Question

Statement 1: There is an ionic bond in sodium chloride compounds.
Statement 2: Some electronegativity is shown by sodium and chlorine atoms.
A. False, false
B. True, false
C. True, true
D. False, true
E. Statement 1 and statement 2 are true and statement 2 is the correct explanation of statement 1.

Answer 1

When we look into the structure of $\,NaCl\,$, the six closest neighbours of each negatively charged $\,C{l^ - }\,$ anion are positively charged $\,N{a^ + }$ cations in the sodium chloride crystalline system. Negatively charged $\,C{l^ - }\,$ anions are the six closest neighbours to each positively charged $\,N{a^ + }\,$ cation.

Complete step by step answer:
Let’s analyse the first statement. For that, let us look into the reaction to form $\,NaCl\,$;
$\,2Na(solid) + \,C{l_2}(gas) \to 2NaCl(solid)\,$
In this reaction, we can see the following changes happening;
The atoms in solid $\,Na\,$ are isolated, requiring energy.
Each $\,C{l_2}\,$ molecule is divided into atoms of $\,Cl\,$, which require energy.
To produce a $\,C{l^ - }\,$ion that releases energy, an electron is added to each Cl atom.
In a crystal lattice, all $\,N{a^ + }\,$ cations and one $\,C{l^ - }\,$ anion are assembled in a $\,1:1\,$ ratio to produce $\,NaCl\,$, which releases a very large amount of lattice energy.
From the above explanations, we can understand that, during the formation of sodium chloride, ions are formed and a bond is formed between the compound by transfer of electrons. This bond is known as an ionic bond and hence the first statement is true.
Now, let's look into the second statement;
Electronegativity is the pushing power of another element that an element has on an electron. As we saw earlier, sodium chloride has an ionic bond. However, Sodium and chlorine atoms display distinct electronegativity. Sodium is electro-positve, while chlorine is electronegative. Hence the second statement is not true.

But after the bond formation, an electron has been moved from chlorine to sodium. Sodium has an electronegativity of $\,1.0\,$ and the electronegativity of chlorine is $\,3.0\,$ . That is a difference of $\,2.0(3.0 - 1.0)\,$ in electronegativity, making the bond between the two atoms very, very polar. The general rule of thumb is that if the difference in electronegativity between the two components is greater than around $\,1.5\,$ , the bond would be ionic.

So, the correct answer is Option B.

Note: As electronegativity number is higher, higher is the pull. If the difference between electronegativities is less than $\,0.7\,$ , then the bond is neutral covalent bond. The mutual electron spends more time close to the more electronegative part if it is bigger, rendering it a polar covalent bond. The electron only spends time around the electronegative part making it an ionic bond when it is greater than about $\,1.5\,$ as we saw above.