Predict the nature of the bond in the following molecules. (i) NaCl (b) NaBr (c ) Nal (iv) NaF (v) NaH

Predict the nature of the bond in the following molecules. (i) NaCl (

1.	Predict the nature of the bond in the following molecules. (i) NaCl (b) NaBr (c ) Nal (iv) NaF (v) NaH
Answer» Solution :(i) NaCl : Electronegativity difference between na and Cl is , 3 -1= 2 . That is grater than 1.7, therefore NaCl is iionic bond. (ii) NaBr : Electronegativity difference between Na and Br is , 2.8 - 1 = 1.8. That is greater than 1.7, therefore NaBr is ionic bond. (iii) Nal : Electronegativity difference between Na and I is, 2.5 - 1 = 1.5. That is lesser than 1.7, therefore NAI is COVALENT bond. (IV) NaF : Electronegativity difference between Na and F is , 4-1 = 3. That is greater than 1.7, therefore NaF is ionic bond. (V) NaH : Electronegativity difference between Na and H is, 2.1 -1 = 1.1 That is lesser than 1.7, therefore NaH is covalent bond.