1yeah....no.of protons=no.of electrons=26...& mass.no=60.......where did i go wrong??
The electronic conf. of a tripositive metal ion is [Ar]3d5.If it's mass no. is 60,the total no. of fundamental particles init's neutral atom is....???
-
UP 0 DOWN 0 0 8
8 Answers
62Deleted some of my very stupid messages above.. that causes pinky to think it was out of syllabus.. but it is not..
How to solve.. we know that electrons are taken out from the outermost shell only...
So that should be taken as a hint.. try to solve it now,,,
62http://targetiit.com/a/periodic_table.html
may help as a good tool for this quesiton!?
1the given conf. is after the removal of electrons...so
original no. of electrons=23+3=26. So total no.of fundamental particles =60+26=86.
Am i correct??
162oops that reply was a nail in the coffin..
i am not attempting chemistry anymore :D
lol
Manish was out.. so i though i wud do the helping.. it turns out that i did more
I have deleted 3 posts in this thread.. that goes to show how pathetic my chemistry is :)
1mass no. is no. of proton+ no. of netron
no. of electron= no of proton= 26 in original state.
therefore no. of neutron are 34
now wht u want exactly tell??/
186 is the answer .. the number of neutrons (34)+ protons(26) + electrons(26) ...