City University ofNew York Distance from nucleus Distance from nucleus (r) Dista
ID: 512016 • Letter: C
Question
City University ofNew York Distance from nucleus Distance from nucleus (r) Distance from nucleus (r) Distance from nucleus (r) Answer is b 20) Which of the following statements is wrong? (T12-2) a. If any electrons are found in antibonding orbital(s), it means that no bond will form between atoms of that particular molecule b. When two orbitals overlap in-phase with each other, a bonding molecular orbital forms. When two orbitals overlap out-ofphase with each other, an antibonding molecular orbital forms. d. All antibonding M are higher in energy than the atomic orbitalsof which they are composed 21)The molecule XCis has a square pyramidal shape. Which ofthe following atoms could be X? (T10-12) a. O b. P c. S d. Xe e. At least two of these atomscould be x Spring 2017 Exam 2AExplanation / Answer
Since I have understood by seeing the questions the one which are marked in red are correct. So:
19) The question with electron probablity and distance from nucleus isnt complete so i couldnt answer. But to give yu an idea about electron probablity it is a form of wave function and distance from the nucleus gives the radial probablity of the nucleus. Hope this will help.
20) Molecular orbitals theory are formed by overlapping between atomic orbitals. The more electronegative compounds hold the atom more closely. The modelling of Mo can have when there is comparable energy between atomic orbitals. So the two molecules overlap in two ways depending the relationship between the phase not in the phase.
21) Since the XeF4 forms a square planer structure as Xe is having 6 electrons in outer shell 4 in bonded form and 2 in lone form so it can gain two electron from the other. So this is why to form a square bipyramidial shape two atoms are needed so the two atoms cold be X.
22)In a multiple bond the pi bond is formed from the two orbitals that lie perpendicular to it. This why no hybride is inolved.
23) Magnesium (1s22s22p63s2) comes up in group 2 of the preiodic and has a rising ionisation energy. So to gain the second ionisation energy it has to lose the 3s2 shell so the answer would be [Ne] 3s1--> [Ne] + e-
24) The bond length is the distance between the two bond. The more electrostatic the atom would be the bond length would be lesser this is why the answer would be NO2+< NO2-<NO3-
25) The given object having the same kinetic energy so T= p2/ 2m this is same for all the object. So p will be p=sqrt{2mT} and the debroglis wavelength is lambda= h/p so the larger the p the smaller the lambda, so bowling ball will be the answer
26) Addition of the two electron in the B2 that gives a dianion and this results in bond order to be 2. Since their is no unpaired electron left so their will be no effect on the bond strength.
27) First ionisation energy usually increases going across the period 3. And the ionisation energy I.E1 for phosphorus is 1012 kJ/mol