the bond between which pair of elements is the least ionic in character ?

(1) H-F
(2)H-Cl
(3) H-I
(4) H-O
please help

To determine which pair of elements has the least ionic bond character, we need to compare their electronegativity values. Electronegativity is the tendency of an atom to attract a bonding pair of electrons.

The greater the difference in electronegativity, the more ionic the bond character becomes. Lower electronegativity differences indicate covalent bond character.

Let's find the electronegativity values for each element using a periodic table or electronegativity chart:

(1) H-F:
Electronegativity of Hydrogen (H) is 2.2
Electronegativity of Fluorine (F) is 3.98

The electronegativity difference between H and F is 1.78.

(2) H-Cl:
Electronegativity of Chlorine (Cl) is 3.16

The electronegativity difference between H and Cl is 0.96.

(3) H-I:
Electronegativity of Iodine (I) is 2.66

The electronegativity difference between H and I is 0.56.

(4) H-O:
Electronegativity of Oxygen (O) is 3.44

The electronegativity difference between H and O is 1.24.

Now, comparing the electronegativity differences, we find that the pair with the smallest electronegativity difference is H-I, which means it has the least ionic character.

So, the correct answer is (3) H-I.

To determine the bond that is the least ionic in character among the given options, we need to compare the electronegativity difference between the two elements.

Electronegativity is a measure of an atom's ability to attract electrons towards itself in a chemical bond. The greater the electronegativity difference between two atoms, the more ionic the bond becomes.

Now, let's find the electronegativity difference for each option:

1) H-F:
- Electronegativity of H = 2.1
- Electronegativity of F = 3.98
Electronegativity difference = 3.98 - 2.1 = 1.88

2) H-Cl:
- Electronegativity of H = 2.1
- Electronegativity of Cl = 3.16
Electronegativity difference = 3.16 - 2.1 = 1.06

3) H-I:
- Electronegativity of H = 2.1
- Electronegativity of I = 2.66
Electronegativity difference = 2.66 - 2.1 = 0.56

4) H-O:
- Electronegativity of H = 2.1
- Electronegativity of O = 3.44
Electronegativity difference = 3.44 - 2.1 = 1.34

From the calculations, we can conclude that the bond between hydrogen (H) and iodine (I) (option 3: H-I) has the least electronegativity difference and hence, the least ionic character among the given options.

Look up the electronegativity (EN). The ionic character of the bond can be estimated by subtracting the EN of each element. For example, H is 2.1 and F is 4.0, the difference is 1.9. Do the same for the others. The larger the the difference, the more ionic in character is the bond.