomic level: sodium has one electron in its outermost (valence) shell, while chlorine has seven electrons in its valence shell. When a sodium atom transfers an electron to a chlorine atom, forming a sodium cation (Na+) and a chloride anion (Cl-), both ions have complete valence shells, and are energetically more stable.

The reaction is extremely exothermic, producing a bright yellow light and a great deal of heat energy.

In the following demonstrations, a 2.5 liter bottle is filled with chlorine gas. A coating of sand on the bottom of the bottle absorbs some of the heat energy produced during the reaction, and prevents it from breaking. A small piece of freshly-cut metallic sodium is placed in the flask, and then a small amount of water is added, which reacts with the sodium and causes it to become hot. The hot sodium then reacts with the chlorine, producing a bright yellow light, a great deal of heat energy, and fumes of sodium chloride, which deposits on the walls of the bottle.

In the first video clip, the sodium flares up almost immediately upon reaction with the water, and "burns out" quickly. (Don't blink, or you'll miss it.) In the second, water is added twice, to produce one short flash, followed by a much longer one. (This reaction can also be done with molten sodium, but I've never been brave enough to try that.)

Answer 2

❁ $\sf{ \pink{\Huge{Solution}}}$❁

The electronic configuration of a sodium atom is 2,8,1. It has one electron in excess of the stable electronic configuration of the nearest noble gas i.e., neon, ( 2, 8 ). Therefore, an atom of sodium shows a tendency to give up the electron from its outermost shell, so as to acquire stable electronic configuration of neon.

$\sf{ Na \longrightarrow {Na}^{ + } + {1e}^{ - } }$

Here,

◉ Na - ( 2, 8, 1 ) atom

◉ $\sf{ {Na}^{ + } - (2, \: 8) \: cation}$

However, after giving up one electron, the sodium atom is no more electrically neutral. It has 11 protons in its nucleus but only 10 electrons revolving around it. Therefore, it has a net positive charge of +1. This positively charged atom is called sodium ion and is written as $\sf{ {Na}^{ + } }$ and its electronic configuration resembles that of the noble gas - neon.

The electronic configuration of chlorine is 2,8,7.

It has an electronic configuration with one electron less than that of the nearest noble gas i.e., argon ( 2, 8, 8 ). Therefore, the chlorine atom shows a tendency to acquire an electron to attain octet in its outermost shell.

$\sf{Cl + {1e}^{ - } \longrightarrow {Cl}^{ - }}$

Here,

◉ Cl - ( 2, 8, 7 ) atom

◉ $\sf {Cl}^{ - } - (2, \: 8, \: 8) \: anion$

An atom of chlorine is electrically neutral and it has contains 17 protons in its nucleus and 17 electrons revolving around nucleus. But, after acquiring an electron from the sodium atom, the chlorine atom does not remain electrically neutral. It has one electron more than the number of protons in its nucleus and therefore it has a charge of -1 represented as $\sf {cl}^{ - }$ i.e., chloride ion and its electronic configuration resembles that of the noble gas - argon.

Thus, when an atom of sodium combines with an atom of chlorine, "one" electron is transferred from the sodium atom to the chlorine atom, resulting in the formation of a sodium chloride molecule.

The cation $\sf {Na}^{ + }$ and anion $\sf {Cl}^{ - }$ are attached towards each other due to opposite electrical charges or coulomb force existing between them and form an ionic compound.

Refer the attachment above for orbit structure of electrovalent or ionic bonding in sodium chloride.

━━━━━━━━━━━━━━━━━━━━━━━━