Question

Show the formation of NaCl by transfer of electrons

Answer 1

Explanation:

Na=2,8,1

Cl=2,8,7

After forming a ionic bond,

Na becomes +charge and Cl becomes - charge ..

Answer 2

$\sf\purple{Formation\:of\:NaCl\:by\:transfer\:of\:electrons:-}$

Sodium (Na)

• atomic no. = 11
• electronic configuration= 2,8,1

$\sf\pink{↬}$ By losing one electron of its outermost shell , it acquires the inert gas configuration of neon and changes into sodium Ion .

On the other hand,

chlorine(Cl)

• atomic mass= 17
• electronic configuration= 2 ,8 ,7

$\sf\green{↬}$ Accepts one electron released by Sodium to complete its orbit and acquire stable configuration of argon in this process chlorine is converted into chloride ion .

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\put(1,1){\large\bf Na}\put(1.35,1.6){\circle*{0.1}}\put(2,1){ + }\put(3,1){\large\bf Cl}\multiput(3,1.5)(0.3,0){2}{ \times }\multiput(3,0.6)(0.3,0){2}{ \times }\multiput(3.7,1)(0,0.3){2}{ \times }\put(2.6,0.9){ \times }\qbezier(1.4,1.75)(2,3)(2.7,1.4)\qbezier(2.7,1.4)(2.7,1.4)(2.8,1.6)\qbezier(2.7,1.4)(2.7,1.4)(2.5,1.5)\put(4.3,1.25){\vector(1,0){1}}\put(5.5,1.1){ \large\bf [Na^+] }\put(7.5,1.1){ \large\bf Cl }\multiput(7.4,1.6)(0.3,0){2}{ \times }\multiput(7.4,0.7)(0.3,0){2}{ \times }\multiput(8.1,1)(0,0.3){2}{ \times }\multiput(7.1,1)(0,0.3){1}{ \times }\put(6.8,1.1){ \Bigg[ }\put(8.5,1.1){ \Bigg]^- }\put(7.25,1.5){\circle*{0.15}}\end{picture}$

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$\sf \large \red{Some\:Properties\:of\:NaCl!!}$

• Sodium chloride is a white crystalline solid having a density of 2.17 g/ml.

• It melts at 1080 K (807°C) and boils at 1713 K (1440°C).

• Pure sodium chloride is non-hygroscopic, but behaves as hygroscopic due to the impurities of CaCl2 and MgCl2 in it.

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