Answer
Verified
391.5k+ views
Hint: Hybridization of a molecule is determined using the postulates of the Valence band theory and it is basically the phenomenon of mixing of orbitals.
Complete step by step solution:
Hybridization (mixing of orbitals) when two or more than two types of orbitals having less energy difference are combined and form the same number of new orbitals having similar shape and energy but different orientation.
Formula of hybridization; -
$H=\dfrac{(V+M-C+A)}{2}$
Here,
H= hybridization
V = number of valence shell electron of central atom
M= number of monovalent atom surrounding the atom
C= charge on cation
A= charge on anion
-First of all, we have to find out which is the central Atom here so the nitrogen is the central Atom here.
-Therefore, the electronic configuration of nitrogen atom is 1\[{{s}^{2}}\] 2\[{{s}^{2}}\] 2\[{{p}^{3}}\].
-According to the electronic configuration the first (inner) shell will have two electrons and the outermost shell will have two electrons and the outermost shell will have 5 electrons.
-Number of outermost shell electrons is the number of valence shell electrons of Atom.
So, now we are solving for $NO_{2}^{+}$;
By formula of hybridization
$H=\dfrac{(V+M-C+A)}{2}$
where,
V= 5; C= 1
M= 0; A= 0
Put these values in formula
H =$\dfrac{\left( 5\text{ }+\text{ }0\text{ }-1\text{ }+\text{ }0\text{ } \right)}{2}$
H= $\dfrac{4}{2}$= 2
So, the hybridization = \[sp\]
Now solving for $NO_{3}^{-}$;
Again, by formula
V = 5; M= 0
C= 0; A = 1
Put these values in formula
$H=\dfrac{(V+M-C+A)}{2}$
H= \[\dfrac{\left( 5\text{ }+\text{ }0\text{ }-\text{ }0\text{ }+\text{ }1 \right)}{2}\]
H= $\dfrac{6}{2}$ = 3
So, the hybridization = \[s{{p}^{2}}\]
Now solving for $NH_{4}^{+}$;
Again, by formula
$H=\dfrac{(V+M-C+A)}{2}$
V= 5; M = 4
C = 1; A = 0
Put these values in formula
H= $\dfrac{\left( 5\text{ }+\text{ }4\text{ }-\text{ }1\text{ }+\text{ }0 \right)}{2}$
H=$\dfrac{8}{2}$= 4
So, the hybridization is =\[s{{p}^{3}}\]
So, by solving these the hybridization of atomic orbitals of N in $NO_{2}^{+}$, $NO_{3}^{-}$ and$NH_{4}^{+}$ are respectively \[sp\], \[s{{p}^{2}}\], \[s{{p}^{3}}\].
Therefore, the correct option is option (A) \[sp\], \[s{{p}^{2}}\], \[s{{p}^{3}}\].
Note:
-Check properly the number of valence shell electrons of the central Atom.
-Check properly the number of monovalent atoms.
-Check properly the charge of cation and anions
Complete step by step solution:
Hybridization (mixing of orbitals) when two or more than two types of orbitals having less energy difference are combined and form the same number of new orbitals having similar shape and energy but different orientation.
Formula of hybridization; -
$H=\dfrac{(V+M-C+A)}{2}$
Here,
H= hybridization
V = number of valence shell electron of central atom
M= number of monovalent atom surrounding the atom
C= charge on cation
A= charge on anion
-First of all, we have to find out which is the central Atom here so the nitrogen is the central Atom here.
-Therefore, the electronic configuration of nitrogen atom is 1\[{{s}^{2}}\] 2\[{{s}^{2}}\] 2\[{{p}^{3}}\].
-According to the electronic configuration the first (inner) shell will have two electrons and the outermost shell will have two electrons and the outermost shell will have 5 electrons.
-Number of outermost shell electrons is the number of valence shell electrons of Atom.
So, now we are solving for $NO_{2}^{+}$;
By formula of hybridization
$H=\dfrac{(V+M-C+A)}{2}$
where,
V= 5; C= 1
M= 0; A= 0
Put these values in formula
H =$\dfrac{\left( 5\text{ }+\text{ }0\text{ }-1\text{ }+\text{ }0\text{ } \right)}{2}$
H= $\dfrac{4}{2}$= 2
So, the hybridization = \[sp\]
Shape = linear
Now solving for $NO_{3}^{-}$;
Again, by formula
V = 5; M= 0
C= 0; A = 1
Put these values in formula
$H=\dfrac{(V+M-C+A)}{2}$
H= \[\dfrac{\left( 5\text{ }+\text{ }0\text{ }-\text{ }0\text{ }+\text{ }1 \right)}{2}\]
H= $\dfrac{6}{2}$ = 3
So, the hybridization = \[s{{p}^{2}}\]
Shape = trigonal planar
Now solving for $NH_{4}^{+}$;
Again, by formula
$H=\dfrac{(V+M-C+A)}{2}$
V= 5; M = 4
C = 1; A = 0
Put these values in formula
H= $\dfrac{\left( 5\text{ }+\text{ }4\text{ }-\text{ }1\text{ }+\text{ }0 \right)}{2}$
H=$\dfrac{8}{2}$= 4
So, the hybridization is =\[s{{p}^{3}}\]
Shape = tetrahedral
So, by solving these the hybridization of atomic orbitals of N in $NO_{2}^{+}$, $NO_{3}^{-}$ and$NH_{4}^{+}$ are respectively \[sp\], \[s{{p}^{2}}\], \[s{{p}^{3}}\].
Therefore, the correct option is option (A) \[sp\], \[s{{p}^{2}}\], \[s{{p}^{3}}\].
Note:
-Check properly the number of valence shell electrons of the central Atom.
-Check properly the number of monovalent atoms.
-Check properly the charge of cation and anions
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Which type of bond is stronger ionic or covalent class 12 chemistry CBSE
What organs are located on the left side of your body class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
When people say No pun intended what does that mea class 8 english CBSE