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Bond [772]
3 years ago
10

Which best describes the electron behavior in metallic bonding?

Chemistry
1 answer:
aliya0001 [1]3 years ago
4 0

They are delocalized.

When you think of a bond, you probably think of the "ball and stick" model of bonding where there are two atoms connected by one or more bonds. The bonds between those two atoms connect only those two atoms. Because the bond just stays put, it's referred to as localized covalent bonding.

However, metals do different stuff. If you've got a metal, you've got a great big bunch of atoms all held together in a great big bunch of valence electrons. Rather than two valence electrons forming a single bond between two metal atoms, you have all of the valence electrons from every atom in the element moving around all over the place in delocalized bonds. If you imagine putting a bunch of pieces of Velcro into a pile of yarn, the electrons are represented by the disordered yarn and the metal atoms are represented by the Velcro.

OK.

Got for - M<span>isterguch</span>


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The equilibrium constant for the reaction AgBr(s) Picture Ag+(aq) + Br− (aq) is the solubility product constant, Ksp = 7.7 × 10−
barxatty [35]

Answer:

The reaction will be  non spontaneous at these concentrations.

Explanation:

AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)

Expression for an equilibrium constant K_c:

K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]

Solubility product of the reaction:

K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}

Reaction between Gibb's free energy and equilibrium constant if given as:

\Delta G^o=-2.303\times R\times T\times \log K_c

\Delta G^o=-2.303\times R\times T\times \log K_{sp}

\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]

\Delta G^o=69,117.84 J/mol=69.117 kJ/mol

Gibb's free energy when concentration [Ag^+] = 1.0\times 10^{-2} M and [Br^-] = 1.0\times 10^{-3} M

Reaction quotient of an equilibrium = Q

Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}

\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)

\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])

\Delta G=40.588 kJ/mol

  • For reaction to spontaneous reaction:  \Delta G.
  • For reaction to non spontaneous reaction:  \Delta G>0.

Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations

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alukav5142 [94]

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where :

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Explanation:

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