Question

Daily Problem Select the best of the choices. For full credit. some work must be shown in the space below each part, a) Largest species (circle one): Work /logie: b) Largest atomic radius: Work /logie: Ne c) Highest 2nd (least favorable) IE2 Work: (show electron configurations here!) N electron configuration: Na electron configuration: Ne electron configuration: ion(s) electron configuration(s): ion(s) electron configuration(s): ion(s) electron configuration(s): Starting n value: Ending n value: Starting n value: Ending n value: Starting n value: Ending n value:

Answer 1

Date: a) II" Iyst; Ivrt As the change on atom increases its Size gets smaller & smaller This is because cation are formed on aemoval of electrons from outer. most shell & hence mexe positive change means more no. of electrons are being removed Hence for same no. of protons in all thake Iran Irst Irit ; there are leart no. of electrons in Izet . Same number of protons are attracting less not of elections making the size of cation smallest Hence largest species Irat - Smallest species Iro Iith 15thlath group. Nos) Periody AL p ce - As we move along the period from - left to right size decrealle This is because no of Valence e. معما حللللعطواخسیمکالمختنكحيل

Date: increases - Nucleus - Shell the attraction between & the electrons hence the shrinks towards Nucleus largest Radius : Al © Highest and IE. JE It is energy required to remene the outermost election from the gaseous species 2nd J.E : Energy required to sempre second election. N Atomic No. 7 15 20 20 Na Atomie No-ld Is as ap 31 Ne: Atomic No lo stes ap > Stable electronic configuration (fall shall For Na after removal of Ist election - we get Nat with el configuration

Date: Which is Noble gas configuration & hence enllgy required to remove second election from Na is highest Nat 15 25 26 → highest 2nd IE [ N : IS . ง. Is alp Ne: 152 L Net 1599.4 I Ne : 1s as ap