L37 : Numericals-1 - Structure of Atom, Class 11 Video Lecture

hello friends this video instruction of
atom part 37 is brought to you by exam
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no more fear from exam before watching
the video please make sure that you have
watched part 1 - part 36 it take some
example in the first one Caesar you
write electronic conviction of H minus
sign so H - we'll have Amon electrons to
ne plus we'll have sodium as 1111 - 110
po + oxygen has 8 8 - 1 7 and F -
Laurine has 9 9 you add 1 because they
get it shot channel correct so let's
start with H - H - will have 2 electrons
so 1s will be sufficient for this guy so
one has to will be the electronic and
revision measurements NE + as 10 so 1
has sub cell - s sub cells and to P sub
cell is also my correct 10 so let's
attribute first second third comes here
fourth fifth sixth seventh eight nine 10
so the electronic and region will be 1 s
2 second s cell has 2 2 P cell has 6
like this is the electronic curriculum
and it is
similarly o+ has 7 electrons so 1s 2s 2p
so 1s will have 1 2 then third fourth
fifth six seven this is how the
electronic and region o+ will look like
it will be 1s2 2s2 2p3 if you see o+ is
stable because this halfway you see f -
f - has 10 so let's again write 1s to 1s
2s and 2p right so if it is here or if
you see this verse 10 and this first 10
the electronic contribute to be all the
same just I will just show you
electrons first comes here second comes
here comes here for comes here
fifth six seven eight nine
so it will be 1s2 us sub cell has two
and two piece of synthesis these are
electronic configurations up all this
ions second question is what is the
atomic number of element is outermost
electrons represented by 3s one please
note if the outermost is 3 s 1 that
means the electronic continues to be
something like this align c 1s is the
lowest energy then I have 2 s then I
have to pee then I have 3 s right like
this goes up so it is 3 s 1 that means
the electron II can finish be something
like this 1 s 2 2 s will be all filled
will be all filled and then I have 3 s 1
it's correct this will be the electronic
condition of this element a so the total
atomic number is 2 plus 2 4 plus 6 10
plus 1 11 so the 11 is the atomic number
for a let's take B what weed is 2 B 3
this guy so before 2 P 3 I have 1s and
2s they should be all filled so I get to
be 3 if you add 2 + 2 4 + 3 7 so this
guy's sin atomic number similarly 3 D 5
3 D 5 is or stupefy to be 5 Ashlyn's
2 P Phi so 1s2 2s2 2p5 so before this I
acted fill everything
so this becomes 1 s 2 2 s 2 n you be
fine electronic generation is become 5 +
2 + 2 is 9 the second equation which
other items are indicated by the
following electronic configuration
helium 2 s 1 so if you see helium is
nothing but 1 s 2 that's all right
so helium 2 s 1 will be
helium to s 1 will be nothing but 1s2
2s1
so what is the atomic number 2 plus 1 3
so atomic number 3 is what lithium that
is my answer this is des second is neon
3 s 2 2 P 3 so neon I know is or you can
directly say neon is what is the
electron atomic number for neon 10 10
plus 2 12 plus 3 15 so neon 3s2 3p3
atomic number is 15 correct
atomic number 15 is what phosphorous ii
is argon 4s 2 3 D 1
so argon atomic number is 18 in 18 you
add three so I Tomic number becomes 18
plus 3 that is what 21 in 21 atomic
number is 4 and that is my answer easy
only which is the lowest value of n that
allows G orbit to exist the question is
which is the lowest value of n that
allows G orbit to exist see I have s
orbit P or will be F G R right for s
orbit L is nothing but 0 this is 1 2 3
and 4 and we know that n has to be 1
more than L right so for the end to be 0
then minimum value of n is 1 because
it's our from n for getting a p-orbital
the minimum value of n is 2 right it's
art with 2 P 1 for getting a D orbital
the mean value of n is 3 why because it
has to be one more than l4f the minimum
value I saw was 4 so 4 5 minimal of e4g
the minimum error will be fine so the
minimum value of n that allows G to our
existence n because n is always Elvis
month correct an electron is in one of
the three D orbital get the possible
value of n L and L so I am saying my n
is equal to 3 for n is equal to 3 I have
3 option L can be 0 1 n 2 if it is 0 it
is s it is 1 it is P it is 2 it is deep
that's how it is the tree structure
which I told for n is equal to 3 I can
have 3 values of L and 0 the city by is
s once a device t sub cell and
prevents deceptions right the question
is an electron is in one of the 3d
orbitals so we see for this D is equal
to two eigenharp how many orbitals minus
2 minus 1 0 plus 1 and plus 2 hope you
understand see for any value of L they
can have this ml to be minus L 2 plus L
so minus u 2 plus 2 so one electron is
in one of these orbitals correct I have
to find the possible values of n L and M
L so in all these KN is free always L is
2 always but my ml can be minus 2 minus
1 0 plus 1 and cluster and there is nice
hope you understand see the three
structure what we did told was for for a
given value of n I can have lot of value
of n so for n is equal to 3 I got three
values of L 4 L is going to do I got
this five values of ml structure I got
the quotient says that electron is in
one of these orbitals of 3 D sub shells
right so sorry yeah 3 D substance so
this can have each any of these amalgams
here or neutral atom actually because it
is telling atoms all assume it is a
neutral atom is non iron a neutral atom
of element contain 29 electron 35
protons neutrons you defined the number
of protons since it is neutral I can say
that number of electron is going to
number of proton and number of electron
is 29 so the number of protons also 29
first part is all now I defined the
electronic configuration so electronic
configuration depends on the number of
electrons 29 right so I'll write the
electronic and region volt 29 so again
let me write the order of increasing
order of energy a light 1s
2s 2p 3s 3p GD then I have 4s 4p 4d so
I'll take first 1s cause then 2 s 2 P 3
years then the 3p 4s like this
so let me write first though increasing
order it is like this 1s then I have 2 s
then I have to pee right then I have 3
years then I have 3p 4s then I am 3d and
then I have for something like this just
29 so this would be 1s then I'll have to
S then I'll have to pee I'll have 3 yes
the left 3t flat 4s 3d right D will have
how many 10 5 5 5 orbitals 10 electron
3d now for B so it will fill the
electrons they are 29 electrons 1 2 3 4
5 6 7 getting start a 9 10 11 12 13 14
15 16 17 18 19 20 21 22 23 24 25 26 27
2000 Li I'm not doing correct 20 is fine
then after that let me remove this just
3d orbital the clearing won't happen
first so 20 is done 21 22 23 24 25 26 27
28 29 29 is done but now you see this
guy's 9 electrons if it gets one
electron it becomes table so what it
will do it will take one electron form
for us right yet to come here and this
is half filled so this will also be 0 so
the electronic contribution rate
one s will have two 2's will have 2 2 P
will have 6 3 a subset will have to this
was I think three Nia then three P sub
cell will have six four S sub cell will
have one and 3d subshell will have 10
electron and this is what the electronic
configuration will look like correct
because the half else already the
electron jump thank you
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