Chemistry, 25.12.2021 23:30, chels6058

Define electronegativity. A neutral atom has high electronegativity. Describe what happens to this atom during ionic bond formation

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Electronegativity is basically how strongly an atom is attracted to electrons and a neutral atom has a high electronegativity because it has yet to achieve its stable state of 8 valence electrons.

During an ionic bond formation, an atom gives or takes electrons to have a valence shell of 8 electrons. For example, a simple ionic compound such as NaCl has Na (Sodium) which has only one valence electron and wants to get rid of it and Cl (Chlorine) which has 7 electrons and needs just one. So Chlorine takes Sodium's one electron and they form an ionic compound.

answered: miguel454545

Define Electronegativity :

Electronegativity, symbolized as χ, is the tendency of an atom to attract shared electrons to itself.

A neutral atom has high electronegativity. Describe what happens to this atom during

ionic bond formation.

Thus, the term “ionic bonding” is given when the ionic character is greater than the covalent character – that is, a bond in which a large electronegativity difference exists between the two atoms, causing the bonding to be more polar (ionic) than in covalent bonding where electrons are shared more equally.

2. What happens to a neutral atom with high electronegativity during an ionic bond?

When ionic bonds form, a metal donates one or more electrons, due to having a low electronegativity, to form a positive ion or cation. The non-metal atom has a high electronegativity, and therefore readily gains electrons to form a negative ion or anion.

Explanation:

What happens to a neutral atom during ionic bond formation?

A cation (a positive ion) forms when a neutral atom loses one or more electrons from its valence shell, and an anion (a negative ion) forms when a neutral atom gains one or more electrons in its valence shell. ... The properties of ionic compounds shed some light on the nature of ionic bonds.

How does electronegativity affect ionic bond formation?

Electronegativity of the atoms involved in a compound affects the ionic bonds. More electronegative elements have the tendency to form higher ionic character bonds with other elements. Elements that have high electronegativity, will form a bond with any atom that has more ionic character.

answered: Guest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.