
The hybridization of \[N\] atom in \[NO_3^ - \] ,\[NO_2^ - \] and \[NH_4^ - \] respectively is:
A.\[s{p^2},s{p^3},sp\]
B.\[sp,s{p^3},s{p^2}\]
C.\[sp,s{p^2},s{p^3}\]
D.\[s{p^2},sp,s{p^3}\]
Answer
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Hint: We need to know that the orbital hybridization in chemistry is the concept of combining atomic orbitals into new hybrid orbitals suitable for pairing electrons in the theory of valence bonds to form chemical bonds.
As we know that when an atom binds using electrons from both the orbitals s and p, hybridization happens, producing an imbalance between the electrons' energy levels. The s and p orbitals involved are combined to form hybrid orbitals to equalize these energy levels.
Complete step by step answer:
The N atom hybridization in \[NO_3^ - \], \[NO_2^ - \] and \[NH_4^ - \] is \[s{p^2},sp,s{p^3}\], respectively.
In \[NO_3^ - \] there are 3 bonding domains in the central N atom (one single bond and two double bonds) and zero lone electron pairs. We can draw the structure of this compound as,
In \[NO_2^ - \], there are 2 bonding domains in the central N atom (one single bond and one double bond) and zero lone electron pairs. We can draw the structure of this compound as,
In \[NH_4^ - \] , there are 4 bonding domains (4 single bonds and zero lone electron pairs in the central N atom. We can draw the structure of this compound as,
Option D is the correct option as the given hybridization is correct for the given atoms respectively.
Note:
With the exception of the first three noble gases, helium, neon, and argon, and some of the very short-lived elements after bismuth, nitrogen binds to almost all the elements in the periodic table, producing an enormous range of binary compounds with varying properties and applications. Many binary compounds are known: these are usually labelled n, with the exception of nitrogen hydrides, oxides, and fluorides.
We have to know that the orbital hybridization (or hybridization) in chemistry is the process of combining atomic orbitals into new hybrid orbitals (with different energies, sizes, etc., than atomic orbitals) suitable for pairing electrons in valence bond theory to form chemical bonds.
As we know that when an atom binds using electrons from both the orbitals s and p, hybridization happens, producing an imbalance between the electrons' energy levels. The s and p orbitals involved are combined to form hybrid orbitals to equalize these energy levels.
Complete step by step answer:
The N atom hybridization in \[NO_3^ - \], \[NO_2^ - \] and \[NH_4^ - \] is \[s{p^2},sp,s{p^3}\], respectively.
In \[NO_3^ - \] there are 3 bonding domains in the central N atom (one single bond and two double bonds) and zero lone electron pairs. We can draw the structure of this compound as,
In \[NO_2^ - \], there are 2 bonding domains in the central N atom (one single bond and one double bond) and zero lone electron pairs. We can draw the structure of this compound as,
In \[NH_4^ - \] , there are 4 bonding domains (4 single bonds and zero lone electron pairs in the central N atom. We can draw the structure of this compound as,
Option D is the correct option as the given hybridization is correct for the given atoms respectively.
Note:
With the exception of the first three noble gases, helium, neon, and argon, and some of the very short-lived elements after bismuth, nitrogen binds to almost all the elements in the periodic table, producing an enormous range of binary compounds with varying properties and applications. Many binary compounds are known: these are usually labelled n, with the exception of nitrogen hydrides, oxides, and fluorides.
We have to know that the orbital hybridization (or hybridization) in chemistry is the process of combining atomic orbitals into new hybrid orbitals (with different energies, sizes, etc., than atomic orbitals) suitable for pairing electrons in valence bond theory to form chemical bonds.
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