Of the molecules below, the bond in is the most polar. of the molecules below, the bond in is the most polar. hi hbr h2 hf hcl

The compound with the highest electronegativity is HF, Fluor being the most electronegative element that exists.

Explanation:

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To calculate the electronegativity of the compounds, the difference in electronegativity of each of the elements is calculated.

The data we need are:

Fluorine = 4.0

Chlorine = 3.16

Hydrogen = 2.1

Bromine = 2.96

Iodo = 2.66

The electronegativity of the compounds is:

HI: 2.66-2.1 = 0.56

HBr: 2.96-2.1 = 0.86

H2: 2.1-2.1 = 0

HF: 4-2.1 = 1.9

HCl: 3.16-2.1 = 1.06

After the calculations, we can conclude that the compound with the highest electronegativity is HF, Fluor being the most electronegative element that exists.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

increased stability in the completion of fluorine's outer shell.