Oxides of Nitrogen Video Lecture - NEET

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the next important compounds the
nitrogen families oxides of nitrogen so
in this will be studying dinitrogen
oxide will be shutting my trojan
monoxide will be studying nitrogen
dioxide nitrogen dioxide nitrogen
tetroxide nitrogen pentoxide let's study
the first oxides of nitrogen that is
dying nitrogen oxide by nitrogen oxide
is nothing but n 2o oxidation state of
nitrogenous +1 here it is colorless and
neutral s is colorless gas it is also
neutral correct so to prepare my n 2o
what we do is we take NH 4 no.3 and we
heat this we get n 2o and some water
molecule if you draw the structure of
this and to oh this is something like
this this n n 1 is 1 1 3 B committer
and and Oh bond is 1 1 9 you go with and
the shape is linear
let us draw the letter structure of this
n 2 N and O how many valves electron 5 5
& 6 this 5 plus 5 plus 6 that is 16
minutes left on will draw these
electrons will form a bond so we have
used 1 2 3 4 electrons we are left with
2 electrons now now these two electrons
we have to dispute among this nitrogen
and oxygen so that all of these satisfy
a quantity so you can actually draw this
you can draw 3 different structures for
this so in one case 6 2 + bi by 2
for six let's say 8 10 12 this nitrile
oxygen has one two three four five six
and eight octants complete this nitrogen
has six plus seven plus eight octane is
complete this nitrogen has only four so
this nitrogen has only four what we can
do is we can now our typical bond here
between nitrogen and oxygen so that my
nitrogen octet is also complete so in
that case I will get some structure
something like this okay
these electrons moves four electrons
move so in this case nitrogen has 1 2 3
4 into 2 8 electrons this nitrogen has
eight electrons this oxygen has one two
and six eight electrons this is one
possible option the other possible
options we have is you can actually get
too many transform this into a crunch
from this so in that case we get
something like this here we'll have 1 2
3 4 5 6 & 8 electrons nitrogen 8
electrons 4 and here you will have
folded across this is another possible
option another possible option is we'll
get the 4 electrons from this nitrogen
nitrogen 1 so that all will cooperate so
we will get something like this so this
will have 6 electrons you still have 2
electrons now and now if you see this
nitrogen has 1 2 3 4 5 6 7 8 this
nitrogen has 1 2 3 4 1 & 8 electron this
oxygen has 6 plus 2 electrons 8 all
three possible scenarios they but out of
these let's see the charge in this case
nitrogen will have how much charge or
nitrogen has five electrons right 5
minus 1 2 3 4 5 6 7 5 and 7 is minus 2
this one will have 5 minus 1 2 3 4 5
minus 4 s plus 1 this oxygen will have
oxygen valence electron is 6 6 minus 1 2
3 4 5 6 minus 5 is plus 1 so that will
be the charge distribution in this case
we'll have nitrogen is 5 5 minus 1 2 3 4
5
that is minus 1 this nitrogen is 5 - 1 2
3 4 is plus 1 this oxygen will have
charged 6 - 1 2 3 4 by 6 minus 6 then 0
again this nitrogen will have 5 - 1 2 3
4 5 that is 0 charge this nitrogen will
have 5 - 1 2 3 over is plus 1 charge
this oxygen will have 6 - 1 2 3 4 5 6 7
that is minus 1 charge so you see this
is not this is ruled odd because it has
no charge minus 2 charge plus 1 plus 1
uncharged so out of these two these two
are the possible structure for n - OH
actually these two are the possible
structure into Oh correct and now do
these two structures I will say probably
this is more stable oxygen has the
addiction mentally in have positive
charge a way these two are the possible
structure in - let's try to understand
my Trojan War oxide is n o ox addiction
stage of nitrogen is class 2 this is
again a colorless gas and it is also
neutral right to prepare what we have to
do is you have to use any n of 2 and
some vs o 4 and some sulfuric acid react
this you get fe2 so4 3 + n HS o 4 and
some water molecule and you get some
anode this is my hole you want again
write a balanced equation for this so I
will not spend time on balancing this is
how it comes out to be correct
you wanna draw our structure is the
structure n o the bond length is 1 1 5
he committed this is again a linear spin
store the Lewis structure the Lewis
structures is M here is both now in n o
what is the number of this
has 5 this is 6 5 plus 6 is 11 electrons
it want to draw a bond you consume 2
electrons you have 9 electrons left 9
electrons you have to sum or distribute
let's see - good six electrons here 1 2
3 4 5 6 - 6 3 is left 3 we can put here
now if you see we are done with of
electron but oxygen has 1 2 3 4 5 6 7 8
that is fine nitrogen has 1 2 3 4 5 that
is not good right it has to have 8 or at
least 7 so what we'll do we'll transfer
these two electrons these 2 electrons
and these 2 electrons will transparent
form of bond here so it will be
something like this because already
there and these two are transferred
lectrons but when you see right then you
will get bond like this now in this case
oxygen has 1 2 3 4 5 6 7 8 nitrogen has
1 2 3 4 5 6 7 8 11 7 oxygen is a this is
the best you can get for nitrogen so
this will be the structure so na will be
of the structure okay if you want to see
the charge also you can see nitrogen
have will have charge 5 - 1 2 3 4 5 0
oxygen will have charge 6 - 1 2 3 4 5 6
0 so net charge is also 0 in both
nitrogen and oxygen so this is the
structure we have let us talk about
dinitrogen trioxide dinitrogen trioxide
is my blue solid this is how it looks
this is a sedated and this is also clean
and actually we draw the structure of my
nitrogen trioxide this cleaner this is
how dinitrogen trioxide looks this
nitrogen and oxygen bond in this 1 1 4
picometer nitrogen nitrogen bond length
is little bigger 186 Pico meter this
angle is 1 0 5 eb and this angle is 1 30
degree and this nitrogen angle bond
length is
1:21 picometer this is the structure if
you want to prepare this nitrogen
trioxide there is nothing but no.3 what
we can do is you can take nitrogen oxide
if you have just the read react with n 2
O 4 and 1/2 temperature of almost 250
Kelvin you get head into 3 that's how
you prepare in totally correct let's
draw the Lenni structure up in blue oh
three now so we have two nitrogen's and
let's suppose we have three oxygens so
the number of valence electron is they
are three oxygens three into six plus
two nitrogen that is 2 into 5 that is 18
plus 10 that is 28 so we use 2 electrons
the form is 1 2 4 6 8 we are done with 8
electrons minus 8 we have 20 electrons
20 electrons will use 6 electrons each
on these oxygen atoms you can watch my
video on the limit structure which will
give you deep understanding on how to
draw slavey structure listen you're left
with two atoms yay two atoms we can put
here so now let's see how much it is
oxygen has eight nitrogen of all the
oxygen is eight nitrogen this nitrogen
has six electrons and this nitrogen has
one two three four five six both
negative has six electrons each these
two electrons will pull and form a
double bond and these two electrons will
form a double bond so with this what
we'll get is something like this and
this will have six electrons
let me single bone here or electrons
double bond and so this is the structure
we'll get for n 2 O 4
Atlee became daughter's wedding
structure so instead of taking 2
electrons from this oxygen you can take
Learning Sector two electrons from this
oxygen so in this case we'll take our
mode
structure and that will be something
like this correct this is a structure
that is my n 2 O 3 dinitrogen trioxide
the next is nitrogen dioxide that is
no.2 very famous it is a brown gas his
brown gas and as an acid a is acidic
brown gas correct this is no.2 acid II
no gas now we'll draw the structure of
no.2 before drawing the structure let's
see how it is prepared so to prepare
this we use PB no3 old to and we'll heat
this almost 673 Kelvin what you get is
you know two and
structure why not through is something
like this this is my 134 degree and this
length is idle that's the symbol n NN o
is 120 Pico meter let's draw the
structure no.2 is n the active oxygen
nitrogen is 5 oxygen is 6 6 so 5 plus 6
plus 6 that is 17 valence electron we
have we use these four to form bond so
we are left with 3830 let's put 1 2 3 4
5 6 here is oxygen to form octet one two
three four five six here in tox uniform
octet - it is gone one is that so one
will put in this nitrogen so we had done
the electron sharing now I will see this
option is has completed oxide this
oxygen also is has completed octet but
this nitrogen has 1 2 3 4 5 electrons
that is not good so we can transport
these 2 electrons from this oxygen to
this nitrogen so with this will
shaped like this correct
so in this case oxygen has eight this
oxygen has eight electrons this nitrogen
has one two three four five six seven
that is the best we can give for this
nitrogen correct so this will be the
structure actually you can say the ring
structure
why because instead of taking oxygen
from this for electron from this oxygen
we could also take out electron from
this oxygen so in that case you will be
trying getting some other structure
something like this correct
so in this case we took electrons from
this oxygen so these are the two
structures we can get for n 2 O the next
is dinitrogen tetroxide n2o4 okay this n
2 O 4 you actually need you to have 2
no2 it dimerize to form in 204 and we'll
see you know to has a good property of
dimer rising it time wise to comment to
go for there is colourless solid please
note this colorless and it is solid or
sometimes liquid this acid II also okay
this have you have n this is the
structure of no.2 and you take one more
no.2 it dimer is a little form n so do
you one more in here something like this
in 204 okay so this no.2 is brown we
have just seen wrong gas and this is
where my karelis liquid or solid right
this is colorless liquid or solid
correct so if you see this structure of
n2o4 this bonding between n ll n is 175
Pico meter this distance is 120 1 Pico
meter and this angle is 135 million 135
degree so now let's see the level
structure of n2o4 okay so n 2 o 4 let's
draw telling n-o-o-o-o n 2 o 4 2
nitrogen that is doing 2 5 is 10 plus 6
for oxygen it is pouring toxin each
oxygen has 6 valence electron is 24 24
plus 10 is 34 so we will use 1 2 3 4 5 6
7 8 9 10 electrons to form bond that is
used we have 24 it is distribute this 24
electrons around this oxygen so you get
1 2 3 4 5 6 or concocted is complete 1 2
3 4 5 6 6 6 12 1 2 3 4 5 6 12 6 18 18 +
6 is 24 so all the 24 electrons are used
all the oxygen option is complete but
this nitrogen pocket is not complete
this vital it has 6 electrons this
negative also has 6 electrons what to do
so we can pull this electron and form a
bond here we can pull this electron and
form a bond here so this is the
structure you will get is something like
this correct this will have 4 electrons
now this will have as usual 6 electrons
this will have 1 2 3 4 electrons this
will have a 6
electrons this all the elements the
octet is complete correct this is my
structure actually instead of taking
electrons from this and this oxygen you
can take it from two oxygen on the same
side so in that case you will form
structure something like this you take
electrons from the same side correct so
you get structure something like this so
this is again one more resonating
structure thus you can see that either
resonating structure of n2o4 the next is
nine dinitrogen pentoxide it is a 205 so
in this case the oxidation state of
nitrogen is plus five this night
dinitrogen pentoxide is again my
careless solid this is also i say and
form this we use nitric acid and react
with p4o10 to give 4 HP or 3 & 2 this is
what you looking for
correct the structure of n2o5 is
something like this okay so this angle
is 1 1 2 degree and this length is n o
is length is 150 1 picometer and this nu
length is 1 with 9 Pico meter and this
angle is 134 Deeley if you want to draw
the Lewis structure of n2o5 let's draw
this in 205 correct so how many oxygen 5
oxygen 5 into 6 two nitrogen 2 into 5
that is how much 30 plus 10 is 40 HR
into 4 & 2 O 5 is 40 so let us see how
much we can draw
into five so we have to use a 1 2 3 4 5
6 7 8 9 10 11 12 12 reforms ex bonds so
40 minus 12
we are left with the 28 when it gave me
a left with so we have 1 2 3 4 5 6 so 24
we can use in these oxygen nicely so
let's use this one to 3 for by 6 7 8 9
10 11 12 13 14 15 16 17 18 19 20 21 22
23 24 24 we can use minus 24 left with
only 4 electrons all right let it be the
only 4 electrons now so we had to put
this today close we can produce
electrons as opposed here in oxygen hole
sorry 1 2 3 4 left in the way that
everything is done let's see if it is
balance oxygen 80s oxygen 8 years the
softens is also 8 1 2 3 4 5 6 7 8 years
oxygenated oxygen 80s oxygens all
balanced nitrogen HC nitrogen 1 2 3 4 5
6
you want to balance this so let us take
2 electrons from this oxygen put it here
this nitrogen 1 2 3 4 5 6 again let's
take 2 electrons from this oxygen input
here so this what you get is my trillion
the double bond so you see you my triple
bond and oxygen
so now 6 electrons laughs whole electron
is enough one it wrongs in half or in
acknowledged in have 6 electrons and
everything is satisfied so this is one
of the resulting structure in this case
also instead of taking electrons from 2
oxygen on the same side we would have
taken electrons from oxygen on these
sites also there you get oxygen double
bond here and single body this oxygen so
that we can draw more inward resonating
structure for this ok so with this what
you will get is what
enjoy for you something like this
correct so if you take electrons from
these two oxygen in these two oxygens
you will get something like this now
we'll take some numericals on this
nitrogen family the question is why and
not dude I'm area so let's see the
structure of a neutral first and oh wait
nitrogen is 5 plus 6 plus 6 plus 5 plus
6 plus 17 order 17 we have used for to
form form there is 13 13 let's use this
one two three four five six seven eight
nine ten eleven twelve and one like each
other so everything is used now if you
see this nitrogen oxygen has 1 2 3 4 5 6
7 8 all the octave is complete this
oxygen also has a toxic what it is
complete this nitrogen has 1 2 3 4 5 6
sorry 1 2 3 4 5 orchid is not complete
so we'll take two oxygen from this
product here so what you get here is
something like this correct so this
oxygen adopted is complete this oxygen
of it is complete this mechanism has 1 2
3 4 5 6 7 the man that is the best you
can have but this is odd valence
electron now in this case you see this
odd valence electron is not stable this
odd valence electron is not stable thus
it will dimerize to convert into stable
into a force if you see the structure we
had something like this right so this
was the one but it is our 1 so it will
react with one more of an aura into and
then they'll form a bond in these two
right so this it will get something like
this
so now if you see and if you see octet
so everything is stable all oxygen after
this complaint this nitrogen one two
three four eight less nitrogen also
eight so this is again my brown gas and
this is my colorless liquid or solid in
this n beverage so the main reason was
the odd valence electron in no.2 the
next question is what is the core
valence of nitrogen n2o5 let's draw the
structure of n2o5 you've just drawn a
lot repeat you can actually use the same
one so now if you see covalency of
nitrogen here is how much one two three
four in this case one two three four
the covalency of nitrogen is four in a
205 just draw the structure and see the
number of bonds nitrogen has thank you
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