10. Below is a molecular orbital diagram for the hypothetical molecule XY. XY 0
ID: 571969 • Letter: 1
Question
10. Below is a molecular orbital diagram for the hypothetical molecule XY. XY 0 a) What is the first ionization energy of atom Y? eV -4 5 -6 -8 -10 -12 b) What is the first ionization energy of XY? 2s ExY eV What is the bond energy of the reaction: XY+ X, Y (Answer: 4eV) c) d) What is the bond order of XY, XY, XY2+, XY? e) Arrange the following in order of decreasing of the bond dissociation energ y XY, XY*, XY+, XY 11. For a single-electron atom x, Eois--12 eV, Eol -7 eV, and Eis -10 ev. What is AE for the reaction: X2--)r-X ? (Answer: 1 eV)Explanation / Answer
10 (a) As Y contains two electrons the first electron will go to sigma bonding orbital, thus energy shown in diagram will change from -8eV to -10eV thus there is difference of 2eV, therefore the ionisation energy is 2eV.
10 (b) Since Ionisation Energy is the min energy required to remove a electron from the atom, thus to remove a electron from XY, transfer the electron to X, so the energy required is -1-(-4)eV=3eV.
10 (c) Consider the electron removed from XY to make it XY+ goes to Y then the net energy of XY+ is -10+(-10)+(-8) and when dissociation takes place, one electron will be in sigma bonding orbital and rest two in Y's orbital, therefore the net energy is -8+(-8)+(-10)eV. The difference between energy is 2eV but that is for one electron so for two lectron it is 2*2=4eV.
10 (d) Bond Order=(number of bonding electron-number of antibonding electron)/2
therfore the bond order is (1/2),1,(1/2),0.
10 (e) Bond Order is directly proportional to bond dissociation energy
XY-<XY2+=XY<XY+
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